Outline of a development of fractions

Transcription

1 Outlie of a evelopmet of fractios Fractios are a stumblig block for may stuets. I thikig about the reasos, two poits preset themselves as beig particularly relevat: i) To eal successfully with fractios, stuets must revise their coceptio of what a umber is: from a cout or aitive coceptio to a ratio or multiplicative coceptio a escriptio of the relative size of a give quatity, as measure by aother quatity, which fuctios as the uit. I the mathematics eucatio literature, this is sometimes calle the trasitio from aitive thikig to multiplicative thikig. ii) I cotrast to the situatio for whole umbers, we o ot have uique ames for ratioal umbers; the umber iicate by ay give fractio is uchage if both the umerator a eomiator are multiplie by the same factor; a oig computatios with fractios ofte higes o selectig a appropriate represetatio, which may ivolve chagig from the represetatio with which oe is iitially presete. I particular, it is ofte ecessary to replace two fractios with equivalet fractios that have the same eomiator. The Commo Core State Staars i Mathematics attempts to ameliorate the ifficulties i approachig fractios by itroucig uit fractios, a emphasizig that they are ew uits. Cocomitat to this approach is close attetio to the uits attache to umbers. This emphasis o uits ca help with both of the ifficulties metioe above. The goal of this ote is to sketch a evelopmet of the omai of fractios, startig with uit fractios. I: Meetig Fractios A. Itrouce uit fractios: of some quatity (aka the uit, or the whole) is aother quatity, such that copies of it make the origial quatity. It is probably avisable to give may examples. There are iee may examples available from commo practice with measuremet. For example, we have may uits of time. We call 7 of a week, a ay ; we call 24 of a ay, a hour. We call 60 of a hour a miute. We call 60 of a miute a seco. (A though it is ot curret, i Elizabetha plays, you ca fi 60 of a seco - the trice. ) Stuets shoul realize that ay coveiet quatity ca be cosiere to be the uit. Other quatities are the assige umbers escribig how may of the uit they are mae of. For example, i ealig with soa, we have the ca, the 6-pack a the case (4 six-packs). We coul take the uit to be the ca, i which a ca is, the 6-pack is 6 a the case is 24. We coul take the uit to be the six-pack, i which case a ca is 6, a a case is 4. We coul take the uit to be the case, a the a ca is 24, a a 6-pack is 4. The variability or arbitrariess of the uit obliges us, a stuets i particular, to always be specific, a to state the uits to which their umbers refer. Teachers shoul eforce this rule strictly, util satisfie that stuets are quite comfortable with fractios. B: The geeral fractio is efie to be copies of the uit fractio : =. C: Moels May moels a examples of fractios shoul be presete, icluig set moels a circular (pie a pizza) moels, but ot oly these. Especially importat for thikig about fractios are the area moel a the liear moel or umber lie. I the area moel, the uit is a rectagle (aka browie pa, or cor brea), which ca be partitioe ito equal parts (subrectagles) of size by rawig equally space lies parallel to a pair of opposite sies. (If the rectagle has uit area, the the area of each of the subrectagles will have area ; but oe ee ot ivoke the iea of area, sice i the mai applicatios, oe will always be comparig umbers of cogruet rectagles to reach the esire coclusios.) A valuable feature of the area moel is that the uit rectagle ca be subivie ito pieces of size i oe irectio, a pieces of size e i the other irectio, a the iteractio of these two operatios provies valuable isight ito several importat situatios.

2 The liear moel takes place o a ifiite half lie (the umber lie, or umber ray). There is a otio of legth or istace, a i particular, there is a uit of legth. The the poits o the lie are labele by the Measuremet Priciple: The umber labelig a poit tells how far the poit is from the origi/epoit, as a multiple of the uit istace. The epoit itself is labele 0, sice it is at 0 istace from itself. The uit iterval [0, ] has uit legth, a is at istace from 0. The 2 is twice as far from 0 as oe is, so the iterval [0, 2] is compose of two itervals of uit legth, amely [0, ] a [, 2]. A similarly with larger whole umbers. The umber 2 goes i the mile of the uit iterval, so that [0, 2 ] a [ 2, ] have the same legth, which is 2. The label 2 goes with the mile poit, because the the poit is two times as far from 0 as 2 is. Similarly, the poits 3 a 2 3 are the poits that partitio the iterval [0, ] ito three equal subitervals, so that each iterval has 3 the legth of [0, ]. The label 3 goes with the iterval that starts at 0, sice the poit will the be 3 times as far from 0 as is 3. The label 2 3 goes o the other 3 ivisio poit, sice there are two itervals of legth 3 betwee it a 0, or i other wors, it is 2 times as far from 0 as 3 is. Similar reasoig serves to locate fractios with larger eomiators o the umber lie. The placig of fractios o the umber lie is a importat issue a is ot straightforwar for may stuets, so it shoul receive careful iscussio. The placig o the umber lie of the fractios for a fixe eomiator a a umerator that varies through all whole umbers prouces a lovely a compelligly regular image, of equally space poits partitioig the lie ito equal itervals of legth. They look just like the whole umbers, except they are closer together, a of the smaller itervals fit isie each uit iterval. D: A fractio represets a ivisio. Our efiitio of a fractio is as a multiple of a uit fractio. However it is importat for stuets to realize that a fractio also ca be iterprete as the result of a ivisio: =. This shoul be illustrate with several examples. Both liear a area moels ca be use effectively to show why this is true.. Gettig stuets comfortable with this priciple takes substatial attetio to the ieas of measuremet. See Three Pillars of First Grae Mathematics (commocoretools.me/wp-cotet/uploas/202/02/3pillars.pf) for suggestios about how to start. 2

3 . II: Arithmetic of fractios with fixe eomiator Oce stuets are reasoably familiar with the iea of a fractio, they ca begi to stuy their arithmetic. The first stage is to eal with the case of fractios with the same eomiator. A: Aitio The geeral formula is + m = + m. Examples of this rule ca be justifie irectly from the efiitio, a illustrate with various moels, icluig liear a area moels. A attractive feature of the liear moel is that, just as aitio of whole umbers ca be iterprete i terms of legth by placig bars e-to-e, this works equally well for fractios. Thus, although the symbolic calculatio of the sum of fractios may seem quite ifferet from that of whole umbers, the associate geometry is the same. This remais true for the case of aig fractios with uequal eomiators as well. B: Compariso The geeral rule is m if a oly if m. This agai ca be illustrate by example i various moels, a has a atural iterpretatio i terms of legth. C: Multiplicatio by a whole umber The geeral formula is m = m. Similarly to aitio a compariso, this ca be illustrate by example with various moels. It is cosistet with the coceptio of multiplicatio by a whole umber as repeate aitio.. 3

4 III: Relatig fractios with ifferet eomiators The key to workig successfully with fractios is to be able to accommoate fractios with ifferet eomiators. This is base o several key relatioships, which ca be covicigly illustrate with the area moel, a also, sometimes with perhaps more effort, with the liear moel. These ieas may preset ifficulties for some stuets, but sice they are critical for oig a uerstaig calculatios, they shoul be presete carefully a iscusse, with may examples, to promote acceptace. A: Repeate Subivisio As a quatity i its ow right, ca itself be ivie ito fractios of itself. Thus we ca cosier e ( ). This is ot a ew etity, but i fact is a uit fractio of the origial quatity. Precisely, e ( ) = e. This ca be illustrate with the area moel by iviig the uit rectagle ito equal vertical strips, a e equal horizotal strips. The the lies efiig the horizotal strips also ivie each vertical strip ito e equal pieces, which are therefore e ( ). But also, all the small pieces of all the strips are cogruet, a there are e of them i the whole rectagle, so each oe costitutes e of the whole. B: Recostitutio I the same picture as use to emostrate repeate subivisio, we see that the vertical strips of size are compose of e of the small rectagles of size e ; a likewise, each of the horizotal strips of size e is compose of of the small rectagles. This illustrates the relatioships = e e = e e ; a e = e = e. C: Reamig If we multiply the recostitutio ietity by a whole umber, we obtai = = e e = e e = e e. This shows that the umber represete by the fractio ca also be represete (i ifiitely may ways) by other fractios, obtaie by multiplyig both the umerator a eomiator by the same umber. This meas, that i ay situatio where e appears i a calculatio, it ca be replace by e. This process is sometimes calle reamig the fractio. Also, the fractio e e is sai to be equivalet to. The reamig process ca be illustrate a justifie with the area moel by takig the same picture as use to argue for repeate subivisio, a ietifyig the area represetig (which might fill up several uit rectagles if > ) as a uio of the rectagles of size e. D: Reamig to fi a Commo Deomiator I particular, the reamig process shows that, give ay two fractios a m e, we ca reame them both as fractios with the same eomiator: This is essetial for oig arithmetic with fractios. = e e, a m e = m e = m e. 4

5 IV: Arithmetic of geeral fractios A: Aitio The geeral formula for the sum of two fractios is reuce to the alreay kow formula for aitio of fractios with the same eomiator by reamig the fractios to have a commo eomiator: + m e = e e + m e e + m =. e This ca be escribe verbally: multiply the umerator a eomiator of each fractio by the eomiator of the other oe, the a the umerators of the resultig fractios, a put the sum over the prouct of the two eomiators. This ca be illustrate by proucig the area moel for each reame fractio, a otig that the sum of the two is just represete by the uio of the two regios represetig the two fractios. Each fractio shoul be create withi its ow uit rectagle (or uio of several, if it is larger tha ). The rectagles cotaiig each fractio ca be lie up ajacet to each other. This ca also be illustrate with the umber lie. A particularly satisfactory feature of aig fractios o the umber lie is that the priciple that govere whole umber aitio - that you a by juxtaposig itervals with legths equal to the aes, (with oe legth havig its left epoit at the origi), is equally vali for fractios. B: Compariso Give two fractios, a m e, we ca compare them by reamig them as fractios with the same eomiator, a comparig the umerators: = e e m e = m e if a oly if e m. This is ofte calle the cross multiplicatio criterio: multiply the umerator of each fractio by the eomiator of the other, a compare the proucts. Although it is justifie by the reamig process, it ca also be emostrate usig the area moel, subivie as if preparig to reame. C: Multiplicatio Multiplicatio ca also be represete icely usig the picture for repeate subivisio, If the uit rectagle is subivie vertically ito strips of size a horizotally ito strips of size e, the oe sees that that the rectagle forme by vertical strips is also subivie ito equal horizotal strips by the same lies as subivie the whole rectagle. Thus, the itersectio of o of these strips with the rectagle represetig amouts to e of e, so m of these strips costitute m e. But this is just the itersectio of the rectagles represetig a m m e, a it cosists of m of the small rectagles of size e, so amouts to e of the whole. This gives the multiplicatio formula: m e = m e. Of course, this ca be escribe verbally by sayig: the umerator of the prouct is the prouct of the umerators, a the eomiator of the prouct is the prouct of the eomiators; or more telegraphically, multiply the umerators, a multiply the eomiators. Although this reasoig works i geeral, it is easier to execute whe the fractios are less tha ( < a m < e). D: Divisio Divisio is the most ifficult of the operatios to uersta for whole umbers, I fact, it is ot actually a operatio i the strict sese, sice ot all ivisios of whole umber prouce whole umber results. Fractios were ivete to allow ivisio of ay two whole umbers. Oe of the great beefits of exteig the umber system from whole umbers to fractios is that ivisio becomes a true operatio: fractios ot oly give the aswers for iviig oe whole umber by aother, but the result of iviig ay 5

6 fractio by ay fractio is also aother fractio. It is ot ecessary to ivet eve more umbers to allow for ivisio of a fractio by a fractio. More remarkably, the operatios of ivisio a multiplicatio are merge ito a sigle operatio: ivisio by the fractio m e is the same as multiplyig by aother fractio, usually calle the reciprocal of m e. Moreover there is a simple formula for the reciprocal: it is e m. Thus, the formula for ivisio of fractio takes the simple form: m e = e m = e m. This rule is usually escribe as ivert a multiply. Because of the ifficulty associate with uerstaig ivisio, we will give several argumets for the ivert-a-multiply rule. i) Measuremet. The measuremet iterpretatio of whole umber ivisio, whe it is possible, is, how may of quatity fit i quatity? We ca ask the same questios of two fractios: how may copies of m e fit i? To aswer this questio, we shoul express both a m e i the same uits, which meas puttig them over a commo eomiator. Thus, we shoul chage our questio to: how may copies of m e e fit i e? But we ca ow recogize this questio as beig a whole umber ivisio questio, but with the uits beig the uit fractio e. That is, we are askig, how may groups of size m, of uits of size e, fit ito a group of size e of the same uits? A we kow that the aswer to this questio is just the fractio e m. The thikig ivolve i this argumet ca be well illustrate with the area moel. ii) Missig factor. I whole umber situatios, the quotiet a b ca also be uerstoo as a missig a factor: it is the umber x which, if multiplie by b, gives a. I other wors, b is the solutio x to the equatio bx = a. If we take this as efiig the quotiet also whe a a b are fractios, we are lookig for x such that ( m e ) x =. If we use the formula for multiplicatio of fractios, we ca check that e m m e e m = me em = (me) (me) =, satisfies this equatio. Iee, as esire. I this calculatio, we use the commutative a associative rules for multiplicatio (for oriary whole umbers) to rearrage the proucts i the umerator a the eomiator, a the we use the reamig process to simplify the fractio. iii) Uoig multiplicatio. Sometimes ivisio is presete as the uoig of multiplicatio. If we multiply the whole umber m by the whole umber, a the ivie the prouct by m, we just get back to. We ca check by multiplicatio that the prouct of m a m is : m e e m = me em = me me =. It follows from this that multiplicatio by e m oes iee uo multiplicatio by m e : e ( m m e ) = e m m e = e(m) m(e) = (me) (me) =. I this calculatio, we have agai use the formula for multiplicatio, plus the commutative a associative rules for multiplicatio (of whole umbers). This calculatio shows that multiplicatio by e m oes uo multiplicatio by e m, a so shoul be cosiere as ivisio by e m. Thus, ivisio by a fractio is the same as multiplicatio by aother fractio. The fractio e m is kow as the reciprocal of m e. It is obtaie by exchagig the umerator a eomiator e of m. This is what gives rise to the familiar irective: to ivie by e m, ivert a multiply. 6

7 V: Refiemets After stuets are comfortable with the basic aspects of fractios, as sketche above, icluig their arithmetic, several topics that are less cetral but which promote a more refie view of fractios shoul probably be covere. The mai oes are mixe umbers, a reucig fractios, which actually is about fiig the simplest form of a ratioal umber - fiig the smallest umerator a/or eomiator that ca be use to represet the umber. It is ot at all obvious that there is a uique simplest form for a give ratioal umber, but there is, a its existece is relate to some basic umber theory, especially the greatest commo ivisor (GCD), the Eucliea Algorithm a uiqueess of prime factorizatio (which was ubbe the Fuametal Theorem of Arithmetic by Gauss). The Eucliea Algorithm is a geeral metho to fi the greatest commo ivisor of two whole umbers; it was escribe by was give by Eucli i his Elemets. We will iscuss these theoretical issues briefly below. A: Mixe umbers a improper fractios The iscussio above has more or less igore the fact that fractios are usually thought of as beig less tha oe. For a give eomiator, the umerator was freely allowe from the begiig to rage through all whole umbers; a iee, this was a key feature of the umber lie represetatio of the system of all fractios for fixe, as efiig a regular ivisio of the umber lie ito itervals of size However, it is ot straightforwar to tell the size of a fractio. Whe the umerator icreases a the eomiator is fixe, the fractio icreases, but whe is fixe a icreases, the fractio ecreases. These opposite teecies ca make it ot so simple to etermie the size of fractios. For that reaso, amog others, it ca be useful to express a fractio as the sum of a whole umber a a fractio less tha oe. If the whole umber is o-zero, it will costitute the mai part of the fractio, a the part less tha oe will be a relatively small correctio. A fractio betwee 0 a is calle a proper fractio, a the expressio of a fractio as a sum = q + r of a whole umber a proper fractio is calle a mixe umber. The traitioal termiology for a fractio with > is improper fractio, but we o ot avocate for this usage. To obtai the expressio of as a mixe umber, oe performs the ivisio-with-remaier algorithm of by, a fi the DWR quotiet q of by, a the remaier r: = q + r, with r < = q + r. The arithmetic of mixe umbers is somewhat ivolve, which is ot surprisig, sice two mixe umbers ivolves ealig with three ifferet uits: the whole, a the uit fractios correspoig to the two eomiators. However, there is othig coceptually ew ivolve, a the computatios ca be carrie out o the basis of the formulas for arithmetic of part III above, Accorigly, we will ot escribe the etails. B: Simplest form, GCD The fact that fractios have may ames gives rise to a problem: give a fractio, ca we fi aother fractio that equals, but is simpler i the sese that < a <? If we ca, we geerally use i place of, especially for purposes of expositio, sice we te to fi fractios with smaller eomiators easier to iterpret. Thus, we woul fi it easier to process a statemet that 6 of stuets pla to stuy more mathematics, tha that stuets pla to stuy more mathematics. The problem of simplifyig fractios amouts to fiig a commo factor or commo ivisor of the umerator a eomiator; for if f > ivies both a, that is, = f a = f, the = f f = ; that is, ca be simplifie. 7

8 It is a remarkable a coveiet fact that every pair {, 2 } of whole umbers has a greatest commo ivisor (GCD, or GCF). This is a commo ivisor f o, which is greatest ot oly i the sese that it is larger tha ay other commo ivisor f, but is also largest multiplicatively, i the sese that, if f ivies both a 2 exactly, it also ivies f o exactly: f o = ff, for some whole umber f The existece of a greatest commo ivisor guaratees a best possible simplificatio of ay fractio. Give a fractio, if f o is the GCD of a, with = o f o a = o f o, the o is equivalet to = ofo of o, that is, represets the same umber. Moreover, ay other fractio that represets the same umber will have the form ñ = of of for some whole umber f (which is the see to be the GCD of a. Thus, every fractio has a uique simplest form, or reuce form, i which the umerator a eomiator have o commo factors greater tha. We the say that the umerator a eomiator are relatively prime, a that the fractio is reuce. These facts may or may ot be clearly state i K-2 math istructio. Eve if they are clearly state, it is ulikely that they are prove. I particular, it is rarely if ever oe i K-2 mathematics to show the existece of the GCD, outsie of a few examples. The usual approach to fiig GCDs is by meas of prime factorizatio. However, beyo a rather limite rage, prime factorizatio is ifficult to o: o efficiet algorithm is kow. There is, however, a efficiet way to fi the GCD of two umbers. It was escribe by Eucli i Book VII of his Elemets. The Eucliea Algorithm is pretty much igore i K-2 mathematics. (Perhaps it is presete i specialize courses at some elite high schools.) We iscuss it briefly below. C: LCM Whe aig fractios, we ee to express them both i the same uits, which meas reamig them as fractios with a the same eomiator. This is traitioally calle a commo eomiator. Commo eomiators figure also i our iscussio of compariso a ivisio. I iscussig reamig i part II, we showe that, give ay two fractios, the prouct of their eomiators coul always be use as a commo eomiator. However, there are situatios where smaller umbers ca also be use. For example 3 4 = 9 2 a 5 6 = 0 2. The questio thus arises, give two fractios, what is the smallest umber that ca serve as a commo eomiator for fractios equivalet to both of them? If we assume that is i lowest terms, the by part A, we kow that ay fractio equivalet to must have eomiator ivisible by. Similarly, if m e is aother fractio i lowest terms, ay reamig of it must have eomiator ivisible by e. O the other ha, if r = a = be (for whole umbers a a b) is a multiple of both a e, the = a a = a r a m e = bm be = bm r, so r ca serve as a commo eomiator for both fractios. We call the umber r = a = be a commo multiple of a e. The prouct e clearly is a commo multiple of a a b, but it may happe that smaller umbers are also commo multiples. For example, 2 is a commo multiple of 4 a 6, which allowe the reamig above of 3 4 a 5 6 i terms of 2 s. The smallest positive multiple of two whole umbers is calle their least commo multiple (LCM). It ca be show that LCM(, e), the least commo multiple of a e is uique, a is multiplicatively miimal, i the sese that it ivies ay other commo multiple. Moreover, the LCM is relate to the GCD by a elegat formula: LCM(, e) GCD(, e) = e. o Sometimes, whe istructio puts heavy emphasis o puttig fractios i simplest terms, stuets may be tol that whe aig fractios, they shoul fi the LCM of the eomiators. While this may somewhat relieve the computatioal bure i ealig with the sum, fiig the LCM of two umbers is ot a simple process. Iee, it typically is cosierably more work tha fiig the sum. Moreover, the coceptual issues ivolve i fiig the LCM are much more challegig tha the relatively straightforwar formula for the sum of two fractios, as give i part III. Thus, it is ot recommee to raise the issue of LCM while stuets are learig the basics of fractios. It ca be iscusse later, as a refiemet, perhaps for stuets with strog iterest i mathematics. 8

9 VI. Theoretical Aeum: Eucliea Algorithm a Prime Factorizatio A: The Eucliea Algorithm Theoretical justificatio for the existece of the GCD a uiqueess of prime factorizatio is provie by the Eucliea Algorithm (EA). Although it is avoie i school mathematics, we will give a brief iscussio of it here. The EA is a extesio form of the proceure for ivisio-with-remaier, which is ofte calle the ivisio algorithm (DA). The DA expresses a give umber as a multiple of a give umber, plus a remaier r, which is require to be less tha : = q + r, with r <. (DA) The coitio o r is equivalet to sayig that q is the largest umber such that q is less tha or equal to. We ivie ito as may pieces of size as we ca, a the r is what is left, too little for form aother piece of size. This coceptio of ivisio epes strogly o the orerig of the whole umbers. The EA starts with two umbers a 2, with 2 take to be the smaller. It performs the DA of by 2. The remaier is calle 3. The the DA is performe for 2, a 3 (which is guaratee to be less tha 2 by the DA). This process is cotiue util a remaier of 0 is obtaie; a the it stops. Sice the remaiers form a strictly ecreasig sequece, this must happe at some step. Thus, if the EA stoppe at the fourth step, it woul look like this: = q 2 + 3, 2 = q , 3 = q , 4 = q 4 5. (EA) The mai statemet about the EA is this: Theorem: The last o-zero remaier i the EA is the GCD of a 2. We will explai why this is true. The theorem makes two claims: i) The last o-zero remaier ivies the two startig umbers. ii) Ay umber that ivies both of the two startig also ivies the last remaier. These are the efiig properties of GCD. Thus, the last o-zero remaier is the GCD of the two startig umbers; this is what the EA oes. To show ii), oe works forwar usig the successive equatios of the EA. We ee the iea of a itegral combiatio of two umbers r a s. This is a sum ar + bs, where a a b are itegers (whole umbers, perhaps with a - sig). We take as obvious the followig facts: a) A itegral combiatio of whole umbers is a iteger. b) A itegral combiatio of two itegral combiatios of two umbers is agai a itegral combiatio of the origial two umbers. c) If r, s a t are whole umbers, a t ivies r exactly a ivies s exactly, the t also ivies exactly ay itegral combiatio of r a s. Now retur to stuy the EA. The k-th equatio of the EA has the form x k = q k x k+ + x k+2. We ca rewrite this to express the k+2 i terms of k+ a k : x k+2 = x k q k x k+. 9

10 This equatio shows that x k+2 is a itegral combiatio of x k+ a x k. If we work forwar from x a x 2, usig fact b) above, we coclue first that x 3 is a itegral combiatio of x a x 2, a seco that x 4 is a itegral combiatio of x a x 2, a thir that x 5 is a itegral combiatio of x a x 2. Cotiuig like this, we coclue that all the x k are itegral combiatios of x a x 2. I particular, the last o-zero remaier is a itegral combiatio of x a x 2. It therefore follows from statemet c) above that all the remaiers are ivisible by ay umber that ivies x a x 2. I particular, the last o-zero remaier is ivisible by x a x 2. This coclues the proof of claim ii) above. To show i), oe works backwar through the same set of equatios. Observe that the k-th equatio of the EA expresses x k as a itegral combiatio of x k+ a x k+2. If the last o-zero remaier is x l, the the l -th equatio says x l = q l x l, which meas that x l is ivisible by x l. Sice the l 2-th equatio expresses x l 2 as a itegral combiatio of x l a x l, we coclue that x l 2 is also ivisible by x l by fact c) above. The sice x l 3 is a itegral combiatio of x l 2 a x l, we ca agai use fact c) to coclue that x l 3 is ivisible by x l. Workig backwar like this, we evetually coclue that x a x 2 are ivisible by x l. This coclues the proof of claim i) above. So both claims are true. Here is the example of workig backwar i the 4-step algorithm state above; 4 = q 4 5, 3 = q = q 3 (q 4 5 ) + 5 = (q 3 q 4 + ) 5, 2 = q = q 2 ((q 3 q 4 + ) 5 ) + q 4 5 = (q 2 q 3 q 4 + q 2 ) 5 + q 4 5 = (q 2 q 3 q 4 + q 2 + q 4 ) 5, = q = q (q 2 q 3 q 4 + q 2 + q 4 ) 5 ) + (q 3 q 4 + ) 5 = (q q 2 q 3 q 4 + q q 2 + q q 4 ) + (q 3 q 4 + )) 5 = (q q 2 q 3 q 4 + q q 2 + q q 4 + q 3 q 4 + ) 5. A here is the example of workig forwar: 3 = q 2, 4 = 2 q 2 3 = 2 ( q 2 ) = (q + ) 2, 5 = 3 q 3 4 = ( q 2 ) q 3 ((q + ) 2 ) = (q 3 + ) (q q 3 + q + q 3 ) 2. B: Uiqueess of Prime Factorizatio Recall that a prime umber is a whole umber that is ot ivisible by ay umber strictly less tha itself a greater tha. The first so may primes are 2, 3, 5, 7,, 3, 7, 9, 23,... Sice two is the first prime, all primes greater tha 2 are o umbers. The prime umbers are the geerators for the whole umbers uer multiplicatio, as escribe by the followig two facts. i) (Existece of prime factorizatio) Every whole umber ca be expresse as a prouct of prime umbers. This is easy to see, usig the followig priciple, which is a basic property of the whole umbers, a is the basis for mathematical iuctio: Every o-empty set of whole umbers has a smallest elemet. Acceptig this priciple, suppose there are umbers that are ot expressible as proucts of prime umbers, a let o be the smallest oe. The either o is prime, or it is t. If it is prime, it is its ow expressio of itself as a prouct of primes. If it is ot prime, it is expressible as a prouct o = ab, where a (a hece, also b) is a umber strictly betwee a o. So both a a b, beig smaller tha o, are expressible as proucts of primes. The multiplyig these two expressios together gives a expressio for o as a prouct of primes, which is a cotraictio. Hece, the set of umbers ot expressible as a prouct of primes must be empty; i other wors, all umbers are expressible as a prouct of primes, as claime. ii) (Uiqueess of prime factorizatio) A whole umber has oly oe factorizatio ito a prouct of primes, up to reorerig of the primes. Thus = 2 m2() 3 m3() 5 m5()..., 0

11 where all the expoets m p () are uiquely etermie by. The proof agai is a argumet by cotraictio. Suppose there are umbers with more tha oe prime factorizatio, a let o be the smallest oe. The o must ot be prime, or it woul be its ow uique prime factorizatio. Suppose o = p p 2 p 3... p j is oe factorizatio of o ito primes, a o = p p 2 p 3... p k is aother oe. The all the primes i the two factorizatios must be ifferet. For if there were oe prime factor i commo, say p = p, after reorerig for sake of coveiece, the o p = p 2 p 3... p j = p 2 p 3... p k woul be two prime factorizatios of o p. But o p, beig smaller tha o, ca have oly oe prime factorizatio, so the p c must be the same as the p up to reorerig; which meas that the two factorizatios of o were also the same, up to reorerig. So if we have the two prime factorizatios for o, the prime p must ot be equal to ay of the p b. Now write o = p m, where m is the prouct of all the rest of the primes i the seco factorizatio of o. The p oes ot ivie either of p, a it must ot ivie m either, sice m < o has a uique prime factorizatio, which oes ot iclue p. However, this cotraicts the followig Lemma: If a prime p ivies a prouct a b, the it ivies a, or it ivies b. Proof: Suppose that p oes ot ivie a. The the GCD of p a a, sice it must ivie p, a caot be p (sice p oes ot ivie a, by assumptio), ca oly be. By the EA, this meas that is a itegral combiatio of p a a; so we have a equatio rp + sa =, for some itegers r a s. Multiply this equatio by b: brp + sab = b. By assumptio, p ivies ab: ab = cp for a whole umber c. Therefore b = brp + sab = brp + scp = (br + sc)p. This says that p oes ivie b, so the lemma is prove. Sice the coclusio we arrive at by assumig the existece of umbers with more tha oe prime factorizatio cotraicte the lemma, it must be false. So all umbers have uique factorizatios.