# 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, $x \in \N_{> 0}$.

$\Box$

Let $x < 0$.

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

Then $-x > 0$ and so:

 $\ds \paren {-x} y$ $=$ $\ds -\paren {x y}$ Product with Ring Negative $\ds$ $=$ $\ds -\paren {x z}$ $\struct {\Z, +}$ is a group: Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds \paren {-x} z$ Product with Ring Negative $\ds \leadsto \ \$ $\ds y$ $=$ $\ds z$ from above: case where $x > 0$

$\Box$

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

$\blacksquare$

## Proof 2

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

 $\ds y$ $\ne$ $\ds z$ $\ds \leadstoandfrom \ \$ $\ds y - z$ $\ne$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds x \paren {y - z}$ $\ne$ $\ds 0$ Ring of Integers has no Zero Divisorsâ€Ž $\ds \leadstoandfrom \ \$ $\ds x y - x z$ $\ne$ $\ds 0$ 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 $x < 0$.

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

Then $-x > 0$ and so:

 $\ds x y$ $=$ $\ds x z$ $\ds \leadsto \ \$ $\ds \paren {-\paren {-x} } y$ $=$ $\ds \paren {-\paren {-x} } z$ Negative of Ring Negative $\ds \leadsto \ \$ $\ds -\paren {\paren {-x} y}$ $=$ $\ds -\paren {\paren {-x} z}$ Product with Ring Negative $\ds \leadsto \ \$ $\ds -y$ $=$ $\ds -z$ as $-x$ is (strictly) positive, the above result holds $\ds \leadsto \ \$ $\ds y$ $=$ $\ds z$ $\struct {\Z, +}$ is a group: Group Axiom $\text G 3$: Existence of Inverse Element

$\Box$

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.
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