Oakey1w

2022-06-19

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:
$\frac{1}{-16}>\frac{x}{1}-\frac{4}{1}>\frac{1}{16}$

timmeraared

If a and b have the same nonzero sign, then
$a\frac{1}{b}$
(i.e., taking reciprocals reverses the inequality).
If a and b have opposite (nonzero) signs, then
$a
(i.e., taking reciprocals preserves the inequality).
These follow from the fact that the function $f\left(x\right)=1/x$ defined on the nonzero reals is strictly decreasing and sign-preserving on each component $\left(-\mathrm{\infty },0\right)$ and $\left(0,\mathrm{\infty }\right)$ of its domain.
If either of a or b is zero, then you can't take reciprocals.
Finally, compound inequalities like $a should be separated into "$a and $b" and each component considered separately.
I also remark that inverting a sum is not the same as inverting the addends separately.

migongoniwt

It depends if $x$ and $y$ are the same sign.
Case 1: $0 then $0 and $0 and $0 so $0<1/y<1/x$
If both positive, flip.
Case 2: $x<0 then $x/y<0<1$. Then as $x<0$ we flip when we do $x/y\ast \left(1/x\right)>0>1\ast \left(1/x\right)$ so $1/y>0>1/x$ so $1/x<0<1/y$. Don't flip.
Case 3: $x 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 $\frac{1}{1/x-1/4}=\frac{4x}{4-x}$

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