Does $\forall i\in \mathbb Z^+:\left \lfloor log_{2}(i) \right \rfloor = \left \lfloor log_{2}(i+0.999999) \right \rfloor$ Is it true that
$\forall i\in \mathbb Z^+:\left \lfloor log_{2}(i) \right \rfloor = \left \lfloor log_{2}(i+0.999999) \right \rfloor$ ? 
The following is false:
$\forall i\in \mathbb Z^+:\left \lfloor log_{2}(i) \right \rfloor = \left \lfloor log_{2}(i+1) \right \rfloor$ 
(R code):
> for (i in 0:1e9) if ( floor(log2(i)) != floor(log2(i+1))) print(i)
[1] 0
[1] 1
[1] 3
[1] 7
[1] 15
[1] 31
[1] 63
[1] 127
[1] 255
[1] 511
[1] 1023
[1] 2047
[1] 4095
[1] 8191
[1] 16383
[1] 32767
[1] 65535
[1] 131071
[1] 262143
[1] 524287
[1] 1048575
[1] 2097151
[1] 4194303
[1] 8388607
[1] 16777215
[1] 33554431

However
> for (i in 0:1e9) if ( floor(log2(i)) != floor(log2(i+0.999999))) print(i)
[1] 0

Is true for any positive i within the checked range.

Let $$\lfloor\lg(i)\rfloor\le n\land n<\lfloor\lg(i+e)\rfloor$$ for some integer $n$.
This is equivalent to
$$\lg(i)<n+1\land n+1\le\lg(i+e)$$
or, when $i$ is an integer
$$ i\le2^{n+1}-1\land 2^{n+1}\le i+e.$$
By subtraction of the inequalities,
$$1\le e.$$
