Equivalence of Definitions of Absolute Value Function

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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:

\(\ds \paren {-x}^2\) \(=\) \(\ds \paren {-1}^2 x^2\)
\(\ds \) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds +\sqrt {\paren {-x}^2}\) \(=\) \(\ds x\)
\(\ds \) \(=\) \(\ds +\sqrt {x^2}\)

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:

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

$\blacksquare$


Sources

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