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Thread: Inverse Modulus

  1. #1
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    Inverse Modulus

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

    From how I interpret it, it is...

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

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

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

    Then it says that $\displaystyle \overline{a}a\equiv1\quad(mod\; m)$, does it mean $\displaystyle -2\cdot3\equiv1\quad(mod\; m)$ and so when $\displaystyle (-2\times3)\div3$, 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 . . .


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

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

    . . $\displaystyle \text{is a }multiplicative\text{ inverse of }a\text{ (mod }m)$
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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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