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.
 A: 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.
A: 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 $\lfloor 2x \rfloor$?
A: 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 $\rm\:f(\Bbb Z)\subseteq \Bbb Z.$
$\begin{eqnarray}\rm{\bf Proof}\rm\  &&\rm\quad\ \ x &\ge&\rm\ \ \lfloor x\rfloor \\
\Rightarrow\  &&\rm \ \ f(x) &\ge&\rm\:\ \  f(\lfloor x\rfloor)\quad by\ \ f\ \ increasing  \\
\Rightarrow\  &&\rm \lfloor f(x)\rfloor &\ge&\rm\  \lfloor f(\lfloor x\rfloor)\rfloor = f(\lfloor x\rfloor)\ \ \ by\ \ \ f(\Bbb Z)\subseteq \Bbb Z\\
\end{eqnarray}$
A: $\lfloor y\rfloor$ 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?
A: 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 $$2\lfloor x \rfloor\leq \lfloor 2x \rfloor$$.
