# Proving $\lfloor f(\lfloor x\rfloor)\rfloor=\lfloor f(x)\rfloor$

Let $$f:\mathbb{R}\to\mathbb{R}$$ be a continuous increasing function such that $$\forall x\in\mathbb{R} \;f(x)\in\mathbb{Z}\implies x\in\Bbb{Z.}\quad (1)$$

I would like to prove that $$\lfloor f(\lfloor x\rfloor)\rfloor=\lfloor f(x)\rfloor.$$

Denote $$m=\lfloor f(\lfloor x\rfloor)\rfloor$$, if I am not mistaken I just need to prove that $$m\le f(x)

I get that $$m\le f(x):$$ We have $$\lfloor x\rfloor\le x<\lfloor x\rfloor+1$$ so that $$f(\lfloor x\rfloor)\le f(x)\le f(\lfloor x\rfloor+1).$$ By definition of foor function we also have $$m\le f(\lfloor x\rfloor) and therefore $$m\le f(x).$$

I need to prove that $$f(x) Not sur how can I do that, I didn't (yet) the fact that $$f$$ is continuous and the property $$(1).$$

I think you're on the right track.

I would prove $$f(x) by contradiction.

Assume that $$f(x)\geq m+1$$. We also know that $$f(\lfloor x \rfloor) < m+1$$. Putting these two facts together we know (by continuity of $$f$$ and using the intermediate value theorem) that there must exist some $$x_0\in[\lfloor x \rfloor, x]$$ such that $$f(x_0)=m+1$$.

We also know that $$x_0\neq \lfloor x \rfloor$$, since $$f(x_0)\neq f(\lfloor x \rfloor)$$, and we know from the property of $$f$$, that $$x_0$$ is an integer. We now separate two options:

1. $$x_0 = x$$, in which case $$x$$ is an integer, and $$\lfloor x \rfloor = x$$, a contradiction since we know $$x_0\neq \lfloor x \rfloor$$
2. $$x_0 \neq x$$, which means $$x_0$$ is integer between $$\lfloor x\rfloor$$ and $$x$$, a contradiction.

In other words, when we increase the value of $$x$$ in the expression $$f(x)$$ (starting from $$\lfloor x\rfloor$$), we cannot hit the value $$m+1$$ before the input $$x$$ increases to the next integer.

• I try to use IVT as well but not to $m+1,$ thanks! – user575807 Sep 25 '18 at 12:09
• +1. Is there a complete reference including almost all about the floor function? Regards. – mrs Sep 25 '18 at 12:12

Let us prove that $$f(x) < m+1$$. If $$f(y) < m+1$$ for all $$y \ge x$$, then we are done. Note that we can suppose that $$x \notin \mathbb{Z}$$, since the statement is true, if $$x$$ is an integer. In the other case, there exists a first $$y \ge x$$ with $$f(y) = m+1$$. The condition implies that $$y =n$$ for some $$n \in \mathbb{N}$$. Because we have assumed that $$x$$ is not an integer, we must have $$x . On the other hand $$y$$ was chosen minimal with $$f(y) = m+1$$. Thus $$f(x) < f(y)= m+1$$.

Another way to see the proof :

Let $$x \in \mathbb{R}$$. The function $$f$$ is increasing and continuous, so $$f(]\lfloor x \rfloor, x ]) = ]f(\lfloor x \rfloor), f(x) ]$$. If this interval contains an integer, that means that there exists $$y \in ]\lfloor x \rfloor, x ]$$ such that $$f(y) \in \mathbb{Z}$$, so by the property of your function, $$y \in \mathbb{Z}$$. That's impossible because $$]\lfloor x \rfloor, x ]$$ contains no integer.

So $$]f(\lfloor x \rfloor), f(x) ]$$ contains no integer, so $$\lfloor f(\lfloor x \rfloor) \rfloor = \lfloor f(x) \rfloor$$.

• Nicely done, thank you. – user575807 Sep 25 '18 at 12:10