Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

May I know the standard proof technique to prove such kind of inequalities.

$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1$


share|cite|improve this question
Hint: What is $\lfloor 2x\rfloor$, when A) $x$ is in the interval $[n,n+1/2)$, B) when $x$ is in the interval $[n+1/2,n+1)$? – Jyrki Lahtonen Aug 29 '11 at 9:22
Floor(x) = x - frac(x) may help. – barrycarter Aug 29 '11 at 14:42
up vote 2 down vote accepted

By Hermite's identity, we know that $ \lfloor x \rfloor + \lfloor x \rfloor \le \lfloor x \rfloor + \lfloor x + \frac 12 \rfloor = \lfloor 2x \rfloor \le \lfloor x \rfloor + \lfloor x + 1 \rfloor$. Alternatively, as already mentioned, you can use casework on $\{x\} := x - \lfloor x \rfloor$, in particular when $0 \le \{x\} < 1/2$ and when $1/2 \le \{x\} < 1$.

share|cite|improve this answer
Hermite's Identity is a nice way to solve this. Thanks. But I am not able to understand the other method.. – Maverickgugu Aug 29 '11 at 8:28
I can get case 1 when $0 \le \{x\} < 1/2$. but am not able grasp the other case.. – Maverickgugu Aug 29 '11 at 8:36
I'd imagine you figured it out by now, but suppose $x = n + r$ where $n \in \mathbb{Z}$ and $1/2 \le r < 1$. Then $2\lfloor x \rfloor = 2n, \lfloor 2x \rfloor = 2n+1, 2\lfloor x \rfloor + 1 = 2n+1$. – azjps Aug 30 '11 at 7:23
Thanks!! It looks so obvious now.. :) – Maverickgugu Aug 31 '11 at 13:12

Hint: let $n = \lfloor{x\rfloor}$, so $n \le x < n+1$. What about $2x$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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