So the question is in the title. And I mean dense in positive real numbers ofcourse. Somehow I cannot grasp if this is very trivial or not. The prime numbers aren't that dense, but are there enough of primes to make the fractions constructed from them dense?
Density of Fractions – Prime Numerator and Denominator
number theoryreal-analysis
Best Answer
You can establish this with the prime number theorem, which can essentially be seen to state that, where $p_n$ is the $n^{th}$ prime number $$\lim_{n\rightarrow\infty}\frac{p_n}{n\log(n)}=1.$$ Using this, suppose that we wish to show that there is a fraction with prime numerator and denominator in the interval $\left(\frac{1}kx,kx\right)$ for $k>1$. Choose some $k-1>\varepsilon>0$ which we will later use for some wiggle-room. Let $N$ be such that, for all $n>N$ we have $$\left(\frac{p_n}{n\log(n)}\right)^2\in\left(\frac{1+\varepsilon}k,\frac{k}{1+\varepsilon}\right)$$ which is possible due to the limit and since squaring is continuous and the right interval is open and contains $1$. Then, we merely need to find a pair $n_1$ and $n_2$ greater than $N$ satisfying $$\frac{n_1\log(n_1)}{n_2\log(n_2)}\in\left(\frac{x}{1+\varepsilon},x(1+\varepsilon)\right)$$ which should be easy enough to find given the identity that $$\lim_{n\rightarrow\infty}\frac{x k \log(x k)}{k \log(k)}=x$$ meaning we can choose $n_1$ and $n_2$ to have ratio near $x$, and so long as they are sufficiently large, this means that the appropriate ratio of $n\log(n)$ too will be near $x$. Then we will have that the ratio of $\frac{p_{n_1}}{p_{n_2}}$ differs from $\frac{n_1\log(n_1)}{n_2\log(n_2)}$ by a factor in $\left(\frac{1+\varepsilon}k,\frac{k}{1+\varepsilon}\right)$, and hence the ratio of the two primes would be in $\left(\frac{1}kx,kx\right)$, as desired. This works on any interval, hence the fractions with prime terms are dense.