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Showing two certain sets have no elements in common

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Let x and y be irrational numbers such that x-y is also irrational.
    Let A={x+r|r is in Q} and B={y+r|r is in Q}
    Prove that the sets A and B have no elements in common.

    2. Relevant equations



    3. The attempt at a solution

    Since x and y are in A and B, then x+r[itex]_{1}[/itex]-y-r[itex]_{2}[/itex] is irrational.

    Using contradiction, assume their sum is z and z is an element of A[itex]\cap[/itex]B

    x-y=z

    Then x+r[itex]_{1}[/itex]-y-r[itex]_{2}[/itex]=z

    Since z is in both A and B, then z can also be said to be x+r[itex]_{1}[/itex]+y+r[itex]_{2}[/itex]

    Then x+r[itex]_{1}[/itex]-y-r[itex]_{2}[/itex]=x+r[itex]_{1}[/itex]+y+r[itex]_{2}[/itex]

    Solving for the right side,

    0=2(y+r[itex]_{2}[/itex])

    But since the sum is irrational and 0 is not irrational, A and B cannot contain the same elements.


    Am I thinking about this correctly?
     
  2. jcsd
  3. Feb 18, 2012 #2

    HallsofIvy

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    x and y are NOT in A and B, [itex]x+ r_1[/itex] and [itex]y+ r_2[/itex] are. And, from that, it follows that they are rational numbers so the difference is rational, not irrational.

     
  4. Feb 18, 2012 #3
    No, if z is in both A and B, then z=x+r1=y+r2; then you take the subtraction ...
     
  5. Feb 19, 2012 #4
    Did I show that x+r1-(y+r2)=x+r1-(y+r2)?
     
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