Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the conversion from fractions to $\mathbb Z_7$ elements, like it's kind of obvious, it only reports $\frac{1}{3}=5$, $\frac{2}{3} = 10 = 3$, $\frac{32}{9}=2$. Will you please break it down for me?

Remember that $\frac{1}{x}$ means "the element which, when multiplied by $x$, gives $1$."
Alternatively, $\frac{a}{b}$ is just shorthand for "the solution to the equation $bx=a$."
So, in $\mathbb{Z}_7$, the fraction $\frac{1}{3}$ is shorthand for "the solution to the equation $3x=1$", which, when interpreted in $\mathbb{Z}_7$, is just "the solution to the integer congruence $3x\equiv 1\pmod{7}$."
Similarly, $\frac{2}{3}$ means "the solution to the integral congruences $3x\equiv 2\pmod{7}$", and $\frac{32}{9}$ means "the solution to the integral congruence $9x\equiv 32\pmod{7}$."
Since $\gcd(3,7)=\gcd(9,7)=1$, each of these congruences has a unique solution modulo $7$. $3(5)\equiv 1\pmod{7}$, so $5 = \frac{1}{3}$. Hence, $\frac{2}{3} = 2\frac{1}{3}=2(5) = 10 \equiv 3\pmod{7}$, so $\frac{2}{3}$ is $3$. And since $32\equiv 4\pmod{7}$, for $\frac{32}{9}$ you have $4(5)(5) = 100\equiv 2\pmod{7}$. 
