Non-Zero Integers are Cancellable for Multiplication/Proof 3

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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.

$\blacksquare$