A famous result: Two positive integers chosen at random will be coprime with probability .
One has to say what one means by this, since there is no uniform distribution on the positive integers. What we mean is that if we choose and uniformly and independently at random from the integers , then and will be coprime with probability tending to as .
Here is one classic proof: Consider the points such that . If we scale this picture by , we will have exactly the points such that . Therefore, whatever the limiting probability of being coprime is, the probability of having is exactly . The law of total probability then implies
Another way of imagining a random element on a noncompact group is to look at the profinite completion , which is always compact so always possesses a uniform distribution. Now by the Chinese Remainder Theorem,
so choosing an element at random from is the same as choosing an element at random from for each . The probability that a random element of is divisible by is precisely , so the probably that two elements from are coprime is precisely
This does not imply the asymtotic result in mentioned above, but it is a nice way of thinking about it.
EDIT (28/4/12): As pointed out to me by Freddie Manners, the “classic proof” suggested above suffers from the fatal defect of not proving that the limit exists. Here is an alternative proof which does prove that the limit exists. More generally, let and consider positive integers chosen uniformly at random. Let be the number of such that , and observe that the number of such that is precisely . Then
for all . Inverting this relationship,
Thus, the probability that randomly and uniformly chosen integers are all coprime is exactly
The th term here is bounded by , so by the dominated convergence theorem .