In the group \((\Z_{11}^\otimes,\otimes)\) where \(a\otimes b=(a\cdot b)\fmod 11\) we compute
\begin{equation*}
\gexp{6}{8}{\otimes}=6\otimes 6\otimes6\otimes6\otimes6\otimes6\otimes6\otimes6\text{.}
\end{equation*}
We use two approaches.
-
We directly follow the definition, that is, we repeatedly apply \(a \otimes b=(a\cdot b)\fmod 11\text{.}\) We first compute
\begin{equation*}
\gexp{6}{2}{\otimes}=6\otimes 6=(6\cdot 6)\fmod 11=36 \fmod 11= 3
\end{equation*}
We now compute the other powers up to
\(\gexp{6}{8}{\otimes}\) making use of the previous result. In every step we apply
Theorem 15.1.2 Item 1.
\begin{align*}
\gexp{6}{3}{\otimes}\amp =\gexp{6}{2}{\otimes}\otimes 6=3\otimes 6=(3\cdot 6)\fmod 11=18\fmod 11=7\\
\gexp{6}{4}{\otimes}\amp =\gexp{6}{3}{\otimes}\otimes 6=7\otimes 6=(7\cdot 6)\fmod 11=42\fmod 11=9\\
\gexp{6}{5}{\otimes}\amp =\gexp{6}{4}{\otimes}\otimes 6=9\otimes 6=(9\cdot 6)\fmod 11=54\fmod 11=10\\
\gexp{6}{6}{\otimes}\amp =\gexp{6}{5}{\otimes}\otimes 6=10\otimes 6=(10\cdot 6)\fmod 11=60\fmod 11=5\\
\gexp{6}{7}{\otimes}\amp =\gexp{6}{6}{\otimes}\otimes 6=5\otimes 6=(5\cdot 6)\fmod 11=30\fmod 11=8\\
\gexp{6}{8}{\otimes}\amp =\gexp{6}{7}{\otimes}\otimes 6=8\otimes 6=(8\cdot 6)\fmod 11=48\fmod 11=4
\end{align*}
We have computed \(\gexp{6}{8}{\otimes}=4\text{.}\)
-
We compute \(6^8\) in the integers and then compute the result \(\fmod 11\text{:}\)
\begin{equation*}
\gexp{6}{2}{\otimes}=6^8\fmod 11=1679616\fmod 11=4
\end{equation*}
Note that we can easily conduct the computations in
Item 1 by hand, but we would not want to compute
\(6^8\) without the help of a calculator. When bases and exponents are larger, the second approach is not feasible anymore as the numbers become to large for most calculators.
