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Math Help - Prove that 3^(1/2) does not belong to the set Q[2^(1/2)]

  1. #1
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    Prove that 3^(1/2) does not belong to the set Q[2^(1/2)]

    Hi, I have this problem which I do not know how to solve... thanks for any help you can give me! And please, if you can solve it, I would really appreciate it if you could tell me the theorem you have used since I am feeling so lost in abstract algebra

    "Let Q[2^{1/2}] be the set consisting of all numbers of the form
    a+b(2)^{1/2}

    where a and b belong to Q (the set of all rational numbers). Prove that 3^{1/2} does not belong to Q[2^{1/2}]."
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If \sqrt{3}=a+b\;\sqrt{2} with a.b\in\mathbb{Q} then, taking squares 3=a^2+2b^2+2\sqrt{2}\;ab . Now you can easily prove that \sqrt{2} would be a rational number (contradiction) .


    P.S. There another way to prove it considering \mathbb{Q}[\sqrt {2}]=\mathbb{Q}[x]/(x^2-2)
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  3. #3
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    Thanks a lot!
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