# Let $x\in \mathbb{R}$. Show that $2⌊x⌋ ≤ ⌊2x⌋ ≤ 2⌊x⌋+1$

So far I got the below and is unsure where to go from there...

By definition, $$\left\lfloor x\right\rfloor =m$$ such that

$$m\leq x< m+1\text{ }\forall x\in\mathbb{R}\wedge \forall m\in\mathbb{Z}$$ $$\iff m+n\leq x+n $$\therefore\left\lfloor x+n\right\rfloor = m+n=\left\lfloor x\right\rfloor+n$$ $$\forall x\in\mathbb{R},\exists\left\lfloor x\right\rfloor\in\mathbb{Z}:\left\lfloor x\right\rfloor\leq x < \left\lfloor x\right\rfloor+1$$

Edit after receiving hints:

Let $$⌊x⌋=m$$ with $$m\in \mathbb{Z}$$. Then there exists $$\varepsilon \in [0,1)$$ such that $$x=m+\varepsilon$$. Thus, we have: $$2⌊x⌋=2m ≤ ⌊2x⌋=⌊2m+2\varepsilon⌋=2m+⌊2\varepsilon⌋$$, which proves the first inequality.

2⌊x⌋+1= 2m+1, and ⌊2ε⌋≤1 for ε∈[0,1). Thus,⌊2x⌋= 2m+⌊2ε⌋≤2⌊x⌋+1= 2m+1, which proves the second inequality.

• Have you tried writing $x = \lfloor x \rfloor + y$ for some $y \in [0,1)$? – Eric Towers Oct 3 '20 at 19:13
• Also, "\$\lfloor x \rfloor\$" gives $\lfloor x \rfloor$. – Eric Towers Oct 3 '20 at 19:14
• Let ⌊x⌋=m with m∈Z. Then there exists ϵ∈[0,1) such that x=m+ϵ.Thus, we have: 2⌊x⌋=2m ≤ ⌊2x⌋=⌊2m+2ϵ⌋.This proves the first inequality right? – Anon Oct 3 '20 at 19:22
• – Shaun Oct 3 '20 at 19:28
• you can get integers out of the floor function $\lfloor 2m+2\epsilon\rfloor=2m+\lfloor 2\epsilon\rfloor$. Now which interval does $2\epsilon$ belongs to ? – zwim Oct 3 '20 at 19:28

Hint: Show that $$\lfloor 2m+2\varepsilon\rfloor=2m+\lfloor2\varepsilon\rfloor$$. What values can $$\lfloor2\varepsilon\rfloor$$ take?