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Math Help - Modular solutions question

  1. #16
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    Quote Originally Posted by flyingsquirrel View Post
    For example : x \equiv 0\,[2] \Rightarrow 2x \equiv 0\,[2] but the other way is wrong : 2x \equiv 0\,[2] \not \Rightarrow x \equiv 0\,[2]

    If we multiply the modulo and the congruence by 2, both ways are correct : x \equiv 0\,[2] \Leftrightarrow 2x \equiv 0\,[4]
    Because 2 has no inverse in Z/2Z.

    Here, we multiplied by the inverse of a number, so it'll always work.
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    Quote Originally Posted by flyingsquirrel View Post
    For example : x \equiv 0\,[2] \Rightarrow 2x \equiv 0\,[2] but the other way is wrong : 2x \equiv 0\,[2] \not \Rightarrow x \equiv 0\,[2]

    If we multiply the modulo and the congruence by 2, both ways are correct : x \equiv 0\,[2] \Leftrightarrow 2x \equiv 0\,[4]
    Quote Originally Posted by Moo View Post
    Because 2 has no inverse in Z/2Z.

    Here, we multiplied by the inverse of a number, so it'll always work.
    It has nothing to do with inverses in the forward direction. If a\equiv b(\bmod n) then it means n|(a-b) then certainly n|c(a-b) so n|(ac-bc) thus ac\equiv bc(\bmod n).
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  3. #18
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    Quote Originally Posted by ThePerfectHacker View Post
    It has nothing to do with inverses in the forward direction. If a\equiv b(\bmod n) then it means n|(a-b) then certainly n|c(a-b) so n|(ac-bc) thus ac\equiv bc(\bmod n).
    I think we were more talking about the reversal of the implication lol, a little off-topic
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