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Thread: proof help

  1. #1
    Mar 2008

    proof help

    Prove that the square root of 6 is irrational.

    Sorry i don't know how to do the square root in here.
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  2. #2
    Mar 2008

    Proof by contradiction

    I would use the proof by contradiction method for this.

    So let's assume that {\sqrt {6}} is rational.

    By definition, that means there are two integers a and b with no common divisors where:

    {\frac{a}{b}} = {\sqrt {6}}

    So let's multiply both sides by themselves:

    ({\frac{a}{b}})({\frac{a}{b}}) = ({\sqrt {6}})({\sqrt {6}})

    {\frac{a{^2}}{b{^2}}} = 6

    a{^2} = 6b{^2}

    This last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a{^2} must be even. But any odd number times itself is odd, so if a{^2} is even, then a is even.

    Since a is even, there is some integer c that is half of a, or in other words:

    2c = a.

    Now let's replace a with 2c:

    a{^2} = 6b{^2}

    (2c){^2} = (2)(3)b{^2}

    2c{^2} = 3b{^2}

    But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

    Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so {\sqrt {6}} cannot be rational.
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