Extending greatest common across the rationals

Grant Boudreaux & Scott Beslin Nicholls State University, Louisiana, USA

he purpose of this article is to examine one possible extension of greatest Tcommon (or highest common factor) from elementary properties. The article may be of interest to teachers and students of the Australian Curriculum: , beginning with Years 7 and 8, as described in the content descriptions for Number and Algebra. The senior secondary curriculum makes no specific mention ofgreatest common divisor, but the article is nevertheless a good resource for revisiting with students at this level the concepts of greatest common divisor and lowest common multiple in greater depth, and with a view to critical thinking. Certain concepts and problems can be used even in post-secondary instruction. In particular, teachers may find it useful in designing projects for guided self-discovery or collaborative learning. The article is written as a hybrid: part guided discovery, and part exposition of interesting results and applications. Teachers who enjoy factorisation of positive and the concepts of divisor and multiple will hopefully find this content useful and meaningful in making connections of those concepts with fractional . Sample problems and exercises are presented at the end of the article as self-tests and as vehicles for student investigations. Innovative teachers can also formulate their own conjectures and examples. The following word problem (e.g., Pirnot, 2007, p. 252, Problem 77) is typical of a collection of problems found in courses having a component unit in elementary , especially with regard to number relationships Australian Senior Mathematics Journal vol. 27 no. 2 among the positive integers.

Problem A: The Tiling Problem What is the size of the largest possible square tile which can be used to cover the floor of a room 8 metres long and 6 metres wide, without altering or cutting any tiles?

This problem may be used as a simple application of greatest common divisor (GCD) of two positive integers. The greatest common divisor (highest common factor), or GCD, of two positive integers a and b is the largest

55 56 Australian Senior Mathematics Journal vol. 27 no. 2 Boudreaux & Beslin (the set (rational) divisoroftwopositiverationalnumbers fromtheGCD,orgreatest other non-integerdimensionworks? What iftheroomdimensions changeto8.1and8 whole number, andwearenotallowedtoaltertiles.Butsurethatno room to80decimetresby58decimetres.Nowwehavepositiveintegersagain, tempted toaskfurtherquestions,suchasthefollowing. this particularcase,since the GCDof8and6is2,wewouldbeconfidentin that withthedimensions proposedinthequestion,8and5.8= not resultinaninteger. respectively. Whenwehaveaccomplished this extension, wordproblemssimilar that to Problem A involving GCD can be handled without number conversion or to ProblemAinvolvingGCDcanbehandledwithoutnumberconversionor of dimension2.5metres,forexample,willnotworkbecause8÷isa example, other hand, common divisor, oftwopositiveintegers.Alsos positive integerthatisafactorofboth feasible? Cantheproblemstillbesolved? giving ananswerofatilemeasuring2metresbymetres. and theGCDapplies. But supposewedonotwishtomakeconversions,or are there other rational divisors? Is alteration. Belowisaproposed definitionfortheGCDoftwofractionalnumbers. suppose insomeapplicationsconversionsarenoteasilyseenor  • • 1 5 + rationallydividesboth8 and5.8,so and Question 2(roomdimension):Whatifthedimensionsofroomare Question 1(tiledimension):Mustthedimensionoftilebegivenin With closerexaminationoftheproblem,teachersandstudentsmightbe We usetheabbreviationGCRDtodistinguish greatestcommon The greatestcommonrationaldivisoroftwopositivenumbersx of GCD from pairs of positive integers On thisnote,weextendthedomainofGCDfrompairspositiveintegers In thecaseoftiledimensionquestionwithregardtoProblemA,a Let usbrieflyreturntoProblemAQuestion1(room dimension).Notice For theroomdimensionquestion,wecanconvertdimensionsof whole numbers?Forexample,couldasquaretilewithsidelength2.5 not given in whole numbers? For example, suppose the room measured 8 metres longby5.8wide.Canwe,andhowdoapplytheconcept metres work? of GCDinthiscase? written GCRD(x and s t isapositiveintegerwitht r y arepositiveintegers. + +

denote the sets of positive integers and positive rational numbers, denotethesetsofpositiveintegersandrationalnumbers, 4 1 ×

|  3 2 7 2 + becausetheresultofdividing doesnotdivide ), topairsoffractionsorrationalnumbers(theset , y ) isthelargestpositiverationalnumberr 3 5 ands , written 1 5 rationalbutnotnecessarilyinteger. For the greatest common rational divisor? andb 7 2 ∤ 1 5 3 5 metreisaplausibleanswer. But 3 2 , becausedividing |t by (seeBurton,2002,p.21).In (s 3 1

4 1 metres? rationally dividest istheinteger6.On suchthatboth 2 5 9 + , thefraction

3 5 × by

 andy ) means + ). Here ). Here 7 2 does x r ,

Extending greatest common divisors across the rationals Australian Senior Mathematics Journal vol. 27 no. 2 57

, . b b a a is is | d k d k b b and 2 3 , think of we start with start we still equal to ) c d . We would like . We 5 such that gives identical such that 1 ⋅ c d 6 , 0 0 2 2 ⋅ ⋅ 5 1 4 6 and ( b a D and = with positive integers with positive integers ) C . b c a d 9 G ' ' 2 ⋅ c d 1 , 5 5 ⋅ = ⋅ 8 1 . Is . c d ( and D 0 5 2 1 C , with GCD(8, 9) = 1. But does b a and establish a property of G G ) 9 c = b and , ⎞ ⎟ ⎠ = as ) well-defined for positive rational for positive ) well-defined is a common rational divisor of is a common rational d ' ' 9 y 2 5 3 4 a 1 d 2 b metre. a n 1 ( b , , 1 d 8 1 D 5 1 b = ÷ ⎛ ⎜ ⎝ C b a 3 4 G D . Then by definition we have that k | bc. Then by definition we have that R and ), which can be shown directly from the C 6 4 ) = b G and d , = , and find the largest k 8 . To find the GCRD of GCRD the find To n. c ) d = 1 b 8 , . 2 1 b 1 5 , where each variable represents a positive integer. , where each variable represents a positive integer. . So: , ' , ' ÷ | ad and k ') whenever c d 8 'd 2 3 ( ) as the numerator and denominator, respectively, of respectively, ) as the numerator and denominator, ⋅ GCD(a =

D ', d c d . For example, our candidate for the GCRD of . For example, our candidate for are positive integers. is a common rational divisor of is a common rational R ) c ', c b d C 1 , b d G a d and d ( ', b b ) = m and , c, d D ' ' b ? C b a (a ⋅ . Indeed, , b G

and cd' = c = 1 2 b a m = is , defined by G(a and y ) ) = G c d a 9 , G , or equivalently k 8 ⋅ , d 2 (

c d 1 (likewise c D | for any natural any for C , c c G d d k and b c and d. Clearly , b b a For Problem A, notice that we propose For Problem A, notice that we propose En route to a formulation for GCRD by means of the function G Consider the two positive rational numbers Consider the two two positive rational numbers have a common divisor? have a common rational numbers words, can two positive In other We begin with the question: Is GCRD(x with the question: begin We To verify this result, we use the fact that the GCD of positive To To now substantiate our choice for GCRD in general, we consider the we consider general, in GCRD for choice our substantiate now To ? Direct calculation shows, “Yes!” Result 1 further below will explain why. Result 1 further below will explain ? Direct calculation shows, “Yes!” b, may be alternately expressed as expressed alternately may be is (a and b 2 1 3 4 2 3 1 a . Result 1 below states that the function G Result 1 below states a rational number. also a common divisor. This means we seek the largest integer k also a common divisor. and and definition of greatest common divisor of integers. function integers is ‘distributive’: that is, for every positive integer m output on pairs of equivalent . of calculate GCD(80, 58) = 2 decimetres = calculate GCD(80, 58) = 2 decimetres our answer change if the given fractions are not in lowest terms? For example, our answer change if the given fractions and bc. Thus, our candidate for the GCRD the greatest common divisor of ad the common rational divisor to note that, unlike with integers, the set of common divisors of any two positive of any two divisors of common set the integers, unlike with to note that, rational numbers. rational numbers. numbers x numbers rational numbers is an infinite set, since rational numbers For suppose first give a discovery-based a give first positive two of GCRD the finding for argument GCD(m

In that case, does a largest such divisor exist? To answer these questions, we answer these a largest such divisor exist? To In that case, does One way of confirming this speculation is to indeed convert to decimetres and One way of confirming this speculation The numbers a Then, ab' = a'b Result 1 a a, G 58 Australian Senior Mathematics Journal vol. 27 no. 2 Boudreaux & Beslin (pairs offractionalnumbers)as G Result 2 We firstshow If a

A formula for rationalGCD A formula Result 1nowenablesustothinkofG Result 1pointsoutthatifquadruplesareconsideredastwo‘pairs’,thenG which isaninteger. Therefore we showthereisnolargercommondivisor. unchanged whenthosepairsrepresentequivalentfractions. of is lessthanorequalto integers Similarly, also aninteger, sothat isthegreatestcommonrationaldivisorofanytwopositivenumbers. The functionG Suppose a a b , b and , c andd e and d c . Now we show every common positiverationaldivisorof . Nowweshowevery , b G arepositiveintegers,then C f , c . Then D b ( d a andd d , b hasbeendefinedonquadruplesofpositiveintegers. c ) isacommonrationaldivisorofboth G ⎝ ⎜ ⎛ e f G arepositiveintegers.Then G ( C a

C D ⋅ D , b ( b b ( m d a d a ⎝ ⎜ ⎛ ⎝ ⎜ ⎛ , d d G G c = , , b C C b , c c d D D ) ⎝ ⎜ ⎛ ) ⎝ ⎜ ⎛ ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ b b ( ( G d a ) b c | d d a a a . Assume b G ⎝ ⎜ ⎛ C ⎠ ⎟ ⎞ ⎠ ⎟ ⎞ d d ======d c and D , , b b ⎠ ⎟ ⎞ ( G b ( ( ( c c G G G G d a a ) ) b . So ac a ⎝ ⎜ ⎛ d ⎠ ⎟ ⎞ ⎠ ⎟ ⎞ ( , C C C C d c , ' a ab b = = = = c a asa‘well-defined’functionon ) ) c D D D D ' ) ' ⋅ , ' = d a ) b G G ⎠ ⎟ ⎞ c b ' G ( ( ( ( e b f | b ⋅ C C G d ⋅ a a ad ad ab ⋅ ⋅ ⎝ ⎜ ⎛ ' G

G a D D b C ab c C , G ' ab G G ⋅ c ' d C a c d ⎠ ⎟ ⎞ ( b ( D d e f C C C d D c a a ' , , D ' ' n C b ' . ⋅ ( ' d d c D b b D ' , ' d b a | ' ( ' c ( c D ⎠ ⎟ ⎞ d , d b c , , b = c c a ( ( d R ' b b d d d b a d a a d d d c c , ) ) ( b ' d d ) ) D , ' ' ) a c b b ' ' ad , ' , , c d c b b ) a d c c c ) and ( ' isacommonrationaldivisor ) ) ) ' ' a . Withthiscontext,weshow b , , for some integers b b b , c d c c ' c ) c ) ' = ) ' ) e f G | C D b ( d c d a d a b forsomepositive , b and c ) . d c . Secondly, m a b and and +

×  is n d c + .

Extending greatest common divisors across the rationals Australian Senior Mathematics Journal vol. 27 no. 2 59

. ). ⎞ ⎟ ⎠ y c d , , b a ⎛ ⎜ ⎝ and 4 15 is the 4 45 ), which bc), which GCRD ). = g , n if divides since bde , as defined in the defined , as 1 metre. ) 9 8 5 1 = bc , , ⋅ GCD(ad ⎞ ⎟ ⎟ ⎠

= ad f c . bd dg , are integers. ≤ 1 ⋅ a rational multiples rational smaller and bg 10 ⎛ ⎜ ⎜ ⎝ ) = GCD(m . Additionally, thoughtful . Additionally, GCD( 5 = , which means 4 5,29 15 ⋅ . Likewise, den = cf Likewise, . bde | adf , n 8 4 4 45 45 () does not require rational fractions rational require not does and y rationally divides GCRD(x rationally divides ⋅ = 9 8 GCRD = 12 135 GCD 9 8 4 45 = bcf), or bde and y = GCD(4,10) ⎞ ⎟ ⎠ 8) ⋅ = 5 29 , 9 9,15 and 8 1 ⋅ ⋅ be | af; thus, ⎛ ⎜ ⎝ 15 3 1 = rationally divides rationally GCD(4 GCD(4,10) 4 f e 45 = = 1. In other words: ⋅ = GCRD ⎞ ⎟ ⎠ , ≤ GCD(adf ).) So we can now infer from the proof of Result of Result proof the from infer can now we So bc).) ⎞ ⎟ ⎠ 9 8 4 , 15 . 1 10 4 , 15 = d = 2 3 ⎛ ⎜ ⎝ 1 4 ⎛ ⎜ ⎝ 4 4 45 15 results in 4 ) is an integer, then both x ) is an integer, 45 . (In fact, . (In fact, GCRD , y which implies af which implies ) are positive integers, GCRD(m GCRD GCD(ad, ⋅

bc , ad bd GCRD(8, 5.8) = bem = and n GCD( ) = f bcf) ≤ . Hence, bde bde | bcf. Hence,

number that is a ‘factor’ of both of ‘factor’ is a that rational number f e no other non-integer (or integer) larger than 2 is acceptable because For example, For example, For Question 2 on room dimension, if we do not wish to make conversions, For Question 2 on room dimension, if we do not wish to make conversions, Note that by Result 1 our formula for GCRD GCRD for formula our 1 Result by that Note Let us try another problem from scratch. In other words, when two rational numbers are divided by their GCRD, the GCRD, their divided by are numbers rational two when words, other In It is not difficult to show that (exercise!): It is not difficult to show that (exercise!): When m When GCRD(x by their GCRD = 9 8 gives dimension, we have seen that 2.5 metres will not work. We can now also state also now can We work. will not metres 2.5 that seen have we dimension, implies containers with an equal amount of juice in each one. What is the maximum the is What one. each in juice of amount equal an with containers capacity of such a container? example, or just directly compute: of 2 which work, such as consideration of the formula in Result 2 shows that: consideration of the formula in Result Revisiting Problem A and further illustrations that resulting numbers are relatively prime integers. For example, dividing resulting numbers are relatively litres, 36 litres, and 54.1 litres. The distributor wishes to package the juice in juice the package to wishes distributor The litres. 54.1 and litres, 36 litres, largest we have previously verified directly that . For Question 1 on tile Problem the Tiling Let us now revisit Problem A, 2 that every rational divisor of x common Introduction. GCD(adf, GCD, readily seen when b GCD, readily seen GCRD(8, 6) = GCD(8, 6) = 2. There are, of course, are, There 2. = 6) GCD(8, = 6) GCRD(8, to be in reduced form of formula is a proper generalisation . Furthermore, this to be in reduced To answer this question, we may convert the measurements to decilitres, for answer this question, we may convert the measurements to decilitres, for To Therefore, Therefore, Problem B: CranberryProblem Juice A cellar contains barrels of cranberry juice of three different capacities: 25.2 Answer 60 Australian Senior Mathematics Journal vol. 27 no. 2 Boudreaux & Beslin Answer Answer An unexpectedapplicationofGCRDisanother proofoftheirrationality Problem D:LighthouseBeacon Problem C:Travelling Vice-President Result 3 The beacon of one lighthouse passes a certain point every 12.3 seconds; The beacon of one lighthouse passes a certain point every 18daysforone The vice-presidentofacorporationtravelstoAtlantaevery rational multiple.With(arguably)nosurprise, GCRD appliedtothree(ormore)terms. If A higher-level applicationofGCRD A higher-level LCRM(12.3, 18.2) LCM(18, 24)=LCRM(18,72days. root ofn they dosoagain? two beaconssimultaneouslyflashonthislocation,inhowmanysecondswill the smallestpositiveintegerhavingbotha be repeatedwithminimalefforttogivebirthitstwinLCRM,orleastcommon many moredayswilltheybeinAtlantaagainonthesameday? multiple) orLCM.We recallthattheLCMoftwowholenumbersa every 24 days, also for one day.every If both persons are in Atlanta today, in how integers day. ThemanagerofonethebranchesthiscorporationtravelstoAtlanta .30.)Forexample,LCM(12,18)=36.Theentiredevelopment ofGCRDcan p. another passes the same location every 18.2seconds.Ifatthismomentthe another passesthesamelocationevery 2 n The conceptual ‘twin’ of GCD is Cranberry Juiceproblemaffordsanopportunityforconsidering Note thattheCranberry Suppose thek . This fact comes as a corollary tothefollowingtheorem. . Thisfactcomesasacorollary andk k isaninteger. andn arepositiveintegersand thek GCRD(25.2,36,54.1) . Since th rootofn = LCRM k n

) ( is a positive rational number, there exist relatively 123 10 isapositiverationalnumber forsomepositive , 91 5 = = = = = GCRD CDGCRD GCRD GCRD(GCRD(25.2,36),54.1) 10 1 LCM(123 litre (lowest common th rootofn 50 ) ( ( andb ⋅ 5,10 18 5 , ⋅ LCRM 91) 541 10 = asfactors(seeBurton,2002, ⎝ ⎜ ⎛ 126 11193 5 5 ) ( , 36 isrationalthenthek 1 a b = ⎠ ⎟ ⎞ , , d c 541 2238.6 seconds. 10 = ) LCM( bd ad , bc ) . andb is th Extending greatest common divisors across the rationals Australian Senior Mathematics Journal vol. 27 no. 2 61 , with . Hint: , which k k ) b a bc , = is not an is not = a ad k bd ) = xy. n b a () k , y LCM( is irrational. = = 2 n , or c d k , b a , so that their greatest () 0 th root of n th = a ) ⋅ LCM(x ⋅ 5

–2 , y can also be realised by the LCRM ⋅ 3 4 k k ) 45

b a 3 n , = 1 1 () k 9 8 a = 2 , , k 9 8 , GCD(x k , or equivalently , or equivalently b 4 () b 15 b a = 1. Hence, n () 5 2 k 1 = GCD b GCRD(1, GCRD ≠ and n = = = = and y –1 k ) _ ) = 1. But we also have ) = 1. But n GCRD ⋅ 5

–1 . –1 ⋅ 3

2 5 2 3 1 4 ⋅ 5 GCD(1, = = –2 = = 2 such that such that 4 _ _ ⋅ 3 15 _ 2 , , 7 = 1 which implies b = 1 which implies , 22 k 2 3 _ _ ()()() and b , or b 5 6 , 2.8 k , 3.14, , 1 b 9 2 3 11 7 14 GCRD GCRD GCRD 1.7, 0.23 ) = 1. Clearly GCD(1, n ) = 1. Clearly k

, b k a Find the GCRD of each pair or triple. Fill in the blanks with distinct non-integer rational numbers, if possible. Show by example and then by proof that Use the fact that for integers x (a) (a) (b) (c) (b) (d) (c) Thus, 1 = A statement equivalent to Result 3 is that if the k if that 3 is Result to statement equivalent A can also be used to calculate GCRD. An alternative definition for definition alternative An GCRD. to calculate used also be can algorithms Closing remarks Problems for teachers and students integer exponents. For example, integer exponents. For example, is an integer. then it is irrational. For example, the real number integer, prime integers a prime integers prime ‘factorisations’ compute the GCD of two positive integers. Consequently, by Result 2, these integers. Consequently, compute the GCD of two positive common ‘divisor’ is 2 2. GCRD can be formulated by means of factorisation and use of non-positive non-positive of use and factorisation of means by formulated be can GCRD GCD( 3. With no necessity for factorisation, recursive algorithms can be used to With no necessity for factorisation, 1. 4. Let a, b, c, d, e and f be positive integers. Prove the following: [For part (a), refer to the discussion in A formula for rational GCD.] (a) GCRD a , c =1 if g = GCRD a , c ()bg dg ()b d Boudreaux & Beslin e (b) GCRD e ⋅ a , e ⋅ c = ⋅ GCRD a , c ()f b f d f ()b d (c) GCRD a , a + c = GCRD a , c ()b b d ()b d

5. Marcia is making bracelets of three varieties: ones with 4.5 mm beads, ones with 5.5 mm beads, and ones with 6.2 mm beads. What is the shortest bracelet length so that all three varieties are the same length?

6. Derive a formula for the GCRD of a triple consisting of two positive a integers m and n and one positive rational b ; in other words GCRD m, n, a = ? ()b

References

Burton, D. M. (2002). Elementary number theory (5th ed.). New York, NY: McGraw-Hill. Pirnot, T. L. (2007). Mathematics all around (3rd ed.). Boston, MA: Pearson Addison-Wesley. Australian Senior Mathematics Journal vol. 27 no. 2

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