1
$\begingroup$

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

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

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


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.

$\endgroup$
10
  • 1
    $\begingroup$ Have you tried writing $x = \lfloor x \rfloor + y$ for some $y \in [0,1)$? $\endgroup$ – Eric Towers Oct 3 '20 at 19:13
  • $\begingroup$ Also, "\$\lfloor x \rfloor\$" gives $\lfloor x \rfloor$. $\endgroup$ – Eric Towers Oct 3 '20 at 19:14
  • $\begingroup$ 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? $\endgroup$ – Anon Oct 3 '20 at 19:22
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Oct 3 '20 at 19:28
  • $\begingroup$ 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 ? $\endgroup$ – zwim Oct 3 '20 at 19:28
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.