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Math Help - 2 modular arithmetic proofs

  1. #1
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    2 modular arithmetic proofs

    Prove there is an integer e such that
    (i) [a]modn + [e]modn = [e]modn + [a]modn = [a]modn for all integers a

    (ii) for all integers a, there exists an integer b such that
    [a]modn +[b]modn = [b]modn + [a]modn = [e]modn
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  2. #2
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    Quote Originally Posted by jsteel2 View Post
    Prove there is an integer e such that
    (i) [a]modn + [e]modn = [e]modn + [a]modn = [a]modn for all integers a

    (ii) for all integers a, there exists an integer b such that
    [a]modn +[b]modn = [b]modn + [a]modn = [e]modn
    I'm not used to the [a] notation, but this is simply (i) additive identity and (ii) additive inverse. That is, 0 and -a (mod n).

    If e isn't restricted to the set {0,1,...,n-1} then there in fact exist an infinite number of integers e, of the form {0, n, -n, 2n, -2n, ...}. Similar with integers b in part (ii), integers of the form -a + kn.
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