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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!

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    $\begingroup$ Of course, $|x|\le \sqrt 2$ simply translates to $-\sqrt 2\le x\le \sqrt 2$ (and $|x|\le 17\pi-5$ into $-17\pi+5\le x\le 17\pi-5$; the procedure is the same for any positive right hand side). However, the double sign in $|x|\le\pm\sqrt 2$ makes little sense $\endgroup$ – Hagen von Eitzen Apr 11 '18 at 6:14
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    $\begingroup$ Well, $|x| \leq - \sqrt{2}$ is true for no $x$, since the absolute value of a number is always positive... $\endgroup$ – Thomas Bladt Apr 11 '18 at 6:18
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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.

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  • $\begingroup$ Thanks for your reply. How would this change by adding a constant: $|x| - 1 \leq \sqrt{2}$ ?? $\endgroup$ – pkenneth81 Apr 11 '18 at 6:36
  • $\begingroup$ your new problem is equivalent to $|x| \le 1+\sqrt{2}$, check whether $1+\sqrt{2}$ is nonnegative, if it is, then follow the rule above. $\endgroup$ – Siong Thye Goh Apr 11 '18 at 6:43
  • $\begingroup$ So, that would be: $|x| - 1 \leq \sqrt{2}$ corresponds to: $ -(1+\sqrt{2}) \leq x \leq +(1+\sqrt{2} )$ . Am I correct? -- that is $ -2.41.. \leq x \leq 2.41...$ $\endgroup$ – pkenneth81 Apr 11 '18 at 6:51
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    $\begingroup$ yes, $-(1+\sqrt2) \le x \le (1+\sqrt2)$. $\endgroup$ – Siong Thye Goh Apr 11 '18 at 6:55
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We have

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

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