How many roots are rational? 
If $P(x) = x^3 + x^2 + x + \frac{1}{3}$, how many roots are rational? 

EDIT:
$3x^3 + 3x^2 + 3x + 1 = 0$, if any rat roots then,
$x = \pm \frac{1}{1, 3} = \frac{-1}{3}, \frac{1}{3}$, and none of these work. Complete?

There's none, the proof is basically the same as for $\sqrt 2$ being irrational. Assume that $x=p/q$ where $p/q$ is can't be reduced further. Now
$$ x^3 + x^2 + x + 1/3 = p^3/q^3 + p^2/q^2 + p/q + 1/3 = 0$$
Multiply both sides with $3q^3$:
$$ 3p^3 + 3p^2q + 3pq^2 + q^3 = 0$$
Now we see that $3|q$ so we rewrite $q = 3r$:
$$ 3p^3 + 3p^23r + 3p9r^2 + 27r^3 = 3p^3 + 9p^2r + 27pr^2 + 27r^3 = 0$$
Now divide by $3$:
$$ p^3 + 3p^2r+9pr^2 + 9r^3 = 0$$
So also $3|p$ wich contradicts the assumtion.
