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Rational Sets Intersecting

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Let S={p+q[tex]\sqrt{2}[/tex] : p,q [tex]\in[/tex] Q} and T={p+q[tex]\sqrt{3}[/tex] : p,q [tex]\in[/tex] Q}. Prove that S[tex]\cap[/tex]T = Q.

    2. Relevant equations

    See above.

    3. The attempt at a solution

    I was thinking possible using

    S[tex]\cap[/tex]T=Q
    S + T - S[tex]\cup[/tex]T = Q

    But I have no idea how to combine them? I don't believe it's necessary to first prove S and T are rational individually.
     
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    hmmm... not sure but is there some way to show that

    [tex] p + q\sqrt{2} = s + t\sqrt{3}[/tex]
    iff q=t=0
    (maybe consider the case p = s = 0 first)

    then the intersection reduces to only the rationals?
     
  4. Oct 12, 2009 #3
    Thanks for your response. I was thinking the same thing.

    Would it be a bad idea to transfer this to one of the math discussion threads? It's high-level undergrad and low-level grad type proof writing class.
     
  5. Oct 12, 2009 #4

    lanedance

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    why not try it first... (more volume here & they're not meant for homework)
    say you have
    [tex] p + q\sqrt{2} \in Q}[/tex]
    [tex] s + t\sqrt{3} \in T}[/tex]

    the intersection will be given by elements that satisfy
    [tex] p + q\sqrt{2} = s + t\sqrt{3}[/tex]

    re-arranging
    [tex] p-s = (t\sqrt{3} -q\sqrt{2})[/tex]

    the left is cleary rational, can the righthand side ever be rational? (haven't tried to prove it or find conditions, but probably a good place to start)
     
    Last edited: Oct 12, 2009
  6. Oct 12, 2009 #5

    Dick

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    You should probably try and solve it before you transfer it. lanedance has good suggestions. Square both sides if t and q are both nonzero. Is sqrt(6) rational? Or is sqrt(2) or sqrt(3) if one of t or q happen to be zero?
     
    Last edited: Oct 12, 2009
  7. Oct 12, 2009 #6
    [[tex]\sqrt{6}[/tex] is definitely not rational. I squared both sides and am kind of clueless. It seems like no matter what I do it will always come off as irrational.

    It seems like I need to get t or q to be zero to this to work, huh?
     
  8. Oct 13, 2009 #7

    Dick

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    You want to prove t AND q are both zero. You've got three cases to worry about. t=0, q not zero, q=0, t not zero and q and t both not zero. Yes, you need t=q=0 to make it work. Can you show this?
     
  9. Oct 13, 2009 #8
    Would this be acceptable?

    Suppose 3t2-2[tex]\sqrt{6}[/tex]tq+3q is rational, then q:=0. Hence,

    (p-s)2=3t2

    Both sides are clearly rational. Therefore, S[tex]\cap[/tex]T [tex]\in[/tex] Q.
     
  10. Oct 13, 2009 #9

    lanedance

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    helps if you show what you are doing, but i think thats the plan

    Remember this is the equation that must be sastified by any points in the intersection of S & T.

    so note if u is rational, then so is u^2. So as you're hinting... what limits can you put on t & q, to be in the intersection? will need to consider all the cases as Dick mentioned
     
  11. Oct 13, 2009 #10

    lanedance

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    no - consider the case separately when t,q both non zero, then look at the square
    if one of t,q = 0, there is no need to look at the square

    note that sqrt(2) is not rational but (sqrt(2))^2 is not,
    i think this is because you can show that u^2 irrational implies u is irrational, but u irrational does not imply u^2 is irrational
     
    Last edited: Oct 13, 2009
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