What does the relative primality of $a$ and $n$ say about the existence of an inverse for $[a]$ in $\mathbb{Z}_n$?

Question

I have a potential solution for this problem, though I am unsure it is entirely correct and am positive it is inelegantly stated and could use some help tidying it up (if it is indeed the correct logic):

I started by identifying the definitions of relative primality and the congruence class of $a$ modulo $n$

1. $a$ and $n$ relatively prime means: $\exists$$s,t \in \mathbb{Z}$ $|$ $as + nt = 1$
2. $[a] = \{b \in \mathbb{Z} \mid a \equiv b \pmod n\} = \{a + kn\mid k \in \mathbb{Z}\}$

We can say, if we take $s = 1$ then we have the form $a + tn = 1$ where we can identify the $t$ with the $k$ in def $2.$ since they are both integers and the $1$ with the $[a]$ which tells us that there exists $s,t$ such that $[a] = 1$ or, equivalently, there exists an element $[b] \in \mathbb{Z}_n$ such that $[a]$ is invertible. A couple questions:

• Is this the correct logic?
• If so, how can I state this in a more "mathematically sophisticated" way? I find it difficult, thus far, to contain my proofs to math symbolism and sparse english (when necessary). I feel like I'm writing an argument rather than a proof. Any general advice on improving proof-writing in that regard is also welcome.

Answer 1

Is this the correct logic?

Your defintions are fine, but I do not quite understand your proof. The statement $[a] = 1$ is problematic because $[a]$ is a set, so saying that it is equal to an integer does not make sense.

Perhaps you are trying to show that $1 \in [a]$, but that does not imply the existence of an inverse. Keep in mind that $[a]^{- 1} \in \mathbb{Z}_n$ is the multiplicative inverse of $[a]$ if $[a] [a]^{- 1} = [1]$.

As a side note, if you are trying to prove that $1 \in [a]$, we cannot claim that $a + nk = 1$ by $as + nt = 1$ because $s$ may not necessarily be 1. For example, $11$ and $9$ are relatively prime, but there is no integer solution to $11 + 9 k = 1$ even though there exists $s, t$ so that $11 s + 9 t = 1$.

For a proof, I would do the following:

In general, $[a] [x] = [c]$ has a solution in $\mathbb{Z}_n$ if and only if $\text{gcd} (a, n) \mid c$. I know this result as the Modular Arithmetic Theorem, though I am aware that this name is not conventional.

Nevertheless, in the case of the inverse, if $[a]$ is invertible in $\mathbb{Z}_n$, then there exists a $[a]^{- 1}$ such that $[a] [a]^{- 1} = [1]$ holds. This is only possible if $\text{gcd} (a, n) \mid 1$ and thus $\text{gcd} (a, n) = 1$, which implies relative primality.