The Rational Roots Theorem

Theorem 1 (The Rational Roots Theorem): If $p \in \mathbb{Z}[x]$ with $p(x) = c_0 + c_1x + ... + c_nx^n$ and $\frac{a}{b} \in \mathbb{Q}$ is a root of $p$ with $(a, b) = 1$ then $a | c_0$ and $b | c_n$.

• Proof: Suppose that $\frac{a}{b} \in \mathbb{Q}$ with $(a, b) = 1$ is a root of $p$. Then $p \left ( \frac{a}{b} \right ) = 0$, so:

(1)

\begin{align} \quad 0 &= c_0 + c_1 \frac{a}{b} + ... + c_n \frac{a^n}{b^n} \\ &= c_0b^n + c_1ab^{n-1} + ... + c_na^n \end{align}

• Since $b | c_0b^n, c_1ab^{n-1}, ..., c_{n-1}a^{n-1}b, 0$ we have that $b | c_na^n$. Similarly, since $a | c_1ab^{n-1}, ..., c_na^n, 0$ we have that $a | c_0b^n$. Since $(a, b) = 1$ and since $b | c_na^n$ and [$a | c_0b^n$ we have that $b | c_n$ and $a | c_0$. $\blacksquare$

The rational roots theorem says that if $\frac{a}{b}$ is a rational root in lowest terms of $p(x) = c_0 + c_1x + ... + c_nx^n$ then it must be that the numerator $a$ divides $c_0$ and that the denominator $b$ divides $c_n$. This is extremely useful in determining if a rational number is a root of a particular $p \in \mathbb{Z}[x]$.

For example, let $p(x) = 4 + 4x^2 + 5x^3 + 5x^4$. Suppose that $\frac{a}{b}$ is a rational root of $p(x)$. Then by the theorem above we must have that $a|4$ and $b|5$. So $a = 0, \pm 1, \pm 2, \pm 4$ and $b = \pm 1, \pm 5$. So the possible rational roots of $p$ are:

(2)

\begin{align} \quad 0, \pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5} \end{align}

Corollary 1: The number $\sqrt{2}$ is irrational.

• Proof: Suppose instead that $\sqrt{2} = \frac{a}{b}$ is rational. Consider the polynomial $p(x) = x^2 - 2$. Clearly $p(\sqrt{2}) = 0$. By the rational roots theorem, we must have that $a | 2$ and $b | 1$. So $a = \pm 1, \pm 2$ and $b = \pm 1$. So $\frac{a}{b} = \pm 1, \pm 2$. However, $p(\pm 1) \neq 0$ and $p(\pm 2) \neq 0$ which is a contradiction. Therefore $\sqrt{2}$ must instead be irrational. $\blacksquare$