Does $|f(x)|≥g(x)$ imply $-g(x)≥f(x)≥g(x)$?

Question

Since for ever I thought that $|f(x)|≥g(x)\implies -g(x)≥f(x)≥g(x)$, but is it actually true? And same with $>$ instead of $≥$, does it hold? Because I have been looking for research on google and on math exchange and I was told that no. But then I came to this problem:$|x-5\pi|≥4\pi$ which if you apply so, it does seem to work? I think at least if I'm not mistaken it implies $$-4\pi≥x-5\pi≥4\pi$$ $$\pi≥x≥9\pi$$ which in interval notation it means $$x\in (-\infty, \pi]\cup [9\pi, \infty)$$ which is correct from Wolfram Alpha. What's going on here?

Answer 1

When you write $$ \pi \geq x \geq 9\pi $$ it means $\pi \geq x$ and $x \geq 9\pi$ (simultaneously, for the same value of $x$), which by transitivity of $\geq$ implies that $\pi \geq 9\pi.$ That is not correct, of course, so the original set of inequalities is a mistake.

What actually is true is that $\pi \geq x$ or $x\geq 9\pi.$ That is the answer you finally wrote and the answer Wolfram Alpha gave you. So you have the right result but a very dubious method that relies on thinking something very different from what you are writing.

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Regarding your general question, $\lvert f(x)\rvert \geq g(x)$ implies that $f(x)\leq-g(x)$ or $g(x)\leq f(x).$ Very rarely will both inequalities be true at the same time (for the same value of $x$ — perhaps you can deduce what the one exception is) and if you have $>$ instead of $\geq$ then the two inequalities will never both be true at the same time.

Answer 2

This is a common mistake. People are used to replacing "$|a|\leq b$" with "$-b\leq a \leq b$" and so assume you can do the same thing with the inequality reversed, i.e., that you can replace "$|a| \geq b$" with "$-b \geq a \geq b$". This is just plain wrong.

To find the correct statement, let's analyze the statement more precisely.

• "$|a|\geq b$" is equivalent to "not ($|a|< b$)".

• That is equivalent to "not ($-b<a<b$)".

• The compound inequality "$-b<a<b$" is shorthand for "$-b<a$ and $a<b$".

• So "not ($-b<a<b$)" is equivalent to "not ($-b<a$ and $a<b$)".

• By DeMorgan's laws, that is equivalent to "not ($-b<a$) or not ($a<b$)".

• Again by DeMorgan, that is equivalent to "$-b \geq a$ or $a \geq b$".

So we see that we end up not with "$-b \geq a \geq b$" (which would mean "$-b \geq a$ and $a \geq b$") but rather with "$-b \geq a$ or $a \geq b$".