fraction arithmetic. For example, consider this problem the 1995 TIMSS Trends in International Mathematics and Science Study:

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1 Brringer Fll Mth Cmp November, 06 Introduction In recent yers, mthemtics eductors hve begun to relize tht understnding frctions nd frctionl rithmetic is the gtewy to dvnced high school mthemtics Yet, US students continue to do poorly when rnked interntionlly on frction rithmetic For exmple, consider this problem the 99 TIMSS Trends in Interntionl Mthemtics nd Science Study: Wht is /4 + 8/ + /8? The options were A / B 4/4 C 9/4 D /4 More thn 4 percent of US eighth grders who worked this problem chose option A This percent ws exceeded only by Englnd Singpore, Jpn, nd Belgium ll hd fewer thn 0 percent students with nswer A In the words of comedin Red Skelton: spek louder thn words This is very much shortened version of 0 pge pper tht exmines frctions from n dvnced point of view Anyone interested cn write me for pdf version of the whole pper In the following exercises nd 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 /, /, /, On number line, wht common frction is three-fourths of the wy from / to /4? Solution: /6 Wht is the reciprocl of Solution: 7/ +? Wht common frction is equl to 07? Solution: 68/90 4 Wht is the vlue of n for which n n =? Solution: n = If x = 4, wht is the vlue of x? x x+ Solution: x = 8 Unuthorized reproduction/photocopying prohibited by lw c

2 Brringer Fll Mth Cmp November, 06 6 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 =, m =, you get /, /6, /6, / For n =, m = 4, you get /4, /, /6, /6, /, / Try this for n = nd m = 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 7 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 you know there re severl tricks tht cn help you You often get frctions whose numertors re one less tht their denomintors, for instnce, 4 It s esy to figure out the difference between two such frctions 4 = 4 4 = 4 Another exmple is 7/8 /4 = (8 4)/(8 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 + )( + ) 8 Show tht every unit frction cn be expressed s the sum of two different unit frctions Solution: Note tht n = n+ + n(n+) 9 Sums of unit frctions

3 Brringer Fll Mth Cmp November, 06 () Notice tht /7 is expressible s the sum of two unit frctions: /7 = /4 + /8 But /7 cnnot be so expressed Show tht /7 is not the sum of two unit frctions (b) There is conjecture of Erdös tht every frction 4/n where n cn be written s the sum of three unit frctions with different denomintors Verify the Erdös conjecture for n =, 4, nd 4 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:,, 7, 9,,,, 6, 69 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 (e) Find ll solutions to + + = with b c b c 4 Solution: There re five solutions, (,, ); (,, ); (,, ); (,, ); (,, ) Cn you prove tht these re the only solutions? In the Spce Between () Nme frction between / nd / Give n rgument tht your frction stisfies the condition Solution: < 7 < Of course this is the midpoint or the verge of the two frctions, ( + ) = 4 = 7 < 9 < This is the nswer I got from six yer old Tht he niled the question did not surprise me, but his explntion did He sid, is the sme s which is less tht And is less tht 6 which is the sme s < 8 <

4 Brringer Fll Mth Cmp November, 06 Who provides this nswer? A crpenter or nyone who works with rules grduted in eighths of n inch < < My fvorite nswer is obtined by thinking bout the wter-shring model A pir of bicyclers hs bottle of wter between them Three other cyclers hve two bottles of wter mong them If they ll gree to meet nd shre eqully, how much will ech cycler get? This number is clled the medint of the two frctions We see this nswer lot when fifth grders lern ddition of frctions incorrectly Let s use the wter shring model to help our fifth grders understnd frction ddition nd frction comprisons By the wy, do you ever wonder why 0 yer old boys cn very esily compre the frctions 6 8 nd 7 9 Bsebll! (b) Nme frction between / nd 7/0 How bout between 6/7 nd /? (c) Nme the frction with smllest denomintor between / nd 7/0 Or 6/7 nd /? (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 equl pieces Finlly you decide to cut the bord into 7 + = 0 equl pieces How mny mrks re on ech piece? (e) A bicycle tem of 7 people brings 6 wter bottles, while nother tem of 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 009 Some of the mteril is from the book Algebr by Gelfnd nd Shen You ve just bought the lst tickets for rffle tht hs bskets, ech with $00 gift crd for the sme store One bsket lredy hs 4 tickets in it nd the other lredy hs 0 You cn distribute your tickets between the bskets in ny wy you wish After you re done, exctly one ticket will be drwn t rndom from ech bsket How mny tickets should you put into the first bsket? Solution: Mximize x/(x + 4) + y/(y + 0) where x + y = The best vlue is x nd we get 007 Dividing Horses This problem comes from Dude, Cn You Count?, by Christin Constnd 4

5 Brringer Fll Mth Cmp November, 06 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 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 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 nd rides wy How ws this kind of division possible Solution: The sum of the three frctions is less thn : + + = 7 So 9 8 the strnger s horse helps complete the whole Consider the eqution + b = 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 ( + b) = b

6 Brringer Fll Mth Cmp November, 06 Thus 0 = b b = b b = b( ) 44 ( ) = ( )(b ) 44 Now, fctoring 44 = 4 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) = ( ) = x x x This rtionl function hs single verticl symptote t x = / nd zero t x = 0 So you cn see tht for ech 0 x /, there is y such tht /x + /y = / 4 Suppose {, b, c, d} = {,,, 4} () 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 + = 8 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 6

7 Brringer Fll Mth Cmp November, 06 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 if 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 {,,, 4}, nd b? Solution: There re frctions (b) How mny of the frctions in () re less thn? Solution: There re 6 in this set, /, /, /4, /, /4, /4 (c) Wht is the smllest number of the form + c, where {, b, c, d} = b d {,,, 4}? Solution: There re just three pirs of frctions to consider, A = +, 4 B = +, nd C = + Why is C not cndidte for the lest 4 4 vlue? Note tht A bets B here becuse A = + = + + while B + + = + + Does this resoning work for ny four positive 4 4 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 {,,, 4}? Solution: Agin there re just three pirs of frctions to consider, A = + 4, B = 4 +, nd C = + 4 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? 6 Fbulous I () Find two different deciml digits, b so tht < nd is s close to s b possible Prove tht your nswer is the lrgest such number less thn Solution: 8 This is s lrge s cn be Why? 9 7

8 Brringer Fll Mth Cmp November, 06 (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 < but is otherwise s lrge s possible Prove tht your nswer is correct Then chnge b d the to nd to 4 Solution: The denomintor cnnot be 7 or 6 Why? Trying for frction of the form n, where n = 6, 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 < nd the sum is s lrge s possible Then chnge the to 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 nd to Solution: 7 Note tht if / + /b + /c =, then ech positive integer, b, c is no lrger thn 6 Wht is the lrgest integer d such tht / + /b + /c + /d = with b c d? Wht s next? Solution: Noting / + /4 + / + /6 <, we must hve = Now / = /b + /c + /d We see b =, 4, or since / = /6 + /6 + /6 Cse : When b =, then /6 = /c + /d So (c 6)(d 6) = 6 nd the extreml vlue of d is d = 4 Cse : When b = 4, then /4 = /c + /d So (c 4)(d 4) = 6 nd the extreml vlue of d is d = 0 Cse : When b =, then /0 = /c + /d The only vlue to check is c = 6 which fils For if c = 7, then /7 + /8 < /0 8 Fbulous II () Use ech of the digits,,, 4,, 6, 7, 8, 9 exctly once to build some frctions whose sum is For exmple = + + = Find ll solutions Solution: Think bout the multiples of 9: 9, 8, 7, 6, 4, 4, 6, 7, 8 Note tht they use the sme pir of digits or hve no digits in common 9 Thus, = 8 = 7 = 6 ll hve vlue / Note lso tht there is n

9 Brringer Fll Mth Cmp November, 06 9 odd opertion δ tht preserves frction vlue: δ = 9 This opertor is the key to building solutions to the eqution More lter on this (b) Use ech of the digits,,, 4,, 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 nd the third is 7 If either or 7 is used in denomintor, it cn never be neutrlized Since the lest common multiple of the remining numbers is 7, we cn use /7 s the unit of mesurement Now one of the three frctions must be close to This cn only be /( ) or 7/( 4) In the first cse, we re short units Of this, 7 must come from the third frction so tht the other must come from the first frction This is impossible becuse the first frction hs numertor nd does not divide 7 In the second cse, we re 9 units short In this cse, must come from the second frction nd 4 must come from the third This cn be chieved s shown below = 9 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 + 4 First, the four numertor digits re,,, 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 + 4 = > becuse > cd gh cd gh cd gh cd gh gh cd Now compre with 4 ( = ) 98 = Pretty clerly, = ( 76 87) > 0 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 + 4 = 6 Notice tht 9 9 = 9/ 9/ = 9

10 Brringer Fll Mth Cmp November, 06 Cn you find more pirs of two-digit numbers, with the smller one on top, so tht cncelltion of this type works? Do you hve them ll? Solution: Let us first build n eqution using plce vlue nottion Note tht the eqution cn be written 0 + b 0b + c = c, where, b, c re digits nd < c This leds to c(0 + b) = (0b + c), which we messge to get 0c c = 0b bc This, in turn leds to 9c + bc = c(9 + b) = 0b From this it follows tht either c = or 9 + b is multiple of Cse c = Then 0 + b 0b + = Next, letting =, = 4, etc we find tht = : b + 0 = 0b + from which it follows tht b = 9 = : 00 + b = 0b + 0 from which it follows tht 90 = b nd b = 6 = : 0+b 0b+ = = 4 : 40+b 0b+ = 4 gives 0 + b = 0b + which hs not integer solutions gives rise to 00 + b = 40b + 0 which lso hs not solutions Why do we need not to check ny higher vlues of? Cse 9 + b is multiple of Agin we consider =, =, etc = : 9 + b = 0 or 9 + b = If b = 9 we get c = cse we lredy considered If b = 6 we get c = 4, new solution b = 6, bc = 64, nd /c = /4 = : 8 + b is either 0 or One leds to b = nd other b = 7 nd neither of these works = : This leds to b = 8 which does not produce n integer vlue for c = 4 : b = 4, so b = 9 This leds to c = 8 The four solutions re 9 9 = 9/ 9/ = 6 6 = 6/ 6/ = 0

11 Brringer Fll Mth Cmp November, = 6/ 6/4 = = 49/ 9/8 = 4 8 = Problems with Four These problems cn be very tedious, with lots of checking required They re not recommended for children () For ech i =,,, 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 It cn be chieved when the digit left out is i = or i = 7: /4 + 7/6 + /8 + /9 = /4 + /6 + /8 + /9 = (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 = : 9/ + 7/ + 8/ + /6 = 8/ + 7/ + 9/ + 6/4 = 9/ + 8/ + 7/ + 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 = there re 6 solutions For i =,, 9 we hve 4,,, 4, 6,,, 7, nd 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, /+8/4+7/6+/9 = 6 nd /+7/6+4/8+/9 = 7 both hve frctionl prts sets {/, /, /6} How mny different frctionl prts multisets re there? Solution: There re sets of frctionl prts including the empty set They re ϕ, {/, /}, {/4, /4}, {/, /}, {/4, /4, /}, {/, /4, /4},

12 Brringer Fll Mth Cmp November, 06 {/, /, /6}, {/, /, /6}, {/, /, /, /}, {/6, /4, /4, /}, {/4, /, /, /4} (f) Find ll solutions to b + c d + e f + g h = i where ech letter represents different nonzero digit Solution: There re two solutions with i =, 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 = i where ech letter represents different nonzero digit Solution: There re two solutions with i =, one with i = 6, nd three with i = 7 (h) Find ll solutions to b + c d + e f + g h = i where ech letter represents different nonzero digit Solution: There is one solution with i =, one with i = 4, nd four with i = (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 = nd one with i = 4 (j) Find ll solutions to b + c d + e f + g h = i where ech letter represents different nonzero digit Solution: There is only one solution: / + 7/ + 8/4 + 6/9 = = 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

13 Brringer Fll Mth Cmp November, 06 vlue is It is chieved when i = 4; 9/ + 7/ + 8/ + /6 4 = The resoning goes like this The lrgest possible vlue of + c + e + g is b d f h 9/+8/+7/+6/4 = 6+/, so the lrgest possible integer vlue is 6 This mens tht we need only check the vlues G, G, nd G There re only four wys to get n integer by rrnging the digits,, 4,, 6, 7, 8, 9 in the form + c + e + g, nd the resulting integers re, 6, 6, nd 8 So b d f h G = 8 = 7 In cse i = there re just three wys nd the integers 7, 0, nd result, so G = = 0 Finlly, G is the lrgest of 0,,, nd 4 = It would be enough to show tht we cnnot chieve 6 using ll the nonzero digits except, without the, or 4 without the, nd this does not tke too much work This problem ws sent to me by Fred Gluck Find four different frctions, ech with integer numertor one less thn the denomintor, the sum of which is whole number Find ll the solutions Solution: Ech frction is of the form (n )/n The four frctions, when dded together results in whole number N: + b b + c c + d d = N Fred Gluck s solution first notes tht N = Then he exmines the numbers /, /, /4, 4/, etc essentilly looking t ech set of four such numbers He notes tht / must be one of the numbers, otherwise the next four smllest such frctions hve sum / + /4 + 4/ + /6 tht is strictly between nd 4, so not integrl He finds six solutions to the problem Our solution trnsforms the problem into one tht ppers esier Since + b + c + d =, it follows tht = 4 = So now we re b c d b c d confronted with the new problem of finding four different unit frctions with sum See http : //mthforumorg/librry/drmth/view/68html nd http : //wwwijpmeu/contents/0 8 /7/7pdf To solve our unit frctions problem, first, note tht / must be mong our frctions becuse the next four frctions /, /4, /, /6 sum to number less thn So ll our solutions hve the form / + /b + /c + /d = where < b < c < d If b, then b = 4 becuse /+/+/6+/7 < Thus we cn write /+/+/c+/d = or /+/4+/c+/d = There re exctly four solutions to /c + /d = /6: (c, d) = (7, 4), (8, 4), (9, 8), (0, ), nd exctly two solutions to /c+/d+4: (c, d) = (, 0), (6, ) Therefore there re exctly six solutions to the unit frctions problem nd to Fred Gluck s problem s well

14 Brringer Fll Mth Cmp November, 06 4 Build binry tree of frctions s follows At the top, Then below ech frction +b there re two frctions, nd, s shown in the below This b +b b grph is clled the Stern-Brocot Tree level : level : level : level 4: At wht level do we find the frction 7? Solution: 8 The sequence 8 4 length 8, so the frction is t level is of I got this from Art Benjmin s Mgic of Mth book Let n be n integer between 0 nd 90 Wht is n/9? Compute the deciml equivlent Solution: To find the deciml representtion of n/9, multiple n by, nd subtrct to get the first three digits bc Then compute 000 n to get the next three digits def Then the nswer is 0bcdef Here is why (n = ) (0 ) + 0 n is six digit number which simplifies to n(0 ) = n(999) On the other hnd, 0bcdef = bcdef = bcdef (00 999) So we hve n(999) = n 00 = n 9 6 Notice tht 6 = = 6 nd = 8 = 8 With this in mind, for two positive integers nd b, we cll the number we get from writing the digits of followed 4

15 Brringer Fll Mth Cmp November, 06 by the digits of b the conctention of nd b We write the conctention of nd b s b For exmple = nd 46 = 46 Explore the conjecture If b = c d, then b = c b d Solution: The conjecture is not true: =, but On the other hnd, 0 0 suppose c nd d hve the sme number k of digits Then c = 0 k + c nd b d = 0 k b + d Also, suppose = c = u The = bu, c = du nd we hve b d c = 0k +c = 0k bu+du = u b d 0 k b+d 0 k b+d