Non-Zero Integers are Cancellable for Multiplication

From ProofWiki
Jump to navigation Jump to search

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