How I can simplify this inequality or how I can solve it:

$$\left\lceil\dfrac{\ln(t+2)}{\ln 2}\right\rceil-\left\lfloor\dfrac{\ln(t+1)}{\ln2}\right\rfloor>1$$ where $t$ is a positive integer.

Here $\lceil\cdot\rceil$ and $\lfloor\cdot\rfloor$ are respectively the Ceiling and the Floor functions.

I have no idea to start.


This equality is equivalent to this statement:

There is an integer $n$ strictly between $\log_2(t+1)$ and $\log_2(t+2)$

And this is the same as

$t+1< 2^n<t+2$

This happens if and only if $\lceil t+1\rceil $ is a power of $2$ and $t\notin\Bbb Z$.

  • $\begingroup$ @ ajotatxe: Why you add the last condition in the last line! $\endgroup$ – DER Dec 17 '14 at 18:15
  • 1
    $\begingroup$ @DER If $t$ is integer and $\lceil t+1\rceil=2^n$ then $t+1=2^n$, that is, the power of two is not strictly between $t+1$ and $t+2$. $\endgroup$ – ajotatxe Dec 17 '14 at 18:19

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