If you take the reciprocal in an inequality, would it change the $>/< $ signs?

Example:$$-16<\frac{1}{x}-\frac{1}{4}<16$$

In the example above, if you take the reciprocal of $$\frac{1}{x}-\frac{1}{4} = \frac{x}{1}-\frac{4}{1}$$
would that flip the $<$ to $>$ or not?

In another words, if you take the reciprocal of $$-16<\frac{1}{x}-\frac{1}{4}<16$$ would it be like this: $$\frac{1}{-16}>\frac{x}{1}-\frac{4}{1}>\frac{1}{16}$$

If $a$ and $b$ have the same nonzero sign, then $$a<b \iff \frac1a > \frac 1b$$ (i.e., taking reciprocals reverses the inequality).

If $a$ and $b$ have opposite (nonzero) signs, then $$a<b\iff \frac1a <\frac1b$$ (i.e., taking reciprocals preserves the inequality).

These follow from the fact that the function $f(x)=1/x$ defined on the nonzero reals is strictly decreasing and sign-preserving on each component $(-\infty,0)$ and $(0,\infty)$ of its domain.

If either of $a$ or $b$ is zero, then you can't take reciprocals.

Finally, compound inequalities like $a<b<c$ should be separated into "$a<b$ and $b<c$" and each component considered separately.

I also remark that inverting a sum is not the same as inverting the addends separately.

It depends if $x$ and $y$ are the same sign.

Case 1: $0 < x < y$ then $0 < x(1/y) < y(1/y)$ and $0 < x/y < 1$ and $0 < x/y(1/x) < 1 (1/x)$ so $0 < 1/y < 1/x$.

If both positive, flip.

Case 2: $x < 0 < y$ then $x/y < 0 < 1$. Then as $x < 0$ we flip when we do $x/y*(1/x) > 0 > 1*(1/x)$ so $ 1/y > 0 > 1/x$ so $1/x < 0 < 1/y$. Don't flip.

Case 3: $x < y < 0$ then $x/y > 1 > 0$ and $1/y < 1/x < 0$. Flip if they are the same sign.

But FOR THE LOVE OF GOD!!!!!!! the reciprical of $1/x - 1/4$ is !!!!!!!NOT!!!!!! $x/1 - 4/1$!!!!!!!!

It is $\frac{1}{1/x - 1/4} = \frac{4x}{4 - x}$.