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Math Help - Inverse Modulus

  1. #1
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    Inverse Modulus

    From the definition, it says that Let m\in\mathbb{Z^{+}} and a\in\mathbb{Z}. An integer \overline{a} such that \overline{a}a\equiv1\quad(mod\; m) is an Inverse of a\; mod\; m

    From how I interpret it, it is...

    a\; mod\; m\;= Remainder of a\div m

    Inverse of a\; mod\; m\;=a-m\cdot k, where k\in\mathbb{Z}. Is this right?

    So if I have 3\; mod\;5, The normal 3\; mod\;5=3. Inverse 3\; mod\;5=-2

    Then it says that \overline{a}a\equiv1\quad(mod\; m), does it mean -2\cdot3\equiv1\quad(mod\; m) and so when (-2\times3)\div3, the remainder is 1? But it doesn't look like it.

    I am very confused with the notation and how this whole inverse of modulus work.
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  2. #2
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    Hello, xEnOn!

    There are two inverses: additive and multiplicative.
    You are confusing them.

    Here is a corrected version of what you said . . .


    \text{From the d{e}finition, it says that: }\,\text{Let }m\in\mathbb{Z^{+}}\,\text{ and }\,a\in\mathbb{Z}.

    \text{An integer }\overline{a}\text{ such that }\overline{a}\!\cdot\!a\equiv1\text{ (mod }m)

    . . \text{is a }multiplicative\text{ inverse of }a\text{ (mod }m)
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  3. #3
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    Oh... But what's the difference between additive and multiplicative?
    Was my interpretation additive or multiplicative and was it correct in the first place? Because I actually find my own intepretation pretty confusing too.
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