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Math Help - Proof by contradiction

  1. #1
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    Proof by contradiction

    I need to prove that SqRt of 3 is irrational using a proof my contradiction and the lemma: an interger n is a multiple of 3 iff n^2 is a multiple of 3

    I know to assume the SqRt of 3 is rational. Then it equals a/b, where a,b are integers and b doesnt equal to 0..not too sure where to go after that
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    Let’s prove a more general result: If n is a non-square positive integer then \sqrt{n} is irrational.
    So suppose for the purpose of proof by contradiction that \sqrt{n} is rational.
    Using the well order of the positive integers, there is a first positive integer K such that K\sqrt n  \in \mathbb{Z}^ +  .
    Using the floor function: 0 < \sqrt n  - \left\lfloor {\sqrt n } \right\rfloor  < 1.
    Now 0 < K\sqrt n  - K\left\lfloor {\sqrt n } \right\rfloor  < K and \left[ {K\sqrt n  - K\left\lfloor {\sqrt n } \right\rfloor } \right] \in \mathbb{Z}^ +  . (WHY?)
    However, \left[ {K\sqrt n  - K\left\lfloor {\sqrt n } \right\rfloor } \right]\left( {\sqrt n } \right) \in \mathbb{Z}^ + which is a clear contradiction to K being minimum.
    Last edited by Plato; April 9th 2009 at 04:03 PM.
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    Quote Originally Posted by nikie1o2 View Post
    I need to prove that SqRt of 3 is irrational using a proof my contradiction and the lemma: an interger n is a multiple of 3 iff n^2 is a multiple of 3

    I know to assume the SqRt of 3 is rational. Then it equals a/b, where a,b are integers and b doesnt equal to 0..not too sure where to go after that
    Let \sqrt{3}= \frac{a}{b} where a,b integers and b\neq 0 and in their lowest terms with no common factor.

    Now square both sides of the equation and we get :


     3b^2 =a^2.That indicates that a^2 is multiply of 3 ,since it is equal to 3b^2,hence according to the above lemma a is a multiple of 3,thus a=3c.

    Now substitute that to the above equation and we have:

    3b^2=9c^2 or b^2=3c^2.

    Therefor we see that also b is a multiple of 3,since b^2 is a multiple of 3.

    Thus b=3r ,contrary to our assumption that a,and b have no common factor.

    Hence \sqrt{3} is not rational
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