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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Unfortunately I am stuck on one step of a proof for an algebraic limit theorem, specifically:

Why is it exactly that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$ ?

If this doesn't make sense without more context, please let me know. Otherwise, thank you for your help!

share|cite|improve this question
My recommendation is to first draw a picture of the number line :-). Make that two pictures: one with $b$ positive, and the other with $b$ negative. Where can $b_n$ lie? – Srivatsan Sep 20 '11 at 10:30
up vote 4 down vote accepted

There's a version of the triangle inequality that says $\big| \,|x| - |y| \,\big| \leq |x - y|$ for all $x$ and $y$. So you have $$\big|\,|b| - |b_n|\,\big| \leq |b - b_n| < {|b| \over 2}$$ So in particular you have $$|b| - |b_n| < {|b| \over 2}$$ Rearranging this expression gives what you want.

share|cite|improve this answer
@ghshtalt This version of triangle inequality is often called the reverse triangle inequality. – Srivatsan Sep 20 '11 at 10:34
Thanks to all of you for your help! Part of what I overlooked is that $|b_n - b| = |b - b_n|$... I think I was trying to do this, but I kept getting $\frac{3|b|}{2}$ or something... Anyway, I think it makes sense to me finally :) – ghshtalt Sep 20 '11 at 15:49

Hint: use $|x|\le|y|+|x-y|$ hence $|y|\ge |x|-|x-y|$ for suitable values of $x$ and $y$.

share|cite|improve this answer

You may divide everything by $b$, then this is equivalent to the statement: $|x-1|\lt 1/2$ implies $x\gt1/2$.

In other words: if $x$ is at a distance less than $1/2$ from $1$ then $x$ must be greater than $1/2$.

share|cite|improve this answer
Nice idea, but you should divide by $|b|$ and not $b$. (Also if I should nitpick, what if $b$ happens to be zero? :-)) – Srivatsan Sep 20 '11 at 13:02
@Srivatsan Narayanan: Sure (and well...:) – AD. Sep 20 '11 at 13:10
If $b=0$ the implication is trivially true. – André Nicolas Sep 20 '11 at 16:15
@André Nicolas: Sure but then it is just boring words flying around... :) – AD. Sep 20 '11 at 18:42

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.