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Math Help - multiplicative inverse

  1. #1
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    multiplicative inverse

    Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NikoBellic View Post
    Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].
     \frac{1}{1+\sqrt2} = \frac{1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)} = \sqrt2-1
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NikoBellic View Post
    Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].
    This is kind of overkill but here goes:


    You can also do this using norms.  \mathbb{Z}[\sqrt2] has a norm  N(a+b\sqrt2) = a^2-2b^2 .

    Suppose  (1+\sqrt2)^{-1} = a+b\sqrt2 , then  N(1+\sqrt2)=-1\implies N(a+b\sqrt2) = -1 \implies a^2-2b^2 = -1 .

    So we're looking to solve  a^2-2b^2=-1 .

    Now  (a^2-2b^2)^2=(-1)^2\implies (a^2+2b^2)^2-2(2ab)^2=1\implies (2a^2+1)^2-2(2ab)^2=1 since  2b^2=a^2+1 for the last equality.

    So as you can see to find a solution, we're left to solve a form of Pell's equation, which is solvable i.e. we can find  a and  b . Thus  1+\sqrt2 has an inverse.

    After Pell's equation is solved though, we'll have an answer in terms of  2a^2+1 and  2ab , so this proof needs some polishing at the end. I won't do this however since I've already given you another solution; this posting was just a thought I had.
    Last edited by chiph588@; April 24th 2010 at 11:08 PM.
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