Prove if $n\in N$ is composite then there exists an equivalence class $a\bmod n$ that does not have a multiplicative inverse

Question

I know that the if $n=j*k$ with $j,k<n$ that $j\bmod n$ or $k\bmod n$ probably don't have inverses, but I don't remember the theorem for it.

If $j^{-1}j\equiv1$ and $jk\equiv0$ then $k=1k\equiv j^{-1}j*k\equiv0$ — J. W. Tanner, Nov 07 '19 at 20:10
$0$ is always an equivalence class without an inverse $\pmod N$ even if $N$ is prime. If $N$ is composite then you can find a non-zero equivalence class without an inverse. — lulu, Nov 07 '19 at 20:25
You don't need "to remember the theorem for it" when you start with an example of a composite number $n$, say $n=4$. — Dietrich Burde, Nov 07 '19 at 20:31
@J.W.Tanner I sort of see where you're going with your solution, but how does that prove that k has no inverse mod n? — user8714896, Nov 07 '19 at 20:48
If $k\equiv0$ then $k$ times anything is $0$, so nothing multiplies by $k$ to get $1$ ((unless $0=1$)) — J. W. Tanner, Nov 07 '19 at 21:03
@J.W.Tanner I think there also needs to be something written for when both j and k don't have inverses like n = 4 and j,k=2 then $2*1(mod)4 \equiv 2$, which doesn't have an inverse but mod4 doesn't isn't equivalent to 0 — user8714896, Nov 07 '19 at 21:41
We’re asked to prove there’s an equivalence class without an inverse. If both $j$ and $k$ don’t have inverses, then we’re done — J. W. Tanner, Nov 07 '19 at 21:49

score 0 · Answer 1 · answered Nov 07 '19 at 22:32

Per lulu's comment, $0$ is an equivalence class without a multiplicative inverse $\pmod n$.

Here is a demonstration that there is a non-zero equivalence class

without a multiplicative inverse $\pmod n$ when $n$ is composite.

Say $n=jk$ with $j$ and $k$ strictly between $0$ and $n$.

Then $jk\equiv0\pmod n$ but $j,k\not\equiv0\pmod n$.

Assume, aiming for contradiction, there is $k^{-1}$ such that $kk^{-1}\equiv1\pmod n$.

Then $0\equiv0\times k^{-1}\equiv (jk)k^{-1}\equiv j(kk^{-1})\equiv j\times1 \equiv j.$

This contradiction shows that $k$ is a non-zero equivalence class

without a multiplicative inverse $\pmod n$.

Swapping the role of $j$ and $k$ would show that $j$ is also not invertible $\pmod n$.

1 Answers1