Time complexity of binary search
Let me start with the first question. Since $\lfloor x \rfloor \leq x$,
$$
\left\lfloor \frac{\mathit{high}+\mathit{low}}{2} \right\rfloor - \mathit{low} \leq
\frac{\mathit{high}+\mathit{low}}{2} - \mathit{low} \leq
\frac{\mathit{high}-\mathit{low}}{2}.
$$
To complete the proof, we use the inequality $\frac{x}{2} \leq \lfloor \frac{x+1}{2} \rfloor$, which holds for integer $x$. To see this, notice that for even $x = 2y$, the inequality reads $y \leq y$, and for odd $x = 2y+1$, it reads $y+1/2 \leq y+1$.
Now for the second question. We want to know what is the smallest integer $r$ such that $r &gt; \log n$. Since $\lfloor x \rfloor &gt; x-1$, we see that $\lfloor \log n \rfloor + 1 &gt; (\log n - 1) + 1 = \log n$. On the other hand, since $\lfloor x \rfloor \leq x$, we see that $\lfloor \log n \rfloor \leq \log n$, and in particular, it is not the case that $\lfloor \log n \rfloor &gt; \log n$. Therefore $\lfloor \log n \rfloor + 1$ is smallest integer $r$ such that $r &gt; \log n$.