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Thread: Help me understand this proof.

  1. #1
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    Help me understand this proof.

    http://imgur.com/fuoLUOT


    In proofs of contradiction isn't the whole point to have >P be false and have a false statement Q, which makes the proof true? Why in the proof, do we assume >P is true? kind of confuses me..
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  2. #2
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    Re: Help me understand this proof.

    The proof is incorrect. They assume $\sqrt{2}$ is irrational and then go on to describe it as a rational number.
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  3. #3
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    Re: Help me understand this proof.

    I must have wrote this down wrong..or perhaps he said something about irrationality that I forgot. He also left us to do another problem to do at home.

    exercise: Prove by method of contradiction that square root (3) is not a rational number.
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    Re: Help me understand this proof.

    Quote Originally Posted by math951 View Post
    I must have wrote this down wrong..or perhaps he said something about irrationality that I forgot. He also left us to do another problem to do at home.

    exercise: Prove by method of contradiction that square root (3) is not a rational number.
    basically what's going on is you determine a polynomial who's roots are $\sqrt{2}$ or $\sqrt{3}$ or whatever.

    Then you assume these are rational roots (to be contradicted), and use the rational root theorem on the polynomial to determine what the possible rational roots are.

    You then show that none of these possible rational roots are in fact roots and thus it must be that the roots are irrational.
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  5. #5
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    Re: Help me understand this proof.

    Quote Originally Posted by math951 View Post
    http://imgur.com/fuoLUOT
    In proofs of contradiction isn't the whole point to have >P be false and have a false statement Q, which makes the proof true? Why in the proof, do we assume >P is true? kind of confuses me..
    What you posted is the classical proof for the $\sqrt2$ being irrational.
    Assume $\sqrt2$ is rational. That means there are two relatively prime integers $p~\&~q$ such that:
    $\sqrt2=\dfrac{p}{q}$. So $2=\dfrac{p^2}{q^2}$ squaring both sides.
    But that leads us to $2q^2=p^2$ which implies $p=2k$ i.e. $p$ is even.
    So we get $2q^2=4k^2$ or $q$ is also even. How does that contradict that $p~\&~q$ are relatively prime?
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  6. #6
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    Re: Help me understand this proof.

    except that they assumed $\sqrt{2}$ was irrational which was clearly an error, as far as a proof by contradiction goes.
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