If $a$ and $b$ are odd, prove $\gcd(a,b) = \gcd(\frac {\left| {a-b} \right |} {2}, b)$ Honestly I don't have a strong idea. I don't know where to even begin, I have considered that the $\gcd(a,b)$ is somehow less than $a-b$, but I'm not even sure why that would be the case.. Any help would be great!

Some hints.  Let $g=\gcd(a,b)$.  We want to show that $g\mid{\rm RHS}$.
(1) $g$ is a factor of $a-b$ because. . .
(2) therefore $g$ is a factor of $|a-b|$ because. . .
(3) therefore $g$ is a factor of $\frac{1}{2}|a-b|$ because. . . (this is the hard bit)
(4) and it is obvious that $g$ is a factor of $b$, so
(5) $g$ is a factor of ${\rm RHS}$.
See if you can fill in all the reasons where indicated by dots.
Now you have to do the same kind of thing in the other direction: let $h={\rm RHS}$ and prove that $h$ is a factor of ${\rm LHS}$.
Good luck!
