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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.

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1 Answer 1

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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."

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  • $\begingroup$ Thank you. Fixed. $\endgroup$
    – user559038
    Commented Sep 18, 2018 at 5:12
  • $\begingroup$ I added a bit about the $|f(x)|<|g(x)$ case. $\endgroup$ Commented Sep 18, 2018 at 5:15
  • $\begingroup$ You meant to write $|f(x)\color{red}{|} < |g(x)|~\text{iff}~f^2(x) < g^2(x)$. $\endgroup$ Commented Sep 18, 2018 at 8:27
  • $\begingroup$ @N.F.Taussig Thanks $\endgroup$ Commented Sep 18, 2018 at 16:46

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