1. GCD question

if $m>0$

show that

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

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

2. Originally Posted by gutnedawg
if $m>0$

show that

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

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

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

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

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

3. how do I show d|mc?

4. Originally Posted by gutnedawg
how do I show d|mc?
d divides the LHS of $mc=maz+mbt$; therefore, d|RHS

5. 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?

6. Originally Posted by gutnedawg
if $m>0$

show that

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

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

7. Originally Posted by gutnedawg
if $m>0$

show that

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

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

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

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

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