I'm re-learning real analysis and decided to start from Tao's books (sorry Rudin) and Tao left a remark stating we can prove the sum of two natural numbers is again a natural number by the Peano Axioms; specifically we only need these components:
Axiom 2.1 $0$ is a natural number.
Axiom 2.2 If $n$ is a natural number, then $n++$ is also a natural number.
Definition 2.1.3. We define $1$ to be the number $0++$, $\,2$ to be the number $(0++)++$, and so on.
Axiom 2.5 (Principle of mathematical induction) Let $P(n)$ be any property pertaining to a natural number $n$. Suppose the $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number n.
Definition 2.2.1 (Addition of natural numbers). Let $m$ be a natural number. To add zero to $m$ , we define $0+m:=m$. Now suppose inductively that we have defined how to add $n$ to $m$. Then we can add $n++$ to $m$ by defining $(n++)+m:=(n+m)++$.
Proof.
Let $n$ and $m$ be natural numbers.
This will be a proof by mathematical induction; we shall induct on the variable $n$, with the base case $n=0$. Since $m$ is a natural number by the hypothesis, our only cases are either $n =0$ and $m=0$ $or$ $n=0$ and $m\neq 0$. Suppose first that $n=0$ and $m=0$. Then $n+m = 0+0 = 0$. We know $0$ is a natural number by Axiom 2.1. Now suppose $n=0$ and $m \neq 0$. By Definition 2.2.1, the sum $n+m =0+m=m$. We know $m$ is a natural number by the hypothesis. Since we get natural numbers in both base cases, we have have proved the base case.
Now suppose we have shown inductively that the sum $n+m$ is a natural number. We need to show $(n++)+m$ is also a natural number.
By Definition 2.2.1, $(n++)+m = (n+m)++$, and since $n+m$ is a natural number by the inductive hypothesis, $(n+m)++ = (n++)+m$ is also a natural number by Axiom 2.2. Thus, $n+m$ must be a natural number by the principle of mathematical induction, closing the induction.
Proof complete, with the help of ultralegend5385 and Rob Arthan.
