Is my proof correct? => Exercise 11.4.41: `Show that ⌊log_2(n)⌋ is Θ(log_2(n))`.

Exercise 11.4.41:

Show that ⌊log_2(n)⌋ is Θ(log_2(n)).

Proof:

Let integer n>=1.

⌊log_2(n)⌋ <= log_2(n) < ⌊log_2(n)⌋ + 1            by definition of floor

So, ⌊log_2(n)⌋ <= log_2(n) and log_2(n) < ⌊log_2(n)⌋ + 1

Notice:

⌊log_2(n)⌋ > log_2(n) - 1
          = log_2(sqrt(n)) + log_2(sqrt(n)) - 1
          = log_2(sqrt(n)) + log_2(sqrt(n)) - log_2(sqrt(n))    for n>=4
	  = log_2(n^(1/2)) + log_2(n^(1/2)) - log_2(n^(1/2))
	  = (1/2)log_2(n)  + (1/2)log_2(n)  - (1/2)log_2(n)
	  = (1/2)[log_2(n) + log_2(n)       - log_2(n)]
	  = (1/2)[log_2(n^2)                - log_2(n)]
	  = (1/2)[log_2[(n^2)/n]]
	  = (1/2)log_2(n)
		  
Let A=1/2, B=1, k=4.

Then Alog_2(n) <= ⌊log_2(n)⌋ <= Blog_2(n) for each integer n>=k.

Therefore, ⌊log_2(n)⌋ is Θ(log_2(n)).

QED

---
Is my proof correct?