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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?

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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 primes are dense.

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