# Non-Zero Integers are Cancellable for Multiplication/Proof 3

## 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

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:

 $\displaystyle x y$ $=$ $\displaystyle x z$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {-\paren {-x} } y$ $=$ $\displaystyle \paren {-\paren {-x} } z$ Negative of Ring Negative $\displaystyle \leadsto \ \$ $\displaystyle -\paren {\paren {-x} y}$ $=$ $\displaystyle -\paren {\paren {-x} z}$ Product with Ring Negative $\displaystyle \leadsto \ \$ $\displaystyle -y$ $=$ $\displaystyle -z$ as $-x$ is (strictly) positive, the above result holds $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle z$ $\struct {\Z, +}$ 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.

$\blacksquare$