Is the product of uniformly distributed numbers, uniformly distributed too? My question is simple, I think.
If we took two random natural numbers $a$ and $b$ uniformly distributed in a specific range $[c,d]$, is $ab$ a uniformly distributed too?
What if $a$ and $b$ are not natural numbers, but real numbers?
What if $a$ is a natural number and $b$ a real number?
 A: NOTE: This analysis only considers positive real numbers $c,d$ and independent random variables $a,b$ both uniformly distributed in $[c,d]$.
In this case $ab$ is NOT uniformly distributed in $[c^2,d^2]$. Note that $(a,b)$ will be uniformly distributed in $S=[c,d]\times[c,d]$ whereas $ab=k\in[c^2,d^2]$ is a hyperbola that passes through the point $(\sqrt k,\sqrt k)$ in $S$.
Analysis: The probability of $ab<k$ for $k<cd$ will be defined by the equation
$$
\begin{align}
A\cdot P(ab<k)&=\int_c^{k/c}\left(\frac{k}{x}-c\right)\ dx\\
&=k\ln(k/c^2)-c(k/c-c)\\
&=k\ln(k/c^2)-k+c^2
\end{align}
$$
where $A=(d-c)^2$ is the area of the region $S$ and the integral above calculates the area that lies beneath the hyperbola $yx=k\implies y=\dfrac{k}{x}$ yet in the region $S$. So the probabilty does not depend linearly on $k$ which means that the distribution cannot be uniform on $[c^2,d^2]$.

Note that for $k=cd$ the hyperbola passes through the cornes $(c,d)$ and $(d,c)$ of the region. So for $k>cd$ more is cut off due to the boundaries of the region and thus $P(ab<k)$ starts growing slower.

Analysing the case $[c,d]=[1,2]$ entirely leads to the distribution function
$$
P(ab<k)=
\begin{cases}
k\ln(k)-k+1&k\in[1,2]\\
k-k\ln(k/4)-3&k\in[2,4]
\end{cases}
$$
which is illustrated here

where the cases where $k>cd=2$ are covered by calculating the purple area as
$$
P(ab>k)=\int_{k/2}^2\left(2-\frac{k}{x}\right)\ dx=k\ln(k/4)-k+4
$$
and therefore $P(ab<k)=1-P(ab>k)=k-k\ln(k/4)-3$.
A: If you treat each of them as random variables, then the product is also a random variable. To see this note that if $X_1, \ldots, X_n$ are random variables, then $(X_1, \ldots, X_n)$ is a random vector on $\Bbb R^n$, hence applyi,g the continuous, and therefore measurable function
$$(X_1,\ldots, X_n)\mapsto X_1\ldots X_n$$
We get the product to be the composition of measurable functions, hence measurable. I.e. The product is also a random variable.
Note that the distribution may very well not be the same as before, but whether or not it counts as a random variable is not a question of the distribution type. Indeed, the other answers cover pretty well that it will not be uniform in general, but the power of this approach is that it does not depend on the choice of probability measure for the exact space, so it does not matter if you are talking about the naturals or the reals or something else entirely, nor does it matter what way you choose to,define "randomness."
A: Here's a simple example:  Suppose $X$ and $Y$ are independent and identically distributed uniform random variables in $[0,1]$.  Let $Z = XY$.  Clearly, we must have $0 \le Z \le 1$.  Thus, if $Z$ were also uniform, we would have for instance $\Pr[Z > 1/2] = 1/2$.  But $XY \le X$ and $XY \le Y$, hence $XY > 1/2$ only if $X > 1/2$ and $Y > 1/2$; i.e., $$\Pr[XY > 1/2] < \Pr[(X > 1/2) \cap (Y > 1/2)] = (1/2)^2 = 1/4.$$  
A: No. Say you are choosing between 1 and 4. If you take every possible result you get:
1*1 = 1
1*2 = 2;
1*3 = 3;
1*4 = 4;
2*2 = 4;
2*3 = 6;
2*4 = 8;
3*3 = 9;
3*4 = 12;

The result 4 is more likely appear than any other number and if the input is truly random as sample size increases it will become uniform.
A: This answer is trivially no for natural numbers. 
Let $c=1$ and $d=2$. 
Now $P(ab=1)=\frac14$, $P(ab=2)=\frac12$, $P(ab=3)=0$, and $P(ab=4)=\frac14$.
