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Thread: Multiplicative Inverse

  1. #1
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    Multiplicative Inverse

    This is in regard to proving that the statements (a) $\displaystyle a$ and $\displaystyle n$ are coprime and (c) row $\displaystyle a$ of the multiplication table for $\displaystyle Z_{n}$ includes all of $\displaystyle Z_{n}$

    $\displaystyle ab = kn + 1$

    "this equation implies that $\displaystyle a$ and $\displaystyle n$ are coprime because any common factor of $\displaystyle a$ and $\displaystyle n$ must also be a factor of 1

    I don't see why this is so.
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  2. #2
    Member Sylvia104's Avatar
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    Re: Multiplicative inverse

    Let $\displaystyle d$ be a common factor of $\displaystyle a$ and $\displaystyle n,$ so $\displaystyle a=a_0d,$ $\displaystyle n=n_0d$ for some integers $\displaystyle a_0,n_0.$ Then

    $\displaystyle \begin{array}{rcl} ab &=& kn+1 \\ a_0db &=& kn_0d+1 \\ \left(a_0b-kn_0\right)d &=& 1 \end{array}$

    Thus $\displaystyle d\mid1.$
    Thanks from alyosha2
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  3. #3
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    Re: Multiplicative Inverse

    So because the difference is 1 we know the only factor we can pull out is 1. Good stuff. Thanks for the help.
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    Re: Multiplicative Inverse

    Thank you Sylvia for solving this problem, even i had the same problem and i was also thinking of posting this but i saw Alyosha2 had already posted it and from here i got the solution of the problem and this is the best forum where we get maximum math problem solved.
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