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Q on irrational numbers

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Is [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] rational?

    2. Relevant equations

    If n is an integer and not a square, then [tex]\sqrt{n}[/tex] is irrational

    For a rational number a and an irrational number b,

    a + b is irrational
    a * b is irrational if a is not equal to 0

    3. The attempt at a solution

    Assume that [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] = x, with x being a rational number.

    [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] = x - [tex]\sqrt{5}[/tex]
    => ([tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex])2 = (x - [tex]\sqrt{5}[/tex])2
    => 2 + 2[tex]\sqrt{6}[/tex] + 3 = x2 - 2x[tex]\sqrt{5}[/tex] + 5
    => 2[tex]\sqrt{6}[/tex] = x2 - 2x*[tex]\sqrt{5}[/tex]
    => (2[tex]\sqrt{6}[/tex])2 = (x2 - 2x[tex]\sqrt{5}[/tex])2
    => 24 = x4 - 4x3*[tex]\sqrt{5}[/tex] + 20x2

    - 4x3*[tex]\sqrt{5}[/tex] is irrational because 4x3 is rational.
    x4 - 4x3*[tex]\sqrt{5}[/tex] + 20x2 is thus irrational.

    The left hand side of the equation is rational, as 24 is a rational number.

    This is a contradiction, thus our assumption was false, x cannot be a rational number.

    [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] is thus irrational

    Is this a valid proof, or should the equation be worked out further?
     
  2. jcsd
  3. Jul 25, 2010 #2

    Hurkyl

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    Skimming it, the proof looks good.
     
  4. Jul 25, 2010 #3
    I looked at some threads on the internet, but worked the equation out to an eighth degree equation, so I wasn't sure this way was actually correct. Thanks a lot!
     
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