deletions | additions       diff --git a/bits1.tex b/bits1.tex  index 3ac388f..bc3a78a 100644  --- a/bits1.tex  +++ b/bits1.tex 
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\item the opposite of $0$ is $0$ (and $-0=0$).  \end{itemize}  \\  Now, an arbitrary It is time that we introduce more formalism. We can consider any  set $G$, $G$  endowed with an operation $*$ s.t.: s.t.  for any $a,b$ in $G$ the operation outputs another element $a*b$ of $G$.  If the operation satisfies the following stringent properties  \begin{itemize}  \item[i)] $*$ is associative;  \item[ii)] $*$ is commutative;  \item[iii)] $*$ has a neutral element;  \item[iv)] each element of $G$ has an opposite w.r.t. $*$  \end{itemize}  then $G$ (with the operation $*$)  is called an  \emph{abelian group}. \\  \begin{remark}  The adjective "abelian" only indicates that ii) holds, i.e. that the operation $*$ is commutative. If only i), iii) and iv) hold, $G$ is only a \emph{group}.\\ \emph{group}.  \end{remark}  In our context, from the argument above, we can say that both $\ZZ$ and $\Fb$ are \emph{abelian groups w.r.t. the sum}.  \begin{Exercise}\label{QAbelian}  Is $\QQ$, the set of rational numbers, an abelian group w.r.t. the sum? Give detailed comments. 
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We can formalize this property, saying that, both in $\QQ$ and in $\Fb$, the multiplication is \emph{distributive} w.r.t. the sum \footnote{The operations are commutative, so it is not necessary to multiply to the right}.  In formulas  $$\forall a,b,c \in \QQ (\textrm{resp } \Fb) a\cdot (b+c)=(a\cdot b) + (a \cdot c).$$  \smallskip  Now, a set $G$, endowed with two operations, denoted by $+,\cdot$, is called \emph{field} if  \begin{itemize}