Classroom. We investigate and further explore the problem of dividing x = n + m (m, n are coprime) sheep in

Transcription

1 Classroom I this sectio of Resoace, we ivite readers to pose questios likely to be raised i a classroom situatio. We may suggest strategies for dealig with them, or ivite resposes, or both. Classroom is equally a forum for raisig broader issues ad sharig persoal experieces ad viewpoits o matters related to teachig ad learig sciece. Syed Abbas School of Basic Scieces Idia Istitute of Techology Madi 7500 Himachal Pradesh sabbas.iitk@gmail.com Keywords Greedy algorithm of Fiboacci, uit fractios, Egyptia fractios. Sheep Distributio Problem Through Egyptia Fractios We ivestigate ad further explore the problem of dividig x = + m (m, are coprime) sheep i, ratio. m Let us begi with a aciet story. A old perso had 5 sheep ad i his last wish he asked his two childre to distribute his sheep amog them i the ratio of ad 2. Now the problem was that the sheep could ot be divided i half ad oe third. So they wet to a wise ma askig for his help. The wise ma added oe extra sheep to these 5 sheep, which made the total six. The he gave half of these six which is to oe brother ad oe third which is 2 to the other brother. After distributio the wise ma was still left with sheep which he icluded i his herd. We see that the total sum of the distributio is 5 ad hece the brothers were satisfied with this distributio. Now a atural questio to ask is, whether this is true for ay umber of sheep? Ca we always fid a umber such that if we add this umber ad the distribute the sheep ito half ad oe third, we are left with this extra added umber? This is easy if the umber of sheep is a multiple of 5. Suppose oe has 5x sheep, the we ca always add x such that the 80 RESONANCE September 20

2 total umber of sheep becomes 6x. Nowhalfof6x is x adoethirdof6x is 2x, which make a total of 5x ad we still have the added extra sheep with us. Now suppose, oe wats to divide y = + m (m, are coprime, m, > ; this is possible if y>6) umber of sheep i a ratio of ad. Notice here that y is ot m divisible by ad m. The questio here is, whether oe ca always fid a umber l such that if we add it to y ad divide this ew umber y +l by ad m, wegetthe atural umbers whose sum is y. Writig this statemet i the form ( of a equatio, ) ( we get ) y+l + y+l = y, which m gives y + + l + = 0. As + is m m m ot equal to oe, the above equatio ca be writte as y( + m m) +l( + m) = 0. Thus we obtai l = y(m m) = m m ad hece we ca always (+m) choose a iteger l = m m. It is iterestig to ote that if the umber of sheep is some multiple of +m, say a(+m), the oe eeds to add a(m m ) sheep. Hece it is always possible to add a umber of sheep i the herd so that after distributig it i ad m ratio, we are still left with the added umber of sheep. Oe more similar story is as follows, A farmer dies leavig his 7 sheep to be divided betwee his sos i the followig proportios, oe half to the eldest, oe third to the middle so, ad oe ith to the yougest. This was a difficult calculatio for the sos ad so they asked a itelliget ma for help. The ma added oe sheep to make the total umber 8. The calculatio is easy ow. Oe half of 8 is 9, oe third of 8 is 6, ad oe ith of 8 is 2, givig a total of 7. The ma ca ow take his oe sheep away leavig the three sos happy. The above problems ca be geeralized to ay umber, say, y = + m + k of sheep ad dividig these ito,, parts. It is easy to see that the umber y is m k ot divisible by ay of these umbers. Oe ca prove by usig similar argumets as above that there exists a Suppose oe wats to divide y = +m (m, are coprime, m, > ; this is possible if y > 6) umber of sheep i a ratio of / ad /m. Notice here that y is ot divisible by ad m. RESONANCE September 20 8

3 It is well kow that every ratioal umber admits a proper Egyptia fractio represetatio, ad every positive fractio has ifiitely may expressios as sums of uit fractios. atural umber l such that if we divide y +l the umber by, m ad k, wegetthetotaly. Iterestigly this problem ca be further geeralized to ay fiite umber of divisios. The aswer of the above problems lies i expressig ay umber i terms of a Egyptia fractio. I particular, i the first two cases oe wishes to write ay ratioal umber of the form as a sum of Egyptia fractios. Egyptia fractios were used by aciet Egyptia mathematicias to represet ratioal umbers i terms of uit fractios. A uit fractio is a fractio of the form, where 2. A Egyptia fractio is a expressio of the form m = k, where 0 < m <, k. It is well kow that every ratioal umber admits a proper Egyptia fractio represetatio, ad every positive fractio has ifiitely may expressios as sums of uit fractios. The represetatio is said to be proper if all i,i=, k are distict. Europeas have kow how to compute Egyptia umbers sice the 2th cetury. The proof is give by Fiboacci i his book Liber Abaci writte i 202. The proof ivolves the applicatio of greedy algorithm of Fiboacci [], which trasforms ratioal umbers ito Egyptia fractios. Let us first explore the process by pickig the umber 5. This umber lies betwee ad because 5 =. Subtractig from 5, gives 5 = ad hece 5 = +, which is the Egyptia fractio represetatio of 5. Let us explore 8 this process a little bit more ow pick the umber. As =, it clearly lies betwee ad. Subtractig from this umber, we obtai =. Now we repeat this process for the umber to get uit fractios Clearly =, ad thus it lies betwee ad. By subtractig 5, we get =. Combiig both RESONANCE September 20

4 equalities, we get = + 52 = From the above calculatio, it is clear how to proceed with ay umber, say, m. If m/ is ot of the form /k, oe ca always choose a largest atural umber N such that < m <. Let us subtract from m, where N N N m is ot, which gives m N = mn N. Usig the above iequality, we get If m/ is ot of the form /k, oe ca always choose a largest atural umber N such that /N < m/ < /(N ). mn <0 ad mn m< or mn <m, which implies that 0 <mn <m.hece our umerator is less tha m ad positive, which meas that our ew umber mn is smaller tha the origial umber m adthisiswhatwewat. Now,ifmN is N equal to oe, the we do ot proceed further; otherwise we repeat the same process for the umber mn util N we get oe as the umerator. The algorithm is called greedy because we are takig off the maximum possible fractio at every step, which is a greedy thig to do. Oe ca observe that for fractios of the form 2, the greedy algorithm uses at most two terms ad for the fractio, this algorithm uses at most three terms. We ow geeralize the above method. Suppose m < is writte i its lowest terms; the there is a Egyptia fractio with at most m terms whose sum is m. Deote r s = m [ ], m which gives m = + r, where [ ] deotes the greatest [ m ] s iteger fuctio. If r =, we do ot proceed further; otherwise, repeat the process. Further, repeated applicatios of the equality = + guaratees (+) (+) The algorithm is called greedy because we are takig off the maximum possible fractio at every step, which is a greedy thig to do. RESONANCE September 20 8

5 Egyptia fractios also give the best way to divide, say, m pieces of bread amog workers. that each uit fractio has ifiitely may represetatios. Cosider, for example, the umber 7. The, 8 r = 7 = 8, as [ ]=2. Because r is ot equal to s s oe, we repeat the process agai. Let r s = =, 9 9 hece 7 = + +, which is exactly what we discussed above i the case of 7 sheep. Notice that i this case there ca be ifiitely may ways of distributig 7 sheep to,, 5 or a maximum of 7 people. Moreover, oe ca further write = +, which gives aother represetatio. Also becasue = + ad = which gives aother represetatio 7 = But this represetatio is ot useful i the case of distributig 7 sheep, as 8 is ot divisible by 2; it is oly applicable if 7 + is divisible by all deomiators. Hece, i this case, oly umbers allowed i the deomiator are 2,, 6, 9or8. Egyptia fractios are useful to check whether a give umber is larger tha aother give umber. Suppose we wat to kow whether is larger tha or ot. 5 Represetig both umbers by Egyptia fractios gives = + ad = + +, which clearly tells us that is larger tha. 5 Egyptia fractios also give the best way to divide, say, m pieces of bread amog workers. For example, let us say we have 5 pieces of bread ad wat to equally distribute these amog 8 workers. It is ot difficult to see that each oe should get 5 part. Now if we represet 8 this umber i terms of Egyptia fractios, we get 5 = 8 +. It clearly shows that it is better to divide the 2 8 first pieces each ito 2 portios ad the last piece ito 8 portios, ad the give each oe of them a ad 2 8 portio (Figure ). Figure. Distributio of 5 pieces of bread amog 8 workers. 8 RESONANCE September 20

6 The sum of the uit fractios that adds upto oe is of special iterest. It is a ope problem to fid the smallest possible value of k, say p(k), such that k i= i =, where < 2 < <. For some values oe ca check that this is possible, for example, p() = 6,p() = 2, but, i geeral, it is ot kow. We coclude the article by statig oe very famous cojecture called Erdös-Straus cojecture oegyptiafractios give by Paul Erdös ad Erst G Straus i 98. It states that for all itegers 2, the ratioal umber ca be expressed as the sum of three uit fractios, that is, = For example if =5, the we ca write the followig Egyptia fractio 5 = = I this case there are two solutios. The mai difficulty here is to prove the cojecture for positive triplets (, 2, ). The cojecture has bee verified up to 0 [2] usig computers, but it is still a ope problem for all. The above equatio falls ito the category of Diophatie equatios. I geeral, a Diophatie equatio is a polyomial equatio that allows oly iteger solutios. The questio regardig the existece of a algorithm which tells whether a arbitrary Diophatie equatio has a solutio or ot is the famous Hilbert s 0th problem []. Ackowledgemet The author is thakful to the aoymous reviewer for his/her costructive commets ad suggestios, which helped to improve the article cosiderably. Oe very famous cojecture called Erdös Straus cojecture o Egyptia fractios states that for all itegers 2, the ratioal umber / ca be expressed as the sum of three uit fractios, that is, = Suggested Readig [] Wikipedia,e.wikipedi. orgwiki/greedy-algorithm-foregyptia-fractios [2] Alla Swett, The Erdös Straus Cojecture, math. uidy.edu/swett/esc.htm [] M Davis, Hilbert's Teth Prolem is Usolvable, Amer. Math. Mothly, Vol. 80, pp.2 269, 97. RESONANCE September 20 85