# Absolute Value of Product

## Theorem

Let $x, y \in \R$ be real numbers.

Then:

$\size {x y} = \size x \size y$

where $\size x$ denotes the absolute value of $x$.

## Proof 1

Let either $x = 0$ or $y = 0$, or both.

We have that $\size 0 = 0$ by definition of absolute value.

Hence:

$\size x \size y = 0 = x y = \size {x y}$

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

Then:

 $\ds x y$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x y}$ $=$ $\ds x y$ Definition of Absolute Value

and:

 $\ds x$ $=$ $\ds \size x$ Definition of Absolute Value $\ds y$ $=$ $\ds \size y$ $\ds \leadsto \ \$ $\ds \size x \size y$ $=$ $\ds x y$ $\ds$ $=$ $\ds \size {x y}$

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

Then:

 $\ds x y$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x y}$ $=$ $\ds x y$ Definition of Absolute Value

and:

 $\ds -x$ $=$ $\ds \size x$ Definition of Absolute Value $\ds -y$ $=$ $\ds \size y$ $\ds \leadsto \ \$ $\ds \size x \size y$ $=$ $\ds \paren {-x} \paren {-y}$ $\ds$ $=$ $\ds xy$ $\ds$ $=$ $\ds \size {x y}$

The final case is where one of $x$ and $y$ is positive, and one is negative.

Without loss of generality, let $x < 0$ and $y > 0$.

Then:

 $\ds x y$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x y}$ $=$ $\ds -\paren {x y}$ Definition of Absolute Value

and:

 $\ds -x$ $=$ $\ds \size x$ Definition of Absolute Value $\ds y$ $=$ $\ds \size y$ $\ds \leadsto \ \$ $\ds \size x \size y$ $=$ $\ds \paren {-x} y$ $\ds$ $=$ $\ds -\paren {x y}$ $\ds$ $=$ $\ds \size {x y}$

The case where $x > 0$ and $y < 0$ is the same.

$\blacksquare$

## Proof 2

Let $x$ and $y$ be considered as complex numbers which are wholly real.

That is:

$x = x + 0 i, y = y + 0 i$

From Complex Modulus of Real Number equals Absolute Value, the absolute value of $x$ and $y$ equal the complex moduli of $x + 0 i$ and $y + 0 i$.

Thus $\cmod x \cmod y$ can be interpreted as the complex modulus of $x$ multiplied by the complex modulus of $y$.

$\cmod x \cmod y = \cmod {x y}$

As $x$ and $y$ are both real, so is $x y$.

Thus by Complex Modulus of Real Number equals Absolute Value, $\cmod {x y}$ can be interpreted as the absolute value of $x y$ as well as its complex modulus.

$\blacksquare$

## 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

Follows directly from:

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

$\blacksquare$