Consider the language \(a^*\). We have
\begin{align} a^* &{}= \varepsilon + a + aa + aaa + \ldots\\ &{}= 1 + a + a^2 + a^3 + \ldots~, \tag{since \(\varepsilon\) is the unit of concatenation}\\ &{}= \sum_{k = 0}^\infty a^k \end{align}in which we now immediately recognize the familiar \emph{geometric series} whose closed form is well known to be \(\frac{1}{1 - a}\). In that sense,
\begin{align} a^* = \frac{1}{1 - a}~. \end{align}We have to be a bit careful as the above is actually a non-commutative division (since multiplication of words is non-commutative), so we write
\begin{align} \frac{ {|}\,a~}{~b\,{|}} \quad\text{for}\quad \frac{1}{b} \cdot a \qquad\text{and}\qquad \frac{~a\,{|}}{ {|}\,b~} \quad\text{for}\quad{} a \cdot \frac{1}{b}~, \end{align}whereas \(\frac{ {|}\,1~}{~b\,{|}} = \frac{~1\,{|}}{ {|}\,b~} = \frac{1}{b}\) because \(\varepsilon \cdot x = x \cdot \varepsilon = x\).
Using our new insight – "Kleene star is a fraction" –, we can now conveniently prove, for instance, the well-known identity \(a(ba)^* = (ab)^*a\). While the standard proof is rather lengthy and involves several axioms of Kleene algebra, our proof fits on a beer coaster (it was actually devised on such, see picture below) and goes as follows:
\begin{align} a(ba)^* &{} = \frac{\quad{}~a~\quad{|}}{ {|}\,1 - ba}\\ &{} = \frac{\quad{}1\quad{|}}{ {|}\,\frac{1}{a} - b} \tag{right-expand fraction by \(\frac{1}{a}\)}\\ &{} = \frac{ {|}\quad{}~1~\quad}{\frac{1}{a} - b\,{|}} \tag{by \(\frac{ {|}\,1~}{~x\,{|}} = \frac{~1\,{|}}{ {|}\,x~}\)}\\ &{} = \frac{ {|}\quad{}~a~\quad}{1 - ab\,{|}} \tag{left-expand fraction by \(a\)}\\ &{} =(ab)^*a \end{align}There are "only" two problems with all of the above: neither \({\ldots} - a\) nor \(\tfrac{1}{\ldots}\) make sense at the moment.
Open Problem: How can we formally make sense of additive and multiplicative inverses on words, such that \(a^* = 1 + a + a^2 + {\dots} = \frac{1}{1 - a}\), i.e.\ such that \(a^* \cdot (1 - a) = 1\)?