Is $\lfloor \log_{2}(n-1) \rfloor + 1 = \lceil \log_{2}(n) \rceil$ for any $n \in \mathbb{N}$ , $n \geq 2$?

Question

I am trying to solve a problem where I get in the solution that the answer is $\lfloor \log_{2}(n-1) \rfloor + 1$, but the solution book has as an answer the value $\lceil \log_{2}(n) \rceil$. I was wondering are these two exactly the same?

I plotted their difference on wolframalpha and it seems to be $0$.

score 2 · Accepted Answer · answered Nov 07 '17 at 17:57

First check that the following claim is true.

If $x<y$ and there is no integer in the open interval $(x,y)$, then $\lfloor x \rfloor + 1 = \lceil y \rceil.$

Now convince yourself that there is no integer strictly between $\log_2(n-1)$ and $\log_2(n)$.

Is $\lfloor \log_{2}(n-1) \rfloor + 1 = \lceil \log_{2}(n) \rceil$ for any $n \in \mathbb{N}$ , $n \geq 2$?

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