TOPIC 3: VISUAL EXPLANATIONS (PROOFS) (Pge references to Proof re to Bndll, P R et l, Proof in Mthemtics, KMEP, 2002). 3. The tringulr numbers form the sequence, 3, 6, 0,, 2,... T T 2 T 3 T 4 The m be illustrted b tringulr ptterns of numbers (hence their nme) s shown: From the digrms it is cler tht T 3 = 6 = + 2 + 3, T = = + 2 + 3 + 4 +, nd T n = + 2 + 3 +... + n. 3.2 Activities. Conjecture formul for T n in terms of n. Now prove our conjecture. 2. Choose two consecutive tringulr numbers nd dd. eg. 6 + 0 = 6. Repet for other such pirs. Conjecture? 3. Eplin how to prove the result in () b studing the digrm (A) below. 4. Eplin how to prove the result in (2) b studing the digrm (B) below. A B A P B 3 S Q 3 3 D R C 8. (Proof, p0) ABCD is squre, PQRS re mid-points of sides. Conjecture: the octgonl region outlined in the centre is regulr. (Hint: Use grdients) 6. Mke squre of thin crd 8 units b 8 units. Cut it up into 4 pieces s shown. Now re-ssemble these to mke rectngle 3. The re is now 6 squre units. Where hs the etr squre come from? Wht is the fllc here? 7. Drw digrms to illustrte 2 3 = 3 2 k( + b) = k + kb (+b) 2 = 2 + 2b + b 2 k( -b) = k - kb ( + b)(c + d) = c + bc + d + bd 9
8 How do these digrms illustrte the corresponding sttements? -b b b -b b -b -b b 2 -b 2 = (+ b)( - b) ( - b) 2 = 2-2b + b 2 9. A prticulr cse of Pthgors' theorem: 4+2=29 9 4 2 2 0. (from n ide of D nd B Bll) ² ² 2b 2b c² 2cb 2c `( + b) 2 = 2 +2b + b 2 ( + b + c) 2 = 2 + b 2 + c 2 + 2b + 2bc + 2c ² 2b c² 2cb 2c d² 2dc 2db 2d ( + b + c +d) 2 = 2 + b 2 + c 2 + d 2 + 2b + 2c + 2d +2bc + 2bd + 2cd 20
² b c b bc c cb c² ( + b + c) 2 = 2 + b 2 + c 2 + 2b + 2bc + 2c 3.3 Some other demonstrtions. If + is fied - s + = 0 - then 2 + 2 is lest when =. The points P, Q, R,.. lie on the line + = 0. If P is such point with coordintes (,) then + = 0, nd lso 2 + 2 = OP 2. Clerl OP is lest when P is t N, where ON is perpendiculr to AB - nd then = =. 0 A 2 + =0 C N B R 2. This m be dpted to solve: If 2 + = 0, find the minimum vlue of 2 + 2. The minimum vlue is 20, when =4 nd =2. (For it occurs t N, nd b similr tringles, NR = OR/2, = /2). 3.4 Further emples -see Proof, p3.. If is constnt, then + is lest when = 2. If + is constnt, then is gretest when = 3. The results of 3.3. nd 3.4., 3.4.2 m lso be proved lgebricll s follows: (A) ( + ) 2 + ( - ) 2 = 2( 2 + 2 ) If + = k, 2( 2 + 2 ) = k 2 + (-) 2 nd is lest when (-) 2 = 0. (B) ( + ) 2 - ( - ) 2 = 4. If = k, then ( + ) 2 = 4k + ( - ) 2 nd is lest when (-) 2 is zero. [The remining result follows similrl] 0 + = 0 R Q N P 0 S 4. The identities (A) nd (B), nd the second one of (8), bove, m ll be illustrted in one 'flg' digrm: (8ii) (-) 2 = 2 + 2-2,. [using =( - ) 2 ] (A) (+) 2 + (-) 2 = 2( 2 + 2 ) [overlp is =( - ) 2 ] (B) ) ( + ) 2 - ( - ) 2 = 4; [ = ] [We lso hve (+) 2 = 2 + 2 + 2; = ] 2
3. Another demonstrtion relting + nd (dpted from Y. Kobshi, M Gz. 293, No. 06, Jul 2002) < PQR/2 unless = ie. <(+) 2 /2, unless = So if is given, + is lest when = If + is given, is gretest when = Q fold P R = 3.6 Wh digrms m be indequte Cre is needed in rguing from digrms, especill in nlsis. Thus (i) Rolle's theorem sttes tht if f() is continuous nd differentible in the closed intervl [,b] nd f() = f(b) then with < <b such tht f '( ) = 0 (ii) A theorem of Bolzno sttes; If f() is continuous in the closed intervl [,b], nd f(), f(b) hve opposite signs, then f( ) = 0 for some rel, < < b. b b Both theorems re "obvious" from the digrms" (nd indeed the digrms help to conve the sense of the theorems - the re useful to the intuition. Yet the theorems require proof in course of nlsis, nd the requirement becomes more obvious if we consider the following:- () In (i) tke f() = tn in the intervl [0, ]. Here there is point of discontinuit t = /2. (b) In (i) consider f() = - in [-, ] f is not differentible t = 0. (c) In (ii) consider f() = 2-2 between = 0 nd =2, but suppose the line = 0 to consist onl of rtionl points. (Since these re 'densel clustered', the 'picture' will be the sme. The curve would meet the is where = 2 - but there is now no point of the line = 0 for which = 2). (d) Fermt stted tht 3 + 3 = hs no solutions in rtionl numbers (other thn the obvious ones (.) = (0,) or (,0)). Thus the curve in the (,) plne contins onl these two rtionl points nd so threds its w through ll the other rtionl points without pssing through n of them. ) b) 0 - c) d) ³+³= P 22 +=0
In ech of the cses (), (b), (c) the conditions of the theorem re shown to be importnt, nd the forml proofs of these theorems indicte eplicitl wht is the role pled b these conditions, nd show lso tht there re no other conditions, implicit in the digrm, which we hve omitted to consider, i.e. the conditions s stted re dequte to gurntee the truth of the theorems. 3.7 Some hrder emples (see lso Proof pp2-7).. Use sketch grph of the curve = / to show tht u/(+u) < ln(+u) < u for u > -, (with equlit onl if u=0) [Consider the es cse u>0 first; ln( + u) = When - < u < 0, ll three epressions re negtive] u d = shded re. =/ =/ +u n n+ 2n 2. Let g(n) = /(n+) + /(n+2) +... + /(2n). When n is ver lrge, /(n+) is ver smll (ie. s n, /(n+) 0). Similrl for the other terms. "So g(n) will be ver smll for lrge n nd s n, g(n) 0" This is incorrect! Show b considering the sketch tht g(n) ln(2), nd more precisel tht ln(2) -/(2n) < g(n) < ln(2), so tht g(n) ln(2) = 0.69347... s n 3. ) Prove: sin > 3 /3!, >0 b) Prove e > + + 2 /2! + 3 /3! >0. 4. Show + /2 + /3 + /4 +... + /n ln(n). Which is the greter, nd b pproimtel how much? [See lso Proof, p4, e A3.6 for demonstrtions of n i= n 2 n nd n 3 ]. i= 3.8 Conclusion Though digrms cn be misleding (see bove, nd lso Proof p0, (3.2.2), p6, (3.0) ), the cn lso provide vluble insight into wh result is true (wheres lgebric mnipultion, though convincing, m still leve us without n 'overview' of the sitution). 23
[It is interesting tht so mn of our everd metphors for understnding re connected with 'seeing' - insight, illumintion, look t it this w, in the drk, now I see..., enlightenment, flsh of inspirtion...] Don't be frid of digrms - mthemticins t ll levels use them more often thn ou might think. Like ll the other 'scffolding', flse strts nd blind lles which usull go to mke up piece of mthemtics, the re often removed before work is put on public view! 24