text: 1. 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.
  
  2. Hello, xEnOn!
  
  There are two inverses: additive and multiplicative.
  You are confusing them.
  
  Here is a corrected version of what you said . . .
  
  $\displaystyle \text{From the d{e}finition, it says that: }\,\text{Let }m\in\mathbb{Z^{+}}\,\text{ and }\,a\in\mathbb{Z}.$
  
  $\displaystyle \text{An integer }\overline{a}\text{ such that }\overline{a}\!\cdot\!a\equiv1\text{ (mod }m)$
  
  . . $\displaystyle \text{is a }multiplicative\text{ inverse of }a\text{ (mod }m)$
  
  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.