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I am wondering if I have a valid proof. I saw a proof I actually liked more than mine here, but I'd like to know if I'm thinking about this correctly. Here is my proof:

Let $[a]$ $\in$ $\mathbb{Z}_{n}$. Assume there exists $[b] \in \mathbb{Z}_{n}$ with $[a][b] = [1]$. So we have $a = ns + r_{1}$ and $b = nt + r_{2}$ for $s, t, r_{1}, r_{2} \in \mathbb{Z}$. Then
$$\begin{align} [a][b] &= (ns + r_{1})(nt + r_{2})\\ &= n^2st + nsr_{2} + ntr_{1} + r_{1}r_{2}\\ &= n(nst + sr_{2} + tr_{1}) + r_{1}r_{2}\\ &\equiv [1] \end{align}$$

Then we have $r_{1}r_{2} = 1$, so $r_{1} = r_{2} = 1$. Then $[a] \equiv 1 \pmod n$, so $\gcd(a, n) = 1$.

Does this check out? I'm also concerned with my usage of an equal sign when I say $[a][b] = (ns + r_{1})(nt + r_{2})$ and my usage of the congruent symbol when I say $n(nst + sr_{2} + tr_{1}) + r_{1}r_{2} \equiv [1]$. Did I use these symbols correctly?

Arturo Magidin
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Justin
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    I think you show $r_1r_2\equiv[1],$ but it's not clear to me why this implies $r_1r_2=1$ or $r_1=r_2=1.$ For example, this feels like it won't work if, say, we take $a=r_1=2,,b=r_2=3,,n=5.$ – Derive Foiler Mar 28 '21 at 21:53
  • Ah, I see. Thank you for pointing that out. I think I may still be on the right track then, right? Can I say that we have $r_{1}r_{2} \equiv 1$, so $r_{1} \equiv 1$? If so, then $[a] \equiv ns + [1] \equiv 1modn$ right? – Justin Mar 28 '21 at 22:19
  • It's still not clear to me why $r_1\equiv1$ follows from $r_1r_2\equiv1.$ For example, $2\cdot3\equiv1\pmod5,$ but $2\not\equiv1\pmod5.$ – Derive Foiler Mar 28 '21 at 22:20
  • Oh right. Perhaps this path I'm taking can't lead me to the end goal here... – Justin Mar 28 '21 at 22:25
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    What is your definition of $[x]=[y]$ in $\mathbb{Z}_n$? You have equal signs between $[a][b]$ and the next actual number, but you have a congruence sign $\equiv$ in the last line. That seems wrong. Here you have $[a][b]=[ab]=[1]$, so the key question is “what does it mean for $[x]$ to be equal to $[y]$?” – Arturo Magidin Mar 29 '21 at 00:20
  • Yeah, I knew something seemed off there... $[x] = [y] \implies x \equiv cmodn$ and $y \equiv cmodn$ for some $0 \leq c < n$ – Justin Mar 29 '21 at 00:53
  • First, that’s an implication, not a definition. Second, do you mean “$[x]=[y]$ if and only if $x\equiv y\pmod{n}$”? (Why introduce the extra $c$?) What is your definition of $x\equiv c\pmod{n}$? Does it mean “$n\mid x-c$”? – Arturo Magidin Mar 29 '21 at 01:45
  • Yes, that's what I mean. And yes, it does mean $n | x - c$ – Justin Mar 29 '21 at 03:31
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    Well, then $[a][b]=[1]$ means $n|ab-1$, means there exists $k$ such that $kn=ab-1$ means $ab-kn = 1$, means $\gcd(a,n)=1$. – Arturo Magidin Mar 29 '21 at 20:05

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