# Proof simplification and absolute value

Could someone verify the following absolute value inequalitiy:

$b_k < |\epsilon + L| \iff b_k < \epsilon + L$ and $bk < -(\epsilon+L) \iff -bk > (\epsilon +L)$

All together:

$-b_k > (\epsilon +L) > b_k$

Is there any further way to simplify this?

-
What if $b_k=\epsilon=L=1$? –  David Mitra Dec 11 '12 at 19:31
As David Mitra points out, but more nicely, the end inequality doesn't make sense. Let's get rid of the stuff in the middle, you are saying $b_k \lt -b_k$. Let $b_k=17$. –  André Nicolas Dec 11 '12 at 19:45

I think it is correct. Further simplification would be dependent on whether you are seeking bounds for $L$ or for $\epsilon$:

You have $bk < (L+\epsilon) < -bk$, which is equivalent to

• either $bk -L < \epsilon < -bk-L$ (in other words, $\epsilon \in (bk-L, -bk+L)$)
• or, equivalently, $bk - \epsilon < L < -bk - \epsilon$ (in other words, $L \in (bk-\epsilon, -bk+\epsilon)$).
-