Proof. Let $\alpha$ be a root of the equation $$x^n\!+\!r_1x^{n-1}\!+\!r_2x^{n-2}\!+\cdots+\!r_n = 0,$$ where $r_1$ , $r_2$ , ..., $r_n$ are rational numbers ($n > 0$ ). Let $l$ be the least common multiple of the denominators of the $r_j$ 's. Then we have $$0 = l^n(\alpha^n\!+\!r_1\alpha^{n-1}\!+\!r_2\alpha^{n-2}\!+\cdots+\!r_n) = (l\alpha)^n\!+\!lr_1(l\alpha)^{n-1}\!+\!l^2r_2(l\alpha)^{n-2}\!+\cdots+\!l^nr_n,$$ i.e. the multiple $l\alpha$ of $\alpha$ satisfies the algebraic equation $$x^n\!+\!lr_1x^{n-1}\!+\!l^2r_2x^{n-2}\!+\cdots+\!l^nr_n = 0$$ with rational integer coefficients.

According to the theorem, any algebraic number $\xi$ is a quotient of an algebraic integer (of the field $\mathbb{Q}(\xi)$ ) and a rational integer.