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 $x$ and $y$ uniformly and independently at random from the integers $\{1,\ldots,N\}$, then $x$ and $y$ will be coprime with probability tending to $6/\pi^2$ as $N\to\infty$.

Here is one classic proof: Consider the points $(x,y)\in\{1,\ldots,N\}^2$ such that $\gcd(x,y)=1$. If we scale this picture by $M$, we will have exactly the points $(X,Y)\in\{1,\ldots,MN\}^2$ such that $\gcd(X,Y)=M$. Therefore, whatever the limiting probability $P$ of being coprime is, the probability of having $\gcd = M$ is exactly $P/M^2$. The law of total probability then implies

\[ 1 = \lim_{N\to\infty}\mathbf{P}(\gcd(x,y)\in\mathbf{N}) = P \sum_{M=1} \frac{1}{M^2} = P\,\zeta(2).\]

Another way of imagining a random element on a noncompact group $G$ is to look at the profinite completion $\hat{G}$, which is always compact so always possesses a uniform distribution. Now by the Chinese Remainder Theorem,

\[ \hat{\mathbf{Z}} = \prod_p \mathbf{Z}_p,\]

so choosing an element at random from $\hat{\mathbf{Z}}$ is the same as choosing an element at random from $\mathbf{Z}_p$ for each $p$. The probability that a random element of $\mathbf{Z}_p$ is divisible by $p$ is precisely $1/p$, so the probably that two elements from $\hat{\mathbf{Z}}$ are coprime is precisely

\[ \prod_p (1-1/p^2) = 1/\zeta(2).\]

This does not imply the asymtotic result in $\mathbf{Z}$ 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 $d>1$ and consider positive integers $x_1,\ldots,x_d\leq X$ chosen uniformly at random. Let $G(X)$ be the number of $(x_1,\ldots,x_d)\in[1,X]^d$ such that $\gcd(x_1,\ldots,x_d)=1$, and observe that the number of $(x_1,\ldots,x_d)\in[1,X]^d$ such that $\gcd(x_1,\ldots,x_d) = n$ is precisely $G(X/n)$. Then

\[ \lfloor X\rfloor^d = \sum_{n\geq1} G(X/n) \]

for all $X$. Inverting this relationship,

\[ G(X) = \sum_{n\geq1} \mu(n) \lfloor X/n\rfloor^d. \]

Thus, the probability that $d$ randomly and uniformly chosen integers $x_1,\ldots,x_d\leq X$ are all coprime is exactly

\[P(X) = \sum_{n\geq1} \mu(n) \left(\frac{\lfloor X/n\rfloor}{\lfloor X\rfloor}\right)^d. \]

The $n$th term here is bounded by $n^{-d}$, so by the dominated convergence theorem $P(X)\to\sum_{n\geq1} \mu(n) n^{-d} = \zeta(d)^{-1}$.

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