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Question about a dense set.

  1. Aug 15, 2012 #1
    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. jcsd
  3. Aug 15, 2012 #2
    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.
     
  4. Aug 15, 2012 #3

    Bacle2

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    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].
     
  5. Aug 15, 2012 #4

    jgens

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    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.
     
  6. Aug 15, 2012 #5

    Bacle2

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    Right, my bad. I jumped in too quickly.
     
  7. Aug 15, 2012 #6

    jgens

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    Me too. My 'proof' of the result actually fails.
     
  8. Aug 15, 2012 #7
    could we maybe use the fact that there is prime between n and 2n where n is a natural number.
     
  9. Aug 15, 2012 #8
    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.
     
  10. Aug 16, 2012 #9
    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
    [itex] \frac{1}{n}<b-a [/itex] 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
  11. Aug 16, 2012 #10

    Bacle2

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    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.
     
  12. Aug 16, 2012 #11

    jgens

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    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.
     
  13. Aug 16, 2012 #12

    Bacle2

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    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.
     
  14. Aug 16, 2012 #13

    jgens

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    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.
     
  15. Aug 17, 2012 #14

    Bacle2

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    Thanks for the input, Jgens.
     
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