Results 1 to 2 of 2

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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Mar 2008

    Proof by contradiction

    I would use the proof by contradiction method for this.

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

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

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

    So let's multiply both sides by themselves:

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

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

    $\displaystyle a{^2} = 6b{^2}$

    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 $\displaystyle a{^2}$ must be even. But any odd number times itself is odd, so if $\displaystyle a{^2}$ 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:

    $\displaystyle a{^2} = 6b{^2}$

    $\displaystyle (2c){^2} = (2)(3)b{^2}$

    $\displaystyle 2c{^2} = 3b{^2}$

    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 $\displaystyle {\sqrt {6}}$ cannot be rational.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: Jun 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Jun 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Apr 14th 2008, 04:07 PM

Search Tags

/mathhelpforum @mathhelpforum