ROLE OF GEOMETRY IN DIFFERENT DISCIPLINES OF MATHEMATICS

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1 Bulletin of the Marathwada Mathematical Society Vol. 17, No. 2, December 2016, Pages ROLE OF GEOMETRY IN DIFFERENT DISCIPLINES OF MATHEMATICS S.R. Joshi 8, Karmayog, Tarak Housing Society, Opp. Ramkrishna Mission Ashram, Beed-bypass, Aurangabad And V.V. Ambarwadikar, Department of Mathematics, J.E.S. College, Jalna Abstract Geometry is one of the few basic branches of Mathematics. It is one of the important and useful disciplines in Mathematics. In this paper we discuss and prove some results of different branches of mathematics, with the help of pure and coordinate geometry. We shall prove at least one result in each of the following disciplines of mathematics. (1) Classical Algebra (2) Trigonometry (3) Number theory (4) Analysis (5) Inequalities and (6) Group theory. 1 1 INTRODUCTION Among several branches of mathematics, Geometry is one of the basic and usful branches. In primary schools also it is introduced along with numbers and arithmetic operations. Because of various aspects and concepts in Geometry, it gave birth to other disciplines of mathematics. For example the concept of distance gave birth to Metric Spaces. The concept of line segment and direction gave birth to Vector Analysis. Topology, a very important branch of Analysis, is also known as rubber sheet geometry. The set of rotations and reflections of an regular polygon about certain lines has helped to develop the theory of Groups and so on. It is wonderful to note that some properties of logarithm can also be proved with the help of geometry. 1 Keywords: Inequalities, Logarithm, Countable set, Group. AMS Classification: 11B37, Marathwada Mathematical Society, Aurangabad, India, ISSN

2 14 S.R. Joshi and V.V. Ambarwadikar In this paper we discuss and prove some results of different disciplines of mathematics, with the help of pure and coordinate geometry. We shall prove or state at least one result in each of the following disciplines. of mathematics. (1) Classical Algebra (2) Trigonometry (3) Number Theory (4) Analysis (5) Inequalities and (6) Group Theory. 2 CLASSICAL ALGEBRA By classical algebra we mean that algebra which in general, is taught in high school classes. In this section we illustrate two results related to classical algebra. Our first result gives the sum of the first n natural numbers. Proposition 2.1 If n is any positive integer, then Proof: n k = n(n + 1)/2. (2.1) k=1 We give the proof for n = 9 only. The readers may think how to give a geometrical proof in general. In the following figure there are 10 rows and in each row there are 9 symbols each symbol being a dark circle or a simple circle. Hence there are 90 symbols out of which 45 (half of 90) are dark and half simple. (see figure 2.1) Fiugure 2.1 If we consider only the simple circles we observe that in the i th row there are (i 1) simple circles where 1 i 10. Further there are equal number of simple and dark

3 ROLE OF GEOMETRY IN... OF MATHEMATICS 15 circles. Hence the total number of simple cicles = (9 10)/2. Therefore we conclude that = (9 10)/2. The proof is complete. Our next result is related to the sum of an infinite G.P., for which the common ratio is less than 1. Proposition 2.2 Show that = 1. (2.2) 27 Proof: In this result we have to show that the sum of the G.P. for which the first term is 2/3 and the common ratio is 1/3 is 1. For this refer Figure 2.2 given below. In this figure one rectangular strip ABCD of unit area is considered and it is first divided into two disjoint sub-rectangles APSD and PBCS, such that the areas of these sub-rectangles are in the ratio 2 : 1. Hence their areas are 2/3 and 1/3 square units respectively. Then the strip PBCS is divided into two disjoint sub-rectangles PQRS and QBCR respectively such that, their areas are in the ratio 2:1 as shown in the figure. Again the rectangle QBCR is divided into two disjoint sub-rectangles QEFR and EBCF, such that the areas of these sub-rectangles are in the ratio 2:1. The procedure of dividing the next one third strips is continued up to infinity. Figure 2.2 By our construction and the method of division we have adopted, we observe that Area of ABCD = 2/3 unis, Area of PQRS = (2/3)(1/3) unis, Area of QEFR = (2/3)(1/3)(1/3) unis, and so on. Adding all these areas we get the area of the complete rectangle ABCD, which is one unit. Hence we have The proof is complete = 1. (2.3) 27 Remark 2.1 The method adopted in the above result can be used to prove the following general result. If a and r are two positive numbers with r < 1 then, a + ar + ar 2 + ar 3 + = a 1 r. (2.4) For this purpose we have to replace the division ratio 2:1 by (1 r) : r, and the first term 2/3 by a. Further we have to start with a rectangular strip whose area is a/(1 r) instead of 1.

4 16 S.R. Joshi and V.V. Ambarwadikar 3 TRIGONOMETRY In this section we give a geometrical proof of one result related to Trigonometry. Proposition 3.1 If x and y are any two positive numbers then, For the proof see [4]. tan 1 1 x = 1 tan 1 x + y + y tan 1 x 2 + xy + 1. (3.1) Remark 3.1 By giving different positive values to x and y in the above result we get different trigonometric identities, each of which can also be proved separately with the help of pure plane geometry. For example if we take x = 1 andy = 6, we get the well known Hutton s Formula given below. π 4 = tan tan = 2 tan tan (3.2) Similarly by taking x = 1 and y = 4 we get, what is known as Strassnizky s Formula given by π 4 = tan tan = tan tan tan (3.3) These formulae were asked in the Problem corner section of [2] to solve and their solutions were given in Problem corner section of [3]. 4 NUMBER THEORY In this section we give a geometrical proof of a result related to Fibonacci sequence. Proposition 4.1 If {f n } is the Fibonacci sequence given by, then, f n+2 = f n + f n+1, n 0, f 0 = 0, f 1 = 1, (4.1) 1 + n f k = f n+2, n > 0. (4.2) k=1 Proof: In the following figure (Figure 4.1), the recursion relation f n+2 = f n + f n+1, n 0 is shown with x n = f n and by considering an isosceles trapezium with sides and angles as shown in the figure. In the Figure 4.2, all such trapeziums with sides f 1, f 2,, f 3,,, f 7 are shown, the first trapezium being a triangle with upper side having 0 length. Let ABCD be a trapezium with CD > AB and AC = BD and the Angle ABE = 60 degrees, Where E is a point on CD such thay CE = AB, then the sides AB, AC and CD can be considered as the n th, (n+1) th and (n+2) th terms of the Fibonacci sequence {f n } respectively, for different values of n. This means that CD = CA + AB = DB + AB, i.e. lower side = left side + upper side = right side + upper side. We shall use this property in the proof.

5 ROLE OF GEOMETRY IN... OF MATHEMATICS 17 Figure 4.1 Figure 4.2 In Figure 4.2 the numbers f 2, f 3, f 4, f 5 and f 6, f 7 represent bottom sides or upper sides of the SIX isosceles trapeziums with different bases. Some of them are also coinsidered with side lengths of other trapeziums. We have tken n = 5 just for simplicity. Now,( see Figure 4.3) Figure 4.3 f 7 = A 5 B 5 = A 5 C 5 + C 5 B 5 = f 5 + C 5 B 5 = f 5 + A 4 B 4 = f 5 + A 4 C 4 + C 4 B 4 = f 5 + C 4 B 4 + A 4 C 4 = f 5 + f 4 + A 4 C 4 = f 5 + f 4 + A 3 B 3 = f 5 + f 4 + A 3 C 3 + C 3 B 3 = f 5 + f 4 + f 3 + A 2 B 2 = f 5 + f 4 + f 3 + C 2 B 2 + A 2 C 2 = f 5 + f 4 + f 3 + f 2 + A 1 B 1 = f 5 + f 4 + f 3 + f 2 + C 1 B 1 + A 0 B 0 = f 5 + f 4 + f 3 + f 2 + f This completes the proof of the desired result.

6 18 S.R. Joshi and V.V. Ambarwadikar 5 ANALYSIS Proposition 5.1 (1,4) If Z is the set of all integers then, the set Z Z of all order pairs of integers is countable. Remark 5.1 By omitting the points of the form (x, 0) (points on Y axis) where x is any integer, we see that the set of all rational numbers is also countable because every rational number can be considered as an order pair (x,y) of two integers with y 0. Further if we consider a spiral like figure joinig the points (0,0), (1,0), (-1,0), (2,0), (-2,0), (3,0), (-3,0) etc. in this order, so that all points on the X-axis with integer coordinates are covered one by one; we see that the set Z of integers itself is also countable. 6 INEQUALITIES In this section we shall consider geometrical proofs of two inequalities both being related to positive real numbers. Proposition 6.1 If x is a positive real number then, x + 1/x 2.[5] In [5 ] four proofs of this result are given without giving any explanation. We shall consider only two of them. Proof 1: Let x > 1. Consider a right angled triangle ABC with BC = x 1/x and the hypotenuse = x + 1/x. From this we observe that the length of the third side i.e. AB is 2. ( see Figure 6.1) Note that x 1/x > 0 since x > 1 and 1/x < 1. Now AC > AB implies that x + 1/x > 2 and the proof is complete. The proof of the inequality for x < 1 can be given similarly by replacing x and 1/x by 1/x and x respectively. Further when x = 1 it is clear that 1/x is also 1 and in this case we get an equality instead of an inequality. Figure 6.1 Proof 2: In this proof we have used coordinate geometry instead of pure plane geometry. Consider the point A(1, 1) on the rectangular hyperbola xy = 1, i.e.y = 1/x.

7 ROLE OF GEOMETRY IN... OF MATHEMATICS 19 It can be verified that the equation of the tangent at A(1, 1) is x + y = 2. Take any point P (x, y) on the hyperbola. Draw a line P Q parallel to the x-axis to meet the tangent in Q. Let Q = (a, b). Then, clearly b = y. Also x a (why). Hence x + y a + b i.e. x + 1/x 2, since a + b = 2 and the proof is complete. 7 GROUP THEORY In this section we shall consider a geometrical proof of one result related to Group Theory given below. Proposition 7.1 Let a, b, c represent any three letters or symbols. Let S = {a, b, c} and let G be the set of all one one and onto mappings on the set S. Such mappings are also known as permutations on the set S. There are six different permutations in all. Define a binary operation on G as the usual composition of two mappings in S. Then G is a non commutative group with respect this binary operation. Proof: The Six permutations p 1, p 2, p 3, p 4, p 5, p 6 on S can be described as ( ) ( ) ( ) a b c a b c a b c p 1 =, p a b c 2 =, p b c a 3 =, c a b p 4 = ( ) a b c, p b a c 5 = ( ) a b c, p c b a 6 = ( ) a b c. a c b Hence G = {p 1, p 2, p 3, p 4, p 5, p 6 }. In order to show that G is a group, we first give a geometrical meaning to these permutations Figure 7.1 Imagine an equilateral triangle ABC with O as the centroid. Consider the line L as that line passing through O and perpendicular to the plane of the triangle ABC. In the Figure 7.1 different positions of the vertices of this triangle are shown with corresponding small letters a, b, c respectively. Imagine first three permutations

8 20 S.R. Joshi and V.V. Ambarwadikar p 1, p 2, p 3 as rotations of the triangle ABC about the line L as axis of rotation in clock wise direction through the angles of 0 degree, 120 degrees and 240 degrees respectively. The first permutation p 1 can also be thought as rotation of the triangle about L through 360 degrees. Further imagine the last three permutations p 4, p 5, p 6 as reflections of the triangle about the lines cn, bm and al respectively, where, cn, bm and al are perpendiculars drawn from c on ab, from b on ac and from a on bc respectively. (see Figure 7.1) Figure 7.2 In order to show that G is a group, we have to show that p i op j G, where p i and p j G and by p i op j we mean composition of p i and p j. There are 36 such combinations. In the Figure 7.2 shown above only three combinations are considered. The geometrical meaning of p i op j is that first p i acts on the triangle and then p j acts on the resulted position of the triangle. From mathematics point of view p i op j, this meaning is just opposite as far as the order of operation is concerned. The first two rows of figure 7.2 indicate that p 2 op 4 = p 6 and p 4 op 2 = p 5. Note that when p 4 is applied on the triangle abc (order abc is important) we get a new position of the triangle as bac after reflexion. When reflexion p 4 takes place the vertex c on the right side remains unchanged as far as its position is concerned. When p 2 acts on the new position of the triangle i.e. bac, then rotation of this triangle through 120 degrees in the clockwise direction takes place and we get the position of the triangle as acb. This position is also obtained when only p 6 is applied on the triangle abc. This is so because when p 6 is applied the upper vertex remains unchanged and the remaining two vertices are interchanged in the position. Thus p 2 op 4 = p 6. Similarly from the 2nd row of Figure 7.2, we get p 4 op 2 = p 5. By considering other combinations we can show that p i op j G, whenever p i and p j G. It is also easy to observe that the permutation p 1 plays the role of identity element in G. Further by considering the geometrical meaning of these Six permutations it is simple to verify that p 1 op 1 = p 1, p 2 op 3 = p 1, p 3 op 2 = p 1, p 4 op 4 = p 5 op 5 = p 6 op 6 = p 1. This shows

9 ROLE OF GEOMETRY IN... OF MATHEMATICS 21 that every element in G has an inverse element in G. Lastly because p 2 op 4 = p 6 and p 4 op 2 = p 5, we conclude that the set G is a group and that it is not commutative. References [1] Des Machale, Z Z is a countable set, Vol.77, No. 1. Feb. 2004, pp [2] Executive Editors, Problem Corner, Bull. Marathwada Math. Soc., Vol.15, No. 1, June (2014), [3] Executive Editors, Problem Corner, Bull. Marathwada Math. Soc., Vol.15, No. 2, December(2014), [4] Joshi S.R., Geometrical Proofs of Some Mathematical Results, Bull. Marathwada Math. Soc., Vol.7, No. 2, December(2006), [5] Roger B. Nelsen, The sum of a positive number and its reciprocal is at least two (four proofs), Vol.66, June 1993, pp [6] Vincent Terlini, Logarithm of a number and its reciprocal, Vol. 24, No. 1. Feb.2001, pp

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