, 2014

Assume there exist rational numbers $a_0, a_1, \dots, a_{n-1}$ such that
$$a_0 + a_1 r + a_2 r^2 + \dots + a_{n-1} r^{n-1} = 0.$$
Consider the polynomials
$$f(x) = a_0 + a_1 x + \dots + a_{n-1} x^{n-1} \quad \text{and} \quad g(x) = x^n - 2.$$
We then have $f(r) = g(r) = 0$. By Eisenstein's criterion, $g(x)$ is irreducible
over the rationals, i.e. cannot be written as the product of two polynomials
of positive degree with rational coefficients.

Alternatively, this could be shown directly. Assume $g(x) = u(x)v(x)$ where $u(x)$ and $v(x)$ are monic polynomials with integer coefficients. Then $u(0)$ and $v(0)$ are integers and $u(0)v(0) = g(0) = -2$, hence one of these integers is equal to $\pm 1$, say $u(0) = \pm 1$. This implies that the product of the absolute values of the complex roots of $u(x)$ is equal to 1. Hence, there exists at least one complex number $z$ for which $u(z) = 0$ and $|z| \le 1$. Then $|z|^n \le 1$ in contradiction to $g(z) = u(z)v(z) = 0$ which implies $z^n = 2$. By Gauss' Lemma, $g(x)$ is then irreducible over the rationals, too.

Hence, if $f \neq 0$, $\gcd(f, g)$ can only be 1 or $g$. But the degree of $f$ is smaller
than the degree of $g$, and so we conclude $\gcd(f, g) = 1$. This means that
there exist polynomials $s(x)$ and $t(x)$ such that $s(x)f(x) + t(x)g(x) = 1$.
This is in contradiction with $f(r) = g(r) = 0$, hence such rational numbers
$a_i$ cannot exist unless they are all equal to zero.