I found the proof below in a textbook. I would like to know why it is important for the proof that it uses $\lceil \frac{n}{2} \rceil$ instead of just $\frac{n}{2}$? I know that you can't split into calls which aren't natural numbers, but how do I present this argument formally?The proof:The recursion splits the problem into two sub problems, each with at most $\lceil \frac{n}{2} \rceil$ elements. Therefore, we can apply the Master Theorem with $a = b = 2$. So, $\log_b{a}$ = 1. The cost of splitting is $0$ comparisons, and that of combining is at most $n1$ comparisons. Hence the cost of split/combine is $\Theta(n) = \Theta(n\log_b{a})$, so we are in the second case of the Master Theorem, and therefore the total cost is $\Theta(n\log n)$.
Applying the Master Theorem on Merge sort
algorithm analysis;runtime analysis;sorting;recurrence relation;master theorem