# pfranusic/why-RSA-works

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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. \vspace{2ex} %%%% otimes-15 table \begin{table}[!ht] \begin{center} \input{otimes-15.tex} \caption{$a \otimes b \quad (\mathcal{R}_{15})$} \label{otimes-15} \end{center} \end{table} 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 $a \otimes b \in Z_n$. $a \otimes b = ab - kn$