# Multiplication of Integers

On a page on addition of numbers, I assumed that we know how to add, subtract and multiply whole numbers. I mentioned Peano axioms as the foundation on which these operations are defined and their properties established. Let's see how it can be done.

First of all we assume that there is something to talk about: there exists an entity known as the set of natural numbers N whose properties (explicitly or implicitly) are given by the following

### Peano axioms

1. 1 is a natural number. This says that the set N is not empty. There is at least one natural number. This number is denoted by the symbol 1 (pronounced one or unit.)
2. For every x in N there exists a number x' known as the successor of x. Since x = y means that x and y are one and the same number, x = y implies x' = y'.
3. x' ≠ 1. In other words, 1 is not a successor of any natural number.
4. x' = y' implies x = y. Different numbers have different successors.
5. (Axiom of Induction). Let M be a (sub)set of natural numbers with the following properties:
1. 1 ∈ M
2. x ∈ M implies x' ∈ M
Then M = N. In other words, M contains all natural numbers.

It's not my purpose to pursue this subject in every detail. Just, as an example, to demonstrate the logic of derivation from the axioms and, especially The Axiom of Induction, I'll prove a few basic theorems; after which we'll define addition and multiplication for the natural numbers. (A great many additional examples are considered elsewhere.)

### Theorem 1

 x ≠ y implies x' ≠ y'.

### Proof

Indeed, otherwise we would have x' = y'. Axiom 4 would then lead to x = y. Contradiction.

### Theorem 2

 x' ≠ x.

### Proof

Let M be the set of all x for which x' ≠ x: M = {x: x' ≠ x}. According to Axioms 1 and 3, 1' ≠ 1. Therefore 1 ∈ M.

Furthermore, assume x ∈ M, i.e., x' ≠ x. Then, by Theorem 1, (x')' ≠ x' which exactly means that x' ∈ M. Finally, Axiom 5 implies that M=N. To recapitulate, x' ≠ x for all natural x.

### Theorem 3

 If x ≠ 1 then there exists u such that x = u'.

### Proof

Let M be the set that consists of 1 and all those x for which such u exists. Then 1 ∈ M by definition. Let x ∈ M. Taking x = u we see that x'=u'. Therefore, x' ∈ M. Therefore, M = N and, except for 1, every natural number is a successor of another natural number.

1. For every x define x + 1 = x'.
2. For every x, y define x + y' = (x + y)'.

### Theorem 4

 The number x + y is well defined for all natural x and y.

### Definition 2 (Multiplication)

1. For every x define x·1 = x.
2. For every x,y define x·y' = x·y + x.

### Theorem 5

 The number x·y is well defined for all natural x and y.

### Remark

1. x + y and x·y are called the sum and the product of x and y, respectively.
2. I delight in the E. Landau's remark at the beginning of the proof of Theorem 5: Mutatis mutandis (with obvious changes), the proof follows virtually verbatim that of Theorem 4.
3. Recursive definitions, i.e., definitions that depend on Axiom 5, are at the heart of Arithmetic. This axiom provides a way to prove a statement for an infinitude of numbers (or, if you will, infinitely many statements each holding for a single number) in a finite number of steps (viz., 2.)
4. Associativity and commutativity of both addition and multiplication follow from the Peano axioms. Subtraction and division must be defined separately. Without them, the set of natural numbers is a semigroup with respect to both addition and multiplication.
5. The distributive law (the first one) is suggested by the second clause of Definition 2 which could be read (x + 1)·y = x·y + y. A more general theorem is of course based on Axiom 5.

### Theorem 6 (Distributive Law)

 x·(y + z) = x·y + x·z

### Proof

The proof is by induction based on Axiom 5. Let M be the set of all z for which the Law holds. We just saw that, by Definition 2, 1 ∈ M. Let z ∈ M. Then

 x·(y + z') = x·(y + z)' = x·(y + z) + x = (x·y + x·z) + x = x·y + (x·z + x) = x·y + x·z'

where I have assumed associativity of the addition as proven. ## Reference

1. Edmund Landau, Foundations of Analysis, Chelsea Pub Co, 1960
2. J. A. Paulos, Beyond Numeracy, Vintage Books, 1992 ### What Can Be Multiplied? 