# Absolute Value of Product/Proof 1

## 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

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.

$\blacksquare$