Evaluating Inequality with absolute value and a square root: $|x| \leq \sqrt{2}$

Question

Just looking for some help, since I'm always confused in this case:

So, for cases like $|x| \leq 2 $, it is easy to separate the two possible cases as:

1. For the positive part: $ x \leq 2$, and, 2. for the negative part: $ - x \leq 2$, which implies $x \geq -2$

Therefore, we can obtain the interval: $ -2 \leq x \leq 2 $.

Then, since I'm just confused about considering all the possible cases when we have $sqrt{2}$ instead, that is, considering we have $ |x| \leq \pm \sqrt{2}$.

Appreciate your help, Thanks!

Answer 1

Notice that in your original argument, we did not use any explicit properties of the number $2$, we just use the property that it is nonnegative.

For any nonnegative number $a$, $$|x| \le a$$

can be written as $$-a \le x \le a$$

and $a$ can be chosen to be $\sqrt{2}$.

Perhaps some geometric meaning might help, view $|x| \le a$ as $|x-0|\le a$, that is the distance from the origin is less than equal to $a$ and the corresponding to the interval $[-a,a]$.

Remark: $\{x:|x|\le -\sqrt2\}= \emptyset$ since $|x| \ge 0$, we can't have a distance from the origin which is less than negative.

Answer 2

We have

• for $x\ge 0 \quad |x| \leq \sqrt 2\implies x\le \sqrt 2$
• for $x\le 0 \quad |x| \leq \sqrt 2\implies -x\le \sqrt 2\implies x\ge -\sqrt 2$

Answer 3

Remember that of the two real numbers $x$ (one positive and one negative) such that $x^2=2$, the symbol $\sqrt2$ refers to the positive one. The other one (the other square root) is then the negative number $-\sqrt2$.

So, both $|x|\le 2$ and $|x|\le \sqrt2$ are of the form $$|x|\le \alpha,\quad \alpha>0.$$ Then, the procedure is always the same, and $$|x|\le 2 \quad \text{gives}\quad -2\le x\le2,$$ $$|x|\le \sqrt2 \quad \text{gives}\quad -\sqrt2\le x\le\sqrt2,$$ and even inequalities like $$|x|\le\pi \quad \text{gives}\quad -\pi\le x\le\pi,$$ $$|x|\le\cos(1) \quad \text{gives}\quad -\cos(1)\le x\le\cos(1),$$ since both $\pi$ and $\cos(1)$ are positive real numbers.