deletions | additions       diff --git a/bits1.tex b/bits1.tex  index d589e45..1f02c56 100644  --- a/bits1.tex  +++ b/bits1.tex 
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% Can you prove that $\Fb$ is actually a field?   % \end{Exercise}  From the rules of the sum of bits, we can derive how to compute \emph{multiples} of bits. bits.\\  Let us consider first  $2\cdot 0$. It means $0+0$ then it is $0$. \\  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}  Multiplying a bit by $2$ we always obtain $0$.  \end{quote}  This has a precise meaning from a mathematical point of view, the concept of \emph{characteristic}. \emph{characteristic}.\\  Given a field (in our case $\Fb$), it can happen that some positive\footnote{notice: positive means \emph{different} from $0$!!} integer numbers, multiplied to all the elements of the field, give $0 $ as   result (in our case, we saw that $2$ has this property).\\ 
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Bits can also be seen as a way to represent the logical values TRUE/FALSE. The   usual notation is $0=\FALSE$ and $1=\TRUE$. Stating this notation, \\  %Indeed, we have $1 \cdot 1= 1$ and $0 \cdot 1 = 1 \cdot 0 =0$; this means that   %each element different from $0$ (i.e. $1$) has an inverse.   % In formulas $a \neq 0 \Rightarrow \exists \frac{1}{a}$.  % We remark that $1 +1= 0 $ in $\Fb$, so in this field $-1 = 1$.  %By the point of view of logic, considering $0=\FALSE$ and $1=\TRUE$,  Once fixed this notation,  we have the following  well known truth tables: \begin{table}[!htb]  \centering  \begin{tabular}{c|c|c|c|c}