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

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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 $1/2$.

Answer 4

Rewriting the antecedent, we want to show $\frac{|b|}{2} > |b_n-b| \Rightarrow |b_n| > \frac{|b|}{2}$. \begin{align*} \frac{|b|}{2} \color{red}{+|b_n| - \frac{|b|}{2}} = \color{LimeGreen}{|b_n|} &> |b_n-b| \color{red}{+|b_n| - \frac{|b|}{2}} \\ &= |b-b_n| \color{red}{+|b_n| - \frac{|b|}{2}} \\ &\ge |b-b_n+b_n| - \frac{|b|}{2} \\ &= |b| - \frac{|b|}{2} \\ &= \color{LimeGreen}{\frac{|b|}{2}} \end{align*}