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:

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

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

Beginner2022-06-20Added 22 answers

If a and b have the same nonzero sign, then

$a<b\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{a}>\frac{1}{b}$

(i.e., taking reciprocals reverses the inequality).

If a and b have opposite (nonzero) signs, then

$a<b\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{a}<\frac{1}{b}$

(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 $(-\mathrm{\infty},0)$ and $(0,\mathrm{\infty})$ 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.

$a<b\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{a}>\frac{1}{b}$

(i.e., taking reciprocals reverses the inequality).

If a and b have opposite (nonzero) signs, then

$a<b\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{a}<\frac{1}{b}$

(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 $(-\mathrm{\infty},0)$ and $(0,\mathrm{\infty})$ 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.

migongoniwt

Beginner2022-06-21Added 4 answers

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 $x/y\ast (1/x)>0>1\ast (1/x)$ 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 $\frac{1}{1/x-1/4}=\frac{4x}{4-x}$

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 $x/y\ast (1/x)>0>1\ast (1/x)$ 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 $\frac{1}{1/x-1/4}=\frac{4x}{4-x}$