Rational Root Theorem

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Theorem

Let $\map P x$ be a polynomial whose coefficients are all integers:

$\map P x = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$

where $a_n \ne 0$ and $a_0 \ne 0$.

Let the polynomial equation $\map P x = 0$ have a root which is a rational number expressed in canonical form as $\dfrac p q$.

Then:

the leading coefficent $a_n$ of $\map P x$ is divisible by $q$
the constant term $a_0$ of $\map P x$ is divisible by $p$.


Proof

Recall the definition of Canonical Form of Rational Number:

Let $r \in \Q$ be a rational number.

The canonical form of $r$ is the expression $\dfrac p q$, where:

$r = \dfrac p q: p \in \Z, q \in \Z_{>0}, p \perp q$

where $p \perp q$ denotes that $p$ and $q$ have no common divisor except $1$.


We have by hypothesis that $\dfrac p q$ is a root of $P$.

Hence:

\(\ds \map P {\dfrac p q}\) \(=\) \(\ds 0\) Definition of Root of Polynomial
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds a_n \paren {\dfrac p q}^n + a_{n - 1} \paren {\dfrac p q}^{n - 1} + \cdots + a_1 \paren {\dfrac p q} + a_0\) Definition of $P$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n\) multiplying both sides by $q^n$
\(\ds \leadsto \ \ \) \(\ds -a_0 q^n\) \(=\) \(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1}\) subtracting $a_0 q^n$ from both sides
\(\ds \) \(=\) \(\ds p \paren {a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} }\) Distributive Laws of Arithmetic


By the closure of addition and multiplication over the integers:

$a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} \in \Z$

By Euclid's Lemma, either:

$p \divides q^n$

or:

$p \divides a_0$

where $\divides$ denotes divisibility.

Because $p \nmid q$ we have that:

$p \nmid q^n$

Hence:

$p \divides a_0$

$\Box$


Then:

\(\ds 0\) \(=\) \(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds -a_n p^n\) \(=\) \(\ds q \paren {a_{n - 1} p^{n - 1} + \cdots + a_1 p q^{n - 2} + a_0 q^{n - 1} }\)


Using the same argument as for $p \divides a_0$, it follows that:

$q \divides a_n$

$\blacksquare$


Sources

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