# Proof of ceil(log2(n)) = ceil(log2(n+1))

How do you prove ceil(log2(n)) = ceil(log2(n+1))?

I said:

Let x = log2(n), then 2^x = n. It follows that x is not an integer,
since n is odd and if x were an integer, 2^x would be even. Now,
suppose y = log2(n+1), then y may be even since n+1 is even. So,
either (1) y is an integer with ceil(x) = y or (2) y is not an
integer, and ceil(x) = ceil(y). It follows that ceil(x) = ceil(y)
(since ceil(y) = y anyway). Therefore, ceil(log2(n)) = ceil(log2(n+1)).

Is this correct or an incomplete proof?

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