# Natural Number Multiplication Distributes over Addition/Proof 2

## Theorem

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:

$\forall x, y, z \in \N:$
$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

## Proof

We are to show that:

$\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

From the definition of natural number multiplication, we have by definition that:

 $\ds \forall m, n \in \N: \,$ $\ds m \times 0$ $=$ $\ds 0$ $\ds m \times n^+$ $=$ $\ds \paren {m \times n} + m$

Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $\map P z$ be the proposition:

$\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds \paren {x + y} \times 0$ $=$ $\ds 0$ Definition of Natural Number Multiplication‎ $\ds$ $=$ $\ds 0 + 0$ Definition of Natural Number Addition $\ds$ $=$ $\ds x \times 0 + y \times 0$ Definition of Natural Number Multiplication‎

and so $\map P 0$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.

So this is our induction hypothesis:

$\forall x, y \in \N: \paren {x + y} \times k = \paren {x \times k} + \paren {y \times k}$

Then we need to show:

$\forall x, y \in \N: \paren {x + y} \times k^+ = \paren {x \times k^+} + \paren {y \times k^+}$

### Induction Step

This is our induction step:

 $\ds \paren {x + y} \times k^+$ $=$ $\ds \paren {x + y} \times k + \paren {x + y}$ Definition of Natural Number Multiplication‎ $\ds$ $=$ $\ds \paren {x \times k} + \paren {y \times k} + \paren {x + y}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {\paren {x \times k} + x} + \paren {\paren {y \times k} + y}$ Natural Number Addition is Commutative and Associative $\ds$ $=$ $\ds \paren {x \times k^+} + \paren {y \times k^+}$ Definition of Natural Number Multiplication

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$

$\Box$

Next we need to show that:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

So:

 $\ds z \times \paren {x + y}$ $=$ $\ds \paren {x + y} \times z$ Natural Number Multiplication is Commutative $\ds$ $=$ $\ds \paren {x \times z} + \paren {y \times z}$ from above $\ds$ $=$ $\ds \paren {z \times x} + \paren {z \times y}$ Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

$\blacksquare$