Proving $f(x,n)=\lfloor x \lfloor x \lfloor x \lfloor x ...(\text{n times})\rfloor\rfloor \rfloor \rfloor $ is increasing for $x>0$

Question

$$f(x,n)=\lfloor \underbrace{x \lfloor x \lfloor x \lfloor x \dots\rfloor\rfloor \rfloor \rfloor}_{\text{$n$ times}}$$ with $ n \in \mathbb{N}$. Can we prove this function is increasing for $x>0$? Can it be proved by induction? I tried very hard, but not being able to reach a conclusion.

Answer 1

It can be proved by induction. We have $a \geq b$ gives $\lfloor a \rfloor \geq \lfloor b \rfloor$, which proves the case for $n=1$.

If $f(x,n)$ is increasing, then for $a\geq b$, we have $f(a,n) \geq f(b,n)$, which means $\lfloor f(a,n) \rfloor \ge \lfloor f(b,n)\rfloor$. But the left hand side is $f(a, n+1)$, and the right hand side is $f(b, n+1)$, and hence we conclude that $f(x, n+1)$ is increasing.

Answer 2

Composition of two increasing maps is increasing and $x \mapsto \lfloor x \rfloor$ is increasing. Hence by induction on $n$, $f(x,n)$ is increasing.

Answer 3

Let $a>b$. Hence \begin{align*} & \lfloor a \rfloor \geq \lfloor b \rfloor\\ \implies & a\lfloor a \rfloor>b\lfloor b \rfloor\\ \implies & \lfloor a\lfloor a \rfloor \rfloor \geq \lfloor b\lfloor b \rfloor \rfloor\\ \vdots & \ \ \ \vdots \ \ \ \ \vdots \ \ \ \ \vdots \ \ \ \vdots \ \ \ \ddots \\ \implies & \underbrace{\lfloor a\lfloor a \cdots \lfloor a \lfloor a \lfloor a\rfloor\rfloor \cdots \rfloor\rfloor}_{\text{$n$ times}} \geq \underbrace{\lfloor b\lfloor b \cdots \lfloor b \lfloor b \lfloor b\rfloor\rfloor \cdots \rfloor\rfloor}_{\text{$n$ times}} \end{align*}

Hence the function $f(x,n)$ is incresing