Find the multiplicative inverses of each nonzero element of the field $Z/(5), Z/(11), Z/(17).$

Question

For $Z/(5)$, I figured that $[4]$ is a class that has an inverse of its own since $4 \equiv -1 (mod 5)$. Is that correct? Then I tried figuring that $[2]$ is also an inverse of its own since $2 \equiv -3 (mod5)$. Am I doing this correctly?

Answer 1

to find multiplicative inverses for a nonzero elements in field ${\mathbb{Z}_p}$, as for any $a \in {\mathbb{Z}_p}$ find

$b \in {\mathbb{Z}_p}$ such that

$ab \equiv 1\left( {\bmod p} \right)$

Example,in the field ${\mathbb{Z}_5}$

[\begin{gathered} 1 \cdot 1 \equiv 1\left( {\bmod 5} \right) \\ 2 \cdot 3 = 6 \equiv 1\left( {\bmod 5} \right) \\ 4 \cdot 4 = 16 \equiv 1\left( {\bmod 5} \right) \\ \end{gathered} ]

then, the inverse of 1 is 1, inverse 2 is 3, inverse of 3 is 2 and inverse of 4 is 4.

I hope, it may be some helpful

Answer 2

As $5,11$ and $17$ are prime, every non-zero element of $\Bbb Z/p$ will have an inverse.

$1$ and $-1$ are always self-inverse and (for primes $>3$) the other numbers form pairs of inverse elements. As there are only two elements remaining in $\Bbb Z/5$, the inverse table is simple:

\begin{array}{c|l} a & a^{-1} \quad (\Bbb Z/5) \\ \hline 1 & 1 \\ 2 & 3 \\ 3 & 2 \\ 4 & 4 \\ \end{array}

Some inverses for $\Bbb Z/11$ are easy because $11+1$ has small factors and given $a^{-1} = b$ then $(-a)^{-1} = -b$ (although when $p\equiv 1 \bmod 4$ there will be one pair for which $b = -a$)

\begin{array}{c|l} a & a^{-1} \quad (\Bbb Z/11) \\ \hline 1 & 1 \\ 2 & 6 \quad (\text{from }12/2)\\ 3 & 4 \quad (\text{from }12/3)\\ 4 & 3 \\ 5 & 9 \quad (\text{from }\{-6,-2\})\\ 6 & 2 \\ 7 & 8 \quad (\text{from }\{-4,-3\})\\ 8 & 7 \\ 9 & 5 \\ 10 & 10 \\ \end{array}

Then $\Bbb Z/17$ starts to become a little more work.

\begin{array}{c|l} a & a^{-1}\quad (\Bbb Z/17) \\ \hline 1 & 1 \\ 2 & 9 \quad (\text{from }18/2)\\ 3 & 6 \quad (\text{from }18/3)\\ 4 & 13 \quad (\text{from }4^2\equiv -1 \Rightarrow \{4,-4\}) \\ 5 & 7 \quad (\text{from }35/5)\\ 6 & 3 \\ 7 & 5 \\ 8 & 15 \quad (\text{from }\{-9,-2\})\\ 9 & 2 \\ 10 & 12 \quad (\text{from }\{-7,-5\})\\ 11 & 14 \quad (\text{from }\{-6,-3\}\\ 12 & 10 \\ 13 & 4 \\ 14 & 11 \\ 15 & 8 \\ 16 & 16 \\ \end{array}

And for larger values of $p$ inverses in $\Bbb Z/p$ might better be worked out as required with some version of the extended Euclidean algorithm.