# Need help in manipulating rational absolute value inequalities

• bamajon1974
In summary: You can also simplify your work by taking round numbers as bounds, like I showed in my previous post.
bamajon1974
Homework Statement
For context, this question is not for homework but in reference to bounding terms for epsilon-delta limit proofs. I know you can bound the numerator and denominator separately but I really would like to know how manipulate the quotient together as I have forgotten some elementary algebra.
Relevant Equations
How does one manipulate rational absolute inequalities?

For example, I want to transform the absolute value inequality ##|x-3|<1## to ##\frac{|x+3|}{5x^2}<A \ ##, for some number ##\text{A}##, to find an upper and lower bound on the latter term using the constraint in the first term, and not sure what to do with the denominator and changing inequality direction.

I can expand the absolute value inequality as follows: ##|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7. ## How do I introduce the ##5x^2## term in the denominator? Dividing all three sides by ##5x^2## would add a variable to the numbers and possibly change signs. Not sure where to go from here.

For context, this question is not for homework but in reference to bounding terms for epsilon-delta limit proofs. I know you can bound the numerator and denominator separately but I really would like to know how manipulate the quotient together as I have forgotten some elementary algebra.

Thanks!

Last edited:
Your lower bound for ##x## implies an upper bound for ##\frac 1{5x^2}##

bamajon1974 said:
Dividing all three sides by 5x2
Sounds like a bad idea. since ##5x^2## is positive (*), multiplying seems more logical: ##|x+3|<5x^2## and you have to study two cases:$$x\le-3 \ \& \ 5x^2 +x+3>0 \quad \Rightarrow x\le -3$$and$$x>-3 \ \&\ 5x^2-x-3 >0$$which requires solving a quadratic equation

(*) Edit: Oops, have to exclude ##x=0##

##\ ##

PeroK said:
Your lower bound for ##x## implies an upper bound for ##\frac 1{5x^2}##

PeroK said:
Your lower bound for ##x## implies an upper bound for ##\frac 1{5x^2}##
I made a mistake and ##\frac{|x+3|}{5x^2}<A## for some number A to be determined, not 1.

I am guessing you would have to consider the numerator and denominator separately. So for the numerator, ##5<x+3<7## and for the denominator, ##\frac{1}{80}<\frac{1}{5x^2}<\frac{1}{20}.##

Then, can I combine the two expressions, i.e., ##\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20}## and, to find an upper bound? That is, to produce ##-\frac{7}{20}<\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies \frac{|x+x|}{5x^2}<\frac{7}{20}?## It seems arbitrary to separate the numerator and denominator.

Thanks.

BvU said:
Sounds like a bad idea. since ##5x^2## is positive (*), multiplying seems more logical: ##|x+3|<5x^2## and you have to study two cases:$$x\le-3 \ \& \ 5x^2 +x+3>0 \quad \Rightarrow x\le -3$$and$$x>-3 \ \&\ 5x^2-x-3 >0$$which requires solving a quadratic equation

(*) Edit: Oops, have to exclude ##x=0##

##\ ##
I made a mistake in the original post and corrected. What I want to do is use $$|x-3|<1$$ to bound ##\frac{|x+3|}{5x^2}## and equal to some number.

bamajon1974 said:
I made a mistake in the original post and corrected. What I want to do is use $$|x-3|<1$$ to bound ##\frac{|x+3|}{5x^2}## and equal to some number. Does that clarify?

bamajon1974 said:
It seems arbitrary to separate the numerator and denominator.
What's not clear is why this is an issue. Bounding both is precisely what you want to do.

PeroK said:
What's not clear is why this is an issue. Bounding both is precisely what you want to do.
Ok then I should clarify...I understand the process of bounding. The question I have is the algebra involved in bounding, i.e., is it legal to find the bounds f the numerator and denominator separately or as a single quotient. In the end, I found my bounds but the issue was just a side question I asked,

bamajon1974 said:
Ok then I should clarify...I understand the process of bounding. The question I have is the algebra involved in bounding, i.e., is it legal to find the bounds f the numerator and denominator separately or as a single quotient. In the end, I found my bounds but the issue was just a side question I asked,
Remember that ##|a| |b|=|ab|##.

Therefore, ## \displaystyle \left| \frac{1}{5x^2} \right| |x+3| = \left| \frac{x+3}{5x^2} \right| ##.

bamajon1974 said:
is it legal to find the bounds of the numerator and denominator separately or as a single quotient.
Whatever works best. As long as you follow the rules of algebra there's no issue.

By the way, it's often possible to simplify these calculations by taking a round number as the bound. E.g. If we have ##2 < x < 4##, then ##\frac 1 {5x^2} < \frac 1 {20} < 1##.

And, as ##| x + 3| < 7##, we have:$$|x - 3| < 1 \ \Rightarrow \ \frac{|x + 3|}{5x^2} < 7$$Which is just as good as ##\frac 7 {20}##.

SammyS said:
Remember that ##|a| |b|=|ab|##.

Therefore, ## \displaystyle \left| \frac{1}{5x^2} \right| |x+3| = \left| \frac{x+3}{5x^2} \right| ##.
I think I see now. Equivalently, ##|a/b| = |a|/|b|.## In the context of my question, it is ##|a|/b##, since ## |a|=|x+3|## and ##|b|## is ##|5x^2|=5x^2## which always positive. Given that ##|x-3|<1## and ##|a/b| = |a|/|b|.##, I can find upper and lower bounds of the numerator and denominator separately and combine them back into one single absolute value inequality, right?

Maybe something like this?

If $$|x-3|<1,$$ then $$|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7$$ for the numerator and $$\frac{1}{4}<\frac{1}{x}<\frac{1}{2} \implies \frac{1}{16}<\frac{1}{x^2}<\frac{1}{4} \implies \frac{1}{80}<\frac{1}{5x^2}<\frac{1}{20}$$ for the denominator.

Dividing produces $$\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies -\frac{7}{20}<\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies \frac{|x+3|}{5x^2}<\frac{7}{20}.$$

PeroK
Yes, that's the idea.

## What is a rational absolute value inequality?

A rational absolute value inequality is an inequality that involves a rational expression and an absolute value function. It can be written in the form of |f(x)| < g(x), where f(x) is a rational expression and g(x) is a real number.

## How do I solve a rational absolute value inequality?

To solve a rational absolute value inequality, you first need to isolate the absolute value expression. Then, you can split the inequality into two separate inequalities, one with a positive sign and one with a negative sign. Solve each inequality separately and combine the solutions to get the final solution set.

## What are the key properties of rational absolute value inequalities?

The key properties of rational absolute value inequalities include the fact that the absolute value function always returns a positive value, and that the inequality may have more than one solution due to the absolute value symbol.

## Why is it important to manipulate rational absolute value inequalities?

Manipulating rational absolute value inequalities is important because it allows us to find the solutions to the inequality and understand the behavior of the rational expression within the absolute value function. It also helps in solving real-world problems involving rational expressions and absolute values.

## What are some common mistakes to avoid when manipulating rational absolute value inequalities?

Some common mistakes to avoid when manipulating rational absolute value inequalities include forgetting to consider both the positive and negative cases, not simplifying the rational expression before solving, and not checking the solutions in the original inequality to ensure they are valid.

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