# Zero is Zero Element for Natural Number Multiplication

Jump to navigation
Jump to search

## Contents

## Theorem

Let $\N$ be the natural numbers.

Then $0$ is a zero element for multiplication:

- $\forall n \in \N: 0 \times n = 0 = n \times 0$

## Proof

Proof by induction.

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

- $0 \times n = 0 = n \times 0$

### Basis for the Induction

By definition, we have:

- $0 \times 0 = 0 = 0 \times 0$

Thus $\map P 0$ is seen to be true.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis $\map P k$:

- $0 \times k = 0 = k \times 0$

Then we need to show that $\map P {k + 1}$ follows directly from $\map P k$:

- $0 \times \paren {k + 1} = 0 = \paren {k + 1} \times 0$

### Induction Step

This is our induction step:

\(\displaystyle 0 \times \paren {k + 1}\) | \(=\) | \(\displaystyle \paren {0 \times k} + 0\) | Definition of Natural Number Multiplication | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0 + 0\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | Definition of Natural Number Addition |

By definition:

- $\paren {k + 1} \times 0 = 0$

So $\map P k \implies \map P {k + 1}$, and the result follows by the Principle of Mathematical Induction.

Therefore:

- $\forall n \in \N: 0 \times n = 0 = n \times 0$

$\blacksquare$

## Sources

- 1982: Alan G. Hamilton:
*Numbers, Sets and Axioms*... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(g)}$