# Introduction to Mathematical Reasoning, Saylor 111

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1 Frction versus rtionl number. Wht s the difference? It s not n esy question. In fct, the difference is somewht like the difference between set of words on one hnd nd sentence on the other. A symbol is frction if it is written certin wy, but symbol tht represents rtionl number is rtionl number no mtter how it is written. Here re some exmples. The symbol is π frction tht is not rtionl number. On the other hnd 2 is both frction nd 3 rtionl number. Now 0.75 is rtionl number tht is not frction, so we hve exmples of ech tht is not the other. To get little deeper, frction is string of symbols tht includes frction br. Any rel number cn be written s frction (just divide by ). But whether number if rtionl depends on its vlue, not on the wy it is written. Wht we re sying is tht in the cse of frctions, we re deling with syntctic issue, nd in cse of rtionl numbers, semntic issue, to borrow two terms from computer science. For completeness, we sy tht number is rtionl if it CAN be represented s quotient of two integers. So 0.75 is rtionl becuse we cn find pir of integers, 3 nd 4, whose quotient is Here s nother wy to think bout the difference. Consider the question Are these numbers getting bigger or smller? This pprently musing question cn provoke some serious questions bout wht we men by the word number. Indeed, there re two spects of numbers tht often get blurred together: the vlue of number nd the numerl we write for the number. By number we usully men the vlue while the word numerl refers to the symbol we use to communicte the number. So the numbers bove re getting smller while the numerls re getting bigger. This contrst between symbol nd substnce lso explins the difference between rtionl number nd frction. A frction is numerl while rtionl number is number. Rtionl Numbers. The most common wy to study rtionl numbers is to study them ll t one time. Let s begin. A rtionl number is number which cn be expressed s rtio of two integers, /b where b 0. Let Q denote the set of ll rtionl numbers. Tht is, 6. Q = {x x = /b,, b Z, b 0}, where Z denoted the set of integers. The following exercises will help you understnd the structure of Q.. Prove tht the set Q is closed under ddition. Tht is, prove tht for ny two rtionl numbers x = /b nd y = c/d, x + y is rtionl number. Solution: We simply need to write x + y s rtio of two integers. Becuse

2 of the wy we dd frctions, x + y = b + c d = d bd + cb db d + bc =. bd But d + bc nd bd re both integers becuse the integers re closed under both + nd. 2. Prove tht the set Q is closed under multipliction. Tht is, prove tht for ny two rtionl numbers x = /b nd y = c/d, x y is rtionl number. Solution: We simply need to write x + y s rtio of two integers. Becuse of the wy we multiply frctions, x y = b c d = c bd Of course, c nd bd re both integers becuse the integers re closed under. 3. Prove tht the number midwy between two rtionl numbers is rtionl. Solution: Let x = /b nd y = c/d. The midpoint of x nd y is x + y 2 which is quotient of two integers. = d + bc 2bd, 4. For this essy, we ssume the set R of rel numbers is the set of ll positive nd negtive deciml numbers. These decimls hve three forms, those tht terminte, ie. hve only finitely mny non-zero digits, like ; those tht repet like = 4/3, nd those tht do not repet. Prove tht ll rtionl numbers of one of the first two types, nd vice-vers, ny number of the first two types is rtionl. Solution: To see tht ech rtionl number hs either terminting or repeting deciml representtion, suppose x = /b is rtionl with nd b integers. Dividing by b using long division, results in sequence of reminders, nd ech of these reminders is between 0 nd b. If the reminder is ever zero, 2

3 the division process termintes nd the resulting deciml hs only finite number of non-zero digits. If not, then we eventully get repet reminder, nd once this hppens, the reminders reoccur in blocks. The proof in the other direction requires more cre. See the essy on representtion for this proof. 5. Let z be positive irrtionl number. Prove tht there is positive rtionl r number less thn z. Solution: Since z is positive, we cn write z = x 0.x x 2..., where either x 0 or one of the x i is not zero. Let k denote the smllest index such tht x k is not zero. Then x 0.x x 2... x k is rtionl number less thn z. 6. Prove tht the rtionl numbers Q is dense in the set of rel numbers R. Tht is, prove tht between ny two rel numbers, there is rtionl number. Solution: Suppose x n y re ny two rel numbers with x < y. Let x = x 0.x x 2..., nd let y = y 0.y y Suppose k is the smllest subscript where they differ. Then y 0.y y 2... y k is rtionl number between x nd y. In the following problems, we need the notion of unit frction. A unit frction is frction of the form /n where n is positive integer. Thus, the unit frctions re /, /2, /3,..... s Addresses Divide the unit intervl into n-ths nd lso into m-ths for selected, not too lrge, choices of n nd m, nd then find the lengths of ll the resulting subintervls. For exmple, for n = 2, m = 3, you get /3, /6, /6//3. For n = 3, m = 4, you get /4, /2, /6, /6, /2, /3. Try this for n = 3 nd m = 5. Cn you find finer subdivision into equl intervls tht incorportes ll the division points for both denomintors? I got this problem from Roger Howe. Solution: It is probbly best to do fter displying nd scertining some pprecition for the wy the multiples of fixed unit frction divide the number ry into equl intervls. In contrst, when you superimpose two such divisions, the resulting intervls will be quite different in length, nd the sitution my seem chotic. A tke wy would be, you cn, nd the LCM of the denomintors will give such. Also, there will lwys be t lest one intervl whose length is /LCM, nd of course, ll intervls will be multiples of /LCM. 2. Here s problem from Trin Your Brin, by George Grätzer. It is difficult to subtrct frctions in your hed, sid John. Tht s right sid Peter, but 3

4 you know. there re severl tricks tht cn help you. You often get frctions whose numertors re one less tht their denomintors, for instnce, It s esy to figure out the difference between two such frctions = = 4. Simple, right? Cn you lwys use this method? Solution: Yes, this lwys works. Let < b. The b b + + nd b + b+ = ( + )b (b + ) (b + )( + ) = be two such frctions with b + ( + ) (b + )( + ). 3. Show tht every unit frction cn be expressed s the sum of two different unit frctions. Solution: Note tht n = n+ + n(n+). 4. Sums of unit frctions () Notice tht 2/7 is expressible s the sum of two unit frctions: 2/7 = /4 + /28. But 3/7 cnnot be so expressed. Show tht 3/7 is not the sum of two unit frctions. (b) There is conjecture of Erdös tht every frction 4/n where n 3 cn be written s the sum of three unit frctions with different denomintors. Verify the Erdös conjecture for n = 23, 24, nd Solution: = (c) Cn you write s sum of different unit frctions ll with odd denomintors? Solution: Yes, it cn be done. One wy is to dd the reciprocls of ech of the following: 3, 5, 7, 9,, 5, 2, 65, 693. I suspect there re lots of other wys lso. (d) Cn ny rtionl number r, 0 < r < be represented s sum of unit frctions? Solution: You just use the greedy lgorithm: subtrct the lrgest unit frction less thn your current frction. You cn show tht the numertor of the resulting frction is lwys less thn the numertor you strted with, so the process converges. 4

5 5. In the Spce Between () Nme frction between /2 nd 2/3. frction stisfies the condition. Give n rgument tht your (b) Nme frction between /5 nd 7/0. How bout 6/7 nd /3? (c) Nme the frction with smllest denomintor between /5 nd 7/0. Or 6/7 nd /3? (d) First drw red mrks to divide long stright bord into 7 equl pieces. Then you drw green mrks to divide the sme bord into 3 equl pieces. Finlly you decide to cut the bord into = 20 equl pieces. How mny mrks re on ech piece? (e) A bicycle tem of 7 people brings 6 wter bottles, while nother tem of 3 people brings wter bottles. Wht hppens when they shre? Some of this mteril is from Josh Zucker s notes on frctions tken from workshop for techers t Americn Institute of Mthemtics, summer Some of the mteril is from the book Algebr by Gelfnd nd Shen. 6. Dividing Horses This problem comes from Dude, Cn You Count, by Christin Constnd. An old cowboy dies nd his three sons re clled to the ttorney s office for the reding of the will. All I hve in this world I leve to my three sons, nd ll I hve is just few horses. To my oldest son, who hs been gret help to me nd done lot of hrd work, I bequeth hlf my horses. To my second son, who hs lso been helpful but worked little less, I bequeth third of my horses, nd to my youngest son, who likes drinking nd womnizing nd hsn t helped me one bit, I leve one ninth of my horses. This is my lst will nd testment. The sons go bck to the corrl nd count the horses, wnting to divide them ccording to their p s exct wishes. But they run into trouble right wy when they see tht there re 7 horses in ll nd tht they cnnot do proper division. The oldest son, who is entitles to hlf tht is 8 horses wnts to 2 tke 9. His brothers immeditely protest nd sy tht he cnnot tke more thn tht which he is entitled to, even if it mens clling the butcher. Just s they re bout to hve fight, strnger rides up nd grees to help. They explin to him the problem. Then the strnger dismounts, lets his horse mingle with the others, nd sys Now there re 8 horses in the corrl, which is 5

6 much better number to split up. Your shre is hlf he sys to the oldest son, nd your shre is six, he sys to the second. Now the third son cn hve one ninth of 8, which is two, nd there is = left over. The strnger gets on the 8 th horse re rides wy. How ws this kind of division possible. Solution: The sum of the three frctions is less thn : + + = 7. So the strnger s horse helps complete the whole. 7. Consider the eqution + b = 5 2. Find ll the ordered pirs (, b) of rel number solutions. Solution: Thnks to Rndy Hrter for this problem. First, let s try to find ll integer solutions. To tht end, rewrite the eqution s Thus 2( + b) = 5b. 0 = 5b 2b 2 = 5b 2b = b(5 2) 2 44 (5 2) 5 5 = (5 2)(5b 2) 44 Now, fctoring 44 = nd looking t pirs of fctors, we cn find ll the integrl solutions. Wht bout the rest. To this end, replce with x nd b with f(x) nd solve for f(x). We get f(x) = ( 5 2 ) = 2x x 5x 2. This rtionl function hs single verticl symptote t x = 2/5 nd zero t x = 0. So you cn see tht for ech 0 x 2/5, there is y such tht /x + /y = 5/2. 8. Suppose {, b, c, d} = {, 2, 3, 4}. 6

7 () Wht is the smllest possible vlue of + b +. c+ d Solution: We cn minimize the vlue by mking the integer prt s smll s possible nd then mking the denomintors s lrge s possible. Clerly = is the best we cn do, nd then b = 4 is certinly best, etc. So we get (b) Wht is the lrgest possible vlue of + = b +. c+ d Solution: We cn mximize the vlue by mking the integer prt s lrge s possible nd then mking the denomintors s smll s possible. Clerly = 4 is the best we cn do, nd then b = is certinly best, etc. So we get = Smllest Sum. Using ech of the four numbers 96, 97, 98, nd 99, build two frctions whose sum is s smll s possible. As n exmple, you might try 99/ /98 but tht is not the smllest sum. This problem is due to Sm Vndervelt. Extend this problem s follows. Suppose 0 < < b < c < d re ll integers. Wht is the smllest possible sum of two frctions tht use ech integer s numertor or denomintor? Wht is the lrgest such sum? Wht is we hve six integers, 0 < < b < c < d < e < f. Now here s sequence of esier problems tht might help nswer the ones bove. () How mny frctions /b cn be built with, b {, 2, 3, 4}, nd b? Solution: There re 2 frctions. (b) How mny of the frctions in () re less thn? Solution: There re 6 in this set, /2, /3, /4, 2/3, 2/4, 3/4. (c) Wht is the smllest number of the form + c, where {, b, c, d} = b d {, 2, 3, 4}? Solution: There re just three pirs of frctions to consider, A = + 2, 3 4 B = + 2, nd C = + 3. Why is C not cndidte for the lest

8 vlue? Note tht A bets B here becuse A = + 2 = + + while B = + +. Does this resoning work for ny four positive integers < b < c < d? The rgument for four rbitrry positive integers is bout the sme. Let A = + b, B = + b. Then A = + + b, c d d d c d d while B = + + b, nd since c < d, A < B. c d c (d) Wht is the lrgest number of the form + c, where {, b, c, d} = b d {, 2, 3, 4}? Solution: Agin there re just three pirs of frctions to consider, A = 3 + 4, B = 4 + 3, nd C = Why is C not cndidte for the lest vlue? Note tht B bets A. Does this resoning work for ny four positive integers < b < c < d? 0. Simpsons (with thnks to Brt nd Lis shoot free throws in two prctice sessions to see who gets to strt in tonight s gme. Brt mkes 5 out of in the first session while Lis mkes 3 out of 7. Who hs the better percentge? Is it possible tht Brt shoots the better percentge gin in the second session, yet overll Lis hs higher percentge of mde free throws? The nswer is yes! This phenomenon is clled Simpson s Prdox. () Find pir of frctions /b for Brt nd k/l for Lis such tht /b > k/l nd yet, Lis s percentge overll is better. Solution: One solution is tht Brt mkes 6 out of 9 in the second session while Lis mkes 9 out of 4. Now 2/2 > /20. The numbers 2/2 nd /20 re clled medints. Specificlly, given two frctions /b nd c/d, where ll of, b, c, nd d re positive integers, the medint of /b nd c/d is the frction ( + c)/(b + d). (b) Why is the medint of two frctions lwys between them? Solution: Imgine two groups of bicyclists, one with b riders nd bottles of wter mong them, nd the other with d riders nd c bottles of wter mong them. When they meet, they ll gree to shre eqully. Thus we hve b + d cyclists with totl of + c bottles of wter. Is it cler tht +c lies between /b nd c/d? b+d (c) Notice tht the medint of two frctions depends on the wy they re represented nd not just on their vlue. Explin Simpson s Prdox in terms of medints. (d) Define the medint M of two frctions /b nd c/d with the nottion M(/b, c/d). So M(/b, c/d) = ( + c)/(b + d). This opertion is sometimes clled student ddition becuse mny students think this would be good wy to dd frctions. Compute the medints M(/3, 8/9) nd 8

9 M(4/9, 2/2) nd compre ech medint with the midpoint of the two frctions. Now lets see wht the prdox looks like geometriclly on the number line. Here, B nd B 2 represent Brt s frctions, L, L 2 Lis s frctions, nd M B, M L the two medints. L M B M L L 2 B (e) (Brt wins) Nme two frctions B = /b nd B 2 = c/d stisfying 0 < /b < /2 < c/d <. Then find two more frctions L = s/t nd L 2 = u/v such tht i. b t 2 c ii. <, nd d v s+u iii. t+u b+d (f) (Lis wins) Nme two frctions B = /b nd B 2 = c/d stisfying 0 < /b < /2 < c/d <. Then find two more frctions L = s/t nd L 2 = u/v such tht s i. < <, t b 2 u ii. < c <, nd v d +c iii. < s+u. b+d t+v. Fbulous () Use ech of the digits, 2, 3, 4, 5, 6, 7, 8, 9 exctly once to fill in the boxes so tht the rithmetic is correct. + + =. Wht is the lrgest of the three frctions? Solution: We my ssume tht the second numertor is 5 nd the third is 7. If either 5 or 7 is used in denomintor, it cn never be neutrlized. Since the lest common multiple of the remining numbers is 72, we cn use /72 s the unit of mesurement. Now one of the three frctions must be close to. This cn only be 5/(2 3) or 7/(2 4). In the first cse, we re short 2 units. Of this, 7 must come from the third frction so tht the other 5 must come from the first frction. This is impossible becuse the first frction hs numertor nd 5 does not divide 72. In the second B 2 9

10 cse, we re 9 units short. If this, 5 must come from the second frction nd 4 must come from the third. This cn be chieved s shown below =. (b) Find four different deciml digits, b, c, d so tht + c < nd is s close b d to s possible. Prove tht your nswer is the lrgest such number less thn. Solution: 7 + = 7. This is s lrge s cn be. Why? (c) Next find six different deciml digits, b, c, d, e, f so tht + c + e < b d f nd the sum is s lrge s possible. Solution: One is = (d) Find four different deciml digits, b, c, d so tht + c < 2 but is otherwise s lrge s possible. Prove tht your nswer is correct. Then chnge b d the 2 to 3 nd to 4. Solution: The denomintor cnnot be 72 or 63. Why? Trying for frction of the form 2n, where n = 56, we re led to = n Why? (e) Next find six different deciml digits, b, c, d, e, f so tht b + c d + e f < 2 nd the sum is s lrge s possible. Then chnge the 2 to 3 nd to 4. (f) Finlly find four different deciml digits, b, c, d so tht + c > but is b d otherwise s smll s possible. Prove tht your nswer is correct. Then chnge the to 2 nd to 3. Solution: 2. Use exctly eight digits to form four two digit numbers b, cd, ed, gh so tht the sum b + ef is s smll s possible. As usul, interpret b s 0 + b, etc. cd gh Solution: The nswer is First, the four numertor digits re, 2, 3, nd the four denomintor digits re 6, 7, 8, nd 9. Also, if b + ed is s smll cd gh s possible, then < b nd c > d, e < f nd g > h. For convenience, we ssume c < g. Then = > becuse >. cd gh cd gh cd gh cd gh gh cd Now compre 3 3 with 24 ( = ) 98 = Pretty clerly, = 3 ( 76 87) > 3. Next find six different deciml digits, b, c, d, e, f so tht b + c d = e f. Solution: There re mny solutions. One is =

11 4. Problems with Four. These problems cn be very tedious, with lots of checking required. They re not recommended for children. () For ech i =, 2,..., 9, use ll the digits except i to solve the eqution b + c d + e f + g h = N for some integer N. In other words rrnge the eight digits so tht the sum of the four frctions is whole number. For exmple, when i = 8 we cn write = 4. (b) Wht is the smllest integer k such tht + c + e + g = k? Which digit b d f h is left out? Solution: The smllest chievble k is 2. It cn be chieved when the digit left out is i = 5 or i = 7: /4 + 7/6 + 2/8 + 3/9 = 5/4 + /6 + 2/8 + 3/9 = 2. (c) Wht is the lrgest integer k such tht + c + e + g = k? Which digit b d f h is left out? Solution: The lrgest chievble k is 6. It cn be chieved when the digit left out is i = 4 or i = 5: 9/ + 7/2 + 8/3 + 5/6 = 8/ + 7/2 + 9/3 + 6/4 = 9/ + 8/2 + 7/3 + 4/6 = 6. (d) For wht i do we get the gretest number of integers N for which + b c + e + g = N, where S d f h i = {, b, c, d, e, f, g, h}? Solution: For i = 5 there re 36 solutions. For i =,..., 9 we hve 4, 3, 5, 4, 36, 3, 25, 7, nd 2 solutions respectively. (e) Consider the frctionl prt of the frctions. Ech solution of + c + e + b d f g = N belongs to clss of solutions with the sme set of frctionl prts. h For exmple, 5/2+8/4+7/6+3/9 = 6 nd 5/+7/6+4/8+3/9 = 7 both hve frctionl prts sets {/2, /3, /6}. How mny different frctionl prts multisets re there? Solution: There re sets of frctionl prts including the empty set. They re φ, {/2, /2}, {/4, 3/4}, {/3, 2/3}, {/4, /4, /2}, {/2, 3/4, 3/4}, {/2, /3, /6}, {/2, 2/3, 5/6}, {/3, /2, /2, 2/3}, {/6, /4, /4, /3}, {/4, /3, 2/3, 3/4}. (f) Find ll solutions to b + c d + e f + g h = i where ech letter represents different nonzero digit.

12 Solution: There re two solutions with i = 5, one with i = 7, one with i = 8 nd four with i = 9. (g) Find ll solutions to b + c d + e f + g h = 2i where ech letter represents different nonzero digit. Solution: There re two solutions with i = 5, one with i = 6, nd three with i = 7. (h) Find ll solutions to b + c d + e f + g h = 3i where ech letter represents different nonzero digit. Solution: There is one solution with i =, one with i = 4, nd four with i = 5. (i) Find ll solutions to b + c d + e f + g h = 4i where ech letter represents different nonzero digit. Solution: There re two solutions with i = 3 nd one with i = 4. (j) Find ll solutions to b + c d + e f + g h = 5i where ech letter represents different nonzero digit. Solution: There is only one solution: 5/ + 7/3 + 8/4 + 6/9 = 5 2 = 0. (k) Find the mximum integer vlue of b + c d + e f + g h i where ech letter represents different nonzero digit. Solution: Let G i denote the lrgest integer vlue of + c + e + g i b d f h where ech letter represents different nonzero digit. The lrgest possible vlue is 2. It is chieved when i = 4; 9/ + 7/2 + 8/3 + 5/6 4 = 2. The resoning goes like this. The lrgest possible vlue of + c + e + g is b d f h 9/+8/2+7/3+6/4 = 6+2/3, so the lrgest possible integer vlue is 6. This mens tht we need only check the vlues G, G 2, nd G 3. There re only four wys to get n integer by rrnging the digits 2, 3, 4, 5, 6, 7, 8, 9 in the form + c + e + g, nd the resulting integers re 3, 6, 6, nd 8. So b d f h 2

13 G = 8 = 7. In cse i = 2 there re just three wys nd the integers 7, 0, nd 2 result, so G 2 = 2 2 = 0. Finlly, G 3 is the lrgest of 0 3, 2 3, 3 3, nd 4 3 =. It would be enough to show tht we cnnot chieve 6 using ll the nonzero digits except 3, 5 without the 2, or 4 without the, nd this does not tke too much work. 5. Ford Circles. Let r = /b be rtionl number where nd b re positive integers with no common fctors. The Ford circle F r is circle tngent to the x-xis t the point r with dimeter /b 2. For exmple F /2 hs the eqution (x 2 )2 + (y 4 )2 = 64. () Find n eqution for the Ford circles F nd F 2/3. Solution: F is given by (x ) 2 + (y 2 )2 = nd F 4 2/3 is given by (x 2 3 )2 + (y 8 )2 =

14 Shown here re the Ford circles F, F nd F (b) Drwing the Ford circles F nd F, it ppers tht they re tngent to 2 one nother. Prove it. (c) Next, construct the Ford circle F 2 3 tngent to both F nd F 2 (d) Now consider two rbitrry Ford circles F r s, nd notice tht it ppers to be. Prove tht it is. nd F p q, where r s is close to 4

15 p q in the sense tht ps qr =. Show tht the two Ford circles re tngent. 6. Frey Sequences. A Frey sequence of order n is the sequence of completely reduced frctions between 0 nd which, when in lowest terms, hve denomintors less thn or equl to n, rrnged in order of incresing size. Ech Frey sequence strts with the vlue 0, denoted by the frction 0/, nd ends with the vlue, denoted by the frction / (lthough some uthors omit these terms). So F = {0/, /}, F 2 = {0/, /2, /}, nd F 3 = {0/, /3, /2, 2/3, /}. () Build the sequences F 4, F 5 nd F 6. (b) How is ech Frey sequence relted to the one tht cme just before it? 5

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