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Proof involving quantifiers and rationality/irrationality

  1. Jan 22, 2009 #1
    I have no idea how to begin this proof.

    1. The problem statement, all variables and given/known data
    Prove that for every rational number z, there exist irrational numbers x and y such that x + y = z.

    3. The attempt at a solution
    I can't think of even a way to start this proof...it's just quite obvious that the sum of two irrational numbers will equal a rational number, somehow.... Please give me some pointers to start this proof with.
     
  2. jcsd
  3. Jan 22, 2009 #2

    epenguin

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    "it's just quite obvious that the sum of two irrational numbers will equal a rational number" that, which is the statement converse of what you are asked, is not obvious and not true in general.

    For the problem, you do know of a rational number A and an irrational x one that is less than it. Can you prove that (A - x) is also irrational?

    Then you can do it for any pair of rationals.
     
  4. Jan 22, 2009 #3
    Okay, so suppose A is irrational number. Then let x = 1/2z – a. And let y = 1/2z + a. Then x + y = z, and x and y are irrational.
     
  5. Jan 22, 2009 #4

    epenguin

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    Yes, understand you mean x = z/2 – a. And let y = z/2 + a

    But you have to prove explicitly that if z is irrational, so is z/2 + a etc.
     
  6. Jan 22, 2009 #5
    Okay, so:

    Let y = z - x. Then suppose y is rational, then m/n - x = a/b, in which case x = m/n - a/b = (mb - na) / (nb), which is a rational number. But this is a contradiction to the fact that x is an irrational number. Thus the statement (for every rational number z, there exist irrational numbers x and y such that x + y = z) is true.

    I essentially just took your idea and developed it. Thanks for your help. :)
     
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