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Math Help - modular proof

  1. #1
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    Question modular proof


    Prove that if a=bmodn and c=dmodn then a+c=b+dmodn and ac=bdmodn
    My solution for the first part is:

    n|(a-b) so (a-b)=nq and n|(c-d) so (c-d)=np
    so (a-b) +(c-d)=nq+np
    (a+c)-(b+d)=n(q+p)
    So n|[(a+c)-(b+d)]
    So
    a+c=b+dmodn
    I am not sure how to show the second part any help would be great

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  2. #2
    o_O
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    Same kind of proof works.

    a \equiv b \ (\text{mod } n) \ \Leftrightarrow \ a - b = nq
    c \equiv d \ (\text{mod } n) \ \Leftrightarrow \ c - d = np

    So:
    \begin{aligned} ac - bd & = ac - (a-nq)(c - np) \\ & = \cdots \\ & = n \left(anp + cnq - npq\right) \end{aligned}

    which by definition, is what we're looking for.
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