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:
\(\ds x y\) | \(=\) | \(\ds x z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {-\paren {-x} } y\) | \(=\) | \(\ds \paren {-\paren {-x} } z\) | 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.
$\blacksquare$