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Math Help - Simple Irrational Number Proof

  1. #1
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    Simple Irrational Number Proof

    Prove the statement: If x is irrational and y is irrational, the x + y is irrational.

    Contrapositive: If x + y is rational, then x is rational or y is rational.

    My best guess at an approach is to prove the contrapositive. This is how I would start:

    Proof:
    Suppose x + y is rational. Then, by defn., x + y = p/q for some intergers p,q where q ≠ 0...


    I'm not sure what to do from here. How do I deduce that x is rational or y is rational?
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  2. #2
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    Re: Simple Irrational Number Proof

    What you are trying to prove is FALSE:

    $\sqrt{2} + (-\sqrt{2}) = 0$

    and 0 is certainly rational.
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    Re: Simple Irrational Number Proof

    Wow, duh! I think I'm going to throw up. Thanks..
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    Re: Simple Irrational Number Proof

    About the only "non-trivial" assertion we have along these lines is:

    If $x$ is rational, and $y$ irrational, then $x+y$ is irrational.

    Proof: (by contradiction)

    If $x$ is rational and $y$ is irrational, but $x+y$ is rational, we then have $y = (x+y) - x$ is rational (since the rationals form an abelian group under addition), contradiction.
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  5. #5
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    Re: Simple Irrational Number Proof

    Yea, I have to prove that one too. But I was told to prove the contrapositive here. That is, assume x + y is rational and x is rational and then duduce that y is rational. A little less intuitive but I suppose it still gets the job done.
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  6. #6
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    Re: Simple Irrational Number Proof

    It's the same proof:

    If $x+y$ is rational, and $x$ is rational, then $-x$ is rational

    (for if $x = \dfrac{p}{q}$, then $-x = \dfrac{-p}{q}$ and $-p$ is an integer if $p$ is).

    Then, surely, $y = (x+y) + (-x)$ is rational, because the sum of any two rational numbers is rational:

    let $r_1,r_2$ be rational, so:

    $r_1 = \dfrac{a}{b}, a,b,\in \Bbb Z, b \neq 0$

    $r_2 = \dfrac{c}{d}, c,d,\in \Bbb Z, d \neq 0$

    then $r_1 + r_2 = \dfrac{ad + bc}{bd}$ and clearly $ad + bc$ is also an integer, and so is $bd$ and furthermore, $bd \neq 0$.

    In general, any proof by contradiction (of an implication) can be turned into a "straight proof" of the contrapositive, and vice-versa.
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