Is it possible to flip a faction within an inequality? Such that:
$$\frac13 < x < \frac23$$
becomes,
$$3 > \frac1x > \frac32$$
$\endgroup$2 Answers
$\begingroup$This is possible in your case because all parts are guaranteed to be $>0$. In general:
Given a strictly monotone decreasing function $f : A\to\mathbb R$ where $A\subset \mathbb R$ is an interval and an inequality $$a<b$$ where both $a,b\in A$ the inequality implies $$f(a) > f(b)$$
In your case, $A = (0,\infty)$ and $f(x) = \frac1x$. For a non-strict version ($a\le b$) the function $f$ can be monotone (not necessarily strictly monotone).
Using $A = (-\infty, 0)$ and $f(x) = \frac1x$ also works, so $$-\infty < a<x<b<0 \Leftrightarrow \underbrace0_{="\frac1{-\infty}"} > \frac1a > \frac1x > \frac1b > \underbrace{-\infty}_{="\frac10"}$$ and $$0<a<x<b<\infty \Leftrightarrow \underbrace{\infty}_{="\frac10"} > \frac1a > \frac1x > \frac1b > \underbrace0_{= "\frac1\infty"}$$
But we must have that $a,b,x$ have the same sign.
Note that you can see that $\frac10$ cannot be defined here. In the first equation, it becomes a $-\infty$ while in the second it becomes a $+\infty$.
Yes. In general, we have $$0\lt a\lt x\lt b\lt \infty\iff \infty\gt\frac 1a\gt \frac 1x\gt \frac 1b\gt 0.$$
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