2
$\begingroup$

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.

Proof:
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 $\mod m$?

$\endgroup$
3
$\begingroup$

The above proof shows that the set $\{0,x,2x,\dots,(m−1)x\}=\{ax \mid a = 0,\dots,m-1\}$ has distinct elements under $\mod{m}$. Therefore, $|\{0,x,2x,\dots,(m−1)x\}| = m$, therefore, this set under $\mod{m}$ contains all representatives in $\mod{m}$. In particular, there exists $a \in \Bbb{Z}$ such that $ax \equiv 1 \pmod{m}$, so there exists $k \in \Bbb{Z}$ such that $ax = 1 + km$. This should give you $ax - km = 1$, so that $\gcd(a,m) = 1$ by Bezout's Identity.

$\endgroup$
  • 1
    $\begingroup$ The proof should'd mentioned this hint. :-) $\endgroup$ – mathmaniage Feb 24 at 7:55
  • $\begingroup$ The moral of this story is to never overlook a result that counts something. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 24 at 15:37
2
$\begingroup$

The key is in the claim that $0, x, 2x, \ldots, (m-1)x$ are all distinct. If this weren't true then $(a-b)x = km$ for some $a,b<m$ and some integer $k$. This says that $m$ divides $(a-b)x$. Since $x$ and $m$ are relatively prime, $m$ must divide $a-b$.

It might help to think in terms of prime factorization. If $m$ divides $(a-b)x$ then all of the prime powers in $m$ must live in the product $(a-b)x$. $m$ and $x$ being relatively prime says that none of the primes dividing $m$ divide $x$, so all of the prime powers making up $m$ must come from $(a-b)$. This shows that $m$ divides $a-b$. Does the rest of their proof make sense?

$\endgroup$
  • $\begingroup$ thank you for your reply. But, again, you're not stating what does the proof have to do with the GCD being one and as a consequence the multiplicative inverse being existent. You're concluding that m divide a-b, and I also understand that a =! b, but what does that have to do with the multiplicative inverse? $\endgroup$ – mathmaniage Feb 24 at 7:51
  • $\begingroup$ "Relatively prime" means the same thing as having a GCD of 1. The conclusion that $m$ divides $a-b$ comes from the GCD being 1 (in the way I described above). Once you have that $0, x, 2x, \ldots, (m-1)x$ are distinct, you must have that $ax \equiv 1\pmod{m}$ for some $a$. $\endgroup$ – Liam Feb 24 at 7:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.