I am trying to write a proof for-  show that if [b] and [c] are both multiplicative inverses of [a] in Zn, then b congruent c (mod n)   I don't know a lot about multiplicative inverse proofs and any help will be appreciated.
This is my first time posting on Math exchange. I have been self-teaching myself Mathematics and recently started learning some Algebra. I was posed the following questions, in which I would like to answer before I proceed, but having some trouble finding proper solutions. So the format will be question, and my answer.      What name do we give an element $a \in \mathbb{Z}_n$ that has a multiplicative inverse?   A modular multiplicative inverse of an integer $a$ is an integer $x$ such that $ax \cong 1 (\mod m)$. This means that the remainder after dividing $ax$ by integer $m$ is equal to $1$.      Give an example, for some $n \in \mathbb{N}$, of an element of $\mathbb{Z}_n$, that does not have a multiplicative inverse?   This question is throwing me off because - does the $n$ we speak of matter for both $n \in \mathbb{N}$ and $\mathbb{Z}_n$ ot solely $n$? My guess here would be $0$ because we cannot have an inverse of $0$.      Prove that multiplicative inverse are unique; an element $a \in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Suppose that an element $a\in\mathbb{Z}_n$ has an inverse $a'$. Next, suppose that $a^*$ is also an inverse of $a$. Then, $a'$ is a solution to $a \cdot_n x=1$, and similarly $a^*$ has a solution to $a \cdot_n x=1$.   Lemma: Suppose $a'$ is multiplicative inverse of $a$ in $\mathbb{Z}_n$, then for any $b \in \mathbb{Z}_n$ the equation $a \cdot_n x=b$ has a unique solution, such that $x = a' \cdot_n b$.   By Lemma, equation $a \cdot_n x = 1$ has a unique solution, namely $x=a'=a^*$. Therefore, an element $a\in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Thank you for the help. I would love any constructive feedback for these. If you can help me elaborate on some ideas, that would be really helpful also.  Thank you very much!
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