Absolute Value Function is Completely Multiplicative/Proof 1

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Theorem

The absolute value function on the real numbers $\R$ is completely multiplicative:

$\forall x, y \in \R: \left\vert{x y}\right\vert = \left\vert{x}\right\vert \, \left\vert{y}\right\vert$

where $\left \vert{a}\right \vert$ denotes the absolute value of $a$.


Proof

Let $x = 0$ or $y = 0$.

Then:

\(\ds x y\) \(=\) \(\ds 0\)
\(\text {(1)}: \quad\) \(\ds \implies \ \ \) \(\ds \left\vert{x y}\right\vert\) \(=\) \(\ds 0\)

and either $\left\vert{x}\right\vert = 0$ or $\left\vert{y}\right\vert = 0$ and so:

\(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds \left\vert{x y}\right\vert\) from $(1)$ above

$\Box$


Let $x > 0$ and $y > 0$.

Then:

\(\ds \left\vert{x}\right\vert\) \(=\) \(\ds x\)
\(\, \ds \land \, \) \(\ds \left\vert{y}\right\vert\) \(=\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\) \(=\) \(\ds x y\)


and:

\(\ds x y\) \(>\) \(\ds 0\)
\(\ds \left\vert{x y}\right\vert\) \(=\) \(\ds x y\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\)

$\Box$


Let $x < 0$ and $y < 0$.

Then:

\(\ds \left\vert{x}\right\vert\) \(=\) \(\ds -x\)
\(\, \ds \land \, \) \(\ds \left\vert{y}\right\vert\) \(=\) \(\ds -y\)
\(\ds \implies \ \ \) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\) \(=\) \(\ds \left({-x}\right) \left({-y}\right)\)
\(\ds \) \(=\) \(\ds x y\)


and:

\(\ds x y\) \(>\) \(\ds 0\)
\(\ds \left\vert{x y}\right\vert\) \(=\) \(\ds x y\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\)

$\Box$


Let $x < 0$ and $y > 0$.

Then:

\(\ds \left\vert{x}\right\vert\) \(=\) \(\ds - x\)
\(\, \ds \land \, \) \(\ds \left\vert{y}\right\vert\) \(=\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\) \(=\) \(\ds - x y\)


and:

\(\ds x y\) \(<\) \(\ds 0\)
\(\ds \left\vert{x y}\right\vert\) \(=\) \(\ds - x y\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \left\vert{x}\right\vert \, \left\vert{y}\right\vert\)

$\Box$


The same argument, mutatis mutandis, covers the case where $x > 0$ and $y < 0$.

$\blacksquare$