How to know if a variable is an integer when writing proof?

Question

Here is the theorem:

For all integers n, if n is odd then n^2 is odd.

This is the Proof:

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Assume n is any odd integer. By definition of odd, $n = 2r + 1$, for some integer $r$. We must show that the square of $n$ is odd. Then

$n = 2r + 1$

$= (2r+1) ^2$

$= 4r^2+4r+1$

$= 2(2r^2+2r)+1$

Let $2r^2+2r= k$. Notice that $k$ is integer because it’s the product of two integers. Hence $n^2 = 2(k) + 1$ in which $k$ is an integer. It follows by definition of odd, that $n^2$ is odd.

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My problem is this step:

Let $2r^2= k$. Notice that $k$ is integer because it’s the product of two integers.

I want to give a reason as to why K is an integer. I first thought that maybe because it's twice $r$, but then i thought since it's $r$ squared. So, I am confused to which one is valid.

Answer 1

An integer multiplied by an integer, is an integer. This is because the set of integers is closed under multiplication. In your case, we have $2r^2$, because $r$ is an integer we get that $rr=r^2$ is an integer and therefore $2r^2=k$ is an integer. Regardless, your proof is not quite right. For example, you wrote $(2r+1)^2=(4r)^2+1$ which is not true.