# Question about a dense set.

1. Aug 15, 2012

### cragar

if i have this set x/y where x and y can be any prime. is this dense on the real line.
and we will allow x to be negative so we can cover the negative side of the reals.
It seems like it would be. how would i prove that x/y can approach any real?

2. Aug 15, 2012

### ImaLooser

Hmmm. It seems very likely that they are dense. Proving it might not be that easy though: there MIGHT be some exception. The first thing I would try is assume there is such an exception and see whether that leads to a contradiction. Or maybe proving that they are dense on the rationals would be good enough, since the rationals are dense on the reals.

3. Aug 15, 2012

### Bacle2

Well, it seems the only problem you may have is with the fractions 1/k.

Don't you basically get the rationals with the set {x/y: x,y prime} except for {1/k}?

After all, you reduce the rationals {p/q: p,q integer} by eliminating common terms;

you actually eliminate redundancy, but you end up with the same thing: so all you

need, I think , is to deal with density in [0,1/2].

4. Aug 15, 2012

### jgens

Not at all. You actually miss a bunch of rational numbers. For example, you miss all rational numbers of the form p/qn where p,q are distinct primes and 1 < n.

5. Aug 15, 2012

### Bacle2

Right, my bad. I jumped in too quickly.

6. Aug 15, 2012

### jgens

Me too. My 'proof' of the result actually fails.

7. Aug 15, 2012

### cragar

could we maybe use the fact that there is prime between n and 2n where n is a natural number.

8. Aug 15, 2012

### Number Nine

That seems to be the right approach to me. I imagine the best way to go about it would be some sort of quasi-constructive proof, where we take an arbitrary real number and use the above fact (or something like) to show that a prime fraction exists in any neighbourhood of x.

9. Aug 16, 2012

### cragar

we could try something similar to how they prove the rationals are dense in the reals.
given any two reals a and b we pick n large enough so that
$\frac{1}{n}<b-a$ then we take the next prime after n and we call it k.
so now we have a prime on the bottom. ok and now we will pick m such that it puts us in between a and b. m is a natural number but not a prime for sure. But we know there is prime between m and 2m. but we may have skipped over b. but we could make the denominator bigger to give us more options for m.

Last edited: Aug 16, 2012
10. Aug 16, 2012

### Bacle2

Right, I have been thinking about a method to (see if it is possible ) to do this. But, with

my method, the (numbers in the) fractions become way too large. So, say, we want to

approximate 1/2 within 1/100: so my idea is : consider the set {n/2n} (clearly not

a ratio of primes ), and consider, for fixed n, the ratios:Pr:= {(n-1)/(2n-1), (n+1)/(2n+1),

(n+1)/(2n-1)(n-1)/(2n+1)}, }. If both numerator and denominator are prime, we get

a good approximation:

| 1/2 - (n-1)/(2n-1) |= 1/(2n-1) -->0 as n becomes large; similar for other ratios. For

example, with 1/2 itself,

we can consider : 29/59=(30-1)/(60-1), 31/61, 73/37 , 157/79 ,... as approximations.

If we know there are infinitely-many n in the prime-ratio set Pr . Problem is that ,

while there are infinitely-many primes, we cannot guarantee right away ( may need an

additional argument) that the primes beyond a certain point are of this type. I think

this is feasible, but I'm being careful given my previous error.

11. Aug 16, 2012

### jgens

Let (x,y) be an open interval in the real numbers with 0 < x < y. The Prime Number Theorem implies that limq→∞[π(qy)-π(qx)] = ∞ and this means that for a sufficiently large prime q, there exists a prime p such that qx < p < qy. Then x < p/q < y as desired.

Unless I messed up on the computation of limq→∞[π(qy)-π(qx)], then this proof should work fine.

12. Aug 16, 2012

### Bacle2

What do you think about my idea:

What do you think about my argument?

My idea is , e.g., for x=1/2, to construct the set :

{1/2,2/4,.....,2n/4n,.....}

Then, to each term in the sequence , we add/subtract 1 to each numerator and

denominator, so , e.g:

1/2 --> 2/3, 2/1 , 0/1, 0/3

2/4 --> 3/5, 3/3, 1/3, 1/5 ,

etc.

Each of these terms is a potential ratio of primes, and a good approximation to 1/2,

with the caveat that primes of this type must be infinitely-many.

13. Aug 16, 2012

### jgens

I thought about a similar approach to yours yesterday, but I came across too many difficulties trying to make everything work. The problem is that for a fixed rational number p/q it is difficult to ensure that there are infinitely many n for which both np-1 and nq-1 are prime. Using something like Dirichlet's Theorem you should be able to show that there are infinitely many n with np-1 prime and infinitely many n with nq-1 prime, but this (unfortunately) says nothing about n for which both of them are prime. There might be a clever way around this or there might be some theorem which guarantees the existence of infinitely many such n, but with my very limited knowledge I obviously know of neither.

14. Aug 17, 2012

### Bacle2

Thanks for the input, Jgens.