# Absolute Value Function is Completely Multiplicative

## 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 1

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 $y < 0$.

$\blacksquare$

## Proof 2

Follows directly from:

Real Numbers form Ordered Integral Domain
Product of Absolute Values on Ordered Integral Domain.

$\blacksquare$

## Proof 3

 $\ds \size {x y}$ $=$ $\ds \sqrt {\paren {x y}^2}$ Definition 2 of Absolute Value $\ds$ $=$ $\ds \sqrt {x^2 y^2}$ $\ds$ $=$ $\ds \sqrt {x^2} \sqrt{y^2}$ $\ds$ $=$ $\ds \size x \cdot \size y$

$\blacksquare$

## Proof 4

We have the result Multiplicative Group of Reals is Subgroup of Complex.

Therefore, any result applying to all complex numbers will also hold for all real numbers.

The result follows from Complex Modulus of Product of Complex Numbers.

$\blacksquare$