# 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.