# Question about a dense set.

## Main Question or Discussion Point

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?

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?
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.

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].

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

Bacle2
Right, my bad. I jumped in too quickly.

jgens
Gold Member
Right, my bad. I jumped in too quickly.
Me too. My 'proof' of the result actually fails.

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

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

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.

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Bacle2
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.
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.

jgens
Gold Member
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.

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.

jgens
Gold Member
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.

Bacle2