Negative of Absolute Value/Corollary 2

Corollary to Negative of Absolute Value

Let $x, y \in \R$ be a real numbers such that $y > 0$.

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

Then:

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

that is:

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

Proof

Necessary Condition

Let $\size x \le y$.

If $\size x < y$ then from Corollary 1:

$-y < x < y$

Thus:

$-y \le x \le y$

Otherwise, if $\size x = y$ then either $x = y$ or $-x = y$.

Hence the result.

$\Box$

Sufficient Condition

Let $-y \le x \le y$.

If $-y < x < y$ then from Corollary 1:

$\size x < y$

Hence:

$\size x \le y$

Otherwise, if either $-y = x$ or $x = y$ then:

$\size x = y$

Hence the result.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): absolute value: $\text {(iv)}$
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): absolute value: $\text {(iv)}$