Prove the following: If $m$ and $n$ are even integers, then so are $m+n$

Question

I have been asked to prove the following and am having difficulty:

If $m$ and $n$ are even integers, then so are $m+n$ and $mn$.

My professor has hinted to us to use the definition of an even integer in our proof.

This is my proof for $m+n$ thus far:

1. Even integers are defined as being divisible by 2.
2. So, if $m$, is even, using this definition we can rewrite this as $m=2j$.
3. Similarly, we can rewrite $n$ as $n=2k$.
4. We now have $m+n=2k+2j$.
5. Distributivity then allows us to write $2j+2k=2(j+k)$
6. We now have that $m+n=2(j+k)$.
7. I now use associativity to create $m+n=(j+k)2$

8. Next, the definition of divisibility states that 'When $m$ and $n$ are integers, we say $m$ is divisible by $n$ if there exists $j∈ Z$ such that $m=jn$.

9. This allows me to conclude that, since all even numbers are divisible by 2, that $m+n$ must be an even number.

I am unsure about whether or not my last two steps are correct. Our teacher has hinted to us to use the definition of divisibility, but I am having trouble wrapping my head around a. how to use it, and b. how it is accurate to do so?

Any advice would be much appreciated!

Answer 1

This looks mostly right, well done! As noted in the comments, there is a small error in 9. You wish to use the fact that all numbers that are divisible by $2$ are even to conclude that it's even. Instead, you evoke the fact that all even numbers are divisible by $2$.