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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- Jan 27th 2011, 05:38 PM #1

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- Jan 27th 2011, 05:59 PM #2

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$\displaystyle \text{let} \ d=\text{gcd}(ma,mb) \ \ \text{and} \ \ c=\text{gcd}(a,b)$

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

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

Now, all you have to do is show d|mc and you can then say d=mc and you are done.

- Jan 27th 2011, 06:09 PM #3

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- Jan 27th 2011, 06:23 PM #4

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- Jan 27th 2011, 07:50 PM #5

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- Jan 27th 2011, 08:03 PM #6
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.

- Jan 28th 2011, 06:53 AM #7

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