# Prove $2\lfloor x\rfloor \le \lfloor2x\rfloor$

I am trying to prove $$2\lfloor x\rfloor \le \lfloor2x\rfloor$$ which in turn will yield a prove for a homework question. I thought it is a simple prove but I can't figure it out. Maybe it is just a simple thing I overlooked. Any help?

I tried to use $x - \lfloor x\rfloor \ge 0$ to prove $2\lfloor x\rfloor-\lfloor2x\rfloor \le 0$ with no success.

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If $x \ge y,$ then $\lfloor x \rfloor \ge \lfloor y\rfloor$ – cats Feb 18 '13 at 5:06
A generalization is discussed here: math.stackexchange.com/q/514407/11994. – Marnix Klooster Feb 8 '14 at 8:39

So they've all given you significant hints - here's what I was saying: since $x \ge \lfloor x \rfloor,\, 2x \ge 2\lfloor x\rfloor,$ and taking floors of both sides does not change the inequality. Hence, $\lfloor 2x \rfloor \ge 2\lfloor x \rfloor$ since the right hand side is already an integer.

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Hint: For $n \in \mathbb Z$, if $x \in [n,n+\frac 12)$ what is $\lfloor 2x \rfloor$? For $n \in \mathbb Z$, if $x \in [n+\frac 12,n+1)$ what is $\lfloor 2x \rfloor$?

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It's the special case $\rm\:f(x) = 2x\:$ in the following

Theorem $\rm\,\ \lfloor f(x)\rfloor \ge\, f(\lfloor x\rfloor)\$ if $\rm\,f\,$ is an increasing function on $\rm\,\Bbb R\,$ and $\rm\:f(\Bbb Z)\subseteq \Bbb Z.$

$\begin{eqnarray}\rm{\bf Proof}\rm\ &&\rm\quad\ \ x &\ge&\rm\ \ \lfloor x\rfloor \\ \Rightarrow\ &&\rm \ \ f(x) &\ge&\rm\:\ \ f(\lfloor x\rfloor)\quad by\ \ f\ \ increasing \\ \Rightarrow\ &&\rm \lfloor f(x)\rfloor &\ge&\rm\ \lfloor f(\lfloor x\rfloor)\rfloor = f(\lfloor x\rfloor)\ \ \ by\ \ \ f(\Bbb Z)\subseteq \Bbb Z\\ \end{eqnarray}$

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$\lfloor y\rfloor$ is the greatest integer that is no greater than $y$, for any real $y$.

It follows, then, that $\lfloor y\rfloor\leq y<1+\lfloor y\rfloor$ for any real $y$. Try rewriting that inequality chain, subbing in $y=x$ and $y=2x$. Can you see how to manipulate these to get the desired conclusion?

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Write $x = i+\epsilon$, where $i\in \mathbb{Z}$ and $\epsilon\in[0,1)$. Then $$2\lfloor x \rfloor = 2\lfloor i+\epsilon\rfloor = 2i$$ while $$\lfloor 2x \rfloor = \lfloor 2(i+\epsilon)\rfloor = \lfloor 2i+2\epsilon \rfloor = 2i+\lfloor 2\epsilon \rfloor$$ Since $\lfloor 2\epsilon \rfloor = 0$ for $\epsilon\in[0,.5)$, and $\lfloor 2\epsilon \rfloor = 1$ for $\epsilon\in[.5,1)$, we have $$2\lfloor x \rfloor\leq \lfloor 2x \rfloor$$.

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