Correctness of proof of division in ceiling function My professor asked me to present a proof to my fellow students tomorrow that
$$\left\lceil\frac{n}{2^k}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^{k-1}}\right\rceil}{2}\right\rceil$$
I tried to prove it as follows, but I am not sure if my proof is correct.
Proof
The above is equivalent to $$\left\lceil\frac{n}{2^{k+1}}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Let us say that $x = \left\lceil\frac{n}{2^{k+1}}\right\rceil$ for ease of notation.
It is then clear that $$x-1 \lt \frac{n}{2^{k+1}} \le x$$
We will now prove what is asked using contradiction. Assume that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Then there is an integer $a$ such that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \le a \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
This is equivalent to $$\left\lceil\frac{n}{2^k}\right\rceil \le 2*a \lt \frac{n}{2^k}$$ which can obviously not be true.
QED
Do you see any errors in my proof? Is it clear? I'm a bit doubtful about the last step, where I multiply $a$ by 2.
 A: Your reasoning is a bit shaky in a few places. Let me see if I can help you a bit with the rigor.
To further simplify, let's say $x=\left\lceil\frac{n}{2^{k+1}}\right\rceil$ and $y=\left\lceil\frac{n}{2^k}\right\rceil,$ so we must prove that $x=\left\lceil\frac{y}2\right\rceil$. Since $\frac{n}{2^k}\le y,$ then $\frac{n}{2^{k+1}}\le\frac{y}2,$ and so $x\le\left\lceil\frac{y}2\right\rceil$. Now we can assume that $x\ne\left\lceil\frac{y}2\right\rceil,$ from which it follows that $x<\left\lceil\frac{y}2\right\rceil.$ Then indeed, there is an integer $a$ such that $x\le a<\left\lceil\frac{y}2\right\rceil.$ In fact, $x$ is such an integer! More relevant, though, is the fact that $$\frac{n}{2^{k+1}}\le x<\left\lceil\frac{y}2\right\rceil.$$ Since $x$ is an integer less than $\left\lceil\frac{y}2\right\rceil$, then we have $x<\frac{y}2,$ and so $$2x<y.\tag{$\spadesuit$}$$ On the other hand, $$\frac{n}{2^{k+1}}\le x,$$ so $$\frac{n}{2^k}\le 2x,$$ and since $2x$ is an integer, then $$y=\left\lceil\frac{n}{2^k}\right\rceil\le 2x.\tag{$\clubsuit$}$$ By $(\spadesuit)$ and $(\clubsuit),$ we have our contradiction.
