0
$\begingroup$

Let $p_n(t) = c_0 + c_1 t + c_2 t^2 + \ldots + c_n t^n$ with $c_i \in \mathbb{Q}$ and let the roots of $p_n(t) = 0$ be $R = \{r_1, r_2, \ldots r_n \}$. Vieta's formula states that $\sum_{i=1}^n r_i = -\frac{c_{n-1}}{c_n}$. My question is if we know a root - say $r_i$ - to be irrational is it possible that the sum of the roots yields a rational number?

In the case of $n=2$ we would have $r_1 + r_2 = -\frac{c_1}{c_2}$. Assume that $r_1 \in \mathbb{I}$. The rhs is clearly rational, so what about $r_2$? Can that be such that $r_1 + r_2 \in \mathbb{Q}$? I would say that the sum of two irrational numbers (let's stay with $r_1$ and $r_2$) is only rational when the digits of $r_2$ are of the form $q - r_1$ with $q \in \mathbb{Q}$. But since the roots of polynomials span an orthogonal space this can't be but I'm not sure.

And can we say anything about the general case?

$\endgroup$
  • $\begingroup$ Look at $t^2-2$. $\endgroup$ – André Nicolas Nov 5 '14 at 21:24
  • $\begingroup$ $\sqrt{2} - (\sqrt{2}+3)=-3$ even though both are irrational. $\endgroup$ – Ari Nov 5 '14 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.