This is a proof of a theorem from my book, Discrete Mathematics and its Applications
Theorem 1
If $a$ and $m$ are relatively prime integers and $m>1$, then an inverse of $a$ modulo $m$ exists. Furthermore, this inverse is unique modulo $m$. (That is, there is a unique positive integer $\overline a$ less than m that is an inverse of $a$ modulo $m$ and every other inverse of $a$ modulo $m$ is congruent to $\overline a$ modulo $m$.
Proof: By Theorem 6 of Section 4.3, because $\text{gcd}(a,m)=1$, there are integers $s$ and $t$ such that $$sa+tm=1$$
This implies that $$sa+tm\equiv1 \pmod m$$
Here is theorem 6 of Section 4.3:
Theorem 6
Beloit's Theorem: If $a$ and $b$ are positive integers, then there exist integers $s$ and $t$ such that $\text{gcd}(a,b)\equiv sa+tb$.
The first part of the proof, "because $\text{gcd}(a,m)=1$" makes sense because the conditional statement includes the statement that "$a$ and $m$ are relatively prime integers", meaning that their gcd is $1$. I don't get the step that the author uses to get from
$$sa + tm = 1$$
to
$$sa+tm\equiv1 \pmod m$$
Can someone explain how the author got to that step? Can someone give a general overview of inverse of modulo as well? I don't really understand it from my book. I understand modulus: $7 \bmod 3$ is $1$ but what would inverse of $7 \bmod 3$ get you?
