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## An inequality involving logarithms and fractional parts [closed]

I would like to prove that if $n = 2^i + 2^j$, with $i \neq j$, then $\lbrace\lg n\rbrace \leq 1 - \lg 3$, where $\lbrace x\rbrace$ is the fractional part of the real $x$ and $\lg n$ is the binary logarithm of the integer $n$. Or find a counterexample. (I have a long proof that hinges on this "lemma.")

The inequality is equivalent to $2n \leq 3 \cdot 2^{\lfloor\lg n\rfloor}$. The use of the standard inequality $x - 1 < \lfloor{x}\rfloor$ is not conclusive.

The problem is clearly related to the binary notation of $n$, which must contain exactly two 1-bits.

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This seems trivial, just assume $i<j$. Then $\lfloor \lg n\rfloor =j$ and $2n≤2^{j+1}+2^j=3⋅2^j$. – thei Apr 18 2011 at 19:42
I disagree. Something equivalent to the first inequality is (3n/2) < 2^(floor(lg n)). That won't work. Which inequality do you really want? Gerhard "Ask Me About System Design" Paseman, 2011.04.18 – Gerhard Paseman Apr 18 2011 at 20:13
Do you disagree that the second inequality is trivial? – thei Apr 18 2011 at 20:19
The first inequality is presumably {lg n} <= lg 3 -1. – thei Apr 18 2011 at 20:23
The second looks plausible, as does your opinion and analysis of the problem, u(g). Gerhard "Dealing With The Unknown (Google)" Paseman, 2011.04.18 – Gerhard Paseman Apr 18 2011 at 20:52
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## closed as too localized by Pietro Majer, Bill Johnson, Gerry Myerson, Aaron Meyerowitz, Willie WongApr 19 2011 at 12:10

My mistake, the inequality is $\lbrace\lg n\rbrace \leq \lg 3 - 1$. It was indeed trivial, sorry and thanks to "unknown".