# Prove the sum of two natural numbers is again a natural number, using the Peano Axioms.

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.

• Your proof looks good to me. Stylistically, it would be better if you made the distinction between the two parts of the inductive proof clearer (Base case: ... Inductive step: ...). Jul 14, 2021 at 1:07
• thank you for the proof. I am not sure why it is necessary to split the base case ($n=0$) into two sub-cases: (i) $m=0$ or (ii) $m\neq 0$. For the base case, my argument is simply that $0+m:=m$ (by Definition 2.2.1.) and $m$ is a natural number by assumption. But then I am not sure where to use Axiom 2.1 in the proof. To my understanding, $m$ can be $0$ in Definition 2.2.1. Thank you for your help!
– Sean
Nov 4, 2023 at 16:20
• @Sean you are correct - we do not need to split the base case if we just keep $m$ fixed and induct on $n$, and we do not need Axiom 2.1. Dec 4, 2023 at 6:44

You have written a good proof, although as @RobArthan said in comments, you must highlight the parts of your induction, and also saying explicitly that you are using induction on $$n$$, which is unclear as of now. Some suggestions for lines:

[We will use induction on $$n$$ to show this]

[For the base case, when $$n=0$$,]

[We assume, for the sake of induction that]

It is always better to define a statement $$P(n)$$ for induction proofs, but it is your choice after all.

Another important thing, your proof never showed that $$0+0$$ is a natural number. Think about that and edit it accordingly.

Hope this helps. Ask anything if not clear :)

• Thank you for your response. I was working all week...haven't had time to come back here. I edited the proof based on your feedback. Much appreciated. Jul 22, 2021 at 0:49