Non-Zero Integers are Cancellable for Multiplication/Proof 1

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.

$\Box$

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.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.10$