Some true/false questions on absolute values

by RoRi

July 7, 2015

Decide whether the following are true or false.

$x < 5 \implies |x| < 5$ .
$|x-5| < 2 \implies 3 < x < 7$ .
$|1 + 3x| \leq 1 \implies x \geq -\frac{2}{3}$ .
$|x-1| = |x-2|$ is not satisfied by any $x \in \mathbb{R}$ .
For all $x \in \mathbb{R}_{>0}$ there exists a $y \in \mathbb{R}_{>0}$ such that $|2x+y| = 5$ .

False. Let $x = -6$ , then $x < 5$ , but $|x| = |-6| = 6 > 5$ .
True.

$|x-5| < 2 \ \implies \ -(x-5) < 2 \qquad \text{and} (x-5) < 2$

Hence,

$x >3 \qquad \text{and} x < 7.$

This means $3 < x < 7$ .
True.

$|1+3x| \leq 1 \ \implies \ -\frac{2}{3} \leq x \leq 0 \ \implies \ -\frac{2}{3} \leq x.$
False. This equality is true for $x = \frac{3}{2}$ since

$\left| \frac{3}{2} - 1 \right| = \frac{1}{2} \qquad \text{and} \qquad \left| \frac{3}{2} - 2 \right| = \frac{1}{2}.$
False. Take $x = 3$ , then

$|2x+y| = 5 \ \implies \ |6+y| = 5 \ \implies \ y=-11 \text{ or } y = -1.$

Hence, $y < 0$ in either case and so $y \notin \mathbb{R}_{>0}$ .

Stumbling Robot