3. Greatest Common Divisor - Least Common Multiple

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1 3 Greatest Commo Divisor - Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd ab, divisor of a ad b as ( ) Example 32: Fid gcd ( 24,90 ) Solutio: The atural umbers that divide 24 are 1, 2, 3, 4, 6, 12 ad 24 The atural umbers that divide 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30 ad 90 Therefore, gcd 24,90 = 6 ( ) Oe way to compute the greatest commo divisor of two atural umbers a ad b is to write their prime factorizatios For example, i the case of the umbers 24 ad 90 above, 3 2 we have 24 = ( 2) ( 3) ad 90 = ( 2)( 3) ( 5) Notice that we ca iclude all of the prime factors from both umbers by usig a expoet of = ( ) 2 90 = ( 2)( 3) ( 5) Now we take the smallest expoet that occurs for each prime factor ad obtai ( ) = ( ) = ( 2) ( 3) ( 5) = ( 2)( 3) mi(3,1) mi(1,2) mi(0,1) gcd 24, = 6 We geeralize this idea i the theorem below Theorem 33: Let ab, ad suppose p1, p2,, p are prime umbers so that α a= ( p1) ( p2 ) ( p ) ad β1 β2 β b= ( p1) ( p2 ) ( p ) where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) Proof: Let c= gcd( a, b) Sice p1, p2,, p are prime umbers, p divides both a ad b implies that p divides c

2 If p α divides a ad p β divides b, the the smaller of p α ad p β ad b ( 2 2 divides 12 ad 2 divides 30 implies 2 divides gcd(12,30) = 6 ) So, mi( α, β ) p the that divides both a ad b; ad hece c If p does ot divide c But whe we say p does ot divide c Hece, gcd (, ) ( ) 1 1 ( ) 2 2 ( ) p divides both a p divides a but does ot divide b, mi( α,0) 0 p mi( α, β ) mi( α, β ) mi( α, β) ab = p p p = = 1 divides c, we mea Example 34: Fid the greatest commo divisor of 1260 ad 600 Solutio: The prime factorizatio of these umbers is give by = ( 2) ( 3) ( 5)( 7) ad = ( 2) ( 3)( 5) So the combied prime factors are 2, 3, 5 ad 7 Notice that these are all factors of 1260 Also, 600 ca be writte usig these prime factors as The Theorem 33 implies = ( ) ( = ( ) ) ( ) = ( 2) ( 3) ( 5) ( 7) ( ) mi 2,3 mi 2,1 mi 2,1 mi 1,0 gcd 1260, = 60 Remark 35: The TI-83 fuctio gcd ca be used to fid the greatest commo divisor of two atural umbers This fuctio ca be foud by pressig MATH, selectig NUM, ad the scrollig dow to item 9 This will place gcd( o the calculator scree Typig 1260,600) ad pressig ENTER results i the scree shot below Gcd(1260,600) 60 How Do Calculators Compute The Greatest Commo Divisor? Calculators DO NOT use prime factorizatios to compute the greatest commo divisor of two atural umbers Istead, they use a very efficiet process called the Euclidea algorithm We give this process after the defiitio ad theorem below

3 Defiitio 36: Let ab, The iteger part of a/ b is the largest value c chose from { } 0,1, 2,3, so that c a/ b We deote the iteger part of a/ b by ipart(a/b) Remark 37: ipart is a built-i fuctio o the TI-83 You ca access this fuctio by pressig MATH ad selectig NUM The fuctio ipart is item 3 Theorem 38: Let ab, with a ad 0 r < b Proof: c ipart ( a/ b) > b, ad suppose c= ipart ( a/ b) The r = a cb = implies that a c k b = + where 0 k < 1 (otherwise c will ot be the largest iteger which is less tha or equal to a/ b) a c k a b( c k) bc bk a bc bk b = + = + = + = Sice 0 k < 1, we have; 0 bk < b Let r = bk The, r = a cb ad 0 r < b Remark 39: The value r i Theorem 38 is sometimes referred to as the remaider associated with the divisio a/ b, ad the iteger part c of a/ b is sometimes referred to as the quotiet associated with the divisio a/ b Example 310: The iteger part of 2/3 is 0, the iteger part of 9/4 is 2, ad the iteger part of 67/13 is 5 That is, ipart(2/3) = 0, ipart(9/4) = 2 ad ipart(67/13) = 5 Theorem 14 ca be realized with the equatios ad 1= 9 ( 2)( 4) 2= 67 ( 5)( 13) The Euclidea algorithm uses the Theorem 14 to fid the greatest commo divisor of two atural umbers This algorithm was origially proposed i Euclid s Elemets aroud 300 BC It is amazig that this process is still widely used i practice by high speed computig devices The process is give below Euclidea algorithm Let ab, with a > b r1 = a ipart( a/ b) b r = b ipart( b/ r) r r = r ipart( r / r ) r r = r ipart( r / r ) r Cotiue util the first time r = 0 i If i = 1 the gcd ( ab, ) = 1 Otherwise, gcd ( ab, ) ri 1 =

4 Theorem 311: Let ab, with a > b The Euclidea algorithm computes gcd ( ab, ) Proof: Let ab, with a b gcd ab, Suppose the remaider of the divisio of a by b is c The a= qb+ c, where q is the quotiet of the divisio Ay commo divisor of a ad b also divides c (sice c ca be writte as c= a qb); similarly ay commo divisor of b ad c will also divide a Thus, the greatest commo divisor of a ad b is the same as the greatest commo divisor of b ad c Therefore, it is eough if we cotiue the process with umbers b ad c Sice c is smaller i absolute value tha b, we will reach c = 0 after fiitely may steps > We are lookig for ( ) Example 312: Use the Euclidea algorithm to compute gcd ( 1260,600 ) Solutio: You ca verify that r1 = 1260 ipart ( 1260 / 600)( 600) = 1260 ( 2)( 600) = 60 r = 600 ipart 600 / = = 0 So, gcd ( 1260,600) = 60 2 Example 313: Use the Euclidea algorithm to compute gcd( 860,1375 ) Solutio: The algorithm results i r1 = 1375 ( 1)( 860) = 515 r2 = 860 ()( 1 515) = 345 r3 = 515 () = 170 r4 = 345 ( 2)( 170) = 5 r = = 0 Cosequetly, gcd( 860,1375) = 5 5 Example 314: Use the Euclidea algorithm to compute gcd( 1326,741 ) Solutio: You ca verify that ()( ) ()( ) r = = 585 r = = 156 r = = 117 r = = 39 r = = 0

5 Cosequetly, gcd( 1326,741) = 39 A quatity which is closely related to the greatest commo divisor is the least commo multiple Defiitio 315: Let ab, The least commo multiple of a ad b is the smallest atural umber which is a multiple of both a ad b The least commo multiple of a ad lcm ab, b is deoted by ( ) Remark 316: lcm ( ab, ) abwheever ab, Sice the lcm ( ab, ) is the smallest umber that is a multiple of both a ad b, it is also the smallest umber for which both a ad b are divisors As a result, all of the prime factors of a must divide lcm ( ab, ), ad all of the prime factors of b must divide lcm ( ab, ) This leads to the followig theorem Theorem 317: Let ab,, ad suppose the prime factorizatios of a ad b are give by α a= ( p1) ( p2 ) ( p ) ad β1 β2 β b= ( p1) ( p2 ) ( p ) where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p) ( p) ( p ) Proof: Let c= lcm( a, b) If a ad b are both multiples of p, the c is a multiple of too If a is a multiple of bigger of p α max( α, β ) p of p Usig ad p α p β ad b is a multiple of p β, the c is a multiple of the ( sice c is a multiple of both a ad b)hece, c is a multiple of Note that, if a is ot a multiple of p but b is; the c should be a multiple p max(0, β ) β max( α, β ) max( α, β ) max( α, β) = p solves this problem Thus, 1 2 lcm ab, = p p p Example 318: Use Theorem 17 to compute lcm ( 24,15 ) 3 Solutio: We ca see that 24 = ( 2) ( 3) ad 15 ( 3)( 5) commo form as = ( 2) ( 3)( 5) = These ca be writte i p

6 ad As a result, 0 15 = ( 2) ( 3)( 5) ( ( ) = ( ) ) ( ) = ( 2) ( 3) ( 5) ( ) max 3,0 max 1,1 max 0,1 lcm 24, = 120 Example 319: We show the value of lcm ( ab, ), gcd (, ) of ab, ab ad ab for several choices a, b lcm(a,b) gcd(a,b) ab 16, , , , Although it might ot be readily apparet, the table above gives some isight to the lcm, gcd ab, ad ab The table above has bee relatioship betwee ( ab ), ( ) reproduced below, alog with a additioal colum cotaiig the product of lcm ( ab, ) ad gcd ( ab, ) a, b lcm(a,b) gcd(a,b) ab lcm(a,b)gcd(a,b) 16, , , , For the choices of a ad b above, it appears that lcm ( ab, ) gcd ( ab, ) = ab This result is always true Theorem 320: Let, ab The lcm (, ) gcd (, ) ab ab = ab α Proof: Let = ( ) ad ( ) β β β a p1 p2 p b= p1 p2 p where the values α, α,, α, β, β,, β are either atural umbers or 0 The ad 1 2 ( ) = ( ) max( α, β ) max( α, β ) max( α, β) lcm ab, p p p

7 1 2 mi( α, β ) mi( α, β ) mi( α, β) gcd ab, = p p p max( 1, 1) max( 2, 2) mi( 1, 1) mi( 2, 2) ( ab) ( ab) ( p) α β ( p ) α β ( p ) ( p) α β ( p ) α β ( p ) max( α, β) mi( α, β) lcm, gcd, = max( α1, β1) + mi( α1, β1) max( α2, β2) + mi( α2, β 2) max( α, β) + mi( α, β) = ( p1) ( p2) ( p ) If max( α1, β1) = α1 the mi( α1, β1) = β1 Ad if mi( α1, β1) = α1, the max( α1, β1) = β1 Hece, i ay case max( α1, β1) + mi( α1, β1) = α1+ β1 So we have; Exercises α1+ β1 α2+ β2 α+ β ( ab) ( ab) = ( p1) ( p2) ( p ) ( p ) ( p ) ( p ) ( p ) ( p ) ( p ) lcm, gcd, α α α β β β = = ab 1 Use the prime factorizatios of 860 ad 1375 to compute gcd( 860,1375 ) 2 Use the prime factorizatios of 6300 ad 1584 to compute gcd ( 6300,1584 ) 3 Use the prime factorizatios of 1260 ad 2640 to compute gcd( 1260, 2640 ) 4 Use the prime factorizatios of 2373 ad 1374 to compute gcd( 2373,1374 ) 5 Use the divisio algorithm to compute gcd ( 6300,1584 ) 6 Use the divisio algorithm to compute gcd( 1260, 2640 ) 7 Use the divisio algorithm to compute gcd( 2373,1374 ) 8 Use the prime factorizatios of 75 ad 124 to fid lcm( 75,124 ) 9 Use the prime factorizatios of 236 ad 125 to fid lcm(236,125) 10 Use the prime factorizatios of 84 ad 118 to fid lcm ( 84,118 ) ab = ad lcm ( ab, ) = 300 Give gcd ( ab, ) 11 Suppose Let mab,, Show that gcd ( ma, mb) = m gcd ( a, b) ; ie the greatest commo divisor satisfies a distributive property 13 Let abc,, Show that gcd ( gcd( ab, ), c) = gcd ( a,gcd( bc, )) ; ie the greatest commo divisor satisfies a associative property 14 Let a Show that gcd( aa, ) = a; ie the greatest commo divisor is idempotet 15 Let, gcd( ab, ) = gcd ba, ; ie the greatest commo divisor ab Show that ( ) satisfies a commutative property 16 Let ab, Show that lcm( a,gcd( a, b)) = a ad gcd( a,lcm( a, b)) = a 17 Let a Show that lcm( aa, ) = a; ie the least commo multiple is idempotet 18 Let ab, Show that lcm( ab, ) = lcm ( ba, ) ; ie the least commo multiple satisfies a commutative property

8 19 Let,, abc Show that lcm( lcm( ab, ), c) lcm ( a,lcm( bc, )) = ; ie the least commo multiple satisfies a associative property 20 Let,, lcm ma, mb = mlcm a, b ; ie the least commo mab Show that multiple satisfies a distributive property Solutios: = ad 1375 = 5 11 Therefore; mi(0,2) mi(1,3) mi(0,1) mi(0,1) gcd(860,1375) = = ( 2) ( 5)( 11) ( 43) = = ad 1584 = Therefore; ( ) = ( ) ( ) ( ) ( ) ( ) mi(2,4) mi(0,2) mi(0,1) mi(0,1) gcd 6300, = = = ad 2640 = Therefore; ( ) = ( ) ( ) mi(2,4) mi(0,1) mi(0,1) mi(0,1) mi(0,1) gcd 1260, = 2 5 = = ad 1374 = Therefore; ( ) ( ) gcd 2373,1374 = = 3 5 gcd ( 6300,1584 ) 3 ()( ) ( ) r1 = = 1548 r2 = = 36 r = = 0

9 Thus; gcd( 6300,1584) = 36 6 r1 r2 r3 = 120 ( 2) 60 = 0 Cosequetly, gcd ( 1260, 2640) = 60 = = 120 = = 60 7 Hece, gcd( 2373,1374) = 3 7 ()( ) ( ) ()( ) ()( ) ()( ) r1 = = 999 r2 = = 375 r3 = = 249 r4 = = 126 r5 = r6 = = 123 = 3 r = = = 3 5 ad 124 = 2 31 Thus, ( ) = ( ) ( 2) ( 3) ( 5) ( 31) max 2,0 max 1,0 max 2,0 max(1,0) lcm 75, = = = 2 59 ad 125 = 5 Hece, ( ) = ( ) ( 2) ( 5) ( 59) max 2,0 max 3,0 max(1,0) lcm 236, = = = ad 118 = 2 59 Therefore,

10 ( ) = ( ) ( 2) ( 3) ( 7) ( 59) max 2,1 max 1,0 max 1,0 max(1,0) lcm 84, = = Suppose 900 So, ( ab) ab = ad lcm ( ab, ) = 300 We kow that lcm (, ) gcd (, ) ab 900 gcd, = = = 3 The greatest commo divisor is: lcm, 300 gcd ( ab, ) = 3 ( ab) ab ab = ab 12 Let mab,, α Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad b= ( p1) ( p2 ) ( p ), where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) α We have: = ( ) ( ) ad ( ) ( ) ma m p1 p2 p So, the greatest commo divisor is: 1 2 ( ma mb) = ( m) ( p1) ( p2) ( p ) mi( α1, β1) mi( α2, β2) mi( α, β) = m( p ) ( p ) ( p ) mi(1,1) mi( α, β ) mi( α, β ) mi( α, β) gcd, = mgcd( a, b) β β β mb = m p1 p2 p Hece, the greatest commo divisor satisfies a distributive property 13 Let abc,, Suppose 1, 2,, α α α p p p are prime umbers so that a ( p1) ( p2 ) ( p ) β1 β2 β γ1 γ2 γ = ( ) ad ( ) =, b p1 p2 p c= p1 p2 p, where the values α, α,, α, β, β,, β, γ, γ,, γ are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) ( ab) ( p1) ( p2) ( p ) mi( β1, γ1) mi( β2, γ2) mi( β, γ) ( bc) ( p1) ( p2) ( p ) gcd, = ad gcd, = mi(mi( α, β ), γ ) mi(mi( α, β ), γ ) mi(mi( α, β), γ) gcd(gcd ab,, c) = p p p ad

11 ( a ( b c) ) ( p1) ( p2) ( p ) mi( α,mi( β, γ )) mi( α,mi( β, γ )) mi( α,mi( β, γ)) gcd,gcd, = Here, it is importat to observe that mi(mi( α, β), γ) = mi( α, mi( β, γ)) = mi( α, β, γ) The reaso behid this is; while comparig 3 atural umbers, you ca start comparig from whichever you wat If you wat to fid the smallest of 5, 9 ad 6, the arrage them i order; 5 < 6 < 9 The smallest is 5, this does ot chage So, it is ot importat if you compare 5 & 9 first ad the the result with 6; or if you compare 6 & 9 first ad the the result with 5 Therefore, gcd ( gcd( ab, ), c) gcd ( a,gcd( bc, )) = ; ie the greatest commo divisor satisfies a associative property 14 Let a Show that gcd( aa, ) = a; ie the greatest commo divisor is idempotet Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) where the values α1, α2,, α are either atural umbers or 0 The mi( α1, α1) mi( α2, α2) mi( α, α) gcd ( aa, ) = ( p1) ( p2) ( p ) where, of course, mi( α, α ) = α Hece, 1 2 ( aa) = ( p1) ( p2) ( p ) α = ( p ) ( p ) ( p ) = a mi( α, α ) mi( α, α ) mi( α, α) gcd, α α α It is easier to see this result if we restate the questio as what is the greatest umber that divides both a ad a? The aswer is, of course, a itself 15 Let, gcd( ab, ) = gcd ba, ; ie the greatest commo divisor satisfies a commutative property α Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad b= ( p1) ( p2 ) ( p ), where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p1) ( p2) ( p ) ad mi( β1, α1) mi( β2, α2) mi( β, α) gcd ( ba, ) = ( p1) ( p2) ( p ) The coclusio follows from the observatio that mi( α, β) = mi( β, α) gcd( ab, ) = gcd ba, ab Show that ( ) Hece, ( )

12 16 Let ab, Suppose 1, 2,, α α α p p p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad ( ) b= p1 p2 p, where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p1) ( p2) ( p ) ad mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) ( ) max( α,mi( α, β )) max( α,mi( α, β )) max( α,mi( α, β)) lcm( a,gcd( a, b)) = p p p Let s fid max( α1, mi( α1, β 1)) If α1 < β1, the mi( α1, β1) = α1 ad max( α1, mi( α1, β1)) = max( α1, α1) = α1 If β1 < α1, the mi( α1, β1) = β1 ad max( α1, mi( α1, β1)) = max( α1, β1) = α1 Similarly, for ay, max( α, mi( α, β )) = α Therefore; ( 1) ( 2) ( ) α ( p ) ( p ) ( p ) a max( α,mi( α, β )) max( α,mi( α, β )) max( α,mi( α, β)) lcm( a,gcd( a, b)) = p p p = = ( ) mi( α,max( α, β )) mi( α,max( α, β )) mi( α,max( α, β)) gcd( a,lcm( a, b)) = p p p With the reasoig we used above we ca coclude that mi( α, max( α, β )) = α Thus, ( 1) ( 2) ( ) α ( p ) ( p ) ( p ) a mi( α,max( α, β )) mi( α,max( α, β )) mi( α,max( α, β)) gcd( a,lcm( a, b)) = p p p = = Hece, lcm( a,gcd( a, b)) = a ad gcd( a,lcm( a, b)) = a 17 Let a Suppose p1, p2,, p are prime umbers so that ( ) α α α a= p1 p2 p where the values α1, α2,, α are either atural umbers or 0 The 1 2 ( aa) = ( p1) ( p2) ( p ) α = ( p ) ( p ) ( p ) = a max( α, α ) max( α, α ) max( α, α) lcm,

13 Therefore, lcm( aa, ) = a; ie the least commo multiple is idempotet 18 Let ab, Suppose p1, p2,, p are prime umbers so that α = ( ) ad ( ) β β β a p1 p2 p b= p1 p2 p, where the values α, α,, α, β, β,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) ( ab) = ( p1) ( p2) ( p ) ad max( β1, α1) max( β2, α2) max( β, α) ( ba) = ( p1) ( p2) ( p ) Sice =, we ca coclude that lcm( ab, ) lcm ( ba, ) lcm, lcm, max( α, β ) max( β, α ) = 19 Let abc,, Suppose p1, p2,, p are prime umbers so that α = ( 1) ( 2 ) ( ), b ( p1) ( p2 ) ( p ) γ1 γ2 γ ( ) a p p p β β β = ad c= p1 p2 p, where the values α1, α2,, α, β1, β2,, β, γ1, γ2,, γ are either atural umbers or 0 The 1 2 ( ab) ( p1) ( p2) ( p ) 1 2 max( α, β ) max( α, β ) max( α, β) lcm, = ad max( β, γ ) max( β, γ ) max( β, γ) lcm bc, = p p p max(max( α, β ), γ ) max(max( α, β ), γ ) max(max( α, β), γ) lcm(lcm ab,, c) = p p p ad ( a ( b c) ) ( p1) ( p2) ( p ) max( α,max( β, γ )) max( α,max( β, γ )) max( α,max( β, γ)) lcm,lcm, = As we have poited out before, the order at which you start to compare atural umbers does ot matter; max(max( α, β ), γ ) = max( α, max( β, γ )) = max( α, β, γ ) Hece, lcm( lcm( ab, ), c) lcm ( a,lcm( bc, )) satisfies a associative property = ; ie the least commo multiple 20 Let mab,, Suppose 1, 2,, α α α p p p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad ( ) b= p1 p2 p, where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p) ( p) ( p )

14 α We have: = ( ) ( ) ad ( ) ( ) ma m p1 p2 p So, the least commo multiple is: 1 2 ( ma mb) = ( m) ( p ) ( p ) ( p ) 1 2 ( ) β β β mb = m p1 p2 p max(1,1) max( α, β ) max( α, β ) max( α, β) lcm, max( α, β ) max( α, β ) max( α, β) = m p p p = mlcm( a, b) Thus, lcm ( ma, mb) mlcm ( a, b) distributive property = ; ie the least commo multiple satisfies a

15 Graphical Represetatio Of gcd(a,b) = 1 Numbers ab, are said to be relatively prime if ad oly if gcd( ab, ) = 1 It is possible to visualize all of the pairs ( ab, ) for which a ad b are relatively prime by plottig the poits ( ab, ) for a ad b betwee 1 ad 100 We give this plot below, ad the patter is very iterestig! Surprisigly, there are more pairs i this rage that are relatively prime tha those that are ot I fact, of the 10,000 pairs ( ab, ) for a ad b betwee 1 ad 100, there are 6,087 pairs which are relatively prime ( ab, ) for which gcd( ab, ) = 1 for ab, { 1,,100}

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