# Question #6d78f

Dec 3, 2016

$2 \cdot 4 \equiv 1 \text{ (mod "7")}$, thus $4 = {2}^{- 1}$ in ${\mathbb{Z}}_{7}$.

#### Explanation:

(This answer assumes a basic understanding of modular arithmetic)

When working in ${\mathbb{Z}}_{n}$, if $k \in {\mathbb{Z}}_{n}$ is coprime with $n$, that is, if $\text{GCD} \left(k , n\right) = 1$ then there exists a unique ${k}^{- 1} \in {\mathbb{Z}}_{n}$ such that $k \cdot {k}^{- 1} \equiv 1 \text{ (mod "n")}$. We call ${k}^{- 1}$ the multiplicative inverse of $k$ in ${\mathbb{Z}}_{n}$.

To see that ${2}^{- 1} \equiv 4$ in ${\mathbb{Z}}_{7}$, then, we need only show that $2 \cdot 4 \equiv 1 \text{ (mod "7")}$. Indeed,

$2 \cdot 4 \equiv 8 \equiv 1 + 7 \equiv 1 \text{ (mod "7")}$

Thus ${2}^{- 1} = 4$ in ${\mathbb{Z}}_{7}$