If $a | b$, prove that $\gcd(a,b)$=$|a|$.

Question

If $a | b$, prove that $\gcd(a,b)$=$|a|$.

I tried to work backwards. If $\gcd(a,b)=|a|$, then I need to find integers $x$ and $y$ such that $|a|=xa+yb$. So if I set $x=1$ and $y=0$ (if $|a|=a$) or if I set $x=-1$ and $y=0$ (if $|a|=-a$), is this good enough?

Answer 1

If $a \mid b$, then $|a| \mid b$. Moreover, we have $|a| \mid a$. Thus $|a|$ divides both $a$ and $b$. If $q > |a|$, then $q \nmid a$ by definition, so $|a|$ is the greatest common divisor of $a$ and $b$.

Answer 2

If $a|b,\ |a||b\implies |a|\le gcd(a,b)\le a\le |a|\implies |a|=gcd(a,b)$.

Answer 3

That doesn't prove anything, since this relation is true even if $a$ and $b$ are coprime. It is the first step in the Extended Euclid's algorithm.

Just observe that any common divisor of $a$ and $b$ must divide $a$, and by hypothesis $a$ is a common divisor of $a$ and $b$. Hence it ($\lvert a\rvert$) is the greatest of all common divisors.

Answer 4

Hint:-Use Bezout's theorem.Express the GCD in the form $ax+by$ and prove that there is a common factor in the GCD and $a$ & $b$.