I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of looking at Question 1.

Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.

First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$

Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that

$$p\in [rKn, rK(n+1)],$$ $$ q\in [sKn, sK(n+1)].$$

Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$

We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$

Edit: small rewording.

I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of looking at Question 1.

Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.

First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$

Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that

$$p\in [rKn, rK(n+1)],$$ $$ q\in [sKn, sK(n+1)].$$

Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$

We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$

Edit: small rewording.

I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of solving Question 1.

Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.

First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$

Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that

$$p\in [rKn, rK(n+1)],$$ $$ q\in [sKn, sK(n+1)].$$

Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$

We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$

I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of looking at Question 1.

Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.

First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$

Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that

$$p\in [rKn, rK(n+1)],$$ $$ q\in [sKn, sK(n+1)].$$

Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$

We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$

Edit: small rewording.