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Math Help - Unit

  1. #1
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    Unit

    Let R={m+n(2)^(1/2)|m,n are integers}
    Show that m+n(2)^(1/2) is a unit in R if and only if m^2-2n^2=+-1

    I know a we have a unit a if ab=1.
    I was given a hint to try (m+n(2)^(1/2))(x+y(2)^(1/2))=1 then(m-n(2)^(1/2))(x-y(2)^(1/2))=1 and multiply the 2 equations, but I don't understand the reasoning for this hint.
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  2. #2
    Senior Member Tinyboss's Avatar
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    The function f(m+n\sqrt2)=m^2-2n^2 is a multiplicative norm, that is, a function from R into the integers such that for r,s\in R, f(rs)=f(r)f(s). What is the norm of 1? Does that help?
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  3. #3
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    I don't think I've really done anything norms in this section.
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  4. #4
    Senior Member Tinyboss's Avatar
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    Well, the fact that it's called a "norm" isn't really important--the only feature of it that's important to the problem is the one I described, which is straightforward to verify, and the result falls out immediately.
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  5. #5
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    Well f(rs)=f(r)f(s)=m^2-2n^2=1(1)=1
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  6. #6
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Well f(rs)=f(r)f(s)=m^2-2n^2=1(1)=1
    Since f(1)=1, what do you know about f(a) and f(b) if ab=1?
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  7. #7
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    Oh f(a)=f(b)=+-1
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  8. #8
    Senior Member Tinyboss's Avatar
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    You got it!
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  9. #9
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    Ok it's an iff statement, so I would have to go backwards, too right?
    Assume m^2-2n^2=+-1
    Then (m^2-2^(1/2)n)(m^2+2^(1/2)n)=+-1
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  10. #10
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    Let m^2-2n^2=\pm 1. Then m=\pm\sqrt{n^2\pm 1}, which means

    m+n\sqrt{2}=\pm\sqrt{n^2\pm 1}+n\sqrt{2} or \mp\sqrt{n^2\pm 1}+n\sqrt{2}.

    So \sqrt{2n^2\pm1}\mp n\sqrt{2},-\sqrt{2n^2\pm1}\mp n\sqrt{2} are the respective inverses.
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