The multiplication of two even numbers gives an even number

Question

I am given the following proposition:

If $m$ and $n$ are even integers, then $mn$ is also an even integer.

This is my strategy:

An integer $m$ is said to be even if it is divisible by 2 (integer). When $m$ and $n$ are integers, $m$ is divisible by $n$ if there exists $j\in\mathbb Z$ such that $m = jn$. Thus, an even number $m$ could be written as $m = j2$. If $m$ and $n$ are even, then they can be defined as:

$m = j2$ and $n = k2$ where $j,k \in\mathbb Z$

\begin{align*} m\times n &= m\times n\\ m\times n &= j2k2\\ m\times n &= (2jk)2 \end{align*}

Please correct me if I am wrong, but at this point, I am finished because $2jk \in\mathbb Z$, and I have derived the 2. I will perform two multiplications in the parentheses; each multiplication will generate an integer because the multiplication is a binary operation. By definition, a binary operation takes two elements from a set $S$ and generates another element of set $S$ as output. In this case, $S$ is $Z$.

Answer 1

Proof: Let $m=2k$ and $n= 2l$ for some $k$ , $l$ $\in Z$. We now multiply. $m \cdot n$ = $(2k)(2l)$ = $(4kl)$ = $2(2kl)$. Since $k$ and $l$ are integers therefore $kl$ is also an integer. Thus $m \cdot n$ is an even number. This is more formal and simpler to understand. Note after solving a proof try to clean it up and make it more finer. Hope this advise help. Good job by the way.

Answer 2

I assume that you're asked to prove the preposition, and that is what you are asking for help with.

Though correct, your proof is verbose. If you are told that an integer is even, then you already know what even numbers are. If you know what even numbers are, you already know about the properties of multiplication of integers.

Try to get the proof down to two or three lines. You only need to write $m \times n$ once.