Multiplicative inverse questions

Question

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.

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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!

Answer 1

An element with a multiplicative inverse is called a unit.

An element that is not a unit is for instance $0$, you are correct, as soon as $0\neq 1$ which holds whenever $n>1$.

It's kind of cheating to use the lemma for the third question, since the lemma itself is about as difficult to prove as the question. If $a_1$ and $a_2$ are both multiplicative inverses to $a$, compute $a_1\cdot a\cdot a_2$ in two ways.

Answer 2

I believe the second highlighted section could be referring to some group like $\mathbb{Z}_6 = \{0,1,2,3,4,5\}$. Within this group we can see that $2,3,$ and $4$ do not have a multiplicative inverse in $\mathbb{Z}_6$.

To show the uniqueness of the multiplicative inverse of $x\in\mathbb{Z}_n$, let $a_0$ and $a_1$ both be multiplicative inverses of $x$, so that $$\begin{align*} a_0x &= xa_0 = 1\\ a_1x &= xa_1 = 1\end{align*}$$ Then we have $$a_0 = a_0(1) = a_0(xa_1) = (a_0x)a_1 = (1)a_1 = a_1$$ Hence the multiplicative inverse is unique.