# Equivalence of Definitions of Absolute Value Function

## Theorem

The following definitions of the concept of Absolute Value are equivalent:

### Definition 1

Let $x \in \R$ be a real number.

The absolute value of $x$ is denoted $\size x$, and is defined using the usual ordering on the real numbers as follows:

$\size x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$

### Definition 2

Let $x \in \R$ be a real number.

The absolute value of $x$ is denoted $\size x$, and is defined as:

$\size x = +\sqrt {x^2}$

where $+\sqrt {x^2}$ is the positive square root of $x^2$.

## Proof

### Definition 1 implies Definition 2

Let $f: \R \to \R$ be the function defined as:

$\forall x \in \R: \map f x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$

Let $x > 0$.

We have that:

 $\displaystyle \paren {-x}^2$ $=$ $\displaystyle \paren {-1}^2 x^2$ $\displaystyle$ $=$ $\displaystyle x^2$ $\displaystyle \leadsto \ \$ $\displaystyle +\sqrt {\paren {-x}^2}$ $=$ $\displaystyle x$ $\displaystyle$ $=$ $\displaystyle +\sqrt {x^2}$

Thus $\map f x = +\sqrt {x^2}$

$\Box$

### Definition 2 implies Definition 1

Let $f: \R \to \R$ be the function defined as:

$\forall x \in \R: \map f x = +\sqrt {x^2}$

Let $x > 0$.

Then by definition of square and square root:

$+\sqrt {x^2} = x$

Let $x = 0$.

Then $+\sqrt {0^2} = 0$.

Let $x < 0$.

Thus let $x = -y$.

Then:

$+\sqrt {\paren {-y}^2} = y = -x$

$\blacksquare$