Thank you and Skeeter both. Yet for some reason that usual trick still does work for me.
I do see the objection. The inequality is less-than or equal in 11.4.2 but in the one on which it is used.
However, it is true that if $2^k<x<2^{k+1}$ then $k<\log_2(x)<k+1$; the log function is monotone.
The because the is no integer between integers $k~\&~k+1$ we get $\left\lfloor\log_2(x)\right\rfloor=k$.
I am not familiar with this concept. Would you mind fleshing this out for me? I looked on wikipedia and they define monotone with more concepts I don't know.
Usually to be able to apply the floor function and get k, we need something of the form
.
Clearly that is not the case here. I understand your saying there is something special about the log function that allows us to obtain k. Could you explain what that something is.
Monotone, simply means to preserve order. Non-increasing or non-decreasing function.
This really is not special to the log function. It has everything to do with the floor function. We prove that for any real number $t$ there is a unique integer $\left\lfloor t \right\rfloor$ that has the property that $\Large{\left\lfloor t \right\rfloor\le t<\left\lfloor t \right\rfloor+1}$.
There are nine integers from $2^3=8$ to $2^4=16$
thought about it some more, you can switch from < to <= because of a simple mathematical fact,
if something is < then its surely less than or equal. 2 < 6 and its also true that 2 <= 6. So we can apply the theorem with no problems.
Why didn't you just say this? I realized I want to react more strongly to this. You have no business assessing my mathematical capabilities. Furthermore, it is not helpful to me developing my skills. It only served to make me feel discouraged.