Scientific Issues. Catholic University in Ružomberok MATHEMATICA III

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3 Scientific Issues Catholic University in Ružomberok MATHEMATICA III Ružomberok 2009

4 Mathematica III, Scientific Issues Editor: Martin BILLICH Reviewers: Roman FRIČ, Ján GUNČAGA, Graphical Project of a Cover: Martin PAPČO Typesetting by L A TEX: Martin BILLICH c Copyright by VERBUM - Catholic University in Ružomberok Press, 2009 ISBN

5 CONTENTS Billich Martin. On non-isometric distance one preserving mappings 5 Binterová Helena, Činčurová Lenka. Interactive Whiteboard in Basic school Harminc Matúš, Molnár Peter. Influence of visualization on a word problem solving Kopka Jan, Frobisher Leonard, Feissner George. The Strategy of Infinite Descent Kowalczyk Stanisław. Some properties of continuous functions f: R k R n Krech Ireneusz, Płocki Adam, Tlustý Pavel. Stochastic Tree and Construction of Discrete Probability Spaces and Series Summation Kútna Anna, Palásthy Hedviga. Application of information technology in teaching mathematics Lukáč Stanislav. Solving linear optimization problems in mathematics teaching using ICT Major Maciej. Mixed discrete-continuous random variable Major Joanna, Powązka Zbigniew. From research on student difficulties in using the properties of functions while solving equations and inequalities Molnár Peter. Solving of word problems by students in an environment of the recording software Lampášik Muzyczka Zofia, Płocki Adam. Geometrical probability in mathematics for everyone Novikov Sergey. Properties of generalized characteristic multivalued logical functions Pócsová Jana, Semanišinová Ingrid. What do 5th grade pupils know about statistics

6 Powązka Zbigniew. From research on the implementation of Lee Moore s method to forming the idea of function Tlustý Pavel, Šulista Marek. Various Approaches to One Problem of Combinatorics Woźna Bożena, Zbrzezny Andrzej. Verification of Java programs using networks of finite automata with discrete data Zdráhal Tomáš. Geometric solution of the system of linear equations

7 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 ON NON-ISOMETRIC DISTANCE ONE PRESERVING MAPPINGS Martin Billich Department of Mathematics Faculty of Education, Catholic University in Ružomberok Hrabovská cesta 1, Ružomberok, Slovakia Abstract. We deal with mappings between Euclidean spaces preserving the unit distance. Some examples of mappings which are distance one preserving but not isometries are considered. 1. Introduction Let X, Y be two metric spaces and d 1, d 2 the distances on X and Y respectively. A mapping f : X Y is defined to be an isometry if d 2 (f(x),f(y)) = d 1 (x,y) for all elements x,y of X. If f is surjective then the inverse mapping f 1 : Y X is also an isometry. Two metric spaces X and Y are defined to be isometric if there exists an isometry of X onto Y. It thus follows that an isometry is an isomorphism for the metric space structures. Consider the following condition for a mapping f : X Y distance one preserving property (DOPP) (cf. [2], [6]). Let x,y X with d 1 (x,y) = 1. Then d 2 (f(x),f(y)) = 1. (DOPP) In 1970, A. D. Alexandrov posed the following problem: Under what condition is a map of a metric space into itself preserving unit distance an isometry? 2. Non-isometric distance one preserving mappings Let E n denote the n-dimensional real Euclidean space. E. Beckman and D. Quarles [1] proved that if E n E n for 2 n < satisfies the condition (DOPP), then f is an isometry 1. 1 For non-euclidean spaces the Beckman-Quarles property has been derived by the Russian school, especially by A. V. Kuz minykh [4] (see also [7]).

8 6 Martin Billich If f : E n E m preserves some distance, it follows that n m. This is true because E m has equilateral n-dimensional simplices if and only if n m. It remains to examine the case when 1 < n < m <. In the following we explain a method to show how to construct examples to prove that for each n there exists an m and a mapping f : E n E m which is distance one preserving but not an isometry. The following examples illustrate the case of a mapping f : E 2 E 8 (cf. [6]). For this consider partitioning the plane E 2 into squares of unit diagonal (Figure 1). Each square contains the bottom edge, the left edge and the bottom left corner but none of the other corners Figure 1 In this case, there exist sets C 1,C 2,...,C 9 such that (a) E 2 = C 1 C 2... C 9, C i C j =, i j; (b) The ends of each segment in E 2 of unit length belong to distinct sets C i. Let now v 1,v 2,...,v 9 be the vertices of a regular simplex in E 8 with the edge length 1. Put f(x) = v i if x C i, i = 1,2,...,9, then mapping f : E 2 E 8 satisfies the condition (DOPP) but is not an isometry. Remark. A partition of the type (b) will be called colouring of the space by the colours C 1,C 2,.... This idea extends easily to dimension n > 2 (cf. [3]). For this, consider partitioning the space E n into n-dimensional cubes (n-hypercubes) of unit diagonal. The side of n-hypercube has length with condition n 1. We will call it "small" n-hypercube. From these small n-hypercubes we form "big" n-hypercube with the edge length (k + 2) (k N) such that (k + 1) 1. Above conditions for length of small n-hypercubes implies

9 On non-isometric distance one preserving mappings 7 that k = [n] (where [n] denotes the integer part of n). Then the big n- hypercube consists of ([ n] + 2) n small n-hypercubes which interior be coloured by one of ([ n]+2) n colours (use distinct colours for distinct n- hypercubes). Interior colour is transfers to one vertex of small n-hypercube and to edges and faces (i.e. k-hypercubes in E n, k < n) incident with this vertex. Paving E n by the translates of the big coloured cube, one coloures E n. This method of colouring E n may be less difficult, if above conditions for are strong inequalities, i.e. satisfies 1 [ n ]+1 < < 1. n From the condition (a) above it follows that no unit segment can have its ends in one small n-hypercube or in distinct small n-hypercubes of the same colour. As above, it results into a non-isometric distance one preserving mapping f : E n E ([ n]+2) n 1. Lemma. For any natural number n, E n can be partitioned into a finite number of disjoint sets Ω 1,Ω 2,...,Ω k such that (i) Ω i E n (i = 1,2,...,k) : x,y Ω i d(x,y) 1; (ii) E n = Ω 1 Ω 2 Ω k ; Ω i Ω j = for i j. Proof. To be more precise, let us consider the proof with n = 3, k = 3 3, the general case is similar. Partition E 3 (as we did before) into cubes numbered from 1 to 27 with period 3 in each coordinate direction (Figure 2). Figure 2 The side of all cubes (in E 3 ) has lenght with condition 1/2 < ρ < 1/ 3. It is clear to see that if the cubes each have diagonal equal, say to 0.95 (for > 1/2), then the union of all cubes with the number j (j {1,...,27}) is a set Ω j satisfying condition (i).

10 8 Martin Billich We got the following result (see also in [6]). Theorem. For any natural number n, there exists an integer n m such that for each integer N n m there exists a mapping f : E n E N 1 which satisfies the condition (DOPP) but is not an isometry. Proof. We partition E n into the regions E 1,E 2,E 3,... such that each region E i has diameter less than 1. Let any closed n-sphere of radius 1 (in E n ) intersects at most n m of these regions. By the previous lemma we have that regions E 1,E 2,E 3,... can be partitioned into n m disjoint sets Ω 1,Ω 2,...,Ω nm with the condition (i), i.e. if x E i,y E j (i j) and E i,e j belong to the same set Ω k (k {1,2,...,n m }), than d(x,y) 1. Let now v 1,v 2,...,v nm be the vertices of a regular (n m 1)-dimensional simplex in E nm 1 with the edge length 1. Define f : E n E N 1 with N n m as follows: f(x) = v i for x Ω i and i = 1,2,...,n m. Thus for all x,y E n (x y) with d(x,y) = 1 we obtain that both x,y are not in the same set Ω k. It follows that d(f(x),(y)) = 1 and mapping f satisfies the condition (DOPP) but is not an isometry. Some further questions can be asked. For instance: What is the minimum m, 2 n < m, for which a non-isometric distance one preserving mapping E n E m exists? Also, it is still an open problem whether or not there is a distance one preserving mapping f : E 2 E 3 which is not an isometry (cf. [5]). References [1] Beckman, F. S. - Quarles, D. A.: On isometries of Euclidean spaces. Proc. Amer. Math. Soc. 4, , [2] Billich, M.: Transformation preserving unit distance. Acta Fac. Paed. Univ. Tyrnaviensis 5, 11 15, [3] Dekster, B. V.: Non-isometric distance 1 preserving mapping E 2 E 6. Arch. Math. 45, , [4] Kuz minykh, A.V.: On a characteristic property of isometric mappings. Soviet Math. Dokl. 17, 43 45, [5] Rassias, Th. M.: Is a distance one presserving mappings between metric spaces always an isometry?. Amer. Math. Monthly 90, 200, 1983.

11 On non-isometric distance one preserving mappings 9 [6] Rassias, Th. M.: Some remarks on isometric mappings. Facta Univ. Ser. Math. Inform. 2, 49 52, [7] Rassias, Th. M.: Properties of isometric mappings. Journal of Mathematical Analysis and Aplications 235, , [8] Rassias, Th. M.: Isometric mappings and the problem of A.D. Aleksandrov for conservative distance. Proc. Intern. Graz Workshop, World Scientific, Singapore , [9] Rassias, Th. M. - Sharma, C.S.: Properties of isometries. Journal of Natural Geometry 3, No.1, 1 38, [10] Xiang, S. - Rassias, Th. M.: On mappings with conservative distances and the Mazur-Ulam theorem. Univ. Beograd. Publ. Elektrotehn. Fak., Ser. Mat. 11, 1 8, 2000.

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13 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 INTERACTIVE WHITEBOARD IN BASIC SCHOOL Helena Binterová a, Lenka Činčurová b a Department of Mathematics University of South Bohemia Jeronýmova 10, České Budějovice, Czech Republic hbinter@pf.jcu.cz b Department of Applied Mathematics and Informatics University of South Bohemia Studentská 13, České Budějovice, Czech Republic cincurov@ef.jcu.cz Abstract. The article arose as the need of a reaction on problems which occur by the introduction of percents and symmetry at a school using Interactive whiteboard. On the base of a teaching with computer, we demonstrate the method of visualization and geometric interpretation.. 1. Introduction The progress of ICT (Information and Communication Technology) is really fast and we can use it in many spheres of human performance as well as in education. According to (Dostál, 2007), the trend is to integrate the ICT into the teaching process. (Černochová, 1998) writes that a computer improves students creativity and cogitation, it offers a funnier way of teaching, students perceive the subject matter by more senses and they remember it, the teacher gets an immediate feed-back, the use of computers can be adapted to students abilities regarding their own pacing and also lessons can be more schematic. Students are forced to think about given tasks all the time and they have to decide how to accomplish them. If they re not successful they have to search for mistakes and they are able to explain why they have not reached the expected result. Using ICT for teaching brings lots of new possibilities. However, it is necessary to realize that a computer in the teaching process is only an instrument, not the aim of education. It only helps to reach objectives. Teachers can choose from a great amount of various computer programmes suitable for teaching mathematics. We divide them into two

14 12 Helena Binterová, Lenka Činčurová essential groups: CAS (Computer Algebra System) and DGS (Dynamic Geometry System). These programmes represent an incomparable aid in teaching mathematics. Some of them are even available as a freeware so there are no special requirements for a teacher or a school. IWB is a device of last few years. It becomes a common teaching aid in many western countries and gets into Czech schools very fast as well. IWB represents a worth instrument both for teachers and students. It s an excellent visual device helping teachers to present their thoughts in a new and interesting way. It s able to use various sources which are interconnected with a video file, an audio file or a web page. The sources are prepared before the lesson so they re ready to be used immediately. It saves time and keeps the lesson fluent. Teachers can easily change their notes on the board anytime they like, they can move the objects, show new interactions. IWB provides students with a closer contact with a new subject matter; they are engaged in lessons in a more active way. They can present their own thoughts and ideas on the board. Thanks to the visual form of presentations, the students often understand a problem faster. As it results from Moss research, higher demands on teachers, e.g. learning how to use new procedures and technologies can be considered as a disadvantage. Another disadvantage is the time necessary for preparation of a single lesson. Teachers have to make up a new way of interpretation, new examples and new exercises. We can download presentations from various web pages. Their number is rising due to the proportion of the use of IWB. However, not all of the available pages offer presentations of such quality we would like. These presentations have their own imperfections but some of them involve relevant mistakes. Our aim was to make a material for teachers that would offer students another possible way of learning and solving exercises. 2. The worksheets What is the role of technologies in the teaching and learning process, how is it possible to use them in the teaching and learning of mathematics? According to (Kutzler, 2003), it is possible to use them when simplifying, experimenting, for visualisation and concentration. A computer simplifies the execution of arithmetic and algebraic operations, creation of graphs, it is possible to "discover" new pieces of knowledge in the same vain of the contemporary psychological theories of the teaching and learning which consider the learning to be an inductive process, in which experiments play a key role. The role of computers is also important in lessons of geometry. We do not want to say that we get rid of pairs of compasses and rulers. However,

15 Interactive Whiteboard in Basic school 13 the one who taught geometry using a sketchbook appreciates the possibility of visualisation of such situations which can be hardly demonstrated on the board during a single period. Percentage and axial symmetry belong to different disciplines of mathematics. We can demonstrate that it is possible to teach every mathematical matter using a computer and a suitable programme. The worksheets are made in SMART Board programme for IWB. Both themes are divided into several chapters. The main page includes link labels to all of them. Every chapter consists of a motivational exercise that is completely solved. Chapters also contain important information and thesis students should remember a few tasks or a link to another solution, mostly a Microsoft Excel or GeoGebra file. After completing the chapters there is a brush-up at the end of a theme. Worksheet samples We taught mathematics using the worksheets in a secondary school in České Budějovice. Using the IWB really impressed the students, motivated them and showed them the process of teaching from another point of view. Students kept their individual time and the possibility of getting the necessary amount of separated models in term of the process of creating the concepts was really evident. The atmosphere in the class was absolutely extraordinary; students were disposed to cooperate not only with us but with each other as well. 3. Conclusion The worksheets for IWB develop the key competences of students and have in mind the expected outputs according to RVP. The sheets stress the process of creating the concepts, students sense perception and their imagination. We think, we can reach a better motivation using these sheets, make

16 14 Helena Binterová, Lenka Činčurová students interested in the subject and a particular topic and make them participate in lessons in an active way. Why should the pupils be abandoned from using these methods? Because of the lack of time? We are against ourselves. Pupils learn only for an effect on others, for their parents, school and very rarely for themselves and only some of them can use their knowledge effectively in real life as they do not see the mutual coherence. Use of the whiteboard can save a lot of time and open the floor to experimentation and the making of separated models. The change could come with RVP as the teachers will not be bended by a uniform curriculum. However, stereotypes are still prevailing in the Czech education and for this reason the transformation will need some time. References [1] Dostál, J.: Počítač ve vzdělávání, modul 1. Olomouc: Votobia, [2] Černochová, M. - Komrska, T., Novák, J.: Využití počítače ve vyučování: náměty pro práci dětí s počítačem. Praha: Portál, [3] Moss, G. - Jewitt, C. - Levačić, R. - Armstrong, V. - Cardini, A. - Kastle, F.: The Interactive Whiteboards, Pedagogy and Pupil Performance Evaluation: An Evaluation of the Schools Whiteboard Expansion (SWE) Project: London Challenge. London, Institute of Education, [4] Kutzler, B.: CAS as pedagogical tools for teaching and learning mathematics. Univ. S. Boh. Dept. Math. Rep. ser. 11, , 2003.

17 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 INFLUENCE OF VISUALIZATION ON A WORD PROBLEM SOLVING Matúš Harminc a, Peter Molnár b a Institute of Mathematics Faculty of Science, P. J. Šafárik University Jesenná 5, Košice, Slovakia matus.harminc@upjs.sk b Institute of Mathematics Faculty of Science, P. J. Šafárik University Jesenná 5, Košice, Slovakia molpet@centrum.sk Abstract. The influence of the (im)possibility to visualize throughout solving a chosen word problem is presented. We analyze solutions obtained from two groups of students. The first group solved the problem in a software environment not allowing any drawing of pictures or schemes (screen and keyboard), the second one worked conventionally (sheet and pen). 1. Introduction The fact, that solving of word problems compared with solving of other types of mathematical problems is for students much more difficult, is widely accepted by teachers and by students. Therefore it is useful to search for some possibilities (procedures) how solving of such problems can be made easier for students. One advisable possibility is to teach students to create (to draw) visualizations of the situation described by a problem. Visualization is a creative graphical product of a student, which was created in order to understand the problem, to find its solution, or to help with the formulation of an obtained result (Hejný, 2005). Functions of visualization are mentioned in Hejný et al., 2004, Chapter 22. Of course, not every word problem can be visualized in a useful and reasonable way. But it is of great importance, that a student would know about this possibility and he would be also able to use it practically.

18 16 Matúš Harminc, Peter Molnár 2. Description of the research At the turn of April and May 2009 we performed the research, which was aimed at the ability of students to understand an assignment of a word problem. This research was performed at two secondary schools. Twenty Slovak students and thirty one Polish students solved a series of 7 word problems. The word problems were chosen (created) in accordance with the aim of the research. The students had circa 90 minutes for working. It is important that the solving conditions for the groups were not the same. The students from Slovakia solved the word problems on computers, in the environment of some recording software. This software recorded the entire process of solving (written and deleted letters). Thanks to this, a solution can be replayed by a researcher as a movie. It is necessary to emphasize, that the environment of this software did not allow students to draw any pictures. But it was possible to create a table in this environment. The students from Poland solved the word problems conventionally - on the paper. Thus, they had additionally the possibility to visualize (in any kind of way) the situation described by a problem - they could draw pictures and sketch various schemes. The summary of the solving conditions for the both groups is presented by the following Table 1. Country Number of Age of Solving conditions students students SK screen and keyboard, recording software, calculator, impossibility to draw pictures PL writing utensil and paper, calculator, possibility to draw pictures Environment computer classroom ordinary classroom Table 1. In this contribution we direct our attention only on one of the seven word problems. This problem is presented in the following section. 3. Analyzed word problem 3.1. Assignment of the word problem

19 Influence of visualization on a word problem solving 17 The original Slovak assignment Dva sudy obsahujú určité množstvo vody. Ak z prvého nalejeme do druhého práve toľko vody, koľko tam už je, potom z druhého do prvého práve toľko vody, koľko tam už je, bude v každom sude 120 l vody. Koľko litrov vody bolo v každom sude na začiatku? Minimálne koľko litrov vody sa zmestí do každého suda? The Polish translation of the assignment W dwóch beczkach znajduje się woda. Jeżeli z pierwszej do drugiej nalejemy tyle wody, ile już tam jest, a potem z drugiej do pierwszej tyle wody, ile tam już jest, to w każdej beczce będzie 120 litrów. Ile litrów wody było w każdej beczce na początku? Jak najmniejsza może być każda z beczek? The English translation of the assignment Two vats contain a volume of water. If we pour from the first vat to the second one exactly as much of water as there already is, then from the second vat to the first one exactly as much of water as there already is, there will be 120 l of water in each vat. How many liters of water there were in each vat at the beginning? What is the minimal possible volume of each vat? By creating this problem we were inspired by various mathematical schoolbooks. The solution can be found by three different strategies Strategies of solving Strategy 1: Writing an equation The volume of water in both vats changed in accordance with the following table: 1. vat 2. vat At the beginning x y After the 1. pouring x y 2y After the 2. pouring 2(x y) 3y x The corresponding system of linear equations is: 2(x y) = 120 and 3y x = 120. By solving it we find out that x = 150 and y = 90. The volume of water in the first vat changed as follows: 150 l, 60 l and 120 l. In the second one were 90, 180 and 120 liters of water in successive phases. There were in the

20 18 Matúš Harminc, Peter Molnár first vat 150 l and in the second one 90 l of water at the beginning. The minimal possible volume of the first vat is 150 l and of the second one 180 l of water. Note. By forming a system the equation x + y = 240 could also be used. Strategy 2: Working backwards At the end there is 120 l of water in both vats. By the second pouring we have just poured from the second vat to the first one as much of water as there already was. So before the second pouring there had to be in the first vat 60 (half of 120) liters and in the second one 180 ( , that we have just poured to the first vat) liters of water. By using analogy we find out, that before the first pouring (at the beginning) there had to be in the first vat 150 (60 + half of 180) and in the second one 90 (half of 180) liters of water. The volume of water in the first vat changed as follows: 150 l, 60 l and 120 l. In the second one were 90, 180 and 120 liters of water in successive steps. The minimal possible volume of the first vat is 150 l and of the second one 180 l of water. Strategy 3: Guess-check-revise The problem can also be solved by sequential trying of starting values. The values are chosen such that their sum would be 240 liters. To make a new better guess a student additionally uses the information obtained in previous checkings. Note. Also, the fact can be used that there had to be at least 120 l of water in the first vat at the beginning. More about the solving strategies is presented e.g. by J. Kopka in Kopka, 2006, Chapter Crucial parts of the assignment In accordance with the aim of our research (see the Section 2 - Description of the research), the assignment of this word problem contains several parts that can be incorrectly understood by students. By the translation of the original Slovak assignment to the Polish and to the English language it was therefore of great importance to maintain all these crucial parts and not to make new similar parts. Crucial parts of the assignment: - In both vats can be at the beginning a different volume of water. The size of vats can also be different.

21 Influence of visualization on a word problem solving 19 - The word already plays an important role in the assignment. Without this word it would be not clear, to which vat the collocation as there already is is related to. - The collocation as there already is occurs in the assignment twice, but it is always related to another one vat and so a different volume of water is poured. - The second question contains an antisignal. For the correct answer it is necessary to calculate max{150, 60, 120} and max{90, 180, 120}, but not minimums, as the assignment tries to mislead us (see the word minimal in the text of this question). - The word minimal must be understood in the context of the entire assignment and not only in the context of the last sentence (the second question). Otherwise, answers similar to the following one are expected: The minimal possible volume of each vat can be even 1 ml. - The incorrect assumption that the vats have the same size can result in to the following answer to the second question: The minimal possible volume of one vat is 180 l of water. In this case one global maximum - max{150, 60, 120, 90, 180, 120}, instead of two maximums - max{150, 60, 120} and max{90, 180, 120}, is calculated. 4. Influence of the (im)possibility to visualize Note. Students solutions of this word problem are analyzed in this section only from the point of view of the visualization and its influence on the solving process Samples of visualizations in solutions Sample 1 (Writing an equation):

22 20 Matúš Harminc, Peter Molnár Sample 2 (Working backwards): Sample 3 (Guess-check-revise):

23 Influence of visualization on a word problem solving 21 The samples show examples of students visualizations to each of the mentioned strategies. We believe that the visualization helped to understand the problem correctly and it influenced also the choice of the strategy in all the three cases. The analysis of all solutions obtained from the Polish students showed that the visualization strongly affected at least the following two phases: 1. It helped students to understand the assignment correctly in particular by: - Letting them to understand the entire situation described by the assignment - they correctly understood what had happened to water in the vats. - Influencing the correct understanding of the second question. The three Slovak students (of course without any visualization) understood the word minimal only in the context of the sentence, while no Polish student with visualization did that. 2. The visualization markedly contributed to the design (eventually, even to the choice - see the Section 3.2. above) of mathematical model Achieved results of both groups Country Number of students Correct solution Only the first question Success percentage Visualization Appearance of a table Visualization percentage SK % % PL ,4 % ,2 % Table 2. Note. Because two questions were posed by the assignment, the success percentage is calculated as the percentage of correctly answered questions. Note. A table in the solution was not considered as a visualization of the situation described by the problem.

24 22 Matúš Harminc, Peter Molnár The difference between the success percentage and the visualization percentage is relatively small in the Polish group, as well as for the students from Slovakia. In our opinion, this confirms the fact that the (im)possibility to visualize is the decisive factor determining the success of solving this word problem. It is not clear why a table did not occur in the Slovak solutions. The biggest problem of the Slovak students, trying to solve this word problem, was the inability to organize their activities in such way which would help them anyhow. Despite of multiple reading (10-15 times) of the assignment (mostly the second sentence was read repeatedly) a great part of our students was not able to understand what was happening to the water in both vats. They could try to achieve that by: - guessing and checking some concrete starting values (so that their sum would be 240 l), - creating a table (nobody used this possibility in the Slovak group unlikely to the Polish students). But the most natural way - to create (to draw) a visualization of the situation was not disposal to the Slovak students. By solving the remaining six word problems, such striking differences in the success percentage between the students from Slovakia and the students from Poland did not occur. It is important to notice that the situations described in the remaining six assignments were not possible to visualize in any useful or reasonable way. This fact erased the substantial difference in the solving conditions (of remaining word problems) for the both groups. 5. Conclusion Only about 10 % of students in the Polish group solved the problem successfully even without any visualization used in their solutions (77,4 % solved the problem successfully). This corresponds with 10 % of successful students in the Slovak group. Conversely, only about 10 % could not solve the problem despite a (mostly not correct) visualization. Our experiment showed the applicability of creating (drawing) visualizations as a tool (strategy) for word problem solving. It confirmed that it is useful to teach students to visualize the situation described by an assignment for its correct understanding and finding a way to a solution.

25 Influence of visualization on a word problem solving 23 Acknowledgement. The authors thank Marta Pytłak for the Polish translation of the assignment and the students, the teachers and the both schools participating in the research. References [1] Hejný, M. at al.: Dvacet pět kapitol z didaktiky matematiky. UK Praha, Pedagogická fakulta, [2] Hejný, M. at al.: Rozmanitost řešení žáků jako diagnostický nástroj edukačního stylu. Zborník príspevkov z letnej školy z teórie vyučovania matematiky Pytagoras 2005, P-MAT, Bratislava, 19 31, [3] Kopka, J.: Zkoumání ve školské matematice. KU Ružomberok, Pedagogická fakulta, 2006.

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27 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 THE STRATEGY OF INFINITE DESCENT Jan Kopka a, Leonard Frobisher b, George Feissner c a Přírodovědecká fakulta UJEP v Ústí nad Labem České mládeže 8, Ústí nad Labem, Czech Republic kopkaj@sci.ujep.cz b 58 Buckstone Avenue, Leeds LS 17 5HP, England ljfrobisher@btinternet.com c State University of NY, College at Cortland Cortland, New York, 13045, USA gffeisner@hotmail.com Abstract. The strategy of infinite descent was known to the ancient Greeks and remains a powerful tool, especially for proving the impossibility of a construction or the insolubility of an equation. This method is a special case of proof by contradiction. In a problem-solving situation, we have a goal to achieve but may not have immediately the means to attain it. Strategies are tools that are useful in discovering or constructing means to achieve the goal. Some important strategies are, for example: Experimentation, Analogy, Examining a simpler case, Identifying a subgoal, Examining a related problem, Working backward, the Geometric Method, the Algebraic Method, Using indirect reasoning. In this paper we concentrate on the strategy called Strategy of infinite descent. It would appear that this strategy is a characteristic of mathematics that is not appropriate in other spheres of study. It is a method especially suitable for proving negative statements like the insolubility of an equation, or impossibility of a construction. A typical way of using the method is to begin with an assumption that a given problem has a solution S. Starting with S, a never-ending sequence of solutions is constructed; the nature of the problem should indicate that any sequence of solutions must have a final term. This contradiction proves that the problem has no solution: the strategy could be considered to be a special case of proof by contradiction. Strategy of infinite descent is useful in a variety of areas of mathematics. We consider the following geometrical

28 26 Jan Kopka, Leonard Frobisher, George Feissner problem to demonstrate the strategy at work. At the same time we take the opportunity to the problem to show different types of solutions. First we explain the term lattice point. A lattice point in a Cartesian coordinate system is a point whose coordinates are integers. And now the problem: Problem: (a) Is it possible to construct a regular pentagon in a two-dimensional Cartesian coordinate system so that all its vertices are lattice points? (b) Is it possible to construct other regular polygons with this property? Remark: Elementary school children experience the problem when attempting, using rubber bands, to construct a regular pentagon on a squarebased pinboard. The pinboard represents a Cartesian coordinate system with the pins representing lattice points. Figure 1 Young children and older students are often highly motivated by a challenge which can be presented as a claim by a teacher that No one has yet managed to construct (or draw) a regular pentagon on a pinboard. If you have never previously tried to do so, the experience is worthwhile even though you may be conscious that it is impossible. The very act of experimenting, while not necessarily suggesting a proof, enhances one s understanding of the problem and encourages visualization of related images of statements within a proof. Solution a): After some experimentation, we may come to feel that it is impossible to construct such a pentagon, so we conjecture that no such pentagon exists. We present two proofs that our conjecture is correct. Proof 1 (using the strategy of infinite descent): We start with the negation of our conjecture that it is impossible to construct such a regular pentagon, the negation being that it is possible and hence it exists. We thus claim:

29 Problem-Solving Strategies Using Invariants of Transformations 27 There exists in a two-dimensional Cartesian coordinate system a regular pentagon P 1 = A 1 B 1 C 1 D 1 E 1 whose vertices are lattice points (see fig. 2). Now construct the points A 2, B 2, C 2, D 2, and E 2 so that the vector A 1 A 2 equals the vector E 1 D 1. In the same way, we require that B 1 B 2 = A 1 E 1, C 1 C 2 = B 1 A 1, D 1 D 2 = C 1 B 1, and E 1 E 2 = D 1 C 1. We have created another regular pentagon P 2 = A 2 B 2 C 2 D 2 E 2 which also has vertices in a lattice points. Why? (Because coordinates of the vectors on right hand sides of equations are integers, so coordinates of vectors on left hand sides must be also integers - they are the same vectors. And if we add to the point with integer coordinates a vector with integer coodinates we must get point also with integer coordinates.) Figure 2 Similarly, using P 2, we can construct a regular pentagon P 3 whose vertices are lattice points. By iterating this procedure we obtain an infinite sequence of regular pentagons: P 1, P 2, P 3, P 4,... whose vertices are lattice points. Denote by a i the side length of P i for i = 1,2,3,... The squares a 2 1,a 2 2,a 2 3,a 2 4,... form an infinite strictly decreasing sequence of natural numbers. However, any decreasing sequence of natural numbers has a smallest element (this sequence cannot have more terms than has the first one). Hence we have a contradiction. Answer: Thus P 1 cannot be constructed. Our conjecture is confirmed as being true. Comments You may have noticed that during the solution of our problem we used a) the strategy draw a diagram (the geometric method) and

30 28 Jan Kopka, Leonard Frobisher, George Feissner b) the strategy iterating a certain procedure. It is frequently the case that when using a particular strategy to solve a problem or prove a conjecture we resort to sub-strategies within an allembracing strategy. Learners should be reminded constantly of which substrategies are used in order that they become familiar with them, leading to learners incorporating them within their general problem solving skills. We show now an alternative proof of our initial conjecture which, after proof 1, we are able to claim is a mathematical theorem. Proof 2 (also proof by contradiction): Let P 1 = A 1 B 1 C 1 D 1 E 1 be a regular pentagon whose vertices are lattice points. Since P 1 is a bounded set, there are only finitely many lattice points in its interior. Since P 2 is inside P 1 and its vertices are lattice points, there are at least five lattice points inside P 1. But P 3 is inside P 2 and so its vertices, also lattice points, are also inside P 1. Thus, P 1 must have at least 10 lattice points in its interior. Repeating this argument, we obtain an infinite increasing sequence 5,10,15,20,... and so we must conclude that P 1 has infinitely many lattice points in its interior. Again, we have reached a contradiction. Solution b): A little thought shows that we have also made substantial progress on part b) of our original problem generalizing the result to other regular polygons: Theorem: For n > 6, there is no regular n-gon all of whose vertices are lattice points. Proof: For n > 6, the proof is exactly the same as for n = 5! Assuming that such an n-gon exists, we can construct a smaller one inside it, and the smaller one must also have all its vertices at lattice points. The construction is done in the same way as was shown for the pentaton. It is worth the time for the reader to see why the above construction method fails for n = 3,4 and 6. So n-gons for n = 3,4,6 are exceptions from the point of view of our approach. We must treat these three cases separately. The simplest case is n = 4. The square with vertices [0,0], [0,1], [1,0] and [1,1] shows that there is a square all of whose vertices are lattice points. Indeed, if we chose any two lattice points in the plane, drew the line segment connecting them, and then used this line segment as the side of a square, the other two vertices would also be lattice points. An argument involving vectors is perhaps the simplest way to see that this is true.

31 Problem-Solving Strategies Using Invariants of Transformations 29 Remark: This is immediately apparent to those who have constructed squares on a geoboard or pinboard as shown in fig. 3. Figure 3 The situation for n = 3 is not so easy as for n = 4. Remark: With a geoboard and rubber bands it appears reasonably easy to make an equilateral triangle as scalene and isosceles triangle are very easily made. Despite children being told that it is not possible many continue to try to prove such a claim is false. After experimentation we can inductively reason that: Conjecture: For n = 3, there is no regular n-gon, that is an equilateral triangle, whose vertices are lattice points. To prove it we try to use again the strategy of infinite descent, but smaller triangle we must construct in a different way. Proof 1 (using the strategy of infinite descent): Let T 1 = AB 1 C 1 be an equilateral triangle, all of whose vertices are lattice points. If necessary, we can translate the triangle so that vertex A is at the origin. Figure 4 shows the triangle and also shows how the smaller equilateral triangle T 2 = AB 2 C 2 is constructed. Figure 4

32 30 Jan Kopka, Leonard Frobisher, George Feissner Since T 1 is equilateral, it follows that b 2 1 +b2 2 = c2 1 +c2 2 = (b 1 c 1 ) 2 +(b 2 c 2 ) 2 and from these equations, it is easy to show that b 2 1 +b 2 2 = 2(b 1 c 1 +b 2 c 2 ) c 2 1 +c 2 2 = 2(b 1 c 1 +b 2 c 2 ). We now use a strategy from number theory: examining the parity of the numbers. Since the right-hand sides of the equations are even, the left-hand sides must be even also. From this is follows that b 1 and b 2 must have the same parity; they are either both even or both odd. The same is true for c 1 and c 2.Thus, regardless of the parity of b 1 and b 2 and the parity of c 1 and c 2, the quantity b 1 c 1 +b 2 c 2 must be even! Therefore, both b 2 1 +b2 2 and c 2 1 +c2 2 are divisible by 4. It is easy to verify that were b 1 and b 2 both odd, then b 2 1 +b2 2 would be divisible by 2 but not by 4. The same is true of c 1 and c 2. Thus, all four of b 1, b 2, c 1, and c 2 are even. We now construct T 2 by setting B 2 = [b 1 /2,b 2 /2] and C 2 = [c 1 /2,c 2 /2]. Since all four of b 1, b 2, c 1, and c 2 were even, the vertices of T 2 are also lattice points. Now the pattern is clear. We iterate the procedure and obtain an infinite sequence of equilateral triangles T 1,T 2,T 3,T 4,... all of whose vertices are lattice points. Denote by a i the side length of T i for i = 1,2,3,... The squares a 2 1,a2 2,a2 3,a2 4,... form an infinite strictly decreasing sequence of natural numbers. However, any decreasing sequence of natural numbers has a smallest element (this sequence cannot have more terms than has the first one). Hence we have a contradiction. Answer: Thus T 1 cannot be constructed. Our conjecture is confirmed as being true. Comments During construction of proof we used also the sub-strategies: a) draw a diagram (the geometric method), b) iterating a certain procedure, c) parity, and d) analogy.

33 Problem-Solving Strategies Using Invariants of Transformations 31 We offer a second proof of our conjecture (now it is theorem), this time making use of complex numbers. Proof 2 (using complex numbers): We work in the Gauss plane. Consider regular triangle ABC (see fig. 5), for which A[0,0] and B[b 1,b 2 ], where b 1, b 2 are integers and at least one is not zero. We can express points A and B with the help of complex numbers: A = 0+0i and B = b 1 +b 2 i. The third vertex, Figure 5 C[x,y], we can get by rotating vertex B around A with an angle of rotation of +60 or 60. Let us consider the rotation +60. We get it if we multiply number b 1 +b 2 i by complex unit So we have j = cos60 +i.sin60 = 1 2 +i 3 2. (b 1 +b 2 i).j = (b 1 +b 2 i)( i 2 ) = b 1 b 2 3 +i b 1 3+b It means that x = b 1 b and y = b 1 3+b2 2. Because at least one of the numbers b 1, b 2 is not zero, at least one of the numbers x, y is non-integer. We get the similar result if we rotate an angle of 60. Yet a third proof, one not using complex variables, but just a little analytic geometry is possible. We outline it here. Proof 3 (analytic geometry): Consider the equilateral triangle ABC (see fig. 5), for which A = [0,0] and B = [b 1,b 2 ], where b 1, b 2 are integers and at least one is not zero. Vertex C must be on axis of the segment AB and also on the circle with center in point A and radius AB. If you counts coordinates of vertex C, then you find out that both cannot be integral. Result is the same as in proof 2.

34 32 Jan Kopka, Leonard Frobisher, George Feissner For n = 6 situation now is very simple. In a regular hexagon ABCDEF the vertices A, C and E form an equilateral triangle. Since there is no equilateral triangle whose vertices are lattice points, the same applies to a regular hexagon. The general result of b): The only regular polygons whose vertices are lattice points in a two-dimensional Cartesian coordinate system are the squares. The strategy of infinite descent known already in ancient Greece, is a powerful tool for solving some mathematical problems. The method is especially useful for proving negative statements, like the insolubility of an equation, or the impossibility of a construction. Is possible that reader knows other examples of using this strategy for example in the theory of numbers or in mathematical analysis. References [1] Cofman, J.: What to solve? New York, Oxford University Press, [2] Kopka, J.: Výzkumný přístup při výuce matematiky. Ústí nad Labem, UJEP, [3] Frobisher, L., Frobisher, A.: Investigating Problems in the Primary School (book in preparation).

35 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 SOME PROPERTIES OF CONTINUOUS FUNCTIONS f: R k R n Stanisław Kowalczyk Institute of Mathematics Academia Pomeraniensis, Słupsk ul.arciszewskiego 22, Słupsk, Poland stkowalcz@onet.eu Abstract. Given f: R k R n let dist(f,g) = sup f(x) g(x). We study the x R k possibility of approximation of a continuous function f: R k R n by functions which are continuous, not surjective and whose distance from f is arbitrary small. The main goal of the presented paper is a proof that for every n 2 there exist continuous functions f: R n R n such that if dist(f,g) < and g is continuous then g is a surjective function (for n = 1 it is trivial, every function f: R R such that lim x = and lim x = has this property). Let C(X,Y) be the space of all real-valued continuous functions defined on some compact Hausdorff topological space (X,T ) into a metric space (Y,d Y ). Given any functions f,g : X Y, we let dist(f,g) = sup x X d Y (f(x),g(x)). In the proofs that certain continuous binary operations Φ: C(X,Y) C(X,Y) C(X,Y) were not open mappings (see [1,3,4]) a crucial role played the possibility of approximation of f C(X,Y) by a function missing some subsets of the target space with non-empty interior. In [4] this property was studied for continuous functions f: R k R n. We have. Theorem 1. [4] Let f C(R k,r n ) and k < n. Then for each y 0 R n, each ε > 0 and each 0 < λ 1 there exists an h C (R k,r n ) such that [1] dist(f,h) < ε(1+2λ). [2] h(x) y 0 > ε for all x R k. Theorem 2. [4] Let X = (X,τ) be paracompact, dimx k. Fix any natural n > k and let f : X R n be continuous. Then for each ε > 0 and any boundary sets A i R, i = 1,...,n there exists a continuous function h : X R n such that:

36 34 Stanisław Kowalczyk [1] dist(f,h) < ε, [2] h(x) (A 1 A 2... A n ) =. In the presented paper we shall prove that Theorem 1 is not true if k = n or if k > n. Let denote the standard norm in R n. The point (0,0,...,o) R n we simply denote by 0. Throughout the text given x R n and r > 0, we let S n 1 (x,r) denote (n 1)-dimensional sphere in R n centered at x and of radius r. If x = 0 then we simply write Sr n 1 instead of S n 1 (0,r) and if x = 0 and r = 1 then we write S n 1. Definition 1. [2,5] Let (X,T ), (Y,τ) be topological spaces and f,g: X Y be continuous functions. We say that f and g are homotopic if there exists a continuous function F: X [0,1] Y such that F X {0} = f and F X {1} = g. We shall write f g if f and g are homotopic. Obviously, homotopy is transitive: if f g and g h then f h. The following two theorems are very well-known. Theorem 3. [2,5] Let (X,T ) be a topological space and n 1. If f: X S n 1 is continuous and f(x) S n 1 then f is homotopic with constant mapping, f const. Theorem 4. [2,5] For each n 1 an identity Id: S n 1 S n 1, f(x) = x for x X is not homotopic with any constant function const: S n 1 S n 1, const(x) = x 0 for all x S n 1, e.g. Id const. It is easily seen that the last theorem follows from famous Brouwer Fix Point Theory. Lemma 1. Let f: R n R n be continuous and let A = {r > 0: 0 / f(s n 1 r )}. For each r A, define the function F r : S n 1 S n 1 as follows for all y S n 1. Then we obtain F r (y) = f(ry) f(ry) [1] for each r A the function F r is well-defined and continuous, [2] if r 1 < r 2 and [r 1,r 2 ] A then F r1 F r2,

37 Some properties of continuous functions f: R k R n 35 [3] if f(0) 0 then there exists r 0 > 0 such that for each r (0,r 0 ) A we have F r const, [4] if y > f(y) y for all y S n 1 r then F r Id. Proof. Claim (1) is obvious, since the norm is continuous and f(ry) 0 for all y S n 1 and r A. (2) Define H: S n 1 [0,1] S n 1 as follows H(x,t) = f( (r 1 +t(r 2 r 1 ))x ) f ( (r 1 +t(r 2 r 1 ))x ). It is clear that H is well-defined, continuous and H(x,0) = f(r 1x) f(r 1 x) = F r 1 (x) and H(x,1) = f(r 2x) f(r 2 x) = F r 2 (x) for all x S n 1. Therefore F r1 F r2. (3) Suppose that f(0) 0. Let L be a straight line through 0 and f(0), and let l L be a half-line from 0 such that f(0) / l. By continuity off, there existsr 0 > 0 such that whenever x < r 0, then f(0) f(x) < f(0). Therefore for r < r 0 we have f(sr n 1 ) l =. It follows that f(0) f(0) / F r (S n 1 ). Since F r is not surjective we have F r const. (4) Let us define H: S n 1 [0,1] S n 1 as follows Since H(x,t) = rx+t(f(rx) rx) rx+t(f(rx) rx). rx+t(f(rx) r) rx t f(rx) rx r f(rx) rx > 0 for x S n 1, we have that the function H is well-defined and continuous. Furthermore, we obtain and H(x,0) = rx rx = x H(x,1) = rx+f(rx) rx rx+f(rx) rx = f(rx) f(rx) = F r(x). for all x S n 1. Thus Id F r. This finishes the proof. Theorem 5. Let f: R n R n be a continuous function satisfying the following condition limsup f(x) x x < 1. Then the function f is surjective. x

38 36 Stanisław Kowalczyk Proof. Fix any x 0 R n. Suppose that x 0 / f(r n ). Consider the function g: R n R n, g(x) = f(x) x 0 for all x R n. Then g is continuous, g(x) x lim sup x < 1 and 0 / g(r n ). Using notation from Lemma 1, we x obtain A = (0, ). There exists an R > 0 and α < 1 such that f(x) x x < α, whenever x R. It follows that for such x we get f(x) x < α x < x. Then (4) from Lemma 1 implies G r Id for r R, where G r : S n 1 S n 1, G r (x) = g(rx) g(rx). It follows from (2) of Lemma 1 that there is r 0 > 0 such that G r const for r < r 0. On the other hand it follows from (3) of Lemma 1 that G r1 G r2 for r 1,r 2 > 0 whence by transitivity of the homotopy we get Id const. The obtained contradiction shows that f is surjective, i.e. f(r n ) = R n what was to be shown. Theorem 6. Let f: R n R n be a continuous function satisfying the following condition lim sup f(x) x x < 1. Then every function g: R n R n x such that dist(f, g) < is surjective. Proof. First observe that if limsup x then lim sup x g(x) x x limsup x limsup x f(x) x x f(x) x x f(x) x x +limsup x < 1 and dist(f,g) = c < +limsup x Therefore by Theorem 5 g is the surjective function. g(x) f(x) x c x = limsup f(x) x < 1. x x Corollary 1. Let Id: R n R n be identity map Id(x 0 ) = x for some fixed x 0 R n. Then every function g: R n R n such that dist(id,g) < is surjective. For x R k = R n R k n denote x = (ζ x,ξ x ) where x = (x 1,x 2,...,x k ), ζ x = (x 1,x 2,...,x n ) and ξ x = (x n+1,x n+2,...,x k ). Next consider the analogue of Theorem 5 for k > n. Theorem 7. Let f: R k R n, where k > n, be a continuous function (f(x),ξ satisfying the following condition lim sup x) x x < 1. Then the function x f is surjective. Proof. Define f: R k R k by f(x) = (f(x),ξ x ). It follows that f is continuous and for each x R k we get f(x) x) = (f(x),ξ x ) x

39 Some properties of continuous functions f: R k R n 37 whence lim sup x f(x) x x < 0. Thus f: R k R k satisfies the conditions of Theorem 5 and therefore f(r k ) = R k which implies f(r k ) = R n. Corollary 2. Let f: R k R n, where k > n be a continuous function satisfying the following condition lim sup x) x (f(x),ξ x < 1. Then every function x g: R k R n such that dist(f,g) < is surjective. Corollary 3. Let p : R k R n for k > n be the canonical projection, p(x) = (x 1,...,x n ). Then every function g: R k R n such that dist(p,g) < is surjective. References [1] M. Balcerzak, A. Wachowicz and W. Wilczyǹski. Multiplying balls in the spacew of continuous functions on [0, 1]. Studia Math. 170, (2005), [2] M. Greenberg, J. Harper. Algebraic topology. A first course. Springer- Verlag Mathematics Lecture Notes Series, [3] A. Komisarski. A connection between multiplication in C(X) and the dimension of X. Fund. Math. 189, (2006), [4] S. Kowalczyk, S. Ponmonarev. Mappings bypassing holes and unstable sets. Siberian Math. J. [in print]. [5] E. Spanier. Algebraic topology. Springer-Verlag, 1996.

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41 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 STOCHASTIC TREE AND CONSTRUCTION OF DISCRETE PROBABILITY SPACES AND SERIES SUMMATION Ireneusz Krech a, Adam Płocki b, Pavel Tlustý c a Institute of Mathematics Pedagogical University of Cracow ul. Podchorążych 2, Kraków, Poland irekre@tlen.pl b Institute of Mathematics Pedagogical University of Cracow ul. Podchorążych 2, Kraków, Poland adplocki@ap.krakow.pl c Pedagogical Faculty University of South Bohemia Jeronymova 10, Česke Budějovice, Czech Republic tlusty@pf.jcu.cz Abstract. The presented work deals with sums of certain series which are interpreted in context of discrete probability space as consecutive ball draws performed in phases. Stochastic trees represent an unusual tool for determining sums of numerical series. 1. Discrete probability space In this work, the notation N k denotes set {k,k +1,k +2,...}. Definition 1. Let Ω be an arbitrary set with at leats two elements and as much countable as possible. The nonnegative function p : Ω R, satisfying condition p(ω) = 1, ω:ω Ω is called decomposition of probability on set Ω. Example 2. Geometrical series (a n ) n N1, where a n = (1 u) n 1 u and 0 < u < 1, is decomposition of probability on set N 1. This is called geometrical decomposition ([7], p. 9 and 326).

42 40 Ireneusz Krech, Adam Płocki, Pavel Tlustý Definition 3. Let p is a decomposition of probability on the biggest possible countable set Ω and Z = 2 Ω. Function P : Z R given by the functional notation: 0, when A =, p(ω), P(A) = when A = {ω}, p(ω), when A is a set with at least two elements, ω:ω A is a probability, and triple (Ω, Z, P) is probability space considered according to the sense of the axiomatic definition (see [7], p. 124). The triple is called discrete probability space. The elements of class Z are called events. To describe the discrete probability space (Ω,Z,P), it is necessary to describe sufficiently also probability decomposition p on set Ω, and so the discrete probability space can be treated as pair (Ω,p). 2. Discrete random experiment Definition 4. Discrete random experiment is called a real or artificial ([2], p and also [5], p ) experiment δ, which develops and results in accordance with the following conditions: 1) set Ω δ of all results of the experiment is as much countable as possible, 2) for each result, it is possible to determine a priori the probability with which the experiment can end with such a result. Within discrete experiments, we distinguish those, which are performed in phases. They are called multi-phase experiments. Such tosses are n-fold coin tosses. Within multi-phase experiments, we distinguish those in which the number of the phases is random. They are called random experiments with random number of phases. Such a random experiment with a random number of phases is for example when a dice is being thrown until number six is thrown. Definition 5. If Ω δ is a set of results of discrete random experiment δ, and p δ is a function, which assigns to each result of set Ω δ a probability with which experiment δ can end with such a result, then pair (Ω δ,p δ ) is a discrete probability space which is called a model of stochastic drawing experiment δ. 3. Multi-phase experiments and stochastic tree rules In [7] is made a discrete probability space for as drawing experiment as their stochastic model ([6], p. 41). In case of a multi-phase experiment, construction tool of discrete probability space (Ω,p) as its model is stochastic tree

43 Stochastic Tree and Construction of Discrete Probability Spaces and two rules: rule R1: result of an random experiment performed in phases is a progression of results of consecutive phases, such a result is represented by a tree branch; rule R2 (reproduction rule): for each branch of a stochastic tree (and at the same time for each result represented by the branch), there is a corresponding product of numbers assigned to consecutive parts of the branch; the product is called branch weigh ([7], p ); Using rule R1, setωof results of random experiments is determined. The sum of branch weighs of all branches on the stochastic tree is equal to 1. Using rule R2, we determine function p, which is a probability decomposition on set Ω. 4. Numerical series and its sum Definition 6. Let us assume that (a n ) n N1 is an infinite numerical sequence. Let s 1 = a 1 and let s n = a 1 +a 2 +a 3 + a n for n = 2,3,4,... The number s n is called the n-th partial sum of the sequence (a n ) n N1 and an infinite sequence (s n ) n N1 is called the series and is denoted by a n. If the sequence (s n ) n N1 has a finite limit, then we say that the series a n is convergent. If lim s n = s, then number s is called the sum of the sequence n and is denoted by a n = s. n=1 This sum s is understood as an infinite sum a 1 +a 2 +a 3 + that is as a sum of all infinitely numerous terms of the sequence (a n ) n N1 that is a 1 +a 2 +a 3 + = lim n (a 1 +a 2 + +a n ). Definition 7. Let a n = a q n 1 for n N 1, where a and q are established real numbers and q 0. The series a q n 1, as a series created on a base of a geometrical sequence (a q n 1 ) n N1, is called the geometrical series. Theorem 8. If q < 1, then the geometrical series a q n 1 is convergent and a q n 1 = a n=1 1 q. The fraction a 1 q is then the sum of all (infinite number of) terms of geometrical sequence (a q n 1 ) n N1, if q < 1.

44 42 Ireneusz Krech, Adam Płocki, Pavel Tlustý Example 9. If u (0,1), then sequence (a n ) n N1, where a n = (1 u) n 1 u for n N 1 is a geometric sequence, where a = u and q = 1 u, therefore a 1 +a 2 +a 3 + = a n = n=1 u 1 (1 u) = u u = 1. Let Ω = {ω 1,ω 2,ω 3,...}. If p is a probability decomposition on set Ω and p(ω n ) = p n where n N 1, this sum of series p n is equal to Stochastic tree and sums of certain series 1 Let a n = n (n+1) where n N 1. Let us consider series a n also its sum n (n+1) = 1 n=1 In areas of mathematical analysis the sum of this series may be found 1 with the use n (n+1) = 1 n 1 n+1, then s n = ( 1 1 ( 2) ( ( 3) ) n 1 1 n it means that lim s n = lim (1 1 n n n ) = 1. ) = 1 1 n, Definition 10. U b c denotes a box in which there are b white balls and c black balls. Let us consider following multi-phase random experiment δb c P. The first phase is a random ball draw from box U b c. If the randomly drawn ball is black then the experiment ends; if the ball is white, the ball is put back to the box together with another additional white ball. From this new box U (b+1) c another ball is drawn. It is the second phase. If the randomly drawn ball is black then the experiment ends; if the ball is white, the ball is put back to the box together with another additional white ball. From this new box U (b+2) c another ball is drawn. It is the third phase and depending on the result of the draw and the procedure is performed analogically as described. This procedure is performed until a black ball is drawn out. This experiment δb c P with random numerical phases is called Polya s scheme ([2], p. 116). The result of experiment δb c P is unambiguously described by phase number in which a black ball is drawn out first time. Let ω n denote the result: a black ball is drawn out in the n-th phase (n N 1 ). In this case, it is possible to code result using rule R1. Set of results of the experiment δ c is an infinite set Ω P b c = {ω 1,ω 2,ω 3,...}.

45 Stochastic Tree and Construction of Discrete Probability Spaces Example 11. Sum of series 1 a n, where a n = n (n+1) may be found by using stochastic method. Let us consider Polya s scheme δ1 1 P. The figure 1. presents start of fragment stochastic tree this scheme and digraph of function p 1 1 defined by rule R Fig. 1 1 It is important to note that p 1 1 (ω n ) = n (n+1) where n = 1,2,3,.... Expression p 1 1 (ω n ) is the weigh of the tree branch representing result ω n. The fact that the sum of all branch tree weighs is equal to 1 results in the following ω n:ω n Ω P 1 1 p 1 1 (ω n ) = 1 n (n+1) = n=1 + = 1. Example 12. Let us consider Polya s scheme δ1 2 P and serie 1 n(n+1)(n+2) where n N 1 and its sum n(n+1)(n+2) = 1 n=1 Sum of all tree branch weigh of stochastic experiment δ1 2 P is a infinite sum: n 4 4 n 2 n 3 n 1 2 n +. It is sum (n 2)(n 1)n +. The fact that it is the sum of all branch tree weighs, it results in (n 2)(n 1)n + = 1, therefore n=1 1 n(n+1)(n+2) = 1 4.

46 44 Ireneusz Krech, Adam Płocki, Pavel Tlustý Example 13. Let us consider series 1 (2n 1)(2n+1), where n N 1 also Polya s scheme δ2 2 P. It is necessary to say that if p 2 2 is a function defined on set Ω P 2 2 = {ω 1 1,ω 2,ω 3,...} using rule R2, to p 2 2 (ω n ) = (2n 1)(2n+1) where n N 1. It is the branch weight representing the result ω n in the stochastic tree of Polya s scheme δ2 2 P. The sum of all branches of the stochastic tree of the experiment δ2 2 P is equal to: n 3 7 2n 1 2 2n+1 +. It is the series (2n 1)(2n+1) +. Considering that the sum of weigh equals 1, we end up with equation therefore (2n 1)(2n+1) + = 1, 1 (2n 1)(2n+1) = n= = 1 2. In the work, we introduced a method used for determination of sum of some number series using stochastic trees of Polya s scheme: δ P 1 1, δp 1 2 and δ P 2 2. References [1] A. Engel, Wahrscheinlichkeitsrechnung und Statistik. Band 1, Ernst Klett Verlag, Stuttgart, [2] W. Feller, Wstȩp do rachunku prawdopodobieństwa, t. 1. PWN Warszawa, [3] T. Iwiäski, Szeregi nieskończone. WSiP, Warszawa, [4] I. Krech, Rachunek prawdopodobieństwa and sumowanie series. Matematyka 1(2000), [5] A. Płocki, Pravdepodobnost okolo nás. Katolícka univerzita, Ružomberok, [6] A. Płocki, Stochastyka dla nauczyciela. Wydawnictwo Naukowe NOVUM, Płock, 2007.

47 Stochastic Tree and Construction of Discrete Probability Spaces [7] A. Płocki, Dydaktyka stochastyki. Wydawnictwo Naukowe NOVUM, Płock, [8] A. Płocki, P. Tlustý, Pravděpodobnost a statistika pro začátečníky a mírně pokročilé. Prometheus, Praha, [9] P. Trojovský, Patří aritmeticko-geometrická řada na střední školu. Matematika-fyzika-informatika, 1999 (8 10),

48

49 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 APPLICATION OF INFORMATION TECHNOLOGY IN TEACHING MATHEMATICS Anna Kútna a, Hedviga Palásthy b a Department of Informatics Faculty of Education, Catholic University in Ružomberok Námestie Andreja Hlinku 56, Ružomberok, Slovakia : Anna.Kutna@ku.sk b Department of Informatics Faculty of Education, Catholic University in Ružomberok Námestie Andreja Hlinku 56, Ružomberok, Slovakia : hp@ku.sk Abstract. Mathematical sciences require a direct integration of ICT into teaching. This follows from the permanent updating curricula and complementing of graphic displays. It appears that the judicious use of the Internet, various mathematical educational programs and multimedia CD is the ideal. Current communications and rapid dissemination of information should contribute to improving and streamlining the work of the performance of teachers. Introduction Informatics plays an important role in education as it develops the thinking of students, their ability to analyze and synthesize, generalize, to seek appropriate strategies for addressing the problems and verify them in practice. Leads to an accurate expression of ideas and practices, and recorded in the formal record, to serve as a general means of communication. The task of teaching science is to teach students basic concepts, procedures, and informatics resources, build a computer culture, teach effective use of information resources of civilization with respect to legal and ethical principles of the use of information technologies and products. This task should be achieved by the joint operation of objective information and applying information technology (IT) in teaching other subjects and the organization and management of schools. 1. Mathematics and work with information The aim of teaching mathematics in the 1-4 year of primary school is to prepare children for mathematical problems, which everyday life brings

50 48 Anna Kútna, Hedviga Palásthy about. Equally important is to prepare them to receive primary and secondary education in mathematics. In addition to acquiring basic knowledge and skills in the field of arithmetic, algebra and geometry, students are driving to the discovery, understanding and application of appropriate contexts both inside various areas of mathematics, but also between these areas. The solution of verbal tasks, the pupils learn the application of acquired knowledge in the common practice when buying, measurement and so on. Its content and methods of work, the teaching of mathematics helps to develop creative skills, ideas of operations, spatial imagination, the perseverance, diligence, steers student personality characteristics and the like. Suggestions verbal tasks and motivation to adopt some concepts in math and to develop abstract thinking, memory and awareness of the importance of mathematics to the practice. Mathematics develops student s thinking in particular. It teaches them the analysis and synthesis, syllable hypotheses prove and verify their accuracy practice. Bringing their work to the rational, deductive thinking, to a concise formulation of ideas. Mathematics leads students: to develop logical and critical thinking skills, development of argument and in the group in solving the problem, the use of different modes of representation of mathematical content (text, tables, graphs), to the development orientation in the plane and space, algorithmic thinking, the proper use of mathematical symbols, terminology, the independence, perseverance and tenacity. Activating teaching methods such as individual work of students contribute to obtaining budgetary habits and work in pairs and groups. Likewise, the use of suitable demonstration devices, such as geometric models of services and teaching technology in the form of interactive boards. The work is associated with a textbook and a collection of mathematical enrichment of creative activity on the Internet. 2. Information and work with information The objective of computer education is to get familiarized with the computer and the possibilities of its use in everyday life. Using the appropriate topics in other subjects (writing, mathematics, art...) is familiar with the possibilities of drawing, coaching counting, writing and other types of applications, the most representative. Informatics knowledge and skills students

51 Application of IT in teaching mathematics 49 are creating a playful form of the key tools are taught in special education programs for children (e.g. The circus clowns Thomas). With these programs, students are familiar with computers and learn to work with the mouse and keyboard. Informatics has a significant position in education, such as mathematics students develop the thinking, the ability of analysis and synthesis, generalization, seeking appropriate strategies and solutions to the problems of verification in practice. Leads to an accurate expression of ideas and practices recorded in their formal entries, which serve as a general means of communication. The aim of teaching science is making available the basic concepts and techniques used when working with data and to develop algorithms and computational processes. Just as mathematics also computer science in conjunction with IT creates a platform for all other subjects. Education leads to the students acquainted with the concepts of data and information, with different types of data and data collection, storage, viewing, processing and presentation, understanding the concept of algorithm and program. Students should understand that applications are programs that allow dealing with certain classes of problems and issues, understanding the application as a set of closely related algorithms for processing data. Introduced with the main classes of problems and issues addressed by means of information technology. Students should learn to compile algorithms, classify and address the problems presented, evaluated and tested solutions. Contents subject is divided into five logical units that do not separate thematic units. In the whole are listed subtopics, which can be seen as alternatives. Selected subtopic different units, it is necessary to connect to an integrated whole. The teacher selects them according to availability of equipment and addressing the school pupils. It is essential to use the interdisciplinary projects. Thematic units consist of: information about us, communication of information and communication technologies, procedures, problem solving, algorithmic thinking, principles of information and communication technologies, Information Society. For example, in a thematic whole procedure, problem solving and algorithmic thinking solves registration algorithm as a description of the automated handling of data by its features, the method of entry, analysis, correlation with mathematics and programming languages. Then it s an introduction to the algorithm, the solutions to geometric structures, the role

52 50 Anna Kútna, Hedviga Palásthy of logic search algorithms (procedures, guidelines) in everyday life, teaching programming languages (Karel, Baltik and Imagine, etc.), address current challenges in certain algorithms applied to the registration algorithm, features algorithms, programming language, development and analysis of simple algorithms, graphics, work with text, music, animation on the screen. Excellent complements to the teaching are taught programs. Today, there are many programs, even those for which there is no need to pay. 2.1 Imagine and Fractions As an example for educational software suitable for first elementary school level can indicate The Fractions. This program was established in 2006 as a master s thesis student in the KU PF Ružomberok. The project was developed in the programming environment Imagine. Imagine, can be used not only as a tool for teaching programming, but also to create multimedia presentations. We can create a project consisting of multiple pages and each page in addition to text, images and sound can move the animated objects. Imagine supports the work of the Internet and view pages from html files. An interesting novelty is the possibility to publish our projects on the Web, using Imagine a plug-in can run in a web browser finished projects from the network. Fractions program consists of two parts Learning and Instructor (Figure 1). Figure 1. Introduction to Fractions The educational section is divided into fragments theme chapters 5 (Figure 2). In each chapter are explained matters with the image animation explaining matters. Also, there are control issues, whether the pupil to

53 Application of IT in teaching mathematics 51 understand matters. If the answer is correct, program shows the message text Correct! and the button Below. Pupils may continue only if the correct answer (Figure 3). Otherwise, it will not show the button below. Figure 2. Theoretical part Figure 3. Example of 3rd Chapter Practical part is divided into 3 parts: a comparison of fractions, addition and subtraction of fractions. In each section are randomly generated 10 examples, where the student answers correctly, the message Correct appears

54 52 Anna Kútna, Hedviga Palásthy and displays the Continue button. If the student does not supply the correct answer, program display the correct result and the student continues. Finally to view a number of good and bad responses and school mark. This program, we would recommend for the fifth year of elementary school. 3. Conclusion The current school system allows the use of IT in the learning process. It is very closely linked to the material-technical equipment for the various primary schools. Education is a process very difficult and long, which must rest on solid foundations. It is just necessary to recall that the right knowledge, skills and habits acquired in the first years of primary school will form the basis on which to develop and shape the personality pupil (Feszterová, 2001). References [1] Baráth, O., Pomšár, Z.: Rozvíjanie kreativity v prírodovednom a technickom vzdelávaní. In: Zborník Modernizace výuky technicky orientovaných oborech a prědmětech, PF UP, Olomouc, 31 34, [2] Černek, P. a kol.: Učebnice matematiky a pracovné zošity pre ročník. Orbis Pictus Istropolitana, Bratislava, [3] Ďurič, L.: Poznávanie žiakov a rozvíjanie ich tvorivého myslenia. Slovenská pedagogická knižnica a ústav školských informácii, Bratislava, [4] Feszterová, M.: Počítačový program ako učebná pomôcka. [ ] < [5] Turek, I.: Didaktika technických predmetov. SPN, Bratislava, [6] ew&id=3929noveskolstvo.sk/upload/programy/svp_1_priloha.doc [7]

55 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 SOLVING LINEAR OPTIMIZATION PROBLEMS IN MATHEMATICS TEACHING USING ICT Stanislav Lukáč Institute of Mathematics Pavol Jozef Šafárik University in Košice Jesenná 5, Košice, Slovakia Abstract. New information and communication technologies provide means to solve various types of problems in mathematics teaching. Linear optimization problems present an appropriate topic for application of mathematical knowledge about solving systems of linear equations and inequations in solving real-life problems. Optimization problem solving is based on the construction of a mathematical model which allows to find the extremes of functions fulfilling the given conditions. In the paper there are various modelling activities focused on the construction of different kind of models and on the analysis of the possibilities to use them in solving the optimization problem described. Mathematical program Derive and spreadsheet MS Excel are used for modelling and auxiliary calculations. 1. The use of mathematical modelling to problem solving Development of students ability to solve real-life problems is an important objective of mathematics teaching. In real life, people sometimes have to make decisions that take into account specific constraints. As example, we can mention the construction of a production plan in order to achieve the maximal profit consistent with the productive capabilities. Real-life problem solving is often based on mathematical modelling. Any model presents a simplified representation of certain aspects of a real system. Alsina [1] states the following diagram (figure 1) for classical way to the modelling process.

56 54 Stanislav Lukáč Figure 1. A mathematical model is a model created using mathematical concepts such as formulas, equations, inequations, and functions. According to Edwards and Hamson [2] the essential skills for construction a proper model can be summarised as: identifying the key problem variables and making sensible simplifying assumptions, constructing relations between these variables, taking measurements and judging the size of quantities, collecting appropriate data and deciding how to use the data. An objective function expressing the dependence of investigated quantitative measure on input data is an important component of the mathematical optimization model. The extreme of objective function can be found using the mathematical optimization model and appropriate mathematical methods. Results obtained from mathematical model have to be interpreted back in the context of real-life problem. Mathematical modelling can take different form in teaching and learning. The skills needed to be successful in applying mathematics are different from common exercises. Modelling activities can be focused on creation of a model and improvement of model building and model use in order to get outcomes reasonable and compatible in terms of the information given in the original problem. Modelling process can consist of a series of steps which enables students to explore and analyze outcomes using manipulation with the input data and parameters of a model. It is necessary to realize that many different models can be used for tackling the same real problem. Therefore students might develop different simple models based on numerical and graphical representations and investigate their limited range of validity and application to the problem solving.

57 Solving linear optimization problems 55 Various tools of modern ICT can be used for realization of modelling activities in mathematics teaching. Technology can enable students to make patterns visible more readily. Different kinds of mathematical models can usually be converted into computer simulations. Mathematical program Derive offers effective tools for algebraic and graphical solution of systems of linear equations and inequtions. Spreadsheet MS Excel enables students to construct arithmetic models based on formulas expressing relationships between variables in the mathematical model. Iterative calculations can be also used for manipulation with data in arithmetic models. Different iterative methods are implemented in Excel tool called Solver, which also allows optimization problem solving. 2. Graphical representation and solution of linear optimi zation problems Modelling competencies represent important part in the system of key competencies. Even though various modelling activities only play a minor role in every-day mathematics teaching. To change this situation, it is necessary to develop interesting suggestions for integrating modelling in mathematics teaching. For an example we use a word problem describing real-world situation for introducing basic ideas of using mathematical modelling in linear optimization problem solving: Two kinds of wines are made in a winery. The red wine is sold for $5.25 per 0,75 liter bottle, the white wine for $4.00 per one liter bottle. The winery has contracted 200 hours of time on a bottling machine which can fill 200 bottles per hour. It is estimated that 14 (6,5) hours are required to produce 100 l of the red (white) wine for the work in the vineyards and cellar. Staff provides a total of 3640 working hours. How much of the two wines should produce in the winery to get maximum of revenue, given that the demand for the white is limited to at most 250 hl? Two variables can be used for mathematical representation of the given optimization problem: x - number of hl of the white wine, y - number of hl of the red wine. Values of these variables have to be nonnegative. The follow relationships hold for filling the wines to bottles: half an hour is needed to bottle one hl of the white wine and 2/3 hours to bottle one hl of the red wine. The given real situation can be described through inequations. Figure 2 shows the window Algebra of the program Derive, in which these inequations are written.

58 56 Stanislav Lukáč Figure 2. The size of revenue of wine sale is characterized through the objective function f(x,y), which is written in row labelled #6. We use 2D-plot window for graphical solution of the system of inequations. Figure 3. Coordinates of points of a highlighted pentagon in figure 3 fulfill the built-up inequations. For finding the point with a maximum value of the objective function students could add step by step to the figure graphs of objective function for concrete values of revenue. The parallel lines in figure 3 represent values: $50000, $100000, and $ The objective function obtains the maximum value in point M[140, 195]. The sale of 140 hl of the white wine and 195 hl of the red wine returns maximum revenue $ Integer solution of optimization problems After using the graphical representation for solution of the previous problem, students could consider critically limitations of efficiency of this approach to the optimization problem solving. One limitation relates to the number

59 Solving linear optimization problems 57 of variables which determinate the solution of the optimization problem. The graphical method provides only limited possibilities for finding integer solutions of the linear optimization problems. The following problem offers a suitable subject for these considerations. Two kinds of sport boats are made in a workroom. One canoe is made of 55 kg of polyethylene, using 30 minutes of machine time and 3 hours of manual work. The manufacture of kayak consumes 40 kg of polyethylene, 20 minutes work the machine, and 5 hours of manual fitting. The workroom is available to following 3 months 2 tons of polyethylene, 30 hours of machine time and the workers can spend 200 hours for the manual adjustment of boats. The selling price for a canoe is 650 EUR and for a kayak 1000 EUR. Create a production plan to achieve the maximum revenue from the sale of boats. We introduce variables x - number of canoes, and y - number of kayaks. Figure 4 shows the formulated inequations, the objective function, and a result obtained by means of the graphical method. The maximum value of the objective function is /31 EUR for the calculated number of canoes and kayaks. The obtained results cannot be used in real life, because it is not possible to produce 12,9 canoes and 32,258 kayaks. Methods for finding integer solutions of linear optimization problems are based on the stepwise addition of follow-up constraints that allow to eliminate non-integer solutions. Figure 4. Students could carry out additional modelling activities in a spreadsheet environment. Firstly, they can create a simple table for calculating the consumption of polyetylen, machine time, time of manual work, and the revenue for entered numbers of canoes and kayaks. The created table is displayed in figure 5.

60 58 Stanislav Lukáč Figure 5. Summary data are calculated by means of the function SUMPROD- UCT. The formula for calculation of total revenue in cell F5 is displayed in the Formula bar. The given constraints are written in the first row of the table. Students can investigate values of the objective function through a manipulation with input data. The simplest possibility 12 canoes a 32 kayaks does not need to be an integer solution of the optimization problem. This arithmetic model enables students simply data manipulation and evaluation of the fulfilment of the given constraints. It may happen that the experimentation with the numbers of canoes and kayaks does not lead students to resolution of the optimization problem. Therefore, it would be applicable to check systematically all possibilities using a new table. The arithmetic model based on the quantitative relationships between variables in the mathematical model can be used for systematic checking of possibilities for the boat production. Excel formulas represent another form of modelling based on mathematical relations. Arithmetic models can be constructed in the form of tables and they can provide the basis for more sophisticated mathematical models used in algebra. Students should be able to describe real situation through an arithmetic scheme based on the basic operations. The number of canoes is gradually increasing in the first column of the table in figure 6. Figure 6.

61 Solving linear optimization problems 59 The maximum numbers of produced kayaks for entered number of canoes are calculated in columns B, C, D according to the given constraints. For example, the formula in cell B3 for calculating the maximum number of produced kayaks from polyetylen, which remained after production of canoes, is displayed in the Formula bar. The maximum number of produced kayaks in column E for concrete number of canoes is calculated as the minimum value in columns B, C, D. By means of the table in figure 6, students can find out that the maximum revenue EUR is reached by selling 10 canoes and 34 kayaks. The checking of obtained results offers other opportunities for the use of ICT in mathematics teaching. Looking back at the table displayed in figure 5. The mathematical methods implemented in Excel tool Solver can be used for finding the maximum revenue from the sale of boats. Students can enter values 1 into cells B3 and B4 which will be used as starting values for the iterative calculation. The formula in cell F5 for calculating the revenue from the sale of boats can be set to the maximum value by means of Solver. The dialogue window with the corresponding settings is displayed in figure 7. Figure 7. Solver will change integer values in cells B3 and B4 in order to calculate the maximum revenue in cell F5 that takes into account the given constraints. After finishing of the iterative calculation, values 10 and 34 are written in cells B3 and B4. 4. Conclusion Simple linear optimization problems offer the opportunity of making connections between the mathematics incorporated in real-life situations and school mathematics. The use of modelling for their solution promotes the development of key mathematical competencies.

62 60 Stanislav Lukáč References [1] C. Alsina. Less chalk, less words, less symbols... more objects, more context, more actions. In Modelling and Applications in Mathematics Education. Springer, 35 44, [2] D. Edwards, M. Hamson. Guide to Mathematical Modelling. Palgrave, 2001.

63 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 MIXED DISCRETE-CONTINUOUS RANDOM VARIABLE Maciej Major Pedagogical University Institute of Mathematics Podchor ażych Kraków mmajor@up.krakow.pl Abstract. The categories of discrete and continuous random variables do not exhaust the possible types of random variables. The main objective of this article is to provide information about a new category of random variable called mixed discrete-continuous which exhibits the characteristics of a discrete random variable for some events and the characteristics of a continuous random variable for other events. Introduction One of more important concepts of the probability calculus is a random variable that finds applications in many fields like mathematical statistics, economics, theory of insurance, physics and even psychology. Most often the discrete random variables and continuous random variables are considered. There exist also other random variables which distribution functions are neither continuous nor stepwise. These are so called mixed random variables. Variables of this type are little-known but at the same time they are finding application in economics and theory of insurance. The article is devoted to the certain conception of introducing random variables with the use of elementary mathematical means. 1. Basic definitions To begin with, we will quote definitions and theorems which are essential for the more distant part of this work.

64 62 Maciej Major Definition 1. Let (Ω,Z,P) be any probability space. A random variable in this probability space is defined as any function X from the set Ω i R, that satisfies the condition: {ω Ω : X(ω) < x} Z for any x R. (1) Theorem 1. If X is a random variable in probability space (Ω,Z,P) and B is the set of Borel subsets of a straight line and P X is function defined by formula: P X (A) = P({ω Ω : X(ω) A}) for any A B, (2) then the triple (R,B,P X ) is also the probability space. Definition 2. Let X be a random variable in probability space (Ω,Z,P). The function P X defined by formula (2) on the set of Borel subsets of a straight line is called the probability generated on straight line by the random variable X or the distribution of the random variable X and the triple (R,B,P X ) is called the probability space generated on straight line by the random variable X. Definition 3. Function F X defined on R by the formula F X (x) = P X ((,x)) for any x R is called the cumulative distribution function (also cumulative density function or briefly distribution function) of random variable X. Theorem 2. If F X is the distribution function of random variable X, then (i) F X is monotone increasig function, (ii) F X is the left-side continuous function, (iii) lim x F X (x) = 1 and lim x F X (x) = 0. Theorem 3. Random variable X defined on any probability space (Ω,Z,P) is called discrete if in the set of its values X(Ω) there is a subset Ω X of following properties: Ω X ℵ 0, (3) xj Ω X P(X = x j) > 0, (4) x j Ω X P(X = x j ) = 1. (5) The smallest, in the meaning of inclusion, set satisfying the conditions (3), (4) and (5) will be denoted by the symbol Ω X. Definition 4. The function p X defined on the set Ω X by the formula p X (x j ) = P({ω Ω: X(ω) = x j }) for x j Ω X (6) is called the discrete distribution of the random variable X.

65 Mixed discrete-continuous random variable 63 Theorem 4. Function P X defined on B by the formula P X (A) = p X (x j ) for A B x j A is in the meaning of the definition (2) the distribution of the random variable X. Definition 5. The random variable X, for which there exist a nonnegative and integralable function f X defined on R, satisfying the condition F X (x) = x f X (t)dt, is called continuous random variable and its distribution P X is called continuous distribution. The function f X is called the density of the random variable X or the density of distribution P X. Theorem 5. If X is the discrete random variable, then F X (x) = p X (x j ). x j Ω X x j <x 2. Mixed random variable Let (Ω,Z,P) be any probability space and let X, Y be random variables in this space having distribution functions F X and F Y. F αxβy (x) = αf X (x)+βf Y (x). (7) Let α > 0, β > 0 i α+β = 1. Let us define the function F αxβy : R R by the following formula F αxβy (x) = αf X (x)+βf Y (x). (8) It is easy to check that the function F αxβy is meeting conditions (i), (ii), (iii) of the theorem 3, therefore it is a distribution function of a certain random variable. Constants α and β are called the shares of functions F X and F Y in the distribution function F αxβy. Function F αxβy is called the distribution function of the mixed random variables X and Y of distribution functions F X and F Y with shares α and β respectively.

66 64 Maciej Major Let us assume that X and Y are the continuous random variables of densities f X and f Y. In this situation the density f αxβy of the mixed random variables is expressed by the formula f αxβy (x) = αf X (x)+βf Y (x). (9) Let us notice that the distribution function of the mixture of two continuous random variables is a continuous function, so the mixture of two continuous random variables is a continuous random variable. Example 1. The picture below presents the density graphs of the random variables: X: N(0,1) and Y : N(4, 1 2 ) and the density graph of the mixture of the random variables X and Y at shares α = 2 3 and β = 1 3. y f X (x)= 1 e x2 2 2π 2 f Y (x)= π e (x 4) f 23 X 1 3 Y (x)=2 3 f X (x)+1 3 f Y (x) Figure 1. x Let us assume that the random variables X and Y are the discrete random variables. We are going to determine the distribution function p αxβy of the mixture of two discrete random variables of distributions p X and p Y. Let us consider the following cases: 1). If Ω X = Ω Y, then for x j Ω X. p αxβy (x j ) = αp X (x j )+βp Y (x j ) (10) 2). If Ω X Ω Y, then at least one of the sets Ω X \Ω Y and Ω Y \Ω X is not an empty set. In this situation αp X (x j )+βp Y (x j ) for x j Ω X Ω Y, p αxβy (x j ) = αp X (x j ) for x j Ω X \Ω Y, (11) βp Y (x j ) for x j Ω Y \Ω X. Example 2. Let us consider the discrete random variables X and Y of following distributions x j p X (x j ) y j p Y (y j )

67 Mixed discrete-continuous random variable 65 Let us notice that Ω X = Ω Y, so the distribution of the mixture p1 2 X 1 2 Y of two random variables X and Y at shares equal to 1 2 is presented in the below table x i p1 2 X 1 Y(x 1 j) 2 8 Example 3. Let X and Y be discret random variables of the following distributions x j p X (x j ) y j p Y (y j ) 1 3 Let us notice that Ω X Ω Y, so the distribution of the mixture p1 2 X 1 2 Y of two random variables X and Y at shares equal 1 4 and 3 4 should be calculated according to the formula (10). We get x j p1 4 X 3 4 Y(x j) 1 12 In previous discussions the random variables X and Y were of the same type (both discrete or both continuous). Let us consider now the situation when random variable X is discrete and random variable Y is continuous. We have F αxβy (x) = αf X (x)+βf Y (x), (12) where F X is the distribution function of the discrete random variable and F Y of the continuous random variable. It results from the formula (12) and from the distribution function properties that the distribution function of the mixture of random variables X and Y has a finite or countable number of discontinuity points, in addition the sum of jumps at these points is equal to α (and so it s smaller than 1). Let F Z be a distribution function of random variable Z, which is neither discrete nor continuous and satisfies the condition: m L {x R: ε>0 F Z (x+ε) F Z (x ε) > 0} > 0 (13) where m L denotes the Lebesque measure. Let us denote by S the set of discontinuity points of the function F Z. The set S may be only countable because not-decreasing function R R may have only countable number of discontinuity points. (see 4, p. 18). Let ( ) α = x j S lim x x + j F Z (x) F Z (x j ) (14)

68 66 Maciej Major Let us define the function: p X : S R as below: ( p X (x j ) = 1 α lim x x + j F Z (x) F Z (x j ) ) (15) for x j S. It is possible to check that the function p X is the distribution of the discrete random variable X. Let us notice that the function F X : R R given by the formula: F X (x) = p X (x j ) (16) x j S,x j <x is a distribution function of a discrete random variable X. Let F Y (x) = F Z(x) αf X (x) 1 α (17) for x R. Function F Y is a distribution function of a certain continuous random variable Y. From the reasoning carried out it results that it is possible to represent the distribution function of a random variable which is not discrete or continuous and satisfies the condition (13) as a mixture of discrete random function X and continuous random function Y of distribution functions F X and F Y and shares α and β = 1 α. Example 4. In the work [6] the random variable T is being considered (as time of anticipation of person B by person A), defined in the geometrical probability space (Ω,Z,P), where Ω = (0,1) (0,1), Z is the family of subsets of the set Ω having the Lebesqua measure m L, and P is the function defined on Z by the formula P(A) = m L (A). Function T has the following form: { y x dla x < y, T(x,y) = 0 dla x y. It may be checked that the function T is the random variable of the distribution function (see fig. 3) 0 dla x 0, F T (x) = 1+2x x 2 2 dla x (0,1], 1 for x > 1.

69 Mixed discrete-continuous random variable 67 y F T Figure 2. x The random variable T is not the continuous random variable, it isn t also the discrete random variable. Let us notice that the function has only one discontinuity point, S = {0}. Let us determine the share α. We have α = lim x 0 +F T(x) F T (0) = = 1 2. We will determine distribution and distribution function of discrete random variable X. We get p X (0) = 1, F X (x) = { 0 for x 0, 1 for x > 0. Now we determine distribution function and density function of the continuous random variable Y. We get F Y (x) = F T(x) 1 2 F X(x) Finally we get and 0 for x 0, F Y (x) = 2x x 2 for x (0,1], 1 for x > 1. f Y (x) = { 0 for x 0 x > 1, 2 2x for x (0,1]. y F X x Figure 3.

70 68 Maciej Major 1 y F Y 1 Figure 4. x Conclusion It was shown in this article how an advanced term of mixed discretecontinuous random variable may be implemented with the use of elementary mathematical apparatus. The conception suggested above can be used when introducing students to the discussed notion in educational process. References [1] Borowkow, A. A. Rachunek Prawdopodobieństwa. Państwowe Wydawnictwo Naukowe, Warszawa, [2] Krysicki, W., Bartos, J., Dyczka, W., Królikowska, K., Wasilewski, M. Rachunek prawdopodobieństwa i statystyka matematyczna w zadaniach, Rachunek Prawdopodobieństwa 1. Wydawnictwo Naukowe PWN, Warszawa, [3] Krzyśko, M. Wykłady z teorii prawdopodobieństwa. Wydawnictwo Naukowo-Techniczne, Warszawa, [4] Łojasiewicz, S. Wstęp do teorii funkcji rzeczywistych. PWN Warszawa, [5] Mittelhamer, R. Mathematical statistics for economics and business. Springer, [6] Płocki, A. Stochastyka dla nauczyciela. Rachunek prawdopodobieństwa, kombinatoryka i statystyka matematyczna jako matematyka in statu nascendi. Wydawnictwo Naukowe NOVUM, Płock, 2005.

71 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 FROM RESEARCH ON STUDENT DIFFICULTIES IN USING THE PROPERTIES OF FUNCTIONS WHILE SOLVING EQUATIONS AND INEQUALITIES Joanna Major a, Zbigniew Powązka b a Pedagogical University Institute of Mathematics Podchorążych Kraków jmajor@up.krakow.pl b Pedagogical University Institute of Mathematics Podchorążych Kraków powazka@up.krakow.pl Abstract. Solving equations and inequalities containing a root function, rational function or the absolute value is connected with using particular algorithms. However, there are cases in which it is sufficient to arrive at the solution in an easier way, using the properties of the functions occurring in an equation or inequality. The research presented indicates that students did not notice such possibilities or considered them incorrect. As can be observed during various mathematical classes, students of mathematics have difficulties with operative use of mathematical notions. It is particularly noticeable when taking into consideration solving certain task types which contain e.g. equations and inequalities with functions: absolute value, root function, rational function, etc. This fact is confirmed by the results of the research carried out by D. Ciesielska, J. Czaplińska, Z. Powązka (2004), M. Ciosek, B. Pawlik (1998), J. Czaplińska (2003a, 2003b), J. Major, (2006a, 2006b), Z. Powązka (2004a, 2004b, 2006a, 2006b, 2008). Students working on mathematical tasks often approach them in a very schematic way, e.g. they consider cases resulting from applying a definition. Research results indicate that a considerable part of the examined continues to use this strategy even though it very often proves unsuccessful. (see Major, 2006a).

72 70 Joanna Major, Zbigniew Powązka For example, students solving an inequality f (x) (x), (1) where f : D 1 R, g : D 2 R, D 1,D 2 R and D 1 D 2 transform it into a form, f(x) g 2 (x), (2) which in most cases is not equivalent with the inequality (1). In the example (1) let us assume that f(x) = x 2, g(x) = x. In this case the solution to the inequality (1) is x 2, and the solution to the inequality (2) is the set of real numbers. It is possible that the students were forgetting that the necessary condition for the existence of solutions of inequality (1) is D 1 D 2 f(d 1 ) R + {0}. The inequality (1) is a specific case of the inequality h(f(x)) g(x), (3) where f : D 1 R, g : D 2 R, h : D 3 R, D 1,D 2,D 3 R and D 1 D 2 and D 3 f(d 1 ). When the function h has an inverse function and the composition h 1 g is possible that the inequality (3) is transformed into f (x) h 1 (g(x)) if the h function increases, or into the form f (x) h 1 (g(x)) if the function decreases. However, the received solutions of these inequalities require verification with the domains of the functions occurring in these inequalities. The situation proves more complicated when the h function in the inequality (3) does not have the inverse function, e.g. h(x) = x. Many of the surveyed asked to solve equations or inequalities of the (3) type were not able to abandon the algorithm, which consequently resulted in them making mistakes as illustrated in the two tasks below. Task 1. Solve the equation 2x + 1 = x. It should be pointed out that in order to solve the equation 2x+1 = x it is sufficient to notice that x 0, which implies 2x Therefore, using the definition of the absolute value of a real number, the equation in question could be expressed as 2x+1 = x, which means x = 1. It counters the assumption of nonnegativity of x. Therefore, the equation discussed has no solution. Task 2. Solve the inequality 3 x > x+1. There are three strategies which can be used in order to resolve this task:

73 From research on student difficulties 71 S 1 : if we assume that x 0 we can transform the inequality to the form 3 x x 1 > 0, and after we have found the common denominator the received inequality can take the form x(3 x 2 x) > 0. S 2 : if we assume that x 0 we can multiply the inequality by x 2 and transform the inequality received into the form x(3 x 2 x) > 0, S 3 : if we assume that x 0 we can consider two cases x > 0 or x < 0; then, having multiplied by x in both cases, we can resolve a disjunction of conditions: 3 x 2 x > 0 and x > 0 or 3 x 2 x < 0 and x < 0. The S 1 and S 2 strategies directly lead to considering the conjunction of conditions: x(3 x 2 x) > 0 and x 0, whereas when using the S 3 strategy a disjunction of two conjunctions should be resolved: 3 x 2 x > 0 and x > 0 or 3 x 2 x < 0 and x > 0. Finally, the solution of the inequality from Task 2 is contained in the set ( ) ( )., , The results of the research conducted by J. Major show that most of students working on the task 1 were not able to abandon the algorithm method of solving a problem. The surveyed considered cases which were implied by the range of applicability of particular formulas (definition of the absolute value). Furthermore, resolving the tasks by means of the notion method presented above was recognised as incorrect by a large group of the examined (Major, 2006a). Similar difficulties could be observed in solving Task 2. In order to discover whether the reform in the mathematics curriculum brought improvement in secondary-school students skills, they were asked to solve a few tasks, containing the two tasks discussed above. The survey was carried out at the Pedagogical Academia (the Pedagogical University at present) in the academic year 2006/2007. The group subject to the examination consisted of first-year students of mathematics. The objective of the research was to find out how students cope with tasks that are uncomplicated in terms of arithmetic, however require application of necessary definitions and theorems. The group examined consisted of 119 people divided into four groups for the purpose of result analysis. Group 1 included 37 secondary school leavers from classes with mathematical or mathematics and physics profiled classes,

74 72 Joanna Major, Zbigniew Powązka group 2-56 persons from mathematics and information technology profiled classes, group 3-15 people from classes with a general profile, and group 4-11 persons from classes with a humanistic profile or leavers of economic vocational secondary schools. The results of the survey showed that only 11 people (about 9% of the surveyed, 4 persons from group 1, 5 from group 2 and 2 from the remaining groups) resolved Task 1 properly. Interestingly, 8 persons (almost 7% of the examined) on considering correctly the cases: 2x +1 0 or 2x+1 < 0, rejected only one of the two negative solutions. It might be a form of degenerate formalism in applying the definition of the absolute value (Turnau, 1990). Although a great majority of the surveyed (79 persons that is 66% of all students 27 from group 1, 37 from group 2, 8 from group 3 and 6 from group 4) considered the cases 2x+1 > 0 or 2x+1 < 0 properly and resolved the appropriate equations, they did not verify the received solutions using the assumptions made. What needs to be pointed out is that a few mistakes resulted from considering the cases x 0 or x < 0, which can be recognised as a symptom of degenerated formalism mentioned earlier. On analysing the students works it was observed that none of the examined applied the notion method described in this paper. Summing up the analysis of task 1 it is possible to formulate a hypothesis that most of the examined knew the procedures connected with solving equations and inequalities containing the absolute value. Only 9 of the surveyed (7,5%) did not attempt to resolve Task students (about 94%) attempted to resolve Task 2. Simultaneously, only about 22% of the examined (i.e. 25 persons 11 from group 1, 12 from group 2 and 2 from the remaining groups) gave a proper solution to this task, most frequently using the S 1 strategy. It was chosen by 61 persons (23 from group 1, 26 from group 2, 8 from group 3 and 4 from group 4). It might seem that solving the task by means of this method should not cause any arithmetical difficulties as it leads to solving an easy rational inequality. However, the majority of the surveyed who tried to resolve the task using the S 1 strategy finished their work with putting down the inequality or calculating the zero points of the appearing square function. What might have deterred the students from continuing their work on the task was the fact that the solutions of the inequality were irrational numbers. Strategy S 2 was chosen by 18 persons (3 from group 1, 11 from group 2 and 2 from the remaining groups). In this strategy the inequality was transformed to x(3 x 2 x) > 0 as in the S 1 strategy. The biggest number of mistakes appeared on applying the S 3 strategy. This method was chosen by 33 students (8 persons from group 1, 17 from

75 From research on student difficulties 73 group 2, 3 from group 3 and 5 from group 4). One person only considered the necessary cases and started to solve the inequality using solely the assumption x > 0. The remaining students multiplied the inequality by x disregarding the sign of this variable. This procedure could indicate that the examined gained little experience in dealing with rational inequalities in the secondary school and they approached them as an equation associated with a relevant inequality. What confirms this hypothesis is the fact that 22 of the surveyed quitted their calculations with properly assigned roots of the equation. Presented research shows that a lot of students beginning their mathematical studies have problems with operative use of mathematical notions. Even if they know basic definitions and theorems of school mathematics they fail to use them in practice. Students restrict themselves to executing arithmetical procedures but they do not consider requisite assumptions. Moreover, the surveyed are not accustomed to verifying their results, e.g. using function diagrams which appear in discussed tasks. Furthermore, it has to be noticed that students fail to reason in a conceptual manner (Krygowska 1977, Major 2006a). Analysis of task solutions given by different groups surveyed shows that the observed difficulties are independent from curriculums. Their origin might lie in the manner of mathematics teaching at different levels of education. References [1] Ciesielska D., Czaplińska J., Powązka Z.: 2004, Z badań nad egzaminami wstępnymi na studia dzienne na kierunek matematyka w Akademii Pedagogicznej w Krakowie, Roczniki PTM, seria V, Dydaktyka Matematyki 26, [2] Ciosek M., Pawlik, B.: 1998, O trudnościach studentów I roku matematyki w uczeniu się matematyki w świetle analizy ich rozwiązań zadań z geometrii, Roczniki PTM, seria V, Dydaktyka Matematyki 20, [3] Czaplińska J.: 2003a, Przyczynek do badań nad strategiami rozwiązywania zadań matematycznych, Roczniki PTM, seria V, Dydaktyka Matematyki 25, [4] Czaplińska J.: 2003b, Trudności do stosowania pojęć analitycznych przez absolwentów szkółśrednich podczas rozwiązywania zadania egzaminacyjnego, Disputationes Scientificae, Universitatis Catholicae in Ružomberok, Ružomberok, Katolícka Univerzita roč. III, č. 3.

76 74 Joanna Major, Zbigniew Powązka [5] Krygowska Z.: 1977, Zarys dydaktyki matematyki, część 3, WSiP, Warszawa. [6] Major J.: 2006a, Rola zadań i problemów w kształtowaniu pojęć matematycznych na przykładzie bezwzględnej wartości liczby rzeczywistej Roczniki PTM, seria V, Dydaktyka Matematyki 29, [7] Major J.: 2006b, Uwagi dotyczące obrazu pojęcia wartości bezwzględnej liczby rzeczywistej u studentów matematyki, w: Kształcenie matematyczne tendencje, badania, propozycje dydaktyczne, red. Czajkowska M., Treliński G., Wydawnictwo Akademii Świętokrzyskiej, Kielce, [8] Major J., Powązka Z.: 2006, Uwagi dotyczące wartości bezwzględnej liczby rzeczywistej, Annales Academiae Pedagogice Cracoviensis 36, Studia Ad Didacticam Mathematicae Pertinentia I, [9] Powązka Z.: 2004a, Z badań nad rozumieniem treści zadania matematycznego, Roczniki Polskiego Towarzystwa Matematycznego, Seria V, Dydaktyka Matematyki 26, [10] Powązka Z.: 2004b, O kształtowaniu podstawowych pojęć analizy matematycznej na pierwszym roku studiów AP w Krakowie (Conference proceedings XVIII). [11] Powązka Z.: 2006a, O pewnym sposobie opisu procesu rozwiązania zadania, Matematika v škole dnes a zajtra, Zbornik 6. ročnika konferencie s medzinárodnou účast ou, Ružomberok, [12] Powązka Z.: 2006b, Z badań nad wprowadzaniem podstawowych treści analizy matematycznej podczas zajęć na I roku studiów matematycznych, Annales Academiae Pedagogice Cracoviensis Studia Ad Didacticam Mathematicae Pertinentia I, [13] Powązka Z.: 2008, O fałszywych przekonaniach obserwowanych na zajęciach z analizy matematycznej, Academiae Pedagogice Cracoviensis Studia Ad Didacticam Mathematicae Pertinentia II, [14] Siwek, H.: 2005, Dydaktyka matematyki, Teoria i zastosowania w matematyce szkolnej, WSiP, Warszawa. [15] Turnau, S.: 1990, Wykłady o nauczaniu matematyki, PWN, Warszawa.

77 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 SOLVING OF WORD PROBLEMS BY STUDENTS IN AN ENVIRONMENT OF THE RECORDING SOFTWARE LAMPÁŠIK Peter Molnár Institute of Mathematics Faculty of Science, P. J. Šafárik University Jesenná 5, Košice, Slovakia Abstract. In this contribution we present the software Lampášik used for recording of the whole process of solving a word problem by student. Additionally this software offers very valuable information about it, how many times, when (in the process of solving) and how a student read particular parts of an assignment of a word problem. On a concrete example we illustrate the advantages of analyzing a solution of a word problem obtained in such a way. 1. Motivation Anybody, who ever tried to analyze (to correct, to assess, or even to evaluate) student s solution of a word problem knows, how many difficulties it brings. Difficulties arise mostly, when the solution is incomplete, incorrect or incomprehensible (the worst possible case). It is caused by the fact that we simply do not have enough information about the whole process of solving. From the solution written on the paper we can not recognize: - what was lined through, - in what order single parts of the solution were written, - how much time student spent trying to solve the word problem, - etc. Problems of that nature could be solved so, that the student would solve the word problem on a computer and some recording software would record the whole process of solving (as a movie). But this would be only a particular solution of the problem. The main obstruction to successfully solving word problems is the inability of students to understand the problem (the assignment of the problem),

78 76 Peter Molnár to understand the situation described by the problem, or the challenge (the question) given by the problem. [1] Therefore it would be advisable to find out, how the student handled the assignment of a word problem, while trying to solve it. We need to know the answers at least to the following questions: - How many times did the student read the assignment of a problem before he started to solve it? (How many times did he read it globally?) - In what kind of way did the student read single parts of the assignment? - Did he stop by any part of the text? - Did the student, in the process of solving, return back to the assignment of a problem? (Which parts did he return to?) - How many times did he read the question? Software, that would only record the solution of a word problem, can not give us answer to any of these questions (of course, that solution written on the paper can not neither). 2. Software Lampášik For the purpose of detailed analysis of student s word problem solution specialized recording software Lampášik was developed. The author of this software is Veronika Vaneková the graduate of our faculty (see [4]). This software offers, about the whole process of solving a word problem by student, much more information as we could gain from the solution written on the paper or by common recording software (that would only record the solution of a word problem). Functionality of the program The assignment of a word problem is obscured. The student can illume at once only a part of it, by moving a mouse pointer over this part. The program memorizes each mouse move and saves it to a file. Also every change (written or deleted letter) in a solution, made by the student, is recorded. [4] All collected data can be replayed as a movie. Environment sample

79 Solving of word problems by students 77 Thin lines in the obscured area match with the placement and the length of words in the assignment of a word problem. Research with using the software Lampášik In the end of April 2009 we performed a research aimed at the ability of students to understand an assignment of a word problem and their handling of an assignment, while solving a word problem. Twenty students, of the first class of a grammar school, had to solve seven specially chosen word problems. They solved these word problems in the environment of the software Lampášik. For a better understanding of using this software, we present now one of given problems and describe one its obtained solution. 3. Commentary to the mentioned word problem Word problem Mummy asked Annie to go shopping. You will find the money in the kitchen on the table in an envelope she told her. Annie took the envelope and noticed that there is the number 98 written on it. She took the money, put them without counting into the hand-bag and threw away the envelope.

80 78 Peter Molnár In the shop she was asked to pay 90 crowns. When she was going to pay she found out that no 8 crowns are left, but still some few crowns are missing to pay for the commodities. She left the commodities in the shop and run to mummy, if she was not mistaken, when she was counting the money. Mummy told her that she has counted the money correctly and rebuked Annie that she has not paid attention, when she ordered her how much she should buy. What was Annie mistaken at? [3] Model solution Annie made two mistakes. The first one was that she held the envelope upside-down and so she saw the number 98, instead of the number 86. The second one was that she asked in the shop for more, as mummy ordered her. Purpose of enlistment of this word problem By studying (watching) solutions we primarily tried to find out, how the process of monitoring (by students themselves) while reading an assignment of a word problem works. Mainly we were interested in how a student reacts, when he just finds out that he has not understood the perused text, or some of its parts. Secondly we tried to find out, how the process of selecting information (by students) from an assignment works. The selection of information is a volition of information from a text according to some chosen criteria. The criteria reflects a certain value. [2] In the assignment of this word problem some information occur, that are redundant for the whole understanding and successful solving of this problem, e. g.: - the name of the daughter (Annie), - the placement of the envelope (exactly) in the kitchen, - the fact that Annie put the money (exactly) into the hand-bag. In opposite to this, each of the following information is relevant (significant) for the correct understanding and solving of this problem: - Annie took the envelope from the table (she could access it from any side), - she noticed that there is the number 98 written on it (it needed not really to be there, 86 in the view from the opposite side of the table), - Annie did not count the money and threw away the envelope (she could not check the number written on the envelope anymore), - she had not enough money to pay 90 crowns (in the envelope were less than 90 crowns),

81 Solving of word problems by students 79 - she expected that 8 crowns will be left (she thought that in the envelope were 98 crowns), - some few crowns are missing (the number 86 satisfies this condition), - mummy counted the money correctly (mummy gave her the right amount of crowns), - mummy rebuked Annie that she has not paid attention, when she ordered her how much she should buy (she did not send Annie to buy commodities for 90 crowns). 4. Description of the obtained (recorded) solution The student read (by uniform movement) the entire assignment of the word problem. He did not stop by any part of the text. Right after this he began to read the assignment from the beginning. He read twice how much should Annie pay in the shop. After reading the question 15 seconds of silence (the student did not read any part of the text, also he did not write anything) followed. This time he read only the first half of the assignment. He evidently slowed down by numeric data. Another 30 seconds of silence followed. The student began to read the assignment from the beginning. He stopped by the crucial sentence. The word noticed (this is the one of the most important words in the whole assignment) was read very carefully by him. Right after this 5 seconds of examination of the number 98 followed. The student quit reading of the assignment at the end of its first half and wrote the following answer: In my opinion, Annie was holding the envelope upside-down, so 86 and not 98 was written on the envelope. Commentary The student read at first the entire assignment two times. After a short break he read the first half of the assignment. Another break followed and the student read again only the first half. The first half of the assignment was read by him (totally) four times, the second one only two times. His answer did not explain the second part of the story described by the assignment. Source of student s mistake The mistake probably occurred by selecting information from the assignment. It occurred concretely already by choosing a proper criteria. The student considered only those parts of the text to be important, where numeric data were located. All numbers were located in the first half of the assignment.

82 80 Peter Molnár Another possible explanation is that the student, already after the first reading of the assignment, (in his mind) reformulated the question What was Annie mistaken at? to Why did not have Annie enough money, so that she could pay 90 crowns? In that case, he had no reason to verify the real price of commodities that was Annie sent for. 5. Statistical map An important feature of this program is that it can create the statistical map of text reading the assignment of a word problem with its grayed out background is visualized. The more often (and longer) the text was illumed, the lighter (brighter) is the colour in the background of the text. This statistical map confirms the fact that the student spent more time by reading the first half of the assignment (this half is much lighter), as by reading the second half. In the last three lines the colour shades do not change so markedly as by the remaining parts of the text. They were probably read also less thoroughly. The numeric data belong to the brightest parts of the map. To the darkest parts belongs the collocation how much ( koľko in the presented Slovak solution). It corresponds with the fact that the student did not explain the second part of the story. This collocation plays an important role in the entire assignment of our word problem. It points out that Annie asked in the shop for more, as mummy ordered her. 6. Another advantages and possibilities of using this software 1. The possibility to create the statistical map for more students together. This can be efficiently used by an analysis a priori of some word problem. The map will point at the crucial and problematic parts of an assignment. It makes also possible to study, where the basic differences between the statistical maps of successful and unsuccessful solutionists (of some word problem) are. 2. The software as a diagnostic tool (by problems with solving of word problems).

83 Solving of word problems by students Software utilization by an explanation of incomprehensible solutions. In our opinion, this software can be especially helpful by an analysis and an assessment of student s word problem solution. References [1] Hejný, M.: Zmocňování se slovní úlohy. Pedagogika 4, , [2] Matulka, Ž.: Selekcia i synteza informacji w procesie samokstalcenia. Warszawa: PWN, [3] Novoveský, Š. Križalkovič, K. Lečko, I.: 777 matematických zábaviek a hračiek. Bratislava: Slovenské pedagogické nakladateľstvo, [4] Vaneková, V.: Atomárna analýza riešenia slovných úloh. Softwarový projekt PF UPJŠ, Košice, 2004.

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85 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 GEOMETRICAL PROBABILITY IN MATHEMATICS FOR EVERYONE Zofia Muzyczka a, Adam Płocki b a Institute of Pedagogy State Higher Vocational School in Nowy Sącz ul. Chruślicka 6, Nowy Sącz, Poland zmuzyczka@pwsz-ns.edu.pl b Institute of Mathematics Pedagogical University of Cracow ul. Podchorążych 2, Kraków, Poland adplocki@ap.krakow.pl Abstract. The article is about the geometrical aspect of probability in teaching mathematics. It contains some intuitions connected to geometrical probability and errors that occur while introducing it to school textbooks. 1. Geometrical probabilistic space - geometrical probability Let m L mean the Lebesgue s measure on a plain and F L the family of those subsets of the R 2 plain, that have the measure of m L. Let Ω F L and m L (Ω) > 0. Let us assume, that Z is the family of those subsets of the Ω set that have the measure of m L and P is a function from the Z to the R set described by the following: P(A) = m L(A) m L (Ω) for every A Z. It is easy to prove that a (Ω,Z,P) three is a probabilistic space in a sense of the axiom Kolmogorov s definition (see [7], p. 124 and p. 265 see also [6], p. 196). We call this three a geometrical probabilistic space. The elements of the Z family are the events and the P function is the probability in this space. We call this function the geometrical probability. If the A set of the family of Z is a shape having a Jordan s measure (area), it is also the m L measure of this set. Rating the geometrical probability is usually the same as stating the area of two geometrical shapes: Ω and A. Because of this the geometrical probabilistic spaces are introduced in

86 84 Zofia Muzyczka, Adam Płocki school mathematics. Stating the probability becomes an illustration of the integration rule (also known as a fusion rule, see [5], p.76), in the sense of connecting two or more branches of mathematical knowledge while solving a particular problem. At this point we come to a question: what do geometrical objects and stating probability, which in "mathematics for everyone" is understood as stating chances and risk, have in common? Probabilistic space is in this case a model of some chance experiment. Example 1. In [3], an old Polish probabilistic textbook written in the early 20th century, we can find the following: Two people, A and B, wishing to meet between t 0 and t 1 > t 0 hours, agreed that if A comes to the meeting place first, he will wait for α < t 1 t 0 hours, and if B comes first, he will wait for β < t 1 t 0 hours. What is the probability of their meeting successfully? If Ω = (0,1) (0,1), so Ω is a square on a Carthesian plain, the geometrical probabilistic space of (Ω,Z,P) is understood as a model of a chance meeting of A and B if t 0 = 0 and t 1 = 1. If α = 1 4 and β = 1 3, the event A = {the meeting takes place} becomes the shaded shape in the fig. 1, and the probability of A is the area of this shape. The event B = {both A and B come to the meeting place at the same time} is the diagonal of the Ω square, so P(B) = 0, as a line has the area of 0. The event B becomes a counterexample proving that the implication P(B) = 0 = B = is not true (if the probability of B equals 0, the B event is impossible). y B Fig. 1 x A This geometrical space as a model of a chance meeting occurs in many textbooks as an illustration of a geometrical probability "around us". 2. Different aspects of probability The idea of probability (like the idea of a natural number) is a synthesis of different aspects:

87 Geometrical probability in mathematics for everyone classical aspect - probability as a quotient of the number of cases propitious to an event and the total number of equally probable events; 2. statistical aspect - rating the probability of an event is the frequency of the event happening in repeating an experiment many times; 3. geometrical aspect - probability as a proportion of measures like area, length, time, weight etc. This work refers to the geometrical aspect of probability in mathematics for everyone. 3. Geometrical aspect of probability - probability vs. some measures Probability of drawing a red ball from an urn U b c, which contains b white c balls and c red ones is a quotient b+c. Each ball in the urn has an equal chance of being drawn, so it is an equally probable case. Let us consider drawing a color (sector) using a roulette that has white and red sectors in this context. If we take the urn and roulettes from fig. 2, the probability of drawing red equals 1 3. a) b) c) Fig. 2 In case of the U urn probability of 1 3 is the 1 6 fraction meaning number of red balls in the urn total number of balls in the urn. In case of the roulette in fig. 2b) (all sectors are equal) the fraction 2 6 means number of red sectors total number of sectors. In case of the roulette in fig. 2c) the 1 3 probability a quotient of which is the same as angular measure of the red sector angular measure of the perigon, the length of the red sector arc the length of the perimeter of the roulette disc.

88 86 Zofia Muzyczka, Adam Płocki The roulette shows the probability as a part of some integrity in a different way than the urn does. Let us consider drawing using a roulette shown in fig. 3a). Probability of the roulette s hand stopping in a sector (a number in the sector becomes the result of drawing) is a quotient of: length of the sector arc and measure of the perigon, or angular measure of the sector and the perimeter measure a) b a 1 3 b) Fig. 3 Let us assume that the pair (Ω,p) is a model of a random experiment δ ([6], p. 44). Ω is the set of the experiment results and the p function associates a probability of each result to this result. At this point we come to a question: Is it possible for a ω result of the δ experiment that p(ω) = 0? Example 2. Let us consider a special kind of a roulette. On the circle of its disc there is some point marked out. The hand of the roulette may either stop at this point (it is a ω 1 result) or stop somewhere else (it is a ω 0 result). We can consider such an experiment only theoretically. If the (Ω,p) pair is a probabilistic model of the δ experiment, then Ω = {ω 1,ω 2 } and (which is implied by the previous remarks about the roulette) p(ω 0 ) = 1 and p(ω 1 ) = 0. Example 3. Let δ R mean drawing with a second roulette shown in fig. 3. Let us assume there are three possible results of this experience: ω a : the roulette hand stops in section a, ω b : the roulette hand stops in section b, ω c : the roulette hand stops on a line separating the sections. The results of the δ R experience form the set Ω R = {ω a,ω b,ω c }. If p R (ω) is a function which connects a probability to each result of the experience δ R, than: p R (ω a ) = 1 3, p R(ω b ) = 2 3, and p R(ω c ) = 0. The pair (Ω R,p R ) is a model of the δ R experience (it is also a singular, discrete probabilistic space, see [7], p. 17). The singularity of this experience originates from the fact, that there is one result with zero probability.

89 Geometrical probability in mathematics for everyone 87 Example 4. Imagine you have just finished winding a clock. You will not wind it again until it stops. We can ask about the probability of such situations: a) the short hand of the clock stops between half past one and four o clock (rating the probability is counting the quotient of arc length between points 1,5 and 4 on the clock face and the clock face circle length); b) the clock stops at half past five Fig. 4 In case of a) the result of the experience is a point on the circle (pointed to by the clock s short hand); in case of b) the result is a pair (x,y), where x is pointed to by the short hand of the clock and y is pointed to by the long hand. Example 5. A pastry mixed with raisins was then divided into two parts. Each of them was used to form a muffin. Before placing the muffins in the oven it turned out, that one of the raisins is not in the pastry. We come to two questions here: What can we do to place the raisin in the pastry without joining the parts, mixing and dividing again? What can we do if the pastry parts are not equal? Only fortune decides to which muffin our raisin will get. If their masses are equal, each muffin has an equal chance of gaining a raisin. We can sample the muffin for the raisin by tossing a coin. Let us assume, that in case of non-equal parts of the pastry, one was used to make a muffin and the other - to make a croissant. Watching the raisin during mixing and dividing the pastry we have two possible results: ω b - the raisin goes to the muffin, ω r - the raisin goes to the croissant. If p is a function associating probability to each result of the experiment, than: p(ω b ) = muffin weight whole pastry weight p(ω r ) = croissant weight whole pastry weight.

90 88 Zofia Muzyczka, Adam Płocki We use weight - which is some measure - for rating the probability distribution in the set of the experience results. Example 6. A driver is going to turn into a main street which has traffic lights just round the corner. The driver doesn t see the crossroads while turning. The red and green lights are on for 28 seconds each, the yellow is on for 4 seconds. What light will be on when the driver comes to the crossroads is random. So we may consider the light color as a result of a chance experience. There are three possible results: z: the light will be green, c: the light will be red, o: the light will be yellow. Probability of the driver coming to the crossroads with the green light on is in this case a quotient: the duration of the green light period the duration of the whole light color change period This chance experience can be modeled by a (Ω,p) pair, in which Ω = {z,c,o}, p(c) = p(z) = = 7 15 and p(o) = 4 60 = We made this model using some measure again, this time it was time. 4. Shooting and geometrical probability In school mathematics a (Ω, Z, P) geometrical probabilistic space is considered as a model of a δ random experience which is sampling a point on a line or plain. But we do not clear the meaning of sampling a point. What action, real of thought of, is considered here? A mathematician does not have to think about this question. He a priori assumes, that it is a geometrical probabilistic space. In teaching mathematics the procedure of sampling a point needs to be clearly defined. An insight of a circle with a midpoint of a and a radius of r we call the neighborhood of the a point. The substance of geometrical probability is that the probability of hitting the neighborhood of a point in the Ω figure depends only on the neighborhood radius, not on the point itself. We understand shooting the target wheel in the way, that the shooter targets the midpoint, but which point on the wheel will actually be shot depends on the lot. The result of this chance experience is a point that was shot (if the shooter misses, we say the experience has not succeeded).

91 Geometrical probability in mathematics for everyone 89 To determine if the probability of shooting the neighborhood of a point depends on the radius only (that is to verify the H 0 hypothesis stating that a geometrical probabilistic space is a model of a chance experience) we only need do rely on statistical data. We need to analyze the distribution of the points shot if the shooter repeats his shots many times. The distribution reveals the fact, that most of the points shot are placed near the midpoint of the target wheel. The farther from the midpoint, the less points shot. This empirical fact is the basis to lever the H 0 hypothesis. The statistical data is the base for levering the hypothesis Fig. 5 Probability of shooting a point in the j circle of the wheel is not the quotient of areas of the j circle and the whole target wheel. 5. Geometrical probability - errors and mistakes The geometrical probability is introduced to school textbooks (see [8]) and workbooks (see [4]) in the context of throwing darts at the Ω shape. For shapes shown in fig. 6 it is about the probability of hitting the shaded area. This is understood in [8] as a quotient of areas of the shaded part and the whole Ω shape. Rating the probability is resolved to computing areas, which - apparently - makes it close to the idea of functionism. But here a question occurs - what procedure (what chance experience) is this about? In case of shooting a target wheel the procedure was clear (the shooter aims at the midpoint, the chance decides which point is actually shot). What should we mean by throwing a dart? Geometrical probability was in this case introduced in the wrong context. In textbook tasks about geometrical probability we can often find a phrase "sampling a point on a plain". But it is not clear what this phrase

92 90 Zofia Muzyczka, Adam Płocki means. Similarly, we do not know what a phrase "sampling one of the group of s people". In practice the phrase may mean using one of different procedures. In some cases each person in the group has equal chance to be drawn out (for example when we use matches, see [6], p. 26), in other cases it is not so (like in using "the sailor game" - see [6], p. 153). A mathematician would a priori assume that sampling a point on an F shape means a lot experience modeled by an (Ω, Z, P) geometrical probabilistic space, the actual procedure or the action taken in the experience is of no importance to him/her. But in case of teaching mathematics the procedure is very important. a) b) c) Fig. 6 The geometrical aspect of probability should certainly be introduced into school mathematics. The geometrical probability itself however should not, as it is beyond the perceptive capability of pupils (except the capability of computing shape areas). We always rate the probability of some experience result in a particular probabilistic space. A geometrical probabilistic space is non-enumerable, and this issue is not suitable for school children use, because of their psychophysical conditions and their mathematical competence. References [1] T. Hecht, J. Kalas, Matematika pre 4. ročnik gymnázii a SOŠ, Zošit 3. Pravdepodobnostná Štatistika, Orbis Pictus Istropolitana, Bratislava [2] W. Feller, An Introduction to Probability Theory and Its Applicatiions, Vol. 1., John Wiley& Sons, New York, London, Sydney [3] Wł. Gosiewski, Zasady rachunku prawdopodobieństwa, E. Wende i S-ka, Warszawa [4] J. Kováčik, I. Scholtzová, Zbierka prikladov z matematiky pre základné a osemročné gymnáziá, IURA Edition, Bratislava 2002.

93 Geometrical probability in mathematics for everyone 91 [5] A. Płocki, Dydaktyka stochastyki, Wydawnictwo Naukowe NOVUM, Płock [6] A. Płocki, Pravdepodobnosť okolo nás, Katolícka univerzita, Ružomberok [7] A. Płocki, Stochastyka dla nauczyciela, Wydawnictwo Naukowe NOVUM, Płock [8] W. Repáš, P. Černek, Matematika pre 8. ročnik ZŠ, 2. diel, Orbis Pictus Istropolitana, Bratislava 2002.

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95 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 PROPERTIES OF GENERALIZED CHARACTERISTIC MULTI-VALUED LOGICAL FUNCTIONS Sergey Novikov Institute of Computer Science University of Podlasie Senkiewicz Str.51, Siedlce, Poland Abstract. Properties of generalized characteristic functions (v) j i = i for v = j and (v) j i = 0 for v j, v,i,j {0,1,2,...,K 1}, are studied. The algebra of such functions is a Boolean algebra for i = K 1 and it is not a Boolean algebra for i K 1. The efficient formulas for writing of system -intervals in this basis are given. 1. Introduction We will consider the basis {Min(x,y), Max(x,y),(v) j i }, where Min(x,y), Max(x,y) are well-known multi-valued logical functions and (v) j i are generalized characteristic multi-valued logical functions. Our generalized characteristic function (v) j i = i for v = j and (v) j i = 0 for v j, where v,i,j {0,1,2,...,K 1}. Generalized characteristic functions are attractive for the minimization s problem of multi-valued logical functions. The functions (v) j i are efficient tools for writing of system-intervals [1] as formulas (implicants) [2]. The Rosser s characteristic [3] function φ j (v), where φ j (v) = K 1 for v = j and φ j (v) = 0 for v j, and known [4] characteristic function j i (x), where j i (x) = 1 for x = i and j i (x) = 0 for x i, are particular cases of our function (v) j i. It is obvious φ j (v) = (v) j K 1 and j i(x) = (x) i 1. We ll designate the function (v) j K 1 by symbol vj and functions Min(x,y), Max(x,y) by symbols x y, x y for simplification of writing of formulas. 2. Rules of concerned the algebra Our functions (v) j i have interesting properties.

96 94 Sergey Novikov We have for any parameters c,v,j,i {0,1,2,...,K 1} next properties: v j 0 = 0 (1) (v j 0 )0 c = c (2) 0 v j i = 0 (3) (K 1) v j i = vj i (4) c v j i = vj i,if c i (5) c v j i = vj c,if c < i (6) 0 v j i = vj i (7) (K 1) v j i = (K 1) (8) c v j i = c,if c i (9) v j i Justice of this rules for constants follows from definitions of our function and operations N,,. Justice of rules of idempotency x p s xp s = xp s ; (10) x p s xp s = xp s ; (11) follows from definitions of our function v j i and operations,. The justice of rules of addition for i = K 1 x p N(x p ) = K 1; (12) x p N(x p ) = 0 (13) follows from definitions of our function v j = v j K 1 and operations N,,. It is not true rules of addition in our algebra for i K 1. The example of a disturbance rules of addition for i K 1 is showed on Table 1.

97 Properties of GCMVLF 95 x x 3 2 N(x 3 2 ) x3 2 N(x3 2 ) x3 2 N(x3 2 ) Table 1. To prove the commutative property, rules of absorptions and rules of de Morgan we must take at four cases a),b),c),d) of correlations between values x,p,y,q: a) x = p,y = q; b) x = p,y q; c) x p,y = q; d) x p,y q; and two cases 1), 2) of correlations between values s, t : 1) s t; 2)s > t. We ll prove justice of the commutative property x p s y q t = yq t xp s; (14) x p s y q t = yq t xp s; (15) To prove (14) we must prove in the case a) the equality s t = t s. The left and right parts of this equality are equal to t in the case 1) and they are equal to s in the case 2). In the cases b),c) and d) we have from (14) equalities s 0 = 0 s,0 t = t 0 and 0 = 0. The justice of each from them is obvious for cases 1) and 2). In a similar manner we can prove the justice of the equality (15). We ll prove justice of rules of absorptions (x p s y q t ) xp s = x p s; (16) (x p s y q t ) xp s = x p s (17) To prove (16) we must prove in the case a) the equality (s t) s = s. It is obvious that it is true for two cases 1) and 2). We have equalities (s 0) s = s,(0 t) 0 = 0,(0 0) 0 = 0 for cases b),c) and d), which equity is evident for two cases 1) and 2). So our rule (16) is proved for all possible cases. In a similar manner we must prove the justice of four equalities

98 96 Sergey Novikov (s t) s = s,(s 0) s = s,(0 t) 0 = 0,(0 0) 0 = 0 for the proof of the rule (17). The justice of each from them is obvious. So our rules of absorptions are proved for all possible cases. We ll prove justice of rules of de Morgan N(x p s y q t ) = H(xp s) N(y q t ) (18) N(x p s yq t ) = H(xp s ) N(yq t ) (19) We have the equality N(s t) = N(s) N(t) for the case a). The left and right parts of this equality are equal to (K 1) s in the case 1) and are equal to (K 1) t in the case 2). The left and right parts of this equality are equal K 1 for each case b),c),d) and for each case 1) and 2). The proof of the justice (18) is ended. To prove (19) in the case a) we must prove the equality N(s t) = N(s) N(t). If s t, N(s t) = (K 1) t and N(s) N(t) = ((K 1) s) ((K 1) t) = (K 1) t. If s > t, N(s t) = (K 1) s and N(s) N(t) = ((K 1) s) ((K 1) t) = (K 1) s. The left and right parts of the equality N(s 0) = N(s) N(0) in the case b) is equal to (K 1) s, the equality N(0 t) = N(0) N(t) in the case c) is equal to (K 1) t, the equality N(0 0) = N(0) N(0) in the case d) is equal to K 1 in two cases 1) and 2). So our rules de Morgan are proved for all possible cases. To prove properties of associativity and distributivity we must take at eight possible cases of correlations between values x, p, y, q, z, r : A) x = p,y = q,z = r; B) x = p,y = q,z r; C) x = p,y q,z = r; D) x = p,y q,z r;e) x p,y = qz = r;f) x p,y = q,z r; G) x p,y q,z = r;h) x p,y q,z r and four cases of correlations between values s,t,u : 1)s t u;2)s u > t;3)s < t u;4)s < u < t. We ll prove justice of rules of distributivity x p s (y q t ) zr u) = (x p s y q t ) (xp s z r u); (20) x p s (yq t ) zr u ) = (xp s yq t ) (xp s zr u ). (21) To prove (20) in the case A) we must prove the equality s (t u) = (s t) (s u)

99 Properties of GCMVLF 97 for all four cases 1),2),3),4). The left and right parts of this equality is equal to s for cases 1) and 2), is equal to t for cases 3) and is equal to u for cases 4). The left and right parts of (20) is equal to s for cases B),C),D), is equal to t u for case E) and it is equal to 0 for cases F),G),H). To prove (21) in the case A) we must prove the equality s (t u) = (s t) (s u) for all four cases 1),2),3),4). The left and right parts of this equality is equal to t for cases 1), is equal to u for cases 2) and is equal to s for cases 3),4). The left and right parts of this equality are equal to s t for cases B) and to s u for case C) and to 0 for cases D),E),F),G),H). Hence we have proved rules of the distributivity for all A),B),C),D), E),F),G),H) and all four cases 1),2),3),4). In a similar manner we can prove the justice of the associative properties of our operations in our algebra: (x p s y q t ) zr u = x p s (y q t zr u); (22) To prove (22) we must consider equalities: (x p s yq t ) zr u = xp s (yq t zr u ). (23) A) (s t) u = s (t u); B) (s t) 0 = s (t 0); C) (s 0) u = s (0 u); D) (s 0) 0 = s (0 0); E) (0 t) u = 0 (t u); F) (0 t) 0 = 0 (t 0); G) (0 0) u = 0 (0 u); H) (0 0) 0 = 0 (0 0). Corresponding equalities for proving (23) are: A) (s t) u = s (t u); B) (s t) 0 = s (t 0); C) (s 0) u = s (0 u); D) (s 0) 0 = s (0 0); E) (0 t) u = 0 (t u); F) (0 t) 0 = 0 (t 0); G) (0 0) u = 0 (0 u);h)(0 0) 0 = 0 (0 0). The left part of each from 16 equalities is equal to the right part for all cases A),...,H) and all four cases 1),2),3),4) and the validity of the associative properties follows. In fact, this proves the next theorem.

100 98 Sergey Novikov Theorem 1. The algebra of K-valued logical functions (v) j i with two binary operations,, one unary operation N (Lukasevich s negation) and elements K 1,0 is a Boolean algebra for i = K 1 and it is not a Boolean algebra for i K 1. Except properties proved for our generalized characteristic K-valued logical functions (v) j i have follow properties: (v j ) K 1 i = v j i ; (24) c (x p y q ) = (x p y q ) K 1 c = x p c yq c ; (25) c (x p y q ) = (x p y q ) K 1 c = x p c yq c ; (26) (N(v j )) 0 K 1 = v j ; (27) (v j i )0 K 1 = (v j ) 0 K 1 = N(v j ); (28) K 1 j=0 v j i = i; (29) K 1 j=0 K 1 i=0 v j i = vj ; (30) K 1 v j i = i=0 v j i = 0. (31) The validity of formulas can be verified with the help of the corresponding tables. 3. Formulas for analytical recording of any multi-valued logical functions It is obvious what we have true next Statement 1. For any K-valued (K > 2) logical function f(v 1,...,v n ) we have formulas f(v 1,v 2,...,v n ) = α 1 α 2...α n (v α1 1 v α 2 2 vn αn )K 1 f(α 1,...,α ; (32) n)

101 Properties of GCMVLF 99 f(v 1,v 2,...,v n ) = (v 1 ) α1 f(α 1,...,α (v n) 2) α 2 f(α 1,...,α n) α 1 α 2...α n (v n ) αn f(α 1,...,α n). (33) A proof of the completeness of our basis {Min(x,y),Max(x,y),v j i } follows from here. Theorem 2. If f(v 1,...,v n ) is a K-valued function and for β J(α), where J(α) is a interval, α = α 1 α 2...α n, i α i {,0,1,...,K 1}, we have f(β) = j > 0, then the implicant covering f(v 1,...,v n ) on all words from the interval J(α) is equal to P(v) = (v α 1 1 v α 2 2 v αn n )K 1 j, (34) where for α i = a proper factor is absent. Proof. Let β J(α) and f(β) = j > 0, where β = β 1 β 2...β n, β i {0,1,2,...,K 1},α = α 1 α 2...α n,α i {,0,1,...,K 1}. If β J(α), then α i = β i for any component with α i. Hence we have β α i i = K 1 for α i. Our product (Min) of such factors is equal K 1 also. Hence and the definition our function vi K 1 we have P(β) = j and P(γ) = 0, for any γ / J(α). If γ / J(α), we have a component γ i α i and γ α i i = 0. Our function (34) is equal to j for any words β J(α) and it is equal to 0 for any words γ / J(α). Consequently our function (11) is an implicant covering f(v 1,...,v n ) on all words from the interval J(α). Our expression (34) we can write (with the help of formulas from true article) as (v 1 ) α 1 j (v 2 ) α 2 j (v n ) αn j. (35) Example 1. Let for words from the interval J(α) = 34 0 the function f(v 1,...,v 6 ) have sense 3. Then the implicant covering f(v 1,...,v 6 ) on all words from the interval J(α) is equal to It is true next statement. (v 2 ) 3 3 (v 3 ) 4 3 (v 5 ) 0 3. Statement 2. If{P1 1,P1 2,...,P1 d1,...,p(k 1) 1,P(K 1) 2,...,P(K 1) d(k 1) } is the resulting system of implicants for K-valued functionf(v 1,..., v n ) and Pi 1,Pi 2,...,Pi di are implicants covering each of words α, where f(α) = i > 0, then f(v 1,...,v n ) has analytical recording f(v 1,...,v n ) = K 1 di i=1 j=1 Pi j, (36)

102 100 Sergey Novikov where Pi j is written as a formula (34) or (35). Example 2. Let for two functions with n = 6 we have system-intervals: R1 = {(4 {2})}, R2 = {( 4 2 {1}),( 7 {1}),( 7 {2})}, R3 = {( 4{1}),(7 {2}),( 6 {2}),( 0 {1})}, R4 = {( 0 {2}),( 1 {1,2})}, R5 = {( 5{1,2}),( 26 {2}),( 6 {2}),( 0{1})}, R6 = {( 7 {1,2}),(04 {2}),(06 {1}), ( 7 {1}),(0 4 {2})}, R7 = {(7 6 {1,2}),(1 {1})}. Then analytical recording for F = {f 1,f 2 } in the basis {Min(x,y), Max(x,y),v j i } is equal to f 1 = (v 4 3 v2 5 v7 5 )7 2 (v4 6 v0 5 )7 3 (v 5) 1 4 (v5 6 v0 6 )7 5 (v 7 3 v0 1 v6 2 v7 2 )7 6 (v7 1 v6 5 ) v1 1 f 2 = (v 1 ) 4 1 (v 4) 7 2 (v7 1 v6 3 )7 3 (v0 3 v1 5 )7 4 (v5 6 v2 2 v6 3 v6 5 )7 5 (v 7 3 v0 1 v4 2 v0 1 v4 3 )7 6 (v7 1 v6 5 ). References [1] Sergiej Novikov. The Methods of Realization of K-Valued Logical Function s implicants. Mathematica. Proceeding of the XI Slovak - Czech - Polish Mathematical School. Catholic University In Ružomberok. Ružomberok, [2] Sergiej Novikov. Metoda minimalizacji systemów K-znacznych częściowych funkcji logicznych. Studia informatica. Systemy i technologie informacyjne. Volume 2(4), [3] S. B. Rosser, A. R. Turquette. Many-valued logic. Amsterdam, [4] S. V. Jablonski. Vvedenie v diskretnuju matematiku. Moskva, Nauka, 1979.(in russian)

103 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 WHAT DO 5th GRADE PUPILS KNOW ABOUT STATISTICS Jana Pócsová a, Ingrid Semanišinová b a,b Department of Mathematics University of Pavol Jozef Šafárik in Košice Jesenná 5, Košice, Slovakia jana.pocsova@upjs.sk, ingrid.semanisinova@upjs.sk Abstract. The goal of this paper is to present partial results of a research concerning potential implementation of elementary statistics in the fifth grade of primary school. We describe the aims and methods of our investigation. We provide a few examples of treated problems and pupils approaches to solution of these problems. 1. Introduction There is an increasing tendency of introducing elements of statistics into the primary school curriculum as a part of basic literacy in mathematics. On the other hand, the experience of educators and statisticians is, that a large proportion of students, even in college, do not understand many of the basic statistical concepts they have studied. At present very little statistics is typically introduced to students before they enter secondary school. The recent reform of our school system provides a possibility to include elementary statistics and probability in the fifth grade of primary school. Therefore we try to analyse the level of initial statistical understandings of the fifth grade pupils. Our aim is also to find out how difficult for understanding are the elements of statistical reasoning for the fifth grade pupils. 2. Methodology The proposed teaching programme consists of 5 lessons for pupils aged from 10 to 11 years. We try to investigate the level of understanding of basic statistical concepts, such as statistical chart, table of data, average, sample and others, by the pupils of given age category. Our experiment was realised with 14 pupils (3 girls and 11 boys) from one class of 5th grade of primary school. In the report we focus particularly on the following items:

104 102 Jana Pócsová, Ingrid Semanišinová which figures are considered to be statistical charts by pupils, and which statistical charts are understandable for them; how pupils are able to read from statistical charts, what mistakes they do while reading them, what information they can find out in the charts; what can pupils derive from the provided data and what types of statistical charts they create. 3. Experience with realization 3.1. Statistical chart The first item presented above was treated by the following task realised by pupils. The pupils were asked to find out some statistical charts in newspapers and magazines, at internet, etc. Further they had to classify these charts into four groups: histograms, pie charts, polygons and others. The pupils, in general, had no problem to classify standard statistical charts. On the other hand, the pupils have quite a lot of problems to interpret nonstandard charts or they do not consider them as statistical chart at all. For example, pupils think that the table on the figure 1a is histogram and most of them thought that the map was insignificant. The chart on the figure 1b pupils didn t understand at all, they didn t consider it as a statistical chart, they didn t know how to interpret it. a) b) Figure Reading statistical charts In order to find out how pupils are able to read from statistical charts, what mistakes they do while reading them, we show some statistical charts to pupils and ask questions concerning them. We present examples of some questions.

105 What do 5th grade pupils know about statistics 103 Chart 1 (see figure 2a) In which city is the highest rent? In how many monitored cities the rent was inquired? List the cities, where the rent is lower than 450e. Is it true that in more than half of those cities is the rent less than 460e? Can we say that the rent in the west Slovakian cities is higher than the rent in the east Slovakian cities? a) b) Figure 2 Chart 2 (see figure 2b) What were the receipts of the considered company in the period from 2003 to 2007? Try to estimate what is the average amount of receipts during this period. Chart 3 (see figure 3a) Order the following geographical regions according to number of bicycles used by their inhabitants: Europe, U.S., China, Slovakia. a) b) Figure 3 Chart 4 (see figure 3b) What do you think, when (in which year) the police obtained successful

106 104 Jana Pócsová, Ingrid Semanišinová results? Predict the number of stolen cars for the year Chart 5 (see figure 4) How many pupils contributed to the collection in the first grade and in the second grade? Who was more successful in the collection, boys or girls? Explain why. počet žiakov počet žiakov vyzbieraná suma v eurách vyzbieraná suma v eurách Figure 4 The questions were chosen so that we would be able to study the following phenomenon: Do pupils understand different types of charts, could they interpret them? How precisely pupils interpret information from the chart? Were their answers influenced by their own experiences? Do they read and understand the description of the axes? Are the pupils able read and understand implicit information of the chart (estimate the average, made the prediction, chose suitable criterion for a comparison of two groups of data)? We gained the following information: Histogram and pie chart were well understandable for all the pupils. They read explicit information from these charts without significant problems. On the other hand, the reading information from polygon was more complicated for most of pupils. Some pupils did not work properly with the legend of all types of the charts. Most of the pupils did not understand the formulation from...to...(task with interval), did not know what had to be calculated.

107 What do 5th grade pupils know about statistics 105 If the pupils were not able to answer the questions from chart, they complained that the question was incorrectly raised, or their answers were influenced by their own experience. Most of the pupils were able to work with data (estimate the average or made the prediction). Most of the pupils had good visual imagination of the average. Most of the pupils had problems with the selection of criteria for comparing two groups of data. Only some of the pupils chose suitable criterion for comparisons Creating a statistical chart Finally we describe what types of statistical charts pupils create from provided data. We describe what information is important for 5th grade pupils while creating a new diagram. In order to stimulate the pupils to acquire basic statistical data we used the ongoing ice hockey in world championship. As the most of the pupils in this class were boys, selection of the hockey theme was intentional. The task of pupils was to monitor individual matches of Slovak national hockey team and to complete the tables with given structures. We wanted the pupils to monitor the following parameters of a hockey player: height, age and post. They also had to record some information from the match, mainly penalty minutes and the number of won and lost face offs. It turned out that monitoring and recording during the live broadcast is very difficult for the pupils of this age. Thus the required records were made only by a few pupils. For this reason we reduced the monitored attributes only to data concerning the players. This data collection was prepared before the lesson. During the lesson the pupils had to create a statistical chart from the prepared table. Our discoveries can be summarised as follows: Pupils understood the data recorded in the table. Pupils used different types of diagrams for depicting collected data. They preferred histogram; the usage of polygon was incorrect. Pupils had problem to emphasise important information from the table and to omit marginal information. They wanted to depict all the information in the chart. Most of the created charts were just transformations of the table in a visual form (see figure 5). Pupils usually did not know how to divide the intervals on axes.

108 106 Jana Pócsová, Ingrid Semanišinová The understandability of the depiction by other people was not considered to be the essential feature of the statistical diagram. Figure 5 4. Conclusion The results of our experiments we can recapitulate in the following way: Pupils recognizes standard kinds of statistical diagrams (histograms, pie charts and polygons). They understand histograms and pie charts without any serious problem. They can read explicit information provided by the standard diagrams. Most of the pupils are able satisfactory visualise the average in a histogram. The answers of ambiguous questions are often influenced by pupils own experiences. Most of the pupils have problems with a comparison of two groups of data. During the creation of statistical charts the pupils have problem to distinguish important information in the table and to omit marginal information. We think that lessons on statistics might have a positive effect on the children s ability to read and interpret data from statistical charts. According to the reactions of the pupils to the prepared and presented lessons, we can admit that it is possible to include similar tasks of descriptive statistics in the 5th grade of primary school. Such lessons can help pupils understand and work with different statistical information presented by newspapers and media.

109 References What do 5th grade pupils know about statistics 107 [1] J. Garfield and A. Ahlgren: Difficulties in Learning Basic Concepts in Probability and Statistics: Implications for Research Source. Journal for Research in Mathematics Education, Vol. 19, No. 1 (Jan., 1988), [2] E. Fischbein, A. Gazit: Does the Teaching of Probability Improve Probabilistic Intuitions?: An Exploratory Research Study. Educational Studies in Mathematics, Vol. 15, No. 1 (Feb., 1984), 1 24.

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111 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 FROM RESEARCH ON THE IMPLEMENTATION OF LEE MOORE S METHOD TO FORMING THE IDEA OF FUNCTION Zbigniew Powązka Pedagogical University Institute of Mathematics Podchorążych Kraków powazka@up.krakow.pl Abstract. One of the most important ideas in the mathematical analysis is the notion of the function. It can be found in educational programs in many countries around the world. For many years it has played a key role in didactics research to shape the image of this notion correctly in pupils and students conscience. The concept developed by an American mathematician Robert Le Moore ( ) is worth attention. In this paper I would like to show the results of pilot studies aimed at finding at least a partial answer to the question whether Moore s method could be equally effective at our university, and how it was received by the students participating in the experiment. 1. Introduction One of the most important idea in mathematic analysis is notion of function. It can be found in education programs in many countries in the world. For many years it is of prime importance in didactic researches to shape correctly an image of this notion in pupils and students conscience (vide eg. Vinner, 1983, Fulier, 2001, Fulier, Gunčaga, Eisenmann, 2008, Semadeni, 2002, Sajka 2003, 2005). There s a lot of different methods used in teaching of mathematic analysis. The conception of American mathematic Robert Lee Moore ( ) is worth of great attention. It was popularized by his student William S. Mahavier (Mahavier, 1999). The fundament of this conception was Lee Moore s belief that the only way of gaining a knowledge is self-reliant work of student leading to independent discovering of theorems. During his class with group of students consisting of a maximum of 20 persons he applied so called discovery learning reliant on independent

112 110 Zbigniew Powązka solving the tasks set by students. These kits also contained the necessary definitions of terms occurring there. Teacher gave students the consultation and guidance and he monitored and evaluated their work as well. Thus the process of teaching had led to: develop student s ability to work independently with the mathematical text, convince students that they can fully understand the content of mathematical analysis, which was chosen by the course lecturer, learn to prove theorems and present their reasonings, accustom students to use the correct written and spoken mathematical language. 2. Objectives and research organization The experiment was carried out under my direction by Monika Rudowicz with a group of 13 first year students. Our university students participate in lectures and exercises from the analysis, therefore to the survey were selected some specific problems concerning the functions that were not developed in class. During the studies we were observing how students are managing with the independent work with a mathematical text, and what are the difficulties and errors occurring in the solutions of problems. Coordinator has held 4 meetings with the audited group. They were held every week. At the first meeting respondents received a letter indicating how they will work. There have also been fixed forms of permanent contact with the coordinator. In addition, the theoretical material was provided to students: it was a set of definitions and the first three tasks from the list. Students were also informed that to each task has been prepared guidelines to help resolve it. At subsequent meetings, participants reported their solutions, shared their difficulties and received another 4 tasks per each week. During each week, students had to provide written solutions of these tasks. In the event of difficulties they could contact with the leading to obtain guidelines. At the last meeting was carried out short test. Students were also asked to fill out a survey examinating their opinions about the experiment. Below we put the list of tasks which were solving by students: Task 1. Let A = { 1,1,2} and B = { 1,0}. Find all functions f : A B fulfilling following conditions: a) f is constant function.

113 From research on the implementation of Lee Moore s method 111 b) function f has a fix point. Task 2. Let A and B be sets such as in task 1 and f : A B: a) Examine if there exist number a R\{0} such that f(x) = ax for each x A. b) Examine if there exist number a R\{0} such that f(x) = k x for each x A. Task 3. Let A = { 2,2,5}, f(x) = k x, x A, where k 0 is a fixed real number. Adjust the number k which fulfills condition:1 f(a). Task 4. Let A = { 2, 1,0,1,3}. Find a set of fixed points of following functions: a) f(x) = 2x; b) f(x) = ax, a R; c) f(x) = a x, where a > 0. Task 5. Let f : A A, where A = {1,2,3,4,5}; h(x,y) = (f(x),f(y)), where x A i y A and P is set of function h fixed points. Show that if B = (1, 3),(1, 5),(2, 4) is subset of P then f is the identity function on A. Task 6. Find images of relevant sets if: a) A = ( 2,1), f(x) = 3x+2; b) D = [ 2,2], f(x) = 1 x, x 0, f(0) = 0. Task 7. Consider the function f(x) = (a 1)x 2 ax+3, a R. Adjust the number a which fulfills following condition: a) domain of f is the set of real numbers; b) domain of f is the sum of two disjoint ranges. Task 8. Find a value of parameter m such that function f : R R given by formula f(x) = (2m 3)x+3 is positive only for x < 3. Task 9. What will be solution in the task 8 if we skip word "only" in its content? Task 10. Find a value of parameter m such that the solution of inequality x x+x 2 < m is an empty set? Task 11. Let f : A A, where A = (a,b), a,b R and a < b. We define following functions: a) x f(x+6);

114 112 Zbigniew Powązka b) x 2f(x); c) x f( x); d) x 2f(x); e) x 3f(2x). I. Which functions from examples presented above have the same domain as the function f? II. For which of these functions the domain could be disjoint with the set A? III. Which of these functions have codomain equal to codomain of f(a)? IV. Find a set A such that functions from examples a) e) have the same domains and codomains. These tasks come from the book by H. Kąkol and Z. Powązka (Kąkol, Powązka, 1994). It should be noticed that the four last tasks are the most difficult and tasks 8 and 9 are connected. 3. Remarks on researches Students had not major difficulties with the initial tasks from the list but this does not mean that all the written works were drawn correctly. Difficulties appeared in tasks Therefore, we comment their solutions below: Task 7 is the task with parameter. To give a solution a) it should be considered a conjunction of conditions: a 1 > 0 and = a 2 12a+12 < 0 and it should be notice that for a = 1 domain of the function is the set of real numbers x 3. In part b) of this task is sufficient to resolve inequality: a 1 > 0 and = a 2 12a+12 > 0. None of the students participating in the study did not carry out a full discussion, and only 4 persons correctly considered the two inequalities. Others were considering only null-points of the function which the square root was taken of. It shows that most of investigated were treated the expression x (a 1)x 2 ax + 3 as square function. Moreover, even if calculations were correct comments were missing. Therefore, to meet all demands of Moore s method students had to improve and complement their solutions with necessary comments, which decreased the assessment of their work. Considering solutions of tasks 8 and 9, it should be noticed that the given formula describes the set of all linear functions which contain the point (0,3). A solution of the task 8 is only one constant function from

115 From research on the implementation of Lee Moore s method 113 this family which the null-point is x 0 = 3. Conditions from the task 9 are fulfilled by all linear functions of mentioned family which null-points are x 0 3 and by the constant function y = 3, x R. Research shows that the task 8 was easier for students. It has been resolved correctly by the 10 surveyed. In contrast, only 6 persons resolved properly the task 9. Although the content of task 8 differ from the content of task 9 only with one word, errors which occurred indicates that students did not understand the information contained in the task 9. Inequality from the task10 is equivalent to inequalityx x+x 2 > m. To resolve the problem is sufficient to notice that function on the left side of inequality is banded down and its smallest value is x = 5. This proposal can be obtained using the definition of absolute values. Finally, a set of solutions to this inequality is empty if and only if m 5 This task was resolved properly by 4 persons and 9 committed some errors. The most common error was omission of the case m = -5. Others derived from wrongly applied definition of absolute values. Again, it turned out that task with parameter which require analysis of text and correct discussion of all cases gives students difficulties with which they can not handle on their own. This fact was notice in the work of J. Major (Major, 2006a, 2006b, Major, Powązka 2006). Task 11 involves transition of functions graphs. In Poland this issue is discussing in middle school. Putting this in a set of tasks was intended to check on how many students remember these issues. It turned out that 4 people do not have attempted to tackle this task. The others reported solution after receiving tips to consider at first specific functions. Some of solutions includes only drawings without the necessary comments. In the group of surveyed were also some who were not able to generalize their observations of specific cases per case of any function. 4. Summary We can formulate following conclusions based on the experiment: 1. The test showed that the theoretical material has been understood and remembered by the students. 2. Applying the Moore s method requires the selection of tasks in terms of their difficulty. Students participating in a course have not difficulty with easy tasks while the more difficult tasks motivated them to benefit from heuristic guidance. However, the tasks that they seemed to be too difficult deterred them to take the effort to resolve. 3. This method is suitable for working with students who can read mathematical text with understanding and who like to gain knowledge

116 114 Zbigniew Powązka independently. But there are some who prefer the classic lecture and joint resolving of problem. 4. In applying his methods Lee Moore assumed that students do not benefit from any other textbooks, but only from materials prepared by the lecturer. However, our students have noticed that it is a impediment to gaining the knowledge. 5. Half of the participants assessed this method as an interesting way to acquiring knowledge. Thus, despite a major effort, which requires preparation of relevant materials, it is worthwhile to use it as a method of support the process of teaching a group of interested students. References [1] Fulier, J.: 2001, Funkcie a funkčné myslenie vo vyučovaní matematickej analýzy, Univerzita Konštantína Filozofa, Fakulta prírodných vied. Nitra. [2] Fulier, J., Gunčaga, J.: 2006, Modul matematickej analýzy v kurze ďalšieho vzdelávania učiteľov, Matematika v škole dnes a zajtra, Zborník 6. ročníka konferencie s medzinárodnou účasťou, Ružomberok, [3] Kąkol H., Powązka Z.: 1994, Pojacie funkcji cz 1, Wyd. "Dla szkoly", Bielsko-Biala, [4] Mahavier, W., S.: 1999, What is the Moore Method? Primus, vol. 9, < [5] Major J.: 2006a, Rola zadac i problemow w ksztaltowaniu poja matematycznych na przykladzie bezwzgladnej wartolci liczby rzeczywistej Roczniki PTM, seria V, Dydaktyka Matematyki, 29, [6] Major J.: 2006b, Uwagi dotyczące obrazu pojacia wartolci bezwzgladnej liczby rzeczywistej u studentow matematyki, Ksztalcenie matematyczne - tendencje, badania, propozycje dydaktyczne, Wydawnictwo Akademii wiatokrzyskiej, Kielce, [7] Major J., Powązka Z.: 2006, Uwagi dotyczące wartolci bezwzgladnej liczby rzeczywistej, Annales Academiae Pedagogice Cracoviensis 36, Studia Ad Didacticam Mathematicae Pertinentia I,

117 From research on the implementation of Lee Moore s method 115 [8] Sajka, M.: 2003, A secondary school student s understanding of the concept of function - a case study, Educational Studies in Mathematics 53, [9] Sajka M.: 2005, Functional equations as a new tool for researching certain aspects of subject matter knowledge of functions in future mathematics teachers. Proceedings of the 57th CIEAEM Conf., Piazza Armerina, Italy, [10] Semadeni Z: 2002, Trudnolci epistemologiczne związane z pojaciami: pary uporządkowanej i funkcji,roczniki PTM, seria V, Dydaktyka Matematyki 24, [11] Vinner, S.: 1983, Concept definition, concept image and the notion of function,international Journal of Mathematical Education in Science and Technology 14(3),

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119 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 VARIOUS APPROACHES TO ONE PROBLEM OF COMBINATORICS Pavel Tlustý, Marek Šulista Department of Applied Mathematics and Informatics University of South Bohemia České Budějovice, Czech Republic Abstract. In this paper we discuss solutions of students of interesting problems from combinatorics. 1. Introduction The Department of Mathematics of the Pedagogical Faculty of the University of South Bohemia in České Budějovice organizes so called optional seminar of mathematics which is usually a four day meeting of teachers and students of mathematics which focuses on solving various mathematical tasks, interesting problems, games, brain teasers etc. The aim of the seminar is above all to promote and to make mathematics popular among students who do not study mathematics as the main field of their study. Every participant prepares a short presentation of a problem which he/she introduces to the other students and moderates the discussion, directs raised ideas, and with the others looks for the right solution. The seminar is very well evaluated by the participating students. We are glad to see the rising interest among teacher trainees of Year 1-4 and students who do not study mathematics. The survey which is carried out regularly revealed that the students are attracted not only by the friendly atmosphere, informal teaching, but also by the place, where the seminar is organized (Lipno, Horská Kvilda, Bezdrev Lake, etc.). In the contribution, we try to outline in brief the way students tried to find the solution of one concrete mathematical problem. 2. Problem of Combinatorics Problem: On a chess board 6 4 (6 rows, 4 columns), we are supposed to place 12 coins in the way that in each of the columns there are exactly three

120 118 Pavel Tlustý, Marek Šulista coins, and in each of the rows, there are exactly two of them. In how many ways is it possible to do it? (One example of such a placement meeting the given conditions is presented in Figure 1) Fig. 1. Solution 1: Let s start with the first column. There are ( 6 3) = 20 possibilities in total of placing the coins into the first column. We can assume that we have occupied the first three rows of the first column without influencing the generality. Now, the other three columns of the chess board "have been divided" into two parts We have to place in each row one coin in the first part of the board. According to the number of coins in one column, we can distinguish three cases (Figure 2): a) b) c) Fig. 2. a) All three coins are in the same column (Figure 2a). In this case, we have three possibilities of choosing this column. The coins can be placed to the bottom part of the board (3 3) only in one way. In situation a), we have three possibilities. b) In one column, there are two coins, the remaining (third) coin is in a different column (Figure 2b). In this case, we can choose the first column in three ways. It is possible to place the coins into the chosen column again in three ways. For each of them, it is possible to choose a column with one coin in two ways. Therefore, there are = 18 way to place coins into the upper part (3 3) chess board. In how many times, is it possible to fill the bottom (3 3) chess board? Three coins must be placed in the column where there has not been any coin yet. It can be done just in one way. There are three ways of placing

121 Various Approaches to One Problem of Combinatorics 119 one coin in the column with two other coins. Then, the other coins can be placed in one way only. In situation b), we have 18 3 = 54 possibilities. c) All three coins are in different columns (Figure 2c), it can be done in 3 2 = 6 ways. The bottom chess board (3 3) can be filled in 3 2 = 6 ways. In situation c), we have 6 6 = 36 possibilities. There are 20 ( ) = possibilities of meeting the conditions of the problem. Solution 2: There are ( 6 3) = 20 possibilities of placing the coins into each of the columns. According to the placement of coins into the first two columns, we can distinguish four cases (Figure 3): a) b) c) d) Fig. 3. a) No coin from the first column is placed in the same row as the coins in the second column (Figure 3a). The coins can be placed into the first column in ( 6 3) = 20 ways. There is the only possibility of placing the coins in the second column in the way that none of them would lie in the same row as the coins in the first column. Coins in the third column can be placed arbitrarily, this means in ( 6 3) = 20 ways in total. The placement of coins in the fourth column is unambiguous. In the situation a), we have = 400 possibilities. b) One coin in the first column is placed in the same row as the coins in the second column (Figure 3b). We can choose the row in six ways. The other two coins of the first column can be placed in ( 5 2) = 10 ways and there are ( 3 2) = 3 ways to place the other two coins of the second column. Now, we place coins into the third and fourth column. The empty column can be filled in just one way. The other two coins can be placed into the third column in ( 4 2) = 6 ways. The placement of two last coins in the fourth column is unambiguous. In situation b), we have (5 2) ( 2) 1 ( 2) 1 = possibilities.

122 120 Pavel Tlustý, Marek Šulista c) Two coins of the first column lie in the same row as the two coins in the second column (Figure 3c). The rows can be selected in ( 6 2) ways. The remaining coin of the first column can be placed in four ways and the remaining coin of the second column in three ways. Coins can be placed into the third and the fourth column in two ways. In situation c), we have ( 6 2) = 360 possibilities. d) All coins of the first column lie in the same row as the coins in the second column (Figure 3d). Three rows common to the first two columns can be chosen in ( 6 3) = 20 ways. There is one way only of placing the coins ) in the third and the fourth column. In situation d), we have = 20 possibilities. ( 6 3 There are ( ) = possibilities in total of meeting the conditions of the given problem. Solution 3: We ( consider all permissible placements of two coins in one row. There are 4 ) 2 = 6 of them and they are shown in Figure 4. a c e Fig. 4. We can see that corresponding couples (a,b), (c,d) and (e,f) are dual. From the problem assignment, it flows that a+c+e = 3, b+d+f = 3. Considering the duality, it is sufficient to deal with the solution of equation a + c + e = 3. We have three possibilities in total: 1. One of the rows a,c,e is repeated three times. We have three three ways to choose repeated row and ( 6 3) possibilities of placing the row into the chess board. In this situation, we have 3 (6 3) = 60 possibilities in total. 2. One of the rows a,c,e is repeated twice. We have three ways to choose the repeated rows, two ways to choose the third row, ( 6 2) ways to place the repeated row in the board, four ways to place the third row in the board, and 3 possibilities of placing the remaining rows. In this situation, we have 3 2 (6 2) 4 3 = possibilities in total. b d f

123 Various Approaches to One Problem of Combinatorics None of the rows a,c,e is repeated. Each of the rows a,b,c,d,e,f is present once only. There are 6! = 720 possibilities of placing them in the chess board. There are ( ) = possibilities of fulfill the assignment of the problem. Solution 4: We consider arbitrarily permissible placement of the coins into the chess board (Figure 5a). Using the placement, it is possible to obtain four more using mirror symmetry (Figure 5b,c). Axeso 1 and o 2 divide the chess board 6 4 into 4 parts A,B,C,D. o o1 A C a) b) c) d) Fig. 5. Let s denote #A the number of coins in part A. Then, it is true that: #A = #D, #B = #C, (#A+#B) = (#A+#C) = 6 Considering the given equalities and symmetry, the problem is divided into four cases which will be investigated separately. The results are presented in the following tables: possibilities symmetry total 1. #A = #A = #A = #A = References [1] A. Engel. Stochastics. Ernst Klett Verlag, Stuttgart, [2] A. Płocki, P.Tlustý. Pravděpodobnost a statistika pro začátečníky a mírně pokročilé. Prometheus, Praha, B D

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125 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 VERIFICATION OF JAVA PROGRAMS USING NETWORKS OF FINITE AUTOMATA WITH DISCRETE DATA Bożena Woźna, Andrzej Zbrzezny Institute of Mathematics and Computer Science Jan Długosz University of Częstochowa al. Armii Krajowej 13/15, Częstochowa, Poland {b.wozna, Abstract. In the paper we show that automatic verification of Java programs is feasible. In particular, we show that for a given Java program it is possible to build a network of Finite Automata with Discrete Data (FADDs) that models the behaviour of the program under consideration. Such a model enables to verify some interesting properties of the considered program by means of tools like Uppaal and VerICS. 1. Introduction The model checking tools like VerICS [4] or Uppaal [1] accept a description of a network of finite automata with discrete data (FADDs) [7] as input. This FADD formalism is accurate enough to detect concurrency errors and yet abstract enough to make model checking tractable. So, in order to automatically verify Java programs one has to bridge the semantic gap between a non-finite-state system expressed as a Java source code, and those tools input languages. This requires the application of sophisticated transformation techniques, and one of such methods is shown in this paper; it has been implemented as a J2FADD module of the tool VerICS. We illustrate our approach by presenting the automata models of the two well known solutions (written in Java) of the dining philosophers problem (DPP) - an illustrative example of a common computing problem in concurrency. We also provide some experimental results that help to evaluate our method. The subset of Java that can be translated to FADDs contains: definitions of integer variables, standard programming language constructs like assignments, expressions with most operators, conditional statements and

126 124 Bożena Woźna, Andrzej Zbrzezny loops (for, while, do while), instructions break and continue without labels, definitions of classes, (static) objects, constructors and methods, static and non static methods, and synchronisation methods and blocks. There are also standard thread creation constructs recognised and special methods: Thread.wait(), Thread.notify(), and Random.nextInt(int), the methods Thread.join(Thread) and Thread.notifyAll(). For more details we refer to [8]. To the best of our knowledge, there are only two other model checkers for verification of Java programs: JavaPathFinder [5] and Bandera [2]. However, both of them operate on the Java bytecode, and do not produce a low-level models like finite automata with discrete data. 2. Dining Philosophers Problem (DPP) In this section we show FADD models of the two well known solutions of DPP, and we model check it by means of the tools: Uppaal, VerICS. In particular, we search for deadlocks, which can be defined formally as follows [6]: A set of processes is deadlocked if each process in the set is waiting for an event that only another process in the set can cause. FADDs are standard automata augmented to include integer variables over which standard arithmetic and Boolean expressions can be defined. These automata take as an input a set of initialised integer variables and a set of propositional variables, true at particular states. 2.1 Problem description The description of the dining philosophers problem (DPP) we provide below is based on that in [3]. Consider n (n 2) philosophers. Each philosopher has a room in which he engages in his professional activity of thinking. There is also a common dining room, furnished with a circular table, surrounded by n chairs, each labelled by the name of the philosopher who is to sit in it. On the left of each philosopher there is a fork, and in the centre stands a large bowl of spaghetti, which is constantly replenished. Whenever a philosopher eats he has to use both forks, the one on the left and the other on the right of his plate. A philosopher is expected to spend most of his time thinking, but when he feels hungry, he goes to the dining room, sits down on his own chair, and picks up the fork on his left provided it is not used by the other philosopher. If the other philosopher uses it, he just has to wait until the fork is available. Then the philosopher tries pick up the fork on his right. When a philosopher has finished he puts down both his forks, exits dining-room and continues thinking.

127 2.2 Possible solutions Verification of Java programs 125 We have implemented a possible solution of the DPP problem that could lead to a deadlock (see Listing 1). The deadlock can happen, if every philosopher sits down on his own chair at the same time and picks up his left fork. Then all forks are locked and none of the philosophers can successfully pick up his right fork. As a result, every philosopher waits for his right fork that is currently being locked by his right neighbour, and hence a deadlock occurs. The results for the deadlock property are in Table 1. public class College3 { public static void main( String args [ ] ) { Fork fork0 = new Fork ( false ) ; Fork fork1 = new Fork ( false ); Fork fork2 = new Fork ( false ) ; Philosopher p0 = new Philosopher (0, fork0, fork1 ) ; Philosopher p1 = new Philosopher (1, fork1, fork2 ) ; Philosopher p2 = new Philosopher (2, fork2, fork0 ) ; (new Thread (p0 ) ). start ( ) ; (new Thread (p1 ) ). start ( ) ; (new Thread (p2 ) ). start ( ) ; } } class Fork { private boolean unavailable ; public Fork (boolean unavailable ) {this. unavailable = unavailable ;} public synchronized void acquire () { while ( unavailable ) { try {wait ( );} catch( InterruptedException e ){} } unavailable = true ; } public synchronized void release () { unavailable = false ; notify ( );} } class Philosopher implements Runnable { private int nr ; private Fork left, right ; public Philosopher ( int nr, Fork left, Fork right ) { this. nr = nr ; this. l e f t = l e f t ; this. right = right ; } public void run () { while (true ) { l e f t. acquire ( ) ; right. acquire ( ) ; right. release ( ) ; l e f t. rel ease ( ) ; } } } Listing 1: Java source code of DPP (3 Philosophers). The main class. Tools No. Ph sec. MB J2FADD + VerICS J2FADD + Uppaal Table 1: Dining Philosophers. Deadlock.

128 126 Bożena Woźna, Andrzej Zbrzezny Assume now another solution for DPP (see Listing 2), where there is a lackey who ensures that at most n 1 philosophers can be present in the dining room at the same time. This lackey ensures that no deadlock is possible (see Table 2 for the results). public cl a ss Col l ege3l { public static void main ( String args [ ] ) { Fork f0 = new Fork ( false ) ; Fork f1 = new Fork ( false ) ; Fork f2 = new Fork ( false ) ; Lackey s = new Lackey ( 2); Philosopher p0 = new Philosopher (0, f0, f1, s ) ; Philosopher p1 = new Philosopher (1, f1, f2, s ) ; Philosopher p2 = new Philosopher (2, f2, f0, s ) ; (new Thread (p0 ) ). start ( ) ; (new Thread (p1 ) ). start ( ) ; (new Thread (p2 ) ). start ( ) ; } } class Fork { private boolean unavailable ; public Fork (boolean unavailable ) { this. unavailable = unavailable ; } public synchronized void acquire () { while ( unavailable ) { try {wait ( ) ; } catch ( Exception e ){} } unavailable = true ; } public synchronized void release () { unavailable = false ; notify ( ) ; } } cl a ss Lackey { private int m; private int max; public Lackey ( int max) {this.max = max;} public synchronized void acquire () { while (m >= max) { try {wait ();} catch ( Exception e ) {} } } public synchronized void release () { m; notify ( ) ; } ++m; } class Philosopher implements Runnable { private int nr ; private Lackey s ; private Fork left, right ; public Philosopher ( int nr, Fork left, Fork right, Lackey s ) { this. nr = nr ; this. s = s ; this. l e f t = l e f t ; this. right = right ; } public void run () { while (true ) { s. acquire ( ) ; l e f t. acquire ( ) ; right. acquire ( ) ; right. release ( ) ; l e f t. releas e ( ) ; s. release ( ) ; } } } Listing 2: Java source code of DPP (3 Philosophers). The main class. Tools No. Ph sec. MB J2TADD + Uppaal J2TADD + Uppaal Table 2: Dining Philosophers. Absence of deadlocks.

129 Verification of Java programs 127 All of the experiments have been performed on a computer equipped with the processor Intel Core 2 Duo (2 GHz), 2 GB main memory and the operating system Linux. 2.3 Automata models for the solution without lackey In this section we present an automata model for DPP with three philosophers (see Listing 1). The model consists of a network of the following six FADDs: one FADD for each Philosopher, and one FADD for each Fork. v2 e1! e3! e2! e0! e6! z3 = z3 + 1 e5! e4! e7! z3 = z3 + 1 v1 Figure 1: The automaton for Fork f ork0. v18 e14? v8 v22 z1 = 0 v23 z1 == 1 e24? e10? z4 == 0 e33! z4 = z4-1 v16 e4? e10? v17 z1 == 0 e8? v7 v21 e12? v24 e12? v15 z1 = 1 v25 z0 = 1 v14 v19 v20 e2? v10 v9 e25? e6? v13 z0 == 1 e0? z0 == 0 v12 e2? v11 e4? v28 z3 == 0 z3 = z3-1 e36! v26 z0 = 0 v27 Figure 2: The automaton for Philosopher p0. On Figure 2 we show an automaton forphilosopherp0, and on Figure 1 an automaton for the Fork fork0. The automata for philosophers p1 and

130 128 Bożena Woźna, Andrzej Zbrzezny p2 are analogous to that of Philosopher p0, and the automata for forks f ork1 and f ork2 are analogous to that of Fork f ork0. The automaton for a philosopher reflects the method run of the class Philosopher. Namely, the fragment consisting of the locations: v11, v12, v13, v10, v9, v14, v15, v16 reflects the instruction left.acquire(); the fragment consisting of the locations: v16, v17, v18, v8, v7, v19, v20, v21 reflects the instruction right.acquire(); the fragment consisting of the locations: v21, v22, v23, v24, v25 reflects the instruction right.release(); and the fragment consisting of the locations: v25, v26, v27, v28, v11 reflects the instruction left.release(). The automaton for a fork consists of two locations, operates as a binary semaphore, and implements critical sections. 2.4 Automata models for the solution with lackey In this section we present an automata model for DPP with lackey and three philosophers (see Listing 2). The model consists of a network of the following seven FADDs: one FADD for each Philosopher, one FADD for each Fork, and one FADD for the Lackey. On Figure 4 we show an automaton for Philosopher p0, and on Figure 3 automata for the Fork f ork0 and for the Lackey s. The automata for philosophers p1 and p2 are analogous to that of Philosopher p0, and the automata for forks fork1 and fork2 are analogous to that of Fork fork0. v8 e33! z8 = z8 + 1 e34! z8 = z8 + 1 e30! e32! e31! z8 = z8 + 1 e35! e24! e25! v7 z5 = z5 + 1 e6! e2! e0! v2 e3! v1 e27! e28! e26! e29! e1! e4! e5! e7! z5 = z5 + 1 The automaton for Lackey s; The automaton for Fork fork0. Figure 3: The automaton for a philosopher reflects the method run of the class Philosopher. Namely, the fragment consisting of the locations: v15, v16, v17, v14, v13, v18, v19, v20 reflects the instruction s.acquire(); the fragment consisting of the locations: v20, v21, v22, v23, v11, v12, v24, v25 reflects the instruction left.acquire(); the fragment consisting of

131 Verification of Java programs 129 the locations: v25, v26, v27, v28, v10, v9, v29, v30 reflects the instruction right.acquire(); the fragment consisting of the locations: v30, v31, v32, v33, v34 reflects the instruction right.release(); the fragment consisting of the locations: v34, v35, v36, v37, v38 reflects the instruction left.release(); and the fragment consisting of the locations: v38, v39, v40, v41, v15 reflects the instruction s.release(). The automaton for a fork consists of two locations, operates as a binary semaphore, and implements locking a fork. The automaton for a lackey consists of two locations, operates as a bounded semaphore, and implements locking the dining room. z6 = z6-1 v30 v31 v32 e49! v33 e10? z1 = 0 z6 == 0 e12? e12? v29 v9 v10 e36? z1 = 1 e14? e8? v28 z1 == 0 z1 == 1 v27 v25 v26 v14 e10? v12 e6? v24 e4? e37? v22 e33? e39? e38? z0 = 1 v11 z0 == 1 v13 e0? v23 z0 == 0 v17 v18 v19 v20 e24? z3 = z3 + 1 e30? e2? z3 >= z4 v21 z3 < z4 e56! z8 = z8-1 e27? v41 v40 v16 e30? e55! z8 = z8-1 z3 = z3-1 v15 z8 == 0 v39 e2? z5 == 0 v34 v35 z0 = 0 e4? v36 v37 v38 e27? z5 = z5-1 e52! Figure 4: The automaton for Philosopher p0. 3. Conclusions In the paper we have shown that model checking of Java programs by means of toolsuppaal and VerICS that accept a description of a network of FADDs is feasible. We have presented the method using the dining philosophers problem, a classic multi-process synchronisation problem

132 130 Bożena Woźna, Andrzej Zbrzezny The experiments confirm that our approach provides a valuable aid for Java software verification. References [1] J. Bengtsson, K. Larsen, F. Larsson, P. Pettersson, W. Yi, C. Weise. New generation of Uppaal. In Proceedings of the International Workshop on Software Tools for Technology Transfer, [2] J. Corbett, M. Dwyer, J. Hatcliff, Robby C. Pasareanu, S. Laubach, H. Zheng. Bandera: Extracting finite-state models from java source code. In Proceedings of the 22nd International Conference on Software Engineering (ICSE 00), New York, NY, ACM Press. USA, , [3] C.A.R. Hoare. Communicating sequential processes. Prentice Hall, [4] A. Niewiadomski, W. Penczek, A. Pólrola, M. Szreter, B. Woźna, M. Kacprzak, W. Nabialek, A. Zbrzezny. VerICS a model checker for knowledge and real-time. Fundamenta Informaticae, 85(1-4): , [5] C. Pasareanu, W. Visser. Verification of Java Programs Using Symbolic Execution and Invariant Generation. In Proceedings of SPIN 04, volume 2989 of LNCS, Springer-Verlag, , [6] Andrew S. Tanenbaum. Modern Operating Systems. Prentice Hall, Vrije University, Amsterdam, The Netherlands, 2/e edition, [7] A. Zbrzezny and A. Pólrola. Sat-based reachability checking for timed automata with discrete data. Fundamenta Informaticae, 79(3 4): , [8] A. Zbrzezny and B. Woźna. Towards verification of Java programs in VerICS. Fundamenta Informaticae, 85(1-4): , 2008.

133 Catholic University in Ružomberok Scientific Issues, Mathematica III, Ružomberok 2009 GEOMETRIC SOLUTION OF THE SYSTEM OF LINEAR EQUATIONS Tomáš Zdráhal Department of Mathematics Purkyně University in Ústí nad Labem Hoření 13, Ústí nad Labem, Czech Republic Abstract. The contribution is intended as an attempt to help students better understanding of problems concerning of systems of linear equations. The geometric program Cabri 3D is used to show solutions of homogeneous systems of three equations of three variables. Cabri illustrates what is geometrically carried out when the system is solved by Gauss-Jordan elimination. Given the non-homogeneous system of three linear equations with real coefficients a 11 x+a 12 y +a 13 z = b 1 a 21 x+a 22 y +a 23 z = b 2 a 31 x+a 32 y +a 33 z = b 3 By means of scalar product this system can be written as following three equations (a 11,a 12,a 13, b 1 ).(x,y,z,1) = 0 (a 21,a 22,a 23, b 2 ).(x,y,z,1) = 0 (a 31,a 32,a 33, b 3 ).(x,y,z,1) = 0 To solve the original system of linear equations is therefore is equivalent to the problem of finding a vector (x, y, z, 1) that is perpendicular simultaneously to the vectors(a 11,a 12,a 13, b 1 ),(a 21,a 22,a 23, b 2 ), (a 31,a 32,a 33, b 3 ). Further, it is not our purpose to derive known result: If vectors (a 11,a 12,a 13 ), (a 21,a 22,a 23 ), (a 31,a 32,a 33 ) generate 3-dimensional vector space, 2-dimensional vector space, 1-dimensional vector space then the solution of the homogeneous system is 0-dimensional orthogonal complement (point (0, 0, 0)), 1-dimensional orthogonal complement (line), 2- dimensional orthogonal complement (plane) respectively. The solution of

134 132 Tomáš Zdráhal corresponding non-homogeneous system is subsequently a point, a line or it doesn t exist and a plane or or it doesn t exist. On the other hand we use this result to illustrate what is geometrically carried out when the homogeneous system is solved by Gauss-Jordan elimination. Three homogeneous systems will be "solved" in Cabri 3D Geometry. There are two notices on this place to serve reader to better understand following text, actually following figures: 1. We don t draw any plane a i1 x + a i2 y + a i3 z = b i in Cabri 3D, we draw vectors (a i1,a i2,a i3 )! 2. Reader should study figures carefully to really understand what vectors, planes, lines and points mean. There is one recommendation with this respect: Try to draw similar figures in Cabri 3D yourself! (Once you can use different colors and animations, you better understand this mixture of lines...) Now, let us start with three concrete homogeneous systems of three linear equations. (To enable to "study" figures comfortably, there is detailed working description directly under each figure as well.) 1. Solve the system i.e. solve these three equations x+y +z = 0 x+y z = 0 2x+2z = 0 (1,1,1).(x,y,z) = 0 (1,1, 1).(x,y,z) = 0 (2,0,2).(x,y,z) = 0 Let us find thus the vector(s) (x,y,z) perpendicular simultaneously on the vectors (1, 1, 1), (1, 1, 1) and (2, 0, 2). It is again obvious, that the vectors perpendicular on (1,1,1) lie in the plane x+y +z = 0, on (1,1, 1) lie in x+y z = 0 etc. The result vector is of the form (0,0,0) and generates the trivial vector space - orthogonal complement of the vector space given by the vectors (1,1,1), (1,1, 1) and (2,0,2). To solve the system by Gauss- Jordan elimination means to find this orthogonal complement of this vector space given by the "suitable" basis i.e. the basis as "simple" as possible (echelon reduced matrix is replaced by an identity matrix). Here the basis is as follows: [(1,0,0),(0,1,0),(0,0,1)]. 2. Solve the system x+y +z = 0

135 Geometric solution of the system of linear equations 133 i.e. solve these three equations x+y z = 0 2z = 0 (1,1,1).(x,y,z) = 0 (1,1, 1).(x,y,z) = 0 (0,0,2).(x,y,z) = 0 Let us find thus the vector(s) (x,y,z) perpendicular simultaneously on the vectors (1,1,1), (1,1, 1) and (0,0,2). It is again obvious, that the vectors perpendicular on (1,1,1) lie in the plane x+y+z = 0, on (1,1, 1) lie in x+y z = 0 etc. The result vectors are of the form (x, x,0) and generate the 1-dimensional vector space - orthogonal complement of the vector space given by the vectors (1,1,1), (1,1, 1) and (0,0,2). To solve the system by Gauss-Jordan elimination means to find this orthogonal complement of this vector space given by the "suitable" basis i.e. the basis as "simple" as possible (echelon reduced matrix is replaced by an identity matrix). Here the basis is as follows: [(1, 1, 0),(0, 0, 1)]. 3. Solve the system i.e. solve these three equations x+y +z = 0 2x+2y +2z = 0 3x+3y +3z = 0 (1,1,1).(x,y,z) = 0 (2,2,2).(x,y,z) = 0 (3,3,3).(x,y,z) = 0 Let us find thus the vector(s) (x,y,z) perpendicular simultaneously on the vectors (1,1,1), (2,2,2) and (3,3,3). It is obvious, that the vectors perpendicular on (1,1,1) lie in the plane x + y + z = 0, on (2,2,2) lie in 2x + 2y 2z = 0 i.e. x + y + z = 0 etc. The result vectors are of the form (x,y, x y) and generate the 2-dimensional vector space - orthogonal complement of the vector space given by the vectors (1,1,1), (2,2,2) and (3, 3, 3). To solve the system by Gauss-Jordan elimination means to find this orthogonal complement of this vector space given by the "suitable" basis i.e. the basis as "simple" as possible (echelon reduced matrix is replaced by an identity matrix). Here the basis is as follows: [(1,1,1)]. Corresponding 3 figures of above solved 3 homogeneous systems are drawn in Cabri 3D Geometry in what follows.

136 134 Tomáš Zdráhal

137 Geometric solution of the system of linear equations 135 References [1] Hotová, E.: Řešené příklady k učebnici Geometrie 1 pro distanční vzdělávání a vizualizace geometrických pojmů. Diplomová práce, Olomouc, [2] Příhonská, J.: Berličky z algebry. Technická univerzita v Liberci, Fakulta pedagogická. Liberec, [3] Rygal, G., Bryll, G.: About definition of a periodic functions. Matematyka XII, Prace naukowe, Akademia im. Jana Dlugosza w Czestochowie, , [4] Zdráhal, T.: Systems of linear equations and the theory of vector spaces. Teaching mathematics: innovation, new trends, research. Scientific Issues, Catholic University in Ružomberok (2008),