deletions | additions       diff --git a/bits1.tex b/bits1.tex  index c0572d9..0ab559a 100644  --- a/bits1.tex  +++ b/bits1.tex 
...
with the sum. Unfortunately, this cannot always be done, as we will see from now on.   \smallskip  Given $2 \in \ZZ$, we know that we \emph{cannot} find any integernumber  $b$ such that $2 \cdot b = b \cdot 2 =1$; in order =1$.  It is still possible  to find such an element, a number $b$, but  we need to move to \emph{fractions} i.e. consider the set of \emph{rational numbers} $\QQ$ instead of $\ZZ$. The set $\QQ$ is not so different from $\ZZ$, in particular  all the properties previously stated for $\ZZ$ also hold for $\QQ$. Still, if we consider $2 \in \QQ$ we can easily find $b= \frac{1}{2} \in \QQ$, for which $2 \cdot \frac{1}{2} = \frac{1}{2} \cdot 1 =1$.  \\  Notice that all the properties that we stated for $\ZZ$ also hold for $\QQ$ and consider $2 \in \QQ$. In this case, we We  can easily find $b= \frac{1}{2} \in \QQ$, a similar rational for each \emph{nonzero} element of $\QQ$ (while we cannot  for which $2 \cdot \frac{1}{2} = \frac{1}{2} \cdot 1 =1$. $0$).  \\  This can be easily done for each \emph{nonzero} element of $\QQ$, while it It  is false for $0$. of utmost interest to observe that bits behave like rational numbers rather than integers.   Indeed, in $\Fb$ $1 \cdot 1=1$, but we cannot find any bit $b$ such that $1 \cdot b = b \cdot 1 =1$.  \\  Wepoint out that exactly the same holds for bits: $1 \cdot 1=1$ but no elements in $\Fb$ can be multiplied by zero, getting $1$ as results.  \\  We  can formalize this property, saying that each nonzero element of $\QQ$ (resp $\Fb$) has a \emph{inverse} w.r.t. multiplication, \emph{multiplicative inverse},  i.e. $$\forall a \in \QQ\setminus \{0\}\, (\textrm{ resp } \Fb\setminus \{0\}),\, \exists a^{-1} \in \QQ\setminus \{0\}\, (\textrm{ resp } \Fb\setminus \{0\}) \textrm{ s.t. } a\cdot a^{-1} =a^{-1} \cdot a =1. $$  In $\Fb$, the (multiplicative)  inverse of $1$ is $1$ and $0$ has no inverse. (multiplicative) inverse  (but keep in mind that the opposite of $0$ exists: in $\Fb$ $-0=0$).  \\  Finally, we show how sum and multiplication interact. Taken $2,3,4\in \QQ$, we have that   $$2 \cdot(3+4)= 2\cdot 7 =14 = (2\cdot 3)+(2\cdot 4) = 6+8. $$