greatest common divisor proof

Question

Suppose $\gcd(a,b)=1$. Does this necessarily imply that $\gcd(a,2b)=1$?

If $\gcd(a,b)=1$ then $\exists x,y \in \Bbb{Z} (ax+by =1)$

I don't see how to manipulate this equation to give me $ax+2by \neq 1$ I'm assuming that's what I have to do?

Answer 1

Let $a=2$, $b=3$ then $\gcd(a,b)=1$ but $\gcd(a,2b)=\gcd(2,6)=2$, thus the implication is not true.

Answer 2

No. If $a$ is even and $\textrm{gcd}(a,b)=1$ then $\textrm{gcd}(a,2b) = 2$.