I basically don't understand the way the proof is given in my text as well as some references online. They basically do the same thing.
THEOREM: Let $m,x$ be positive integers such that $\gcd(m, x) = 1$. Then $x$ has a multiplicative inverse modulo $m$, and it is unique, $\mod m $.
Now, this is a sufficient and necessary condition, but let's just look at the proof that it is the sufficient condition which is given in my texts.
Consider the sequence of m numbers $0,x,2x,...,(m−1)x$. We claim that these are all distinct,$\mod m$. Since there are only $m$ distinct values modulo $m$, it must then be the case that $ax = 1 \mod m$, for exactly one $a (\mod m)$. This a is the unique multiplicative inverse. To verify the above claim, suppose that $ax = bx \mod m$ for two distinct values $a,b$ in the range $0 ≤ a,b ≤ m−1$. Then we would have $(a−b)x = 0 \mod m$, or equivalently, $(a−b)x = km$ for some integer $k$ (possibly zero or negative). But since $x$ and $m$ are relatively prime, it follows that $a−b$ must be an integer multiple of $m$. This is not possible since $a,b$ are distinct non-negative integers less than $m$.
What I don't see particularly in this proof is the connection betweent the $\gcd$ being one and the existence of the multiplicative inverse. What I do see instead is that the proof that the multiplicative inverse is unique.
What is the connection of the $\gcd$ being $1$ and the existence of the multiplicative inverse $\mod m$?