Solutions

The natural numbers have been defined as:

1. The number 1 belongs to the natural numbers
2. If n belongs to the natural numbers, then n + 1 also belongs to the
   natural numbers

Thus, so long as the union of odd and even numbers possess the same characteristics, then we can claim that the natural numbers are either even or odd.

Let us define the set of even numbers:

$$\begin{aligned} \mathbb{N}^{even} = \{ 2n: n \in \mathbb{N}\} \end{aligned}$$

Also, let us also define the set of odd numbers:

$$\begin{aligned} \mathbb{N}^{odd} = \{ 2n - 1 : n \in \mathbb{N}\} \end{aligned}$$

Now, to prove that our our two sets respect the first criterion, notice that we can obtain the element 1 from the odd numbers by setting $n = 1$.

$$\begin{aligned} 2n - 1 = 2 \cdot 1 - 1 = 1 & \implies 1 \in \mathbb{N}^{odd} \\ & \implies 1 \in \mathbb{N}^{even} \cup \mathbb{N}^{odd} \end{aligned}$$

Next, and finally, we need to prove that if $n$ belongs to the union of our two sets, then $n + 1$ also belongs to it.

$$\begin{aligned} n \in \mathbb{N}^{odd} \cup \mathbb{N}^{even} & \implies n \in \mathbb{N}^{odd} \vee n \in \mathbb{N}^{even} \\ & \implies [\; \exists k \in \mathbb{N}: 2k - 1 = n \; ] \; \vee \; [\; \exists j \in \mathbb{N}: 2j = n \;] \\ & \implies [\; \exists k \in \mathbb{N}: 2k = n + 1 \; ] \; \vee \; [\; \exists j \in \mathbb{N}: 2j + 1 = n + 1 \;] \\ & \implies [ \; n + 1 \in \mathbb{N}^{even} \; ] \vee [ \; n + 1 \in \mathbb{N}^{odd} \; ] \\ & \implies n + 1 \in \mathbb{N}^{even} \cup \mathbb{N}^{odd} \end{aligned}$$