if a|b and a>0 then gcd(a,b)=a

Question

Is this proof right?

Prove: if a|b and a>0 then gcd(a,b)=a.

Let a|b and the GCD(a,b) = m, then b=aq for some integer q and the GCD(a,b) can be expressed as a linear combination with some integer x and y,

       ax + by = m .

Substituting in b we get,

      ax + (aq)y = m
      a(x + qy) = m

But x + qy = GCD(1,q) =1. Thus,

      a(1) = m

Hence a = GCD(a,b), which is what we needed to show.

Thanks for all the comments :)

Update 1: I did not think I said that x and y were any integer I said that there was some integer x and some integer y so that ax +by = m Since I am supposing that the GCD(a,b) = m and a|b, then by substituting b into the GCD(a,b), then

      GCD(a,b)=GCD(a,aq)

              =a*GCD(1,q) Which is a property of GCD

But the GCD(1,q)=1. So, GCD(a,b) = a

Thanks for the comments :)

Answer 1

$a|b \Rightarrow b=aq$, for some integer $q$.

GCD means greatest common divisor.

If $ a\geq 1$, then what is the largest integer which divides both $a$ and $aq$?

Answer 2

$a|a $ and $a|b $ so $a $ is a common divisor of $a $ and $b $.

Anything bigger than $a$ cannot divide $a$. So $a $ is the greatest common divisor.

That's it.