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

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!

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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

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.

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@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

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 $1/2$.

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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

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

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