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Rational and Irrational Number Set proof.

  1. Jul 1, 2007 #1
    Hello, here is my problem:

    how can i prove that if [tex]a\in\mathbf{Q}[/tex] and [tex]t\in\mathbf{I}[/tex], then [tex]a+t\in\mathbf{I}[/tex] and [tex]at\in\mathbf{I}[/tex]?

    My original thought was to show that neither a+t or at can be belong to N, Z, or Q, thus they must belong to I. However i'm not certain if that train of thought is correct.

    Also, i have a question that says given two irrational numbers s and t, what can be said about s+t and st.

    My original thought he was that nothing can be shown, since it is possible to create numbers that belong to N, Z, Q, or I.

    thanks for clarification.
    Last edited: Jul 1, 2007
  2. jcsd
  3. Jul 1, 2007 #2

    matt grime

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    The rational numbers are a field. Oh, and I is not standard notation, by the way.

    As for the second one, then you can't say anythingabout s or t's rationality. Just construct some examples.
  4. Jul 1, 2007 #3
    whoa, thanks, i would have never gotten that.
  5. Jul 1, 2007 #4


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    For a moment I thought you were trying to prove that the sum of a rational number and an integer was an integer!
  6. Jul 2, 2007 #5
    The first set of problems are standard proofs by contradiction.

    Suppose a is rational and t is irrational and at is rational and a+t is rational.

    Since at is rational, at=m/n for appropriate integral m & n.

    Then, t=m/na, which is rational. But t is irrational by our hypothesis. Therefore, at cannot be rational, hence it is irrational.

    The proof for a+t is similar.
    Last edited: Jul 2, 2007
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