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Thread: GCD question

  1. #1
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    GCD question

    if $\displaystyle m>0$

    show that

    $\displaystyle m(a,b)=(ma,mb)$

    I'm told that $\displaystyle (ma,mb) = max + mby = m(ax + by) = m(a,b)$
    isn't sufficient... why?
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  2. #2
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    Quote Originally Posted by gutnedawg View Post
    if $\displaystyle m>0$

    show that

    $\displaystyle m(a,b)=(ma,mb)$

    I'm told that $\displaystyle (ma,mb) = max + mby = m(ax + by) = m(a,b)$
    isn't sufficient... why?
    $\displaystyle \text{let} \ d=\text{gcd}(ma,mb) \ \ \text{and} \ \ c=\text{gcd}(a,b)$

    $\displaystyle d=max+mby \ \ x,y\in\mathbb{Z} \ \ \ mc=maz+mbt \ \ z,t\in\mathbb{Z}$

    $\displaystyle d|ma \ \ \text{and} \ \ mc|ma\Rightarrow mc|d$

    Now, all you have to do is show d|mc and you can then say d=mc and you are done.
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  3. #3
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    how do I show d|mc?
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  4. #4
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    Quote Originally Posted by gutnedawg View Post
    how do I show d|mc?
    d divides the LHS of $\displaystyle mc=maz+mbt$; therefore, d|RHS
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  5. #5
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    but we have shown that mc|d not d|mc so I'm not sure how you come to this conclusion

    never mind: since d=(ma,mb) thus d|max+mby for any x,y in Z correct?
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  6. #6
    Senior Member roninpro's Avatar
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    Quote Originally Posted by gutnedawg View Post
    if $\displaystyle m>0$

    show that

    $\displaystyle m(a,b)=(ma,mb)$

    I'm told that $\displaystyle (ma,mb) = max + mby = m(ax + by) = m(a,b)$
    isn't sufficient... why?
    Maybe to answer your original question, the bad part of your proof is here: $\displaystyle m(ax+by)=m(a,b)$. The only thing that you can conclude about the number $\displaystyle ax+by$ is that it is a multiple of $\displaystyle (a,b)$, which is a little bit too weak to prove what you want.
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  7. #7
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    Quote Originally Posted by gutnedawg View Post
    if $\displaystyle m>0$

    show that

    $\displaystyle m(a,b)=(ma,mb)$

    I'm told that $\displaystyle (ma,mb) = max + mby = m(ax + by) = m(a,b)$
    isn't sufficient... why?

    Let $\displaystyle d=(a,b)$ be the greatest common divisor of $\displaystyle a $ and $\displaystyle b $. So, $\displaystyle d $ is the smallest positive integer that can be expressed a linear combination of $\displaystyle a $ and $\displaystyle b $.(Theorem.)

    Consider the following linear combinations : $\displaystyle (ma)x+(mb)y=m(ax+by)$, where $\displaystyle x,y $ are integers.

    From the equation we can see that $\displaystyle (ma)x+(mb)y$ is the smallest positive linear combination of $\displaystyle ma $ and $\displaystyle mb $ precisely when $\displaystyle ax+by$ is the smallest positive linear combination of $\displaystyle a $ and $\displaystyle b $. Applying the theorem we have $\displaystyle (ma,mb)=m(a,b)$.
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