A proposition from algebra: How would Pythagoras phrase it?

Transcription

1 A proposition from lgebr: How would Pythgors phrse it? Moshe Stupel, Gordon Acdemic College of Eduction, Hif, Isrel & Shnn Acdemic College of Eduction, Hif, Isrel Avi Sigler, Shnn Acdemic College of Eduction, Hif, Isrel Abstrct: The uthors present three methods from different brnches of mthemtics to solve the sme tsk. Originlly n lgebric problem, the first solution presented utilizes n lgebric method. Subsequently, the uthors trnslte the question into geometric lnguge nd explore wys Pythgors might phrse nd solve the problem. Lstly, they shre geometric nd trigonometric solutions deling with specil tringle whose ngles re α, α, nd 4α - geometric progression. Solution of the sme problem using different methods deepens student mthemticl understnding while promoting mthemtics s field composed of intertwining brnches. Keywords. Rich tsks, problem solving, connections, proof 1 Introduction In recent times, much considertion hs been given to problem solving nd teching methods tht connect vrious res of mthemtics into more unified whole. Eductors in the field of mthemtics gree tht using more thn one pproch to solve the sme problem promotes the development of mthemticl resoning (Poly, 1973; Schoenfeld, 1985; Ntionl Council of Techers of Mthemtics, 000). The solution of problems using different methods encourges flexibility nd cretivity (Tll, 007; Leiken & Lev, 007). Moreover, proving result or solving problem using methods from vrious brnches in mthemtics (e.g., geometry, trigonometry, nlyticl geometry, vectors, complex numbers) deepens mthemticl understnding. Our pproch of presenting multiple proofs of the sme problem s tool for constructing mthemticl connections is supported by Poly (1973, 1981), Schoenfeld (1988), Ersoz (009), The Ntionl Council of Techers of Mthemtics (000), nd Levv-Wynberg nd Leiken (009). Similr to the ide of one problem with mny solutions/proofs, is the ide of multiple solution tsks presented by Leiken & Lev (007), Leiken (009), Levv-Wynberg nd Leiken (009), nd Stupel & Ben-Chim(013). Leiken (009) points out tht differences between proof methods my be explined long four dimensions, nmely: (1) different representtions of mthemticl concept; () different properties (definitions or theorems) of mthemticl concepts from prticulr mthemticl topic; (3) mthemticl tools nd theorems from different mthemticl brnches; Pge Ohio Journl of School Mthemtics 75

2 (4) different theorems nd tools from different subjects (not necessrily from mthemtics). In our cse, we pply the third type of difference: in this pper, we present different solutions using tools nd theorems from Eucliden geometry, nlyticl geometry, trigonometry, vectors, nd complex numbers. We present geometric proof for n lgebric proposition tht students typiclly prove using lgebric techniques. Our geometric proof is composed of four uxiliry propositions. The proposition If, b, nd c re positive numbers, nd (1) nd () hold: + c = b (1) b + b = c () then we my conclude + bc = c (3) 3 An lgebric solution The solution presented here is one of severl possible methods ccessible to students in erly secondry level courses (i.e., typicl yer olds). From (1) we eliminte nd obtin c = b (4) Next, we substitute the resulting expression for c in (), nd obtin b + b = ( b ) b( + b) = (b ) (b + ) Since nd b re positive, b + 0, therefore b = (b ) (b+). We expnd, collect similr terms, nd obtin b 3 b b + 3 = 0 (5) We denote + bc c s M nd prove tht M = 0 (which proves the proposition). We substitute b for c from (4), nd b + b for c from () into the expression for M, obtining M = + b(b ) (b + b) From (5) we obtin: M = 3 +b 3 b b b + bc = c. = 3 b b +b 3 = 0 nd, hence, it follows tht Ohio Journl of School Mthemtics 75 Pge 3

3 4 How would Pythgors hve formulted the problem, hd he lived tody? Now we recst the problem in geometric context, sking how Pythgors my hve formulted the proposition. In ncient Greece, ment the re of the squre whose side length is, nd bc ment the re of the rectngle whose side lengths re b nd c. Therefore, is probble tht Pythgors would hve given the problem geometricl interprettion such s the following. 4.1 The proposition in its geometricl form Recsting the originl proposition in geometric context, we re given three segments whose lengths re, b, c, s shown in Figure 1. Fig. 1: Geometric interprettion of the originl proposition. In corresponding mnner, we prove tht the geometric equivlent holds, s shown in Figure. Fig. : Geometric interprettion of the originl result. Note tht the nottion inside the squres is modern script. The Greeks would likely hve only used the drwings of the qudrilterls. Pge 4 Ohio Journl of School Mthemtics 75

4 5 A geometric proof The proof we offer here is comprised of 4 uxiliry propositions (lemms), fter which we present proof of the min proposition. Lemm 5.1. The following inequlity holds: c > b > Proof. From (1) it follows tht b >, nd from () it follows tht c > b. Lemm 5.. The following inequlity holds: + b > c. Proof. It is given in () tht b + b = c. From proposition (1) it follows tht c > b, therefore it is cler tht + b > c, becuse ( + b)b = c. From the segments, b, c, one cn form tringle (by combining propositions (1) nd ()). Lemm 5.3. If in the tringle ABC, whose sides re, b, c, nd whose ngles re α, β, nd γ, there holds (1) + c = b, then β = α. Fig. 3: Geometric interprettion of Lemm 5.3. Proof. We extend CB to D, so tht BD = AB (see Figure 3).Therefore DC = + c. But from the dt there holds ( + c) = b, therefore +c b = b DC, in other words AC = AC BC. Therefore, from the first theorem of similrity, ABC DAC, nd therefore D = α. But DB = AB (the uxiliry construction), nd hence β = α. Lemm 5.4. If in the tringle ABC whose sides re, b, c, nd whose ngles re α, β, nd γ, there holds + c = b nd b + b = c, then γ = 4α (see Figure 4). Fig. 4: Geometric interprettion of Lemm 5.4. Proof. From Lemm 5.3, β = α. We extend AC to E, so tht BC = CE = (see Figure 5). From the dt, there holds b + b = c, therefore b( + b) = c. nd hence it follows tht b c = c +b. In the sme mnner (s in the proof of Lemm 5.3), we hve ABC AEB. Therefore, E = α, but the tringle BCE is n isosceles tringle, nd hence γ = 4α. Ohio Journl of School Mthemtics 75 Pge 5

5 Fig. 5: Extending AC to E. 5.1 Conclusion from Lemms If, b, nd c re three segments tht sttisfy the reltions + c = b nd b + b = c, then from the segments, b, nd c, one cn form ABC with mgnitudes of ngles A, B, nd C equl to α, α, nd 4α, respectively. Therefore, 7α = 180, nd α = (see Figure 6). Fig. 6: Angle mesures in ABC. We now return to the min proposition nd prove it. Theorem 5.5. To show: + c = b b + b = c = + bc = c Proof. See Figure 7. We mrk off the segment AD on AB, so tht, AD = AC = b(c > b). Therefore the length of the segment BD is c b. The tringle ADC is isosceles with vertex ngle of α, nd therefore ech of the bse ngles is 3α (since it ws proved tht α = ), nd lso C = 4α, nd therefore DCB = α. Hence, it follows tht BCD BAC, nd there holds BC = BD BA, in other words = (c b)c, nd it follows tht + bc = c. 6 A trigonometric proof In order to present mthemtics s n extensive field comprised of vrious interconnected brnches, we present trigonometric proof of the tsk. Theorem 6.1. Given is, b, c > 0 nd + c = b (6) b + b = c (7) Pge 6 Ohio Journl of School Mthemtics 75

6 Fig. 7: Visul confirmtion of the reltionship. We prove the following. + bc = c (8) Proof. As in the geometric proof, we note tht c > > b nd + b > c nd therefore the lengths of the segments, b, nd c represent segments tht form tringle. From the Lw of Sines, we obtin from (6) tht sin ( A) + sin( A) sin( C) = sin ( B) = (9) sin( A) sin( C) = sin ( B) sin ( A) (10) = (sin( B) sin( A))(sin( B) + sin( A)) (11) Using the trigonometric identities for the sum nd difference of ngles, we obtin the following. B A B + A B + A B A sin cos sin cos = sin( A + B) sin( B A) (1) Since sin C 0, we obtin. = sin C sin( B A) (13) sin A = sin( B A) (14) Since this is tringle, we hve A = B A = B = A. In the sme mnner, from (7) we obtin C = B. From the reltions obtined between the ngles, we hve C = B = 4 A, nd since the sum of ngles in the tringle is A + B + C = 180, we obtin From which there follows: And therefore A + A + 4 A = 7 A = 180 = C = A (15) C = A = C A = 180 C (16) sin( C A) sin B = sin B sin C (17) sin C sin A = sin B sin C = + bc = c (18) Ohio Journl of School Mthemtics 75 Pge 7

7 7 Concluding remrks nd implictions for teching We constructed proofs of our proposition using combintion of severl fields in mthemtics. In our work, geometry cme to the id proof in lgebr while lgebr provided the source of our chllenging problem. We used trigonometry to construct nother proof. Through it ll, it ws pprent to us how beutiful mthemtics is. In the words of Hrdy (1940), the mthemticin s ptterns, like the pinter s or the poet, must be beutiful; the ides, like the colors or the words, must fit together in hrmonious wy. Beuty is the first test: there is no permnent plce in this world for ugly mthemtics. The three different methods we employed resulted in the sme reltions between the sides of the tringle. Geometry is goldmine for multiple solution tsks. Proofs my be derived by pplying different methods within the specific topic of geometry or within other mthemticl res. The multiple solutions tht were presented herein for one problem demonstrte the connectivity between different res of mthemtics. Multiple proofs foster both better comprehension nd incresed cretivity in mthemtics for the student/lerner nd chllenge for the techer. References Ersoz, F. A. (009). Proof in different mthemticl domins. ICME Study, 1, Hrdy, G. H. (1940). A mthemticin s pology. Cmbridge, UK: Cmbridge University Press, 40. Levv-Wynberg, A., & Leikin, R. (009, Jnury). Multiple solutions for problem: A tool for evlution of mthemticl thinking in geometry. In Proceedings of CERME (Vol. 6, pp ). Leikin, R., & Lev, M. (007, July). Multiple solution tsks s mgnifying glss for observtion of mthemticl cretivity. In Proceedings of the 31st interntionl conference for the psychology of mthemtics eduction (Vol. 3, pp ). Seoul, Kore: The Kore Society of Eductionl Studies in Mthemtics. Leikin, R. (009). Multiple proof tsks: Techer prctice nd techer eduction. ICME Study, 19(), Leikin, R., & Lev, M. (007, July). Multiple solution tsks s mgnifying glss for observtion of mthemticl cretivity. In Proceedings of the 31st interntionl conference for the psychology of mthemtics eduction (Vol. 3, pp ). Seoul, Kore: The Kore Society of Eductionl Studies in Mthemtics. Ntionl Council of Techers of Mthemtics (Ed.). (000). Principles nd stndrds for school mthemtics (Vol. 1). Reston, VA: Author. Poly, G. (014). How to solve it: A new spect of mthemticl method. Princeton University Press. Schoenfeld, A. H. (1985). Metcognitive nd epistemologicl issues in mthemticl understnding. Teching nd lerning mthemticl problem solving: Multiple reserch perspectives, Schoenfeld, A. H. (1988). When good teching leds to bd results: The dissters of well-tught mthemtics courses. Eductionl Psychologist, 3(), Stupel, M., & Ben-Chim, D. (013). One problem, multiple solutions: How multiple proofs cn connect severl res of mthemtics. Fr Est Journl of Mthemticl Eduction, 11(), 19. Tll, D. (007, Mrch). Techers s Mentors to encourge both power nd simplicity in ctive mthemticl lerning. In Plenry t The Third Annul Conference for Middle Est Techers of Science, Mthemtics nd Computing (Vol. 3, pp ). Pge 8 Ohio Journl of School Mthemtics 75

8 Moshe Stupel, is professor nd pre-service techer eductor in two cdemic colleges, Shnn Acdemic College nd Gordon College. He hs published nd presented 40 ppers on mthemtics nd mthemtics eduction. Recently, his reserch is focused on vrious methods of problem solving nd on vrince nd invrince property in dynmic geometricl environments (DGEs) in mthemtics. Dr. Avi Siegler hs studied mthemtics eduction in secondry schools for lmost four decdes. In recent yers he hs worked s senior lecturer in vrious colleges of Techer Eduction. In ddition, he hs published mny rticles in the field of geometric constructions, with prticulr emphsis on specil fetures tht exist in different geometric shpes. Ohio Journl of School Mthemtics 75 Pge 9