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While solving the problem in my other question, I've come across an identity, which I've empirically found to be true, but can't seem to prove.

Here I've simplified the problem a bit, so that the formulas don't look too bulky. So, define a function:

$$f(n)=\begin{cases} \left\lceil\frac n{1+\alpha}\right\rceil+\left\lfloor\alpha\left\lceil\frac n{1+\alpha}\right\rceil\right\rfloor & \mathrm{if}\;\; \left\lceil\frac n{1+\alpha}\right\rceil<\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil,\\ \left\lceil\frac{\alpha n}{1+\alpha}\right\rceil+\left\lfloor\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil\right\rfloor& \mathrm{otherwise.} \end{cases}$$

The statement to prove is: if $\alpha\in\mathbb R \setminus \mathbb Q$ and $n\in\mathbb N$, then $$f(n)=n.$$

How can I prove this?

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Consider the first case:

$$\left\lceil\frac n{1+\alpha}\right\rceil<\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil.$$

From this we can say

$$\left\lceil\frac n{1+\alpha}\right\rceil+\left\lfloor\alpha\left\lceil\frac n{1+\alpha}\right\rceil\right\rfloor<\left\lceil\frac n{1+\alpha}\right\rceil+\left\lceil\frac {\alpha n}{1+\alpha}\right\rceil.$$

In the second case with

$$\left\lceil\frac n{1+\alpha}\right\rceil>\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil$$

we'll have

$$\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil+\left\lfloor\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil\right\rfloor<\left\lceil\frac {\alpha n}{1+\alpha}\right\rceil+\left\lceil\frac n{1+\alpha}\right\rceil.$$

Thus, for $\left\lceil\frac n{1+\alpha}\right\rceil\alpha\ne\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil$ (which is always the case for irrational $\alpha$) we have that

$$f(n)<\left\lceil\frac {\alpha n}{1+\alpha}\right\rceil+\left\lceil\frac n{1+\alpha}\right\rceil\le\left\lceil\frac {\alpha n}{1+\alpha}+\frac n{1+\alpha}\right\rceil+1=n+1,$$

i.e. we have an upper bound on $f$: $$f(n)<n+1.\tag1$$

On the other hand, in any case we can say for the upper expression in definition of $f$, just using the relations between floor and ceiling functions:

$$\left\lceil\frac{n}{1+\alpha}\right\rceil+\left\lfloor\alpha\left\lceil\frac{n}{1+\alpha}\right\rceil\right\rfloor \ge \left\lceil\frac{n}{1+\alpha}\right\rceil+\left\lceil\alpha\left\lceil\frac{n}{1+\alpha}\right\rceil\right\rceil-1\ge\\ \ge \left\lceil\frac{n}{1+\alpha}+\alpha\left\lceil\frac{n}{1+\alpha}\right\rceil\right\rceil-1 > \left\lceil\frac{n}{1+\alpha}+\alpha\frac{n}{1+\alpha}\right\rceil-1=n-1,$$

where we get strict inequality from the fact of irrationality of $\frac n{1+\alpha}$.

Similarly for lower expression we have

$$\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil+\left\lfloor\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil\right\rfloor \ge \left\lceil\frac{\alpha n}{1+\alpha}\right\rceil+\left\lceil\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil\right\rceil-1 \ge \left\lceil\frac{\alpha n}{1+\alpha}+\frac1\alpha\left\lceil\frac{\alpha n}{1+\alpha}\right\rceil\right\rceil-1 > \left\lceil\frac{\alpha n}{1+\alpha}+\frac1\alpha\frac{\alpha n}{1+\alpha}\right\rceil-1=n-1.$$

Thus, we now also have a lower bound on $f$:

$$f(n)>n-1.\tag2$$

Combining $(1)$ and $(2)$, we have

$$n-1<f(n)<n+1$$

Since $f(n)$ and $n$ are integral, we can rewrite this as

$$n\le f(n)\le n,$$

which implies

$$f(n)=n,$$

Q.E.D..

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