# Proofs involving the addition of natural numbers

This article contains mathematical proofs for some properties of addition of the natural numbers: the additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers.

## Definitions

This article will use the Peano axioms for the definition of natural numbers. With these axioms, addition is defined from the constant 0 and the successor function S(a) by the two rules
| A1: | a + 0 = a |
| A2: | a + S(b) = S(a + b) |
For the proof of commutativity, it is useful to give the name "1" to the successor of 0; that is,
    1 = S(0).
For every natural number a, one has
| | S(a) |
| = | S(a + 0) | [by A1] |
| = | a + S(0) | [by A2] |
| = | a + 1 | [by Def. of 1] |

## Proof of associativity

We prove associativity by first fixing natural numbers a and b and applying induction on the natural number c.
For the base case c = 0,
    (a + b) + 0 = a + b = a + (b + 0)
Each equation follows by definition [A1]; the first with a + b, the second with b.
Now, for the induction. We assume the induction hypothesis, namely we assume that for some natural number c,
    (a + b) + c = a + (b + c)
Then it follows,
| | (a + b) + S(c) |
| = | S((a + b) + c) | [by A2] |
| = | S(a + (b + c)) | [by the induction hypothesis] |
| = | a + S(b + c) | [by A2] |
| = | a + (b + S(c)) | [by A2] |
In other words, the induction hypothesis holds for S(c). Therefore, the induction on c is complete.

## Proof of identity element

Definition [A1] states directly that 0 is a right identity. We prove that 0 is a left identity by induction on the natural number a.
For the base case a = 0, 0 + 0 = 0 by definition [A1]. Now we assume the induction hypothesis, that 0 + a = a. Then
| | 0 + S(a) |
| = | S(0 + a) | [by A2] |
| = | S(a) | [by the induction hypothesis] |
This completes the induction on a.

## Proof of commutativity

We prove commutativity (a + b = b + a) by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. we prove that 0 and 1 commute with everything).
The base case b = 0 follows immediately from the identity element property (0 is an additive identity), which has been proved above: a + 0 = a = 0 + a.
Next we will prove the base case b = 1, that 1 commutes with everything, i.e. for all natural numbers a, we have a + 1 = 1 + a. We will prove this by induction on a (an induction proof within an induction proof). We have proved that 0 commutes with everything, so in particular, 0 commutes with 1: for a = 0, we have 0 + 1 = 1 + 0. Now, suppose a + 1 = 1 + a. Then
| | S(a) + 1 |
| = | S(a) + S(0) | [by Def. of 1] |
| = | S(S(a) + 0) | [by A2] |
| = | S((a + 1) + 0) | [as shown above] |
| = | S(a + 1) | [by A1] |
| = | S(1 + a) | [by the induction hypothesis] |
| = | 1 + S(a) | [by A2] |
This completes the induction on a, and so we have proved the base case b = 1. Now, suppose that for all natural numbers a, we have a + b = b + a. We must show that for all natural numbers a, we have a + S(b) = S(b) + a. We have
| | a + S(b) |
| = | a + (b + 1) | [as shown above] |
| = | (a + b) + 1 | [by associativity] |
| = | (b + a) + 1 | [by the induction hypothesis] |
| = | b + (a + 1) | [by associativity] |
| = | b + (1 + a) | [by the base case b = 1] |
| = | (b + 1) + a | [by associativity] |
| = | S(b) + a | [as shown above] |
This completes the induction on b.