Proof involving inverses and modulo

Question

I'm working through an exercise which states:

Let a' be the inverse of a modulo m and let b' be the inverse of b modulo m. Prove that a'b' is the inverse of ab modulo m.

So far what I have is:

We have $a*a'\equiv 1 (mod\ m)$ and $b*b' \equiv 1(mod\ m)$ So $mk = a*a' - 1$ and$ml = b*b' - 1$ for some $k,m \in \mathbb{Z}$

I have no idea where to go from here or if I've even started in the right direction. Could someone nudge me in the right direction?

Thanks!

Answer 1

You simply have to show that $(ab)(a'b')\equiv 1\pmod{m}$.

$$(ab)(a'b')\equiv (aa')(bb')\equiv 1\cdot 1\equiv 1\pmod{m}$$