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?

Yes, you can prove this by induction.
Claim. $2\nmid 1$.
Proof. For positive integers, it holds that $k\mid n$ implies $k\leq n$, since $1<2$, we have that $2\nmid 1$. Furthermore, $1$ is the successor of $0$ and $2$ is the successor of $1$, therefore the only non-negative numbers smaller than $2$ are $0,1$ and so they are the only possible remainders when dividing a positive integer $k$ by $2$. $\square$
Lemma. If $k$ is even then $k+1$ is odd, and if $k$ is odd then $k+1$ is even.
Proof. If $k$ is even then $2\mid k$, and then $2\nmid k+1$ because $2\nmid 1$; therefore $k+1$ is odd. If $k$ is odd then $2\nmid k$ and so the remainder of $k$ when divided by $2$ is $1$, so $2\mid k+1$, and it is even. $\square$
Claim. Every $n\in\Bbb N$ is either odd or even.
Proof. We prove using complete induction. Suppose that for every $k<n$ either $k$ is odd or $k$ is even. In particular this is true for $k=n-1$. If $k$ is even then $k+1=n$ is odd; and if $k$ is odd then $k+1$ is even. $\square$
Claim. Every $k\in\Bbb Z$ is either odd or even.
Proof. By the above claim it holds for $k\geq 0$, and $2\mid k\iff 2\mid -k$, so the claim holds for every integer. $\square$
