# Natural Number Multiplication Distributes over Addition/Proof 2

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## 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:

$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

for all $x, y, z \in \N$.

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

 $\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m \times 0$ $=$ $\displaystyle 0$ $\displaystyle m \times n^+$ $=$ $\displaystyle \left({m \times n}\right) + m$

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

For all $z \in \N$, let $P \left({z}\right)$ be the proposition:

$\forall x, y \in \N: \left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \left({x + y}\right) \times 0$ $=$ $\displaystyle 0$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle 0 + 0$ Definition of Natural Number Addition $\displaystyle$ $=$ $\displaystyle x \times 0 + y \times 0$ Definition of Natural Number Multiplication

and so $P \left({0}\right)$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

$\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$

Then we need to show:

$\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$

### Induction Step

This is our induction step:

 $\displaystyle \left({x + y}\right) \times k^+$ $=$ $\displaystyle \left({x + y}\right) \times k + \left({x + y}\right)$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle \left({x \times k}\right) + \left({y \times k}\right) + \left({x + y}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({\left({x \times k}\right) + x}\right) + \left({\left({y \times k}\right) + y}\right)$ Natural Number Addition is Commutative and Associative $\displaystyle$ $=$ $\displaystyle \left({x \times k^+}\right) + \left({y \times k^+}\right)$ Definition of Natural Number Multiplication

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

$\forall x, y, z \in \N: \left({x + y}\right) \times n = \left({x \times z}\right) + \left({y \times z}\right)$

$\Box$

Next we need to show that:

$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

for all $x, y, z \in \N$.

So:

 $\displaystyle z \times \left({x + y}\right)$ $=$ $\displaystyle \left({x + y}\right) \times z$ Natural Number Multiplication is Commutative $\displaystyle$ $=$ $\displaystyle \left({x \times z}\right) + \left({y \times z}\right)$ from above $\displaystyle$ $=$ $\displaystyle \left({z \times x}\right) + \left({z \times y}\right)$ Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

$\blacksquare$