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Math Help - Closure and MIV

  1. #1
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    Closure and MIV

    Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.

    Q[Sqrt(2)] is define by a+bsqrt(2), a and b in Q.

    MIV is existence of multiplicative inverse Ex: aa^-1=1
    Last edited by JCIR; April 25th 2008 at 05:06 PM.
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  2. #2
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    Quote Originally Posted by JCIR View Post
    Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.
    When you post a question why don't you define the terms?
    I dare say that most of us have no idea what 'MIV' means!
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  3. #3
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    Quote Originally Posted by JCIR View Post
    Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.

    Q[Sqrt(2)] is define by a+bsqrt(2), a and b in Q.

    MIV is existence of multiplicative inverse Ex: aa^-1=1
    To construct MIV:

    \frac1{a+b\sqrt{2}}  = \frac1{a+b\sqrt{2}} \times \frac{a-b\sqrt{2}}{a-b\sqrt{2}} = \frac{a}{a^2 - 2b^2}+\frac{-b\sqrt{2}}{a^2 - 2b^2}

    If a,b \in \mathbb{Q}, so will \frac{-b}{a^2 - 2b^2} and \frac{a}{a^2 - 2b^2}


    Thus (a+b\sqrt{2})^{-1} \in \mathbb{Q}[\sqrt2]

    To prove its existence, start with the constructed form and multiply out and show that the product is 1
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    I suppose showing it is closed under addition is no problem?
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