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It is of utmost interest to observe that bits behave like rational numbers rather than integers.
Indeed, $1 \cdot 1=1$ in $\Fb$, but we cannot find any bit $b$ such that $1 \cdot b = b \cdot 1 =1$.
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We can formalize this property, saying that each nonzero element of $\QQ$
(resp (respectively $\Fb$) has a \emph{multiplicative inverse}, i.e.
$$\forall a \in \QQ\setminus \{0\}\, (\textrm{ resp } \Fb\setminus \{0\}),\, \exists a^{-1} \in \QQ\setminus \{0\}\, (\textrm{
resp respectively } \Fb\setminus \{0\}) \textrm{ s.t. } a\cdot a^{-1} =a^{-1} \cdot a =1. $$
In $\Fb$, the (multiplicative) inverse of $1$ is $1$ and $0$ has no (multiplicative) inverse
(but keep in mind that the opposite of $0$ exists: $-0=0$ in $\Fb$).
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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
$$\forall a,b,c \in \QQ
\; (\textrm{resp \quad (\textrm{respectively }
\Fb)\; \Fb)\quad a\cdot (b+c)\,=\,(a\cdot b) + (a \cdot c)\,.$$
\smallskip
A set $G$, endowed with two operations (sum and multiplication), denoted by $+,\cdot$, is called \emph{field} if
\begin{itemize}
\item[i)] $G$ is an abelian group w.r.t. $+$;
\item[ii)] $G^*$, that is $G$ without let $0$ be the neutral
element w.r.t. $+$, element;
\item[ii)] $G\setminus\{0\}$ is an abelian group w.r.t. $\cdot$;
\item[iii)] $\cdot$ is distributive w.r.t. $+$.
%\item[ii)] $\cdot$ is associative;
%\item[iii)] $\cdot$ is commutative;
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%\item[vi)] $\cdot$ is distributive w.r.t. $+$.
\end{itemize}
According to the above definition, we can say that both $\QQ$ and $\Fb$ are fields, whereas $\ZZ$ is
not \textbf{not} a field.
The notation for the set of bits (that - from now on - we will call the field of bits), reflects this fact. Indeed, $\FF$ stands for "field" and the subscript $2$ stands for the size of the set itself, i.e. the number of its elements.
