Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have seen the following simplification:

$$\left|\frac{1}{(-1-\frac{1}{n})^4 - 1}\right| = \frac{1}{\left|-1-\frac{1}{n}\right|^4 - 1}$$

I really don't have a clue why this is possible...

I am sorry but I can't give any ideas because I don't have any.

Thank you for advice.


share|cite|improve this question

2 Answers 2

up vote 2 down vote accepted

Note that $x \geq 0$ if and only if $|x|=x$. So we have $\frac{1}{(-1-\frac{1}{n})^4 - 1}\geq 0$.

Inequlity $(-1-\frac{1}{n})^4 - 1\geq 0$ implies that $1+\frac{1}{n}\geq 1$. Hence for every natural number $n$, we have $|\frac{1}{(-1-\frac{1}{n})^4 - 1}| = \frac{1}{|-1-\frac{1}{n}|^4 - 1}$.

share|cite|improve this answer
That made it clear -thank you! –  FunkyPeanut Jan 25 '14 at 12:08
You mean "$1+\frac{1}{n}\geq 1$ implies that $(-1-\frac{1}{n})^4-1\leq 0$", right? –  JiK Jan 25 '14 at 12:23
NO, $(-1-\frac{1}{n})^4-1\geq 0$ implies that $1+\frac{1}{n}\geq 1$ –  Babak Miraftab Jan 25 '14 at 12:25
But $(-1-\frac{1}{n})^4-1\geq 0$ is what we want to prove to show the equality in the question. –  JiK Jan 25 '14 at 12:27
@JiK i was intersected in for which n, we have $\left|\frac{1}{(-1-\frac{1}{n})^4 - 1}\right| = \frac{1}{\left|-1-\frac{1}{n}\right|^4 - 1}$. We can move to chat if you want more discussion. –  Babak Miraftab Jan 25 '14 at 13:26

Have you tried $\left|f(x)\right|=\sqrt{f(x)^2}$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.