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\section{Friends}  All %All  properties we have shown so far represent similarities between the operations in $\Fb$ and in $\ZZ$ or $\QQ$. \\  An %\\  %An  important difference between the sum in $\Fb$ and in $\ZZ$ or $\QQ$ is that summing the bit $1$ with the bit $1$, one gets $0$: $$1+1=0.$$  We %$$1+1=0.$$  %We  see now that this difference reflects also in the computation of \emph{multiples} of bits. \\  Let %\\  %Let  us consider first $2\cdot 0$. It means $0+0$ then it is %is  $0$. \\ If %If  we take $2 \cdot 1$ we have $2 \cdot 1=1+1=0$, so we can notice that $0$ is also the \emph{double} of $1$. \begin{quote} %\begin{quote}  %  Multiplying a bit by $2$ we always obtain $0$. \end{quote} %\end{quote}  There exist particular phenomena such the one related with time that have a kind of {\em ciclicity}. We focus on two examples  \begin{enumerate}  \item The day days  of the week, or if you like music the musical notes, that are $7$; \item The hour hours  of a day that we assume are $12$, as in an analog clock. \end{enumerate}  The $2$ clocks in the picture represent exactly these $2$ cases. These examples are similar in some sense. In fact we have that the first day of a week is Monday, the $8^{th}$ is Monday again. The seventeenth is Wednesday. In this case we have $2$ tour tours  around the clock and the remainder is $3$. That is \[  17=2 \times 7 + 3.   \]  The remainder, namely $3$, is the number interesting for hour our  purpose. This is nothing but an application of division algorithm. Focus on the second example and suppose we want to find what is the $26^{th}$ hour. We have also in this case two tours around the clock and then the remainder hors are $2$, that is  \[  26=2 \times 12 + 2.   \]  Obviously the remainder depends on how many are the hours the clock. Hence every number has a unique representation after fixing the number of hours. Because of this reduction we always assume that the labelling of the hours is exactly  \[  0,1,2,\ldots, n-1 \ZZ_n=\{0,1,2,\ldots, n-1\}  \]  where $n$ is the number of the hours of the clock. clock, namely $\ZZ_n$.  Over these clocks we can define a very nice arithmetic that is called the clock arithmetic, or modular arithmetic. That is we can enrich the set of "hours" with addition and multiplication in a natural way. That is Thus  we can add or multiply the numbers in the standard way and then apply the division with respect to the fixed hours and find the remainder, hence the hour. hour or element of $\ZZ_n$.  As for the case $\Fb$ we represent the tables of the two operations. operations on $\ZZ_7$ and $\ZZ_12$.  First of all we consider the sum. In both clocks, $\ZZ_{12}$ and $\ZZ_7$ clocks  the sum is nothing but a rotation of the first row by one element. Now focus on the multiplication
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\[  9x=1 \mod 12.  \]  If you consider each multiple of 9 modulo 12 is in the set of hours $\{0,3,6,9\}$ hence the equation has no solution. In this case we say that the equation do not satisfy the "cancellation" property. In fact in a "standard" equation similar to the first one we can get multiply $9$ by its inverse, getting rid of the $9$, and obtaining the unique solution $0$. In fact by the table we observe that there are not multiple of $9$ that gives $1$. That is $9$ is not invertible and has not a "regular" behaviour. Moreover if we have the equation  \[  7x=0 \mod 12.  \]  there exists the unique solution $x=0$, and with respect to the equation  \[  7x=9 \mod 12.  \]  there is a unique solution, that is $x=3$.  \begin{figure}