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Can you help me to finish up this question?

  1. Oct 19, 2006 #1
    Consider Q[sqrt(2)].
    Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
    Prove if true, and give a counterexample if false.


    My solution:
    sqrt(sqrt(2)) = a + bsqrt(2)

    if I square both sides then I will have :

    sqrt(2) = (a + b*sqrt(2))^2
    = a^2 + 2ab*sqrt(2) + 2b^2

    =======================
    I think the answer should be false. Am I right?

    If I am right. Can you suggest me a counterexample. Thank you very much.

    If I am wrong. Please correct me. Thanks
     
  2. jcsd
  3. Oct 19, 2006 #2

    StatusX

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    Well you're almost done. You need to show no such a and b will work.
     
  4. Oct 19, 2006 #3

    Office_Shredder

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    How about -1?
     
    Last edited: Oct 19, 2006
  5. Oct 20, 2006 #4

    HallsofIvy

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    sqrt(2) = a^2 + 2ab*sqrt(2) + 2b^2
    = (a^2+ b^2)+ 2ab sqrt(2),
    Since every number in Q(sqrt(2)) can be written uniquely as "x+ ysqrt(2)" for rational x, y, what does that tell you about a and b?
     
  6. Oct 21, 2006 #5
    Thank you very much for your opinion. I have solved this question.
     
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