Let x ∈ R. Show that $2\lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2\lfloor x \rfloor + 1 $

Question

I already checked some questions about this statement, however I can't understand why the first inequality is true.

We know that $\lfloor x \rfloor \leq x$, then $2\lfloor x \rfloor \leq 2x$.

In various questions link1 and link2 they say that $2\lfloor x \rfloor \leq \lfloor 2x \rfloor$ because $2\lfloor x \rfloor$ is integer . I don't understand this conclusion that they made.

Answer 1

$\lfloor {2x}\rfloor$ is the largest integer less than or equal to $2x$. So any integer $n$ such that $n \leq 2x$ satisfies the inequality $n \leq \lfloor {2x}\rfloor$. In particular you can take $n=2\lfloor {x}\rfloor$.

Second inequality: if $n \leq x <n+1$ then $2n\leq 2x <2n+2$ which implies either $2n\leq 2x <2n+1$ or $2n+1\leq 2x <2n+2$. So $\lfloor {2x}\rfloor$ is either $2n$ or $2n+1$. In both cases $\lfloor {2x}\rfloor \leq 2n+1=2\lfloor {x}\rfloor+1$.

Answer 2

$\lfloor x \rfloor$ is an increasing function in $x$ and $\forall x~ \lfloor \lfloor x \rfloor \rfloor = \lfloor x \rfloor$, so $\lfloor x \rfloor \leq x \Rightarrow 2 \lfloor x \rfloor \leq 2x \Rightarrow \lfloor 2 \lfloor x \rfloor \rfloor = 2 \lfloor x \rfloor \leq \lfloor 2x \rfloor.$ That gives you the first inequality.

Also, $\lfloor x \rfloor \leq x$ (with equality if and only if $x \in \Bbb Z$.) Thus, $x \lt \lfloor x \rfloor + 1 \Rightarrow 2x \lt 2 \lfloor x \rfloor + 2 \Rightarrow \lfloor 2x \rfloor \leq \lfloor 2 \lfloor x \rfloor + 2 \rfloor = 2 \lfloor x \rfloor + 2.$ Equality cannot hold so $\lfloor 2x \rfloor \leq 2 \lfloor x \rfloor + 1$.