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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 $m\le f(x)<m+1.$

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 $m\le f(\lfloor x\rfloor)<m+1$ and therefore $$m\le f(x).$$

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

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I think you're on the right track.

I would prove $f(x)<m+1$ 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.

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  • $\begingroup$ I try to use IVT as well but not to $m+1,$ thanks! $\endgroup$ – user575807 Sep 25 '18 at 12:09
  • $\begingroup$ +1. Is there a complete reference including almost all about the floor function? Regards. $\endgroup$ – mrs Sep 25 '18 at 12:12
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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 $x <n$. On the other hand $y$ was chosen minimal with $f(y) = m+1$. Thus $f(x) < f(y)= m+1$.

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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 $\lfloor f(\lfloor x \rfloor) \rfloor = \lfloor f(x) \rfloor$.

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

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