deletions | additions       diff --git a/bits3.tex b/bits3.tex  index 172d147..ae01e13 100644  --- a/bits3.tex  +++ b/bits3.tex 
...
Let $\Fb[x]$ where we fix the relation $x^2+x+1=0$, that is $x^2=x+1$. The finite set of polynomials are $\{0,1,x,x+1\}$. The inverse of $1$ is $1$, the inverse of $x$ is $x+1$. In fact $x(x+1)=x^2+x=x+1+x=1$.  Let $\Fb[x]$ where $x^2=0$. Even in this case the set of polynomials are $S=\{0,1,x,x+1\}$, $\{0,1,x,x+1\}$,  but in this case the inverse of $x$ does not exist. In fact multiplying $x$ by each element in $x(x)=x^2=0$ and $x(x+1)=x^2+x=x$. $\{0,1,x,x+1\}$ we have either $0$ or $x$ that is $x$ is not invertible.  \[  x\cdot 0=0, x\cdot 1 =x, x\cdot x=x^2=0, x\cdot (x+1)=x^2+x=x.   \]  %In algebra we call $x$ a $0$-divisor.  \end{Example}