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Mathematical proofs for addition of the natural numbers: additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers.

Definitions

• a + 0 = a [A1]
• a + S(b) = S(a + b) [A2]

It is useful to define another natural number closely related to the successor function, namely "1". We define 1 to be the successor of 0, in other words,

: $1 equiv S\left(0\right),$.

Note that for all natural numbers "a",

:

due to [A1] and [A2] .

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,

: $\left(a+b\right)+0=a+b=a+\left(b+0\right)$

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",

: $\left(a+b\right)+c=a+\left(b+c\right)$

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 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). Clearly, for "a" = 0, we have 0 + 1 = 0 + "S"(0) = "S"(0 + 0) = "S"(0) = 1 = 1 + 0. Now, suppose "a" + 1 = 1 + "a". Then

: "S"("a") + 1: = "S"("a") + "S"(0): = "S"("S"("a") + 0) [by A2] : = "S"(("a" + 1) + 0): = "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): = ("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"

This completes the induction on "b".

*Binary operation
*Commutativity
*Associativity
*Induction
*Proof
*Identity
*Ring

References

*Edmund Landau, Foundations of Analysis, Chelsea Pub Co. ISBN 0-8218-2693-X.

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