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

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

Thanks!

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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
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Floor(x) = x - frac(x) may help. –  barrycarter Aug 29 '11 at 14:42
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2 Answers 2

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 $1/2 \le \{x\} < 1$.

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

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