I am currently reading $\textit{Scientific Computing: An Introductory Survey}$ by Michael Heath. In Section 5.2.1 when talking about the Bisection Method it is said that the number of iterations $n$ required to achieve a tolerance $tol$ is
$$ n = \left\lceil log_{2}\left( \frac{b-a}{tol} \right)\right\rceil $$
where the starting interval is $[ a, b]$.
$\textbf{I would like to prove this claim but can't figure out where to start}$.
I know I need to prove that $$n - 1< log_{2}\left( \frac{b-a}{tol} \right) \leq n $$
and I also know that the length of the interval after $k$ iterations is $\frac{b-a}{2^{k}}$ but I don't know where to go from there.
Could anyone please give me a hint?
After $k$ steps the size of the interval is $\frac{b-a}{2^k}$, and we want to set this as a tolerance. $$tol=\frac{b-a}{2^k}$$ so $$k=\log_2\frac{b-a}{tol}$$ Since the solution of the above equation can be a non-integer, it does not make sense to have something like 5.3 steps. To guarantee that we get within tolerance, we must choose any integer grater than the value above. And the smallest one is $$\left\lceil\log_2\frac{b-a}{tol}\right\rceil$$