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Math Help - Q Field Questions

  1. #1
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    Q Field Questions

    Hello everybody,

    I reached this long section in my book about Q Fields, and after reading through this section, a couple of the examples have me feeling less than enlightened. Here they are:

    1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square. I read this and was like what????

    My work thus far: I don't know what they mean by a/a_bar. I know I did Sqrt[(alpha-a)/b)=d and somehow we force d not to be an integer... perhaps they mean a_bar = alpha-a ? No clue

    2. Does anyone know if every element of Q[Sqrt(2)] have a square root in Q[Sqrt(2)] ? Can this be proven/disproven with a counterexample?

    I just have problems entirely on where to go with that one.

    I really appreciate any and all help! Thank you!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Samson View Post
    1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square?
    do you mean
    an expression for \alpha\cdot \overline{\alpha}
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Samson View Post
    Hello everybody,

    I reached this long section in my book about Q Fields, and after reading through this section, a couple of the examples have me feeling less than enlightened. Here they are:

    1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square. I read this and was like what????

    My work thus far: I don't know what they mean by a/a_bar. I know I did Sqrt[(alpha-a)/b)=d and somehow we force d not to be an integer... perhaps they mean a_bar = alpha-a ? No clue

    2. Does anyone know if every element of Q[Sqrt(2)] have a square root in Q[Sqrt(2)] ? Can this be proven/disproven with a counterexample?

    I just have problems entirely on where to go with that one.

    I really appreciate any and all help! Thank you!
    1. I do not know what a_bar could refer to. My guess is that the alpha symbol looks enough like "a" that you couldn't distinguish them.

    \displaystyle \frac{\alpha}{\overline{\alpha}} = \frac{a+b\sqrt{D}}{a-b\sqrt{D}}=

    \cdots

    \displaystyle \frac{a^2+b^2D}{a^2-b^2D} + \frac{2ab}{a^2-b^2D}\sqrt{D}

    2. (Should be fixed now compared with what I initially wrote) This is pretty easily disproven. Consider that

    (a + b\sqrt{D})^2 = a^2+b^2D + 2ab\sqrt{D}

    Suppose that \sqrt{3+0\sqrt{2}}\in\mathbb{Q}(\sqrt{2}). So

    a^2+2b^2=3

    2ab = 0

    Clearly a and b can't both equal 0. Suppose a = 0, then

    2b^2=3

    b^2=3/2

    b is not rational, contradiction.

    Suppose b = 0, then a is the square root of 3, also not rational, thus contradiction. QED.
    Last edited by undefined; July 7th 2010 at 08:32 AM.
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  4. #4
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    Quote Originally Posted by Amer View Post
    do you mean
    an expression for \alpha\cdot \overline{\alpha}
    Almost, except \alpha / \overline{\alpha}
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