Negative of Absolute Value

Theorem

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

Let $\size x$ denote the absolute value of $x$.

Then:

$-\size x \le x \le \size x$

Corollary 1

$\size x < y \iff -y < x < y$

Corollary 2

$\size x \le y \iff -y \le x \le y$

that is:

$\size x \le y \iff \begin {cases} x & \le y \\ -x & \le y \end {cases}$

Corollary 3

Let $y \in \R_{\ge 0}$.

Let $z \in \R$.

Then:

$\size {x - z} < y \iff z - y < x < z + y$

Proof

Either $x \ge 0$ or $x < 0$.

If $x \ge 0$, then:

$-\size x \le 0 \le x = \size x$

If $x < 0$, then:

$-\size x = x < 0 < \size x$

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.15$