Proving square of nonzero integer is natural number

Question

I am learning proofs with $\mathbb N$ and have this proposition: Let $m \in\mathbb Z$. If $m \ne 0$, then $m^2 \in\mathbb N$. Previously, I have proven: For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.

If $m \ne 0$, then there are two cases to prove:

Case 1: $m \in\mathbb N$ That's straightforward because the product of two natural numbers is a natural number

Case 2: $-m \in\mathbb N$:

\begin{align*} -m \cdot -m \in\mathbb N\\ (-1 \cdot -1) \cdot m \cdot m \in\mathbb N\\ m \cdot m \in\mathbb N\\ \end{align*}

What do you think?

Answer 1

If $m\in Z$,and if $m\ne0$, you've two cases.

Case 1: $m\in Z^+$ $$ => m^2 \in Z^+ => m^2 \in N$$ (Since multiplication within positive integers is closed, and results in a natural number always.)

Case 2: $m\in Z^-$ $$ => m^2 \in Z^+ => m^2 \in N$$ (Since multiplication within negative integers always results in a positive integer as you said by typing out the below :-)\begin{align*} -m \cdot -m \in\mathbb N\\ (-1 \cdot -1) \cdot m \cdot m \in\mathbb N\\ m \cdot m \in\mathbb N\\ \end{align*}

QED

Your proof is perfect (it's essentially the same as mine!); I'd just suggest you say that multiplication operation is closed for numbers picked from positive integers instead of 'That's straightforward because the product of two natural numbers is a natural number' (In case you want to use a bit of terminology).

Otherwise it's completely fine!

Answer 2

You need to prove that: $(-m)(-m) = m\cdot m$.

By definition of $-m$ is that:

$m + (-m) = 0 \to (m+(-m))(m-(-m)) = 0(m+(-m)) = 0 \to m^2 - m(-m) + (-m)m - (-m)(-m) = 0 \to m^2 = (-m)(-m)$.