If \((a, m) = 1\), then \(ax+my=1\) or \(ax=m(q)+1\), which means:
$$ ax ≡ 1 \mod m $$
Here, \(x\) is the inverse. If \(ab_1 ≡ 1\) and \(ab_2 ≡ 1\), then:
$$ ab_1 ≡ ab_2 \mod m ⇒ m|a(b_1 - b_2) $$
Since \((m, a) = 1\):
$$ m|b_1 - b_2 ⇒ b_1 ≡ b_2 \mod m $$
This shows that the inverse is unique. By 'unique', it means there is only one solution in modulo \(m\).