# Inequalities with single sided absolute values versus double sided absolute values

I am trying to understand how the behavior between the absolute value(s) and the inequality sign differ between equations of the form:

\begin{equation} |x| < a \tag{1} \label{1} \end{equation}

for some $a \in \mathbb{R}$ and

\begin{equation} |f(x)| < |g(x)| \tag{2} \label{2} \end{equation}

for $f : D_1 \subseteq\mathbb{R} \to \mathbb{R}$ and $g : D_2 \subseteq\mathbb{R}\to\mathbb{R}$.

Specifically, I know that when one solves an equation like eq. $\ref{1}$, you do the following:

1. Solve $x < a$.
2. Solve $x > -a$.
3. Solve eq. $\ref{1}$ by taking the intersection of the solution found in steps 1. and 2.

I think of step 2. as first writing $-x < a$ (to make the LHS positive for the case when $x < 0$), and then multiplying both sides by $-1$, thereby flipping the inequality sign to get $x > -a$.

To solve eq. $\ref{2}$, you take each case separately, as shown in, for example, this question. However, in each of these cases, as far as I know, you do not switch the inequality sign like in step 2. for solving eq. $\ref{1}$.

Therefore, my question is why, conceptually, is this done for solving eq. $\ref{1}$, but not done in each case (i.e. when the argument of one of the absolute values is negative) when solving eq. $\ref{2}$.

Thanks in advance. I have been avoiding absolute values + inequalities for most of my career in mathematics; it's time I face them.

Note that the absolute value of a number equals its distance from $0$ which is always a non-negative number.

So the statement that $\vert x\vert<a$ tells us that $a$ is a positive number and that $x$ is closer to $0$ than $a$ is. From this we conclude that $x$ lies somewhere between $-a$ and $a$. In terms of inequalities

$$-a<x\text{ and }x<a$$

A pair of inequalities connected by the word "and" [but not the word "or"] can be united in a combined inequality provided both inequalities are in the same direction. So the pair of inequalities above can be written in combined form as

$$-a<x<a$$

which is read as "negative $a$ is less than $x$ and $x$ is less than $a$" or more simply $x$ is between negative $a$ and positive $a$.

Now for the case where $|f(x)|<|g(x)|$.

Note that for non-negative numbers, the function $y=x^2$ is increasing. So it follows that $0\le a<b$ if and only if $a^2<b^2$. Therefore

$$|f(x)|<|g(x)|\text{ iff }f^2(x)<g^2(x)$$

which is equivalent to

$$g^2(x)-f^2(x)>0$$

which is equivalent to

$$[g(x)-f(x)]\cdot[g(x)+f(x)]>0$$

which is true if and only if $g(x)-f(x)$ and $g(x)+f(x)$ are both the same sign--both positive or both negative.

Ultimately, it means that $f(x)$ must always lie between $g(x)$ and $-g(x)$. That is

Either

1. $-g(x)<f(x)<g(x)$ or

2. $g(x)<f(x)<-g(x)$.

So to find the solution set of $|f(x)|<|g(x)|$ you would find the solution of both 1 and 2 and take their union since the two inequalities are connected by the word "or."

• Thank you. Fixed. Sep 18, 2018 at 5:12
• I added a bit about the $|f(x)|<|g(x)$ case. Sep 18, 2018 at 5:15
• Since this is your first question, may not know that you can pick up a couple of reputation points by checking the arrow beside an accepted answer and the person who has their answer accepted also picks up reputation points. But it is optional. Sep 18, 2018 at 5:22
• You meant to write $|f(x)\color{red}{|} < |g(x)|~\text{iff}~f^2(x) < g^2(x)$. Sep 18, 2018 at 8:27
• @N.F.Taussig Thanks Sep 18, 2018 at 16:46