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Thread: GCD Multiplicative Property

  1. #1
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    GCD Multiplicative Property

    i'm reading Introduction to Analytic Number Theory, Tom M. Apostol and i have a doubt about an exercise in chapter 1. Follow the exercise:

    Prove the following multiplicative property of the gcd:

    $\displaystyle (ah,bk)=(a,b)(h,k)\left ( \frac{a}{(a,b)},\frac{k}{(h,k)} \right )\left ( \frac{b}{(a,b)},\frac{h}{(h,k)} \right )$

    i already tried using Bezout identity and $\displaystyle (a,b)=\frac{a\cdot b}{[a,b]}$
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  2. #2
    Super Member fardeen_gen's Avatar
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    Let
    $\displaystyle (a,b)=m$,
    $\displaystyle (h,k)=n$,
    $\displaystyle a=ma_{1}$,
    $\displaystyle b=mb_{1}$,
    $\displaystyle h=nh_{1}$,
    $\displaystyle k=nk_{1}$,
    then $\displaystyle (a_{1},b_{1}), (h_{1},k_{1})=1$

    Then the question becomes:
    $\displaystyle (a_{1}h_{1},b_{1}k_{1})=(a_{1},k_{1})(b_{1},h_{1})$

    $\displaystyle (\frac{a_{1}}{(a_{1},k_{1})}\times\frac{h_{1}}{(b_ {1},h_{1})},\frac{k_{1}}{(a_{1},k_{1})}\times\frac {b_{1}}{(b_{1},h_{1})})=1$, which is obviously true.

    EDIT: Then I recommend waiting for the more experienced forum members to answer.
    Last edited by fardeen_gen; May 8th 2009 at 07:48 PM. Reason: Comments
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  3. #3
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    I think that your way is not formal. A prove different
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  4. #4
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    Quote Originally Posted by streethot View Post
    i'm reading Introduction to Analytic Number Theory, Tom M. Apostol and i have a doubt about an exercise in chapter 1. Follow the exercise:

    Prove the following multiplicative property of the gcd:

    $\displaystyle (ah,bk)=(a,b)(h,k)\left ( \frac{a}{(a,b)},\frac{k}{(h,k)} \right )\left ( \frac{b}{(a,b)},\frac{h}{(h,k)} \right )$

    i already tried using Bezout identity and $\displaystyle (a,b)=\frac{a\cdot b}{[a,b]}$
    it's obvious that if $\displaystyle \gcd(x,y)=\gcd(x,z)=\gcd(t,y)=\gcd(t,z)=1,$ then $\displaystyle \gcd(xt, yz)=1. \ \ \ \ \ \ \ (1)$

    now let $\displaystyle a_1=\frac{a}{\gcd(a,b)}, \ b_1=\frac{b}{\gcd(a,b)}, \ k_1=\frac{k}{\gcd(k,h)}, \ h_1=\frac{h}{\gcd(k,h)},$ and:

    $\displaystyle x=\frac{a_1}{\gcd(a_1,k_1)}, \ y=\frac{b_1}{\gcd(b_1,h_1)}, \ z=\frac{k_1}{\gcd(a_1,k_1)}, \ t=\frac{h_1}{\gcd(b_1,h_1)}.$ it is obvious that $\displaystyle x,y,z,t$ satisfy

    the conditions in $\displaystyle (1)$ and thus we have $\displaystyle \gcd(xt,yz)=1,$ which is exactly your identity!
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