Uniformly random number on $[0,1]$ has zero entropy? I am computing then entropy of the uniform distribution on $[0,1]$:
$$ H(X) = \frac{1}{1-0} \int_0^1 (1 - 0) \log 1 \, dx = 0 $$
Does that mean at $X$ has zero entropy?
 A: A random number uniformly distributed on $[0,1]$ (actually, in any interval of positive measure) has infinite entropy - if we are speaking of the Shannon entropy $H(X)$, which corresponds to the average information content of each ocurrence of the variable. Indeed, in a real number on the interval $[0,1]$ I can code all the information of the wikipedia, math.stackexchange.com and more.
The differential entropy $h(X)$ is another thing. It's not a true entropy (it can be zero or negative), and among other things it depends on the scale (so, say, the -differential- entropy of the height of the humans gives different values if I measure them in centimeters or in inches - which is rather  ridiculous).
So, yes, the differential entropy of your variable is zero. But (because the differential entropy is not the Shannon entropy) that means nothing special - in particular, it does not mean that the variable has no uncertainty -like a constant. Actually, the differential entropy of a constant variable (which would correspond to a Dirac delta density) is $-\infty$.
A: I think you're right.
Shannon's definition of the entropy of a continuous random variable $X$ is
$$
H(X)=-E[ln(f(x))]
$$
where $f(x$) is its probability density function. The probability density function for a uniform distribution is $f(x)=1$ if $x \in [0,1]$ and $0$ otherwise. Of course
$$
H(X) = -\int_0^1 ln(1) f(x) dx = 0
$$
in this case.  It even makes sense in a physical system.  If the temperature of a metal bar is equal everywhere, then there's no heating or cooling anywhere in that system.
A: Using the comment of Clement C, divide the interval into n subintrvals.  Then $H_n(X)=\sum_{k=1}^{n} \frac{-log_2(\frac{1}{n})}{n}=log_2(n)$ so that $H(X)=\infty$
