Probability of a fraction $a/b$ that cannot be simplified Let $a$ and $b$ be random integers chosen independently from the uniform distribution on $\{1, 2,\dotsc, N\}$. As $N \rightarrow \infty$, what is the probability that the fraction:
$$\frac{a}{b}$$
cannot be simplified?
Note: As specified in the comments, the question is the same as this one.
 A: This question is equivalent to ask for the probability that $a$ and $b$ are coprime. If $a$ and $b$ are random integers below $x$, then the probability $P(x)$ that $a$ and $b$ are coprime , satisfies $$\lim_{x\rightarrow\infty} P(x)=\frac{6}{\pi^2}\approx0.6079$$
A: Peter's answer gets to the heart of the matter relatively quickly, but I feel it would also be best to demonstrate where the $6/\pi^2$ comes from, seemingly out of nowhere to the uninitiated.
So, it should be obvious that $a/b$ is in simplest form if and only if $a,b$ are coprime, i.e. $\gcd(a,b) = 1$. Well, what does that mean? It means that $a,b$ share no common prime number factors.
In particular, it means $a,b$ do not share a factor of $2$. For (uniformly randomly chosen) nonzero integers $a,b$ (less than some other number $x$), there is a $1/2$ chance (in the limit $x \to \infty$) each will have a factor of $2$. Thus,
$$P(\text{a,b do not have a mutual factor of 2}) = 1 - \left(\frac 1 2 \right)^2$$
Similarly, it means that they do not share a factor of $3$. There's a $1/3$ chance each will have a factor of three, and thus,
$$P(\text{a,b do not have a mutual factor of 3}) = 1 - \left(\frac 1 3 \right)^2$$
This clearly generalizes. Consider a prime number $p$. There is a $1/p$ chance that $a,b$ each will have it, and in turn
$$P(\text{a,b do not have a mutual factor of p}) = 1 - \left(\frac 1 p \right)^2$$
For $a,b$ to be coprime this needs to be true of all primes $p$. The events are independent, and we accordingly can multiply the respective probabilities for each prime $p$, obtaining
$$P(\text{a,b are coprime}) = \prod_{\text{p prime}} 1 - \left(\frac 1 p \right)^2 = \prod_{\text{p prime}} 1 -  \frac 1 {p^2}$$
This now ties into something known as the Riemann zeta function. There are two formulas typically associated with it: a summation and a product formula. We often focus on the summation formula but can derive the latter; a proof of said derivation can be found here. In any event, we focus on the prime product formula below:
$$\zeta(s) = \prod_{\text{p prime}} \frac{1}{1-p^{-s}}$$
Bearing in mind this is a product, we can do a manipulation:
$$\frac{1}{\zeta(s)} = \prod_{\text{p prime}} 1-\frac 1 {p^{s}}$$
This looks precisely like the formula for our probability of $a,b$ being coprime but with $s$ in lieu of $2$. Indeed, letting $s=2$,
$$P(\text{a,b are coprime}) = \prod_{\text{p prime}} 1 - \frac{1}{p^2} = \frac{1}{\zeta(2)}$$
$\zeta(2)$ is a known value which Euler calculated to be $\pi^2/6$; finding this value is often referred to as the Basel problem. Accordingly,
$$P(\text{a,b are coprime}) = \frac{1}{\pi^2/6} = \frac{6}{\pi^2}$$
The idea also generalizes further. Say you have some group of $n$ integers ($n$ a positive integer). Then the probability that all $n$ are coprime is given by
$$P(\text{all n numbers are coprime}) = \frac{1}{\zeta(n)}$$
