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%%%% why-RSA-works/otimes-operator.tex
%%%% Copyright 2012 Peter Franusic.
%%%% All rights reserved.
When $n$ is small, the $\otimes$ operator can be specified using a table.
Table \ref{otimes-15} specifies the $\otimes$ operator for the ring $\mathcal{R}_{15}$.
The format is the same as Table \ref{oplus-15}. The values, of course, are different.
Note the rose-like pattern visible in the table.
The table is symmetrical about the diagonals.
If we ignore column 0, then
row 1 is the reverse of row 14, row 2 is the reverse of row 13, etc.
%%%% otimes-15 table
\caption{$a \otimes b \quad (\mathcal{R}_{15})$}
Table \ref{otimes-15} specifies the value of $a \otimes b$ for every possible pair of $a$ and $b$.
For example, let $a=11$ and $b=8$.
The value of $11 \otimes 8$ is specified at the intersection of row $11$ and column $8$.
This value is $13$. Therefore $11 \otimes 8 = 13$.
Notice that every element in the table is in the set $Z_{15}$.
This demonstrates the \emph{multiplicative closure} property of rings.
The multiplicative closure property states that for every pair of elements $a$ and $b$ in $Z_n$,
the product $a \otimes b$ is also an element in $Z_n$.
\[ a \otimes b \in Z_n \]
The value of $a \otimes b$ can also be specified using a rule.
To compute $11 \otimes 8$ we first calculate $11 \times 8$ to get 88.
Since 88 is not an element in $Z_{15}$ we subtract a multiple of the modulus, $kn$.
In this case, $kn = 5 \times 15 = 75$. Therefore $88 - 75 = 13$.
And $13 \in Z_{15}$ so 13 is our final result.
In general, the $\otimes$ operator takes two integers $a$ and $b$,
multiplies them together using normal multiplication,
then subtracts some multiple of $n$ such that the final value is in $Z_n$.
In other words, we subtract whichever $kn$ works in order to get closure,
where $a \otimes b \in Z_n$.
\[ a \otimes b = ab - kn \]