# Math Help - inverse of modulo

1. ## inverse of modulo

show that if a' is an inverse of a modulo m and b' is an inverse of b modulo m, then a'b' is an inverse of ab modulo m

2. $a'a \equiv 1 \mod m$
$b'b \equiv 1 \mod m$
hence
$1\cdot 1 \equiv a'ab'b \equiv (a'b')ab \equiv \mod m$

3. $aa' \equiv 1 \mod m \Rightarrow aa' = 1+mt$
$bb' \equiv 1 \mod m \Rightarrow bb' = 1+ms$

$ab(a'b') = aa'bb' = (1+mt)(1+ms) = 1 + mt + ms + m^{2}st = 1 + m(t+s+mst)$ $\Rightarrow ab(a'b') \equiv 1 \mod m$