Bisection method absolute error I know that $\varepsilon \le 2^{-n-1}(b_0 - a_0)$, how to conclude from this that I need $n = \lfloor log_2{\frac{b_0 - a_0}{2\varepsilon}}\rfloor+ 1$? 
Using logs I get $ n \le log_2{\frac{b_0-a_0}{\varepsilon}} +1$ and not $n = \lfloor log_2{\frac{b_0 - a_0}{2\varepsilon}}\rfloor+ 1$ :(

Hint: Try doing the algebra again (it's incorrect). Then, try bounding the expression you get on the right by adding 1, and think about what n means (number of iteration required to guarantee convergence to within epsilon of the actual zero).
