Thread: 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?

Originally Posted by gutnedawg

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.

how do I show d|mc?

Originally Posted by gutnedawg

how do I show d|mc?

d divides the LHS of $\displaystyle mc=maz+mbt$; therefore, d|RHS

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?

Originally Posted by gutnedawg

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.

Originally Posted by gutnedawg

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