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Proove that the cubic root of 2 + the square root of 2 is irrational

  1. Nov 10, 2008 #1

    ehj

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    How do you show that the cubic root of two + the square root of two is irrational? I can easily show that each of these numbers is irrational, but not the sum :/.
     
  2. jcsd
  3. Nov 10, 2008 #2
    If both of the numbers have infitely many decimals, there can never be a terminating digit for their sum.
     
  4. Nov 10, 2008 #3

    ehj

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    Thats not a proof? And besides, doesnt 1 - sqrt(2) and 1 + sqrt(2) both have infinately many decimals, nevertheless their sum is rational, 2.
     
  5. Nov 10, 2008 #4

    Mark44

    Staff: Mentor

    But 1/3 = .333333.... has infinitely many decimal places and it is rational as are 1/2 = .50000... (can also be written as .49999999....) and 1/5 = .200000 (can also be written as .1999999...).
     
  6. Nov 10, 2008 #5

    ehj

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    I figured it out. Posting solution in case sombody might run into the same problem in the future :P

    I assume 2^(1/3) + 2^(1/2) = a , where a is rational

    => 2=(a-2^(1/2))^3 <=> 2 = (a^3 + 6a) + sqrt(2)(-3a^2 -2)

    Which is a contradiction since sqrt(2)(-3a^2 -2) is an irrational multiplied by a non-zero rational, which can be proved to always be irrational, and the sum of a rational (a^3 + 6a) and an irrational can be proved to always be irrational, and above cannot equal 2 since 2 is rational.
     
    Last edited by a moderator: Aug 5, 2015
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