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

Question

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

Answer 1

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.

Answer 2

If you have $$x>y\qquad (1)$$ Then you only have to flip the inequality when you take the reciprocal if $x$ and $y$ are of the same signs. Notice that this is because we can multiply both sides of $(1)$ by $(xy)^{-1}$ to get $$\frac{x}{xy}>\frac{y}{xy}\quad\rightarrow \quad\frac{1}{y}>\frac{1}{x}$$ Which naturally flips the signs, however, if $(xy)^{-1}$ is negative (which occurs when $x$ and $y$ are different signs) then we have to flip the inequality sign back because we multiplied by negative number.

Note, however, in your question you misused the reciprocal. $$\frac{1}{x}-\frac{1}{4}=\frac{4-x}{4x}$$ Which has reciprocal $\frac{4x}{4-x}$. Also you can apply this to a "multistaged" inequality by breaking it into a series of inequalities of the form $(1)$.

Answer 3

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

Answer 4

Here are some ideas for how reciprocals work.

A different kind of reciprocal

From a picture point of view, reciprocals have to do with "turning fractions upside down"— like turning $\frac{a}{b}$ into $\frac{b}{a}$.

From your point of view, which is reasonable, turning $\frac{a}{b} + \frac{c}{d}$ into $\frac{b}{a} + \frac{d}{c}$ makes a certain kind of reciprocal.

But in most algebra classes, people use the term reciprocal to talk about starting with a term like $x$ and replacing it with one-divided-by-$x$. Because $\frac{1}{\frac{a}{b} + \frac{c}{d}}$ is not the same as $\frac{b}{a} + \frac{d}{c}$, your way of talking about reciprocals and the way people talk about reciprocals in algebra classes are different.

You can take reciprocals using only multiplication and division

In algebra, there are rules for what you can do to an equation or inequality to make sure it stays true. For example, you can add the same number to both sides of the equation, or you can multiply both sides of the equation by the same number (except zero).

You can take the reciprocal of an equation just by performing several multiplication and division operations in a row. Here's why:

If you have an equation written like: $$\frac{a}{b} = \frac{c}{d},$$ you can turn it into its reciprocal equation by taking the following steps:

You can multiply both sides by the denominators $b$ and $d$:

$$\frac{a\times b \times d}{b} = \frac{c\times b \times d}{d}$$

Some factors cancel, leaving you with:

$$a \times d = c \times b.$$

You can divide both sides by $a$ and $c$ (the numerators from the original equation), giving you:

$$\frac{a \times d}{a \times c} = \frac{c\times b}{a \times c}.$$

Some factors cancel, leaving you with

$$\frac{d}{c} = \frac{b}{a}.$$

which is the reciprocal of your original equation! Because all we did was perform multiplication and division, we know that whenever the original equation is true $a/b = c/d$, the reciprocal equation is also true $b/a = d/c$.

If you had an inequality $\frac{a}{b} < \frac{c}{d}$, the steps for taking the reciprocal would be the same — but you'd additionally have to keep track of whether $a$,$b$,$c$,and $d$ were negative so you would know when to flip the $<$ sign.

You can't take the reciprocal of a sum by taking the sum of the reciprocals

In general, you can't start with a sum of fractions like $\frac{a}{b} + \frac{c}{d} = \frac{e}{f} + \frac{g}{h}$, and convert it into $\frac{b}{a} + \frac{d}{c} = \frac{f}{e} + \frac{h}{g}$ by following the rules of algebra for manipulating equations.

Of course, you already know that if you have $\frac{a}{b} + \frac{c}{d} = \frac{e}{f} + \frac{g}{h}$, you can safely write:

$$\frac{1}{\frac{a}{b} + \frac{c}{d}} = \frac{1}{\frac{e}{f} + \frac{g}{h}}$$

— but in general, $\frac{1}{\frac{a}{b} + \frac{c}{d}}$ will be a very different number from $\frac{b}{a} + \frac{d}{c}$.