How do we know all numbers alternate between odd and even numbers? Basically, I tried proving that multiplying an odd and even number together gives you an even number to a friend of mine. This is what I said.
Let's first take a random number $k$. It doesn't matter if its odd or even. If we then multiply this by $2$, we get an even number (from the definition of a even number, which is basically a number that can be divided by $2$). Now we know that in the number line, it goes odd, even, odd, even,..., and so if we add $1$ to this, then we get an odd number. We now have an odd number and an even number and so lets multiply them.
$$2k \cdot (2k + 1) = 2k(2k + 1) = 4k^2 + 2k) = 2(2k^2 + k)$$
We now have a random number $(2k^2 + k)$ multiplied by $2$ and so this is an even number (by definition of an even number).
My friend then said that we make the assumption that $2k \pm 1$ is odd because the number line alternates between odd and even numbers, how do know that this is true?
I was thinking how to prove this. Would it be by some form of induction?
EDIT: I think I need to change "number" to "integer" in this post don't I?

Let $a$ be even.
Use the fact that $2\; \not \mid 1$. So if $2 \mid a$, (i.e., if a is even), $2 \not\mid (a + 1)$. Hence $a + 1$ is odd.
Then what about $(a + 1) + 1$, the number immediately following odd $(a + 1)$, which in turn immediately follows even $a$.
Well, $(a + 1) + 1 = a + 2$, and since $a$ is even, it can be expressed as $a = 2k$ for some integer $k$. Then $a + 2 = 2k + 2 = 2(k+1).$ And clearly, $2\mid 2(k+1).$
Hence, $a$ even $\implies (a + 1)$ odd $\implies (a + 2)$ even...
