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Homework Help: Proof about product of reals

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove or disprove: ∀x ∈ R ∃y ∈ R so that xy ∈ Z.

    (R denotes set of all real nuimbers, Z denotes set of all integers)

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how to attack this question. It seems false, but I can't think of a good counterexample.

    Like If I say take pi, I don't think there is any other number you could multiply pi by to make an integer, but I don't know how to formulate this into a proof that makes sense. I also thought of playing with irrationals since maybe they could provide a counter example, but I'm not sure how to prove something like this. Maybe it's just something I'm not seeing.

    Can anyone help get me started? Thanks!
  2. jcsd
  3. Apr 26, 2010 #2


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    Sometimes, starting simpler is helps. Since you want to prove something for all integers (namely, that there exists an x such that for all y, xy is not that integer), you could first try to get ideas by proving it for one specific integer.
  4. Apr 26, 2010 #3
    Well, try two cases.

    Case 1: x = 0.

    Case 2: x =/= 0.

    What if y = 1/x?
  5. Apr 26, 2010 #4
    well if you give me any integer, multiplying it by any integer will produce an integer

    if you give me an irrational, how can you prove that no number can make an integer.

    it's defined on reals, not integers. i don't think that 1/x example works since it says there exists, as long as there exists one number that can be multiplied with x to make an integer, it's true.
  6. Apr 27, 2010 #5
    i'm so stupid, i figured it out, given any real number, multiply it by 0 and 0 is an integer...

    i wasn't thinking simple enough....
  7. Apr 27, 2010 #6


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    If you're given a nonzero, can you multiply it by something to get, say, 1?
  8. Apr 27, 2010 #7
    like l'Hopital said, 1/x * x = 1, so yeah I guess that works.

    Thanks a lot guys :D
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