Take the 2-minute tour ×
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.

Given that $0\leq \epsilon\leq 1$, $a,b>0$, how to prove $$\frac{1}{(1+\epsilon)^2}\leq \frac{a}{b}\leq (1+\epsilon)^2\implies |a-b|\leq 16\epsilon b?$$

share|improve this question
Welcome to math.SE. Given that you're new here, I'll try to give you some advice: 1. If your question is homework, tag it as such. This will hint the users that you need guidance or hints, rather than a full solution. 2. It is good to include your work on the problem and where it comes from, since it will hint users on which tools to use when solving the problem. –  Pedro Tamaroff Apr 10 '12 at 21:50

1 Answer 1

wlog assume $a \ge b$ (Why?)

Then we have

$$1 \le \frac{a}{b} \le 1 + 2\epsilon + \epsilon^2$$


$$0 \le \frac{a}{b} - 1\le 2\epsilon + \epsilon^2$$

For $0 \le \epsilon \le 1$, it is easy to see that $2\epsilon + \epsilon^2 \le 16 \epsilon$

In fact, you can replace $16$ by $3$.

share|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.