Even and Odd Integers form Partition of Integers

Theorem

Let $n \in \Z$ be an integer.

Let $\Bbb O$ be the set of odd integers and $\Bbb E$ be the set of even integers.

$\forall n \in \Z: \exists! q, r \in \Z: n = 2 q + r, 0 \le r < 2$

from which it follows that either:

$n = 2 q \in \Bbb E$

or:

$n = 2 q + 1 \in \Bbb O$

Thus:

$(1): \quad$ each element of $\Z$ is in no more than one of $\Bbb E$ and $\Bbb O$

$(2): \quad$ each element of $\Z$ is in at least one of $\Bbb E$ and $\Bbb O$

and:

$(3): \quad$ setting $q = 0$ it is seen that $0 \in \Bbb E$ and $1 \in \Bbb O$ and so neither $\Bbb E$ or $\Bbb O$ is empty.

Thus $\set {\Bbb E \mid \Bbb O}$ is a partition of $\Z$ by definition.

$\blacksquare$