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We know that $|a|<b$ implies $-b<a<b$. Would that still hold if $-|a|<b$? That is, would that imply $-b<-a<b$? Thanks

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    $\begingroup$ $-|3|<1$ does not imply $-1<-3<1.$ $\endgroup$ – mfl Sep 23 '15 at 13:03
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1) If $b\ge0$ - each $a\neq 0$ is a solution to $-|a|<b$, while $|a|<b$ has a solution $a\in (-b,b)$

2)If $b<0$ then $-|a|<b\Leftrightarrow |a|>-b>0\Leftrightarrow a>-b \,\,\text{or}\,\, a<b$ and becasue $-b>b$ this is $a\in (-\infty,b) \cup (-b,\infty)$ which is not the same as the solution to $|a|<b$ (no real $a$ satisfies it).

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If $b \ge 0$, then $-|x|\le 0 \le b \implies x \in \mathbb R$

If $b \lt 0$, then $-|x| < b \implies |x| \gt -b \implies x \in (-\infty,b) \cup (-b,\infty) $

Please note that, in the second case, $b$ is negative and $-b$ is positive.

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