# Non-Zero Integers are Cancellable for Multiplication

## Theorem

Every non-zero integer is cancellable for multiplication.

That is:

- $\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$

## Proof 1

Let $x y = x z$.

There are two cases to investigate: $x > 0$ and $x < 0$.

Let $x > 0$.

From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.

By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication.

$\Box$

Let $x < 0$.

We know that the Integers form Integral Domain and are thus a ring.

Then $-x > 0$ and so:

\(\displaystyle \left({-x}\right) y\) | \(=\) | \(\displaystyle - \left({x y}\right)\) | Product with Ring Negative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle - \left({x z}\right)\) | $\left({\Z, +}\right)$ is a group: axiom $G3$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({-x}\right) z\) | Product with Ring Negative | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle z\) | from above: case where $x > 0$ |

$\Box$

So whatever non-zero value $x$ takes, it is cancellable for multiplication.

$\blacksquare$

## Proof 2

Let $y, z \in \Z: y \ne z$.

\(\displaystyle y\) | \(\ne\) | \(\displaystyle z\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle y - z\) | \(\ne\) | \(\displaystyle 0\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x \paren {y - z}\) | \(\ne\) | \(\displaystyle 0\) | Ring of Integers has no Zero Divisors | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x y - x z\) | \(\ne\) | \(\displaystyle 0\) | Integer Multiplication Distributes over Subtraction |

The result follows by transposition.

$\blacksquare$

## Proof 3

Let $x y = x z$.

There are two cases to investigate: $x > 0$ and $x < 0$.

Let $x > 0$.

From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.

By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication. {{qed|lemma}

Let $x < 0$ and

We know that the Integers form Integral Domain and are thus a ring.

Then $-x > 0$ and so:

\(\displaystyle x y\) | \(=\) | \(\displaystyle x z\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({- \left({-x}\right)}\right) y\) | \(=\) | \(\displaystyle \left({- \left({-x}\right)}\right) z\) | Negative of Ring Negative | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle - \left({\left({-x}\right) y}\right)\) | \(=\) | \(\displaystyle - \left({\left({-x}\right) z}\right)\) | Product with Ring Negative | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle - y\) | \(=\) | \(\displaystyle - z\) | as $-x$ is (strictly) positive, the above result holds | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle z\) | $\left({\Z, +}\right)$ is a group: axiom $G3$ |

$\Box$

So whatever non-zero value $x$ takes, it is cancellable for multiplication.

$\blacksquare$

## Also known as

Some sources give this as the **cancellation law**, but this term is already in use in the context of a group.

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: I. Basic Concepts*... (previous) ... (next): Introduction $\S 5$: The system of integers