1. ## 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.

2. Hello, xEnOn!

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

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

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

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

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

3. 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.