# Timeless Theorems of Mathematics/Rational Root Theorem

The rational root theorem states that, if a rational number
$\frac{p}{q}$ (where $p$ and $q$ are relatively prime) is a root of a
polynomial with integer coefficients, then $p$ is a factor of the
constant term and $q$ is a factor of the leading coefficient. In other
words, for the polynomial,
$P(x) = a_{n}x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ...+ a_0$, if
$P(\frac{p}{q}) = 0$, (where $a_i\in\Z$ and $a_0, a_n \neq 0$) then
$\frac{(a_0)}{p}\in\Z$ and $\frac{a_n}{q}\in\Z$

## Proof

Let $P(x) = a_{n}x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ...+ a_0$,
where $a_i\in\Z$.

Assume $P(\frac{p}{q}) = 0$ for coprime $p, q\in\Z$. Therefore,
$P(\frac{p}{q})=a_n(\frac{p}{q})^n+a_{n-1}(\frac{p}{q})^{n-1} + a_{n-2}(\frac{p}{q})^{n-2}+...+a_1(\frac{p}{q})+a_0 = 0$
$\Rightarrow a_np^n + a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^2+...+a_1pq^{n-1}+a_0q^n=0$
$\Rightarrow p(a_np^n-1 + a_{n-1}p^{n-2}q+a_{n-3}p^{n-2}q^2+...+a_1q^{n-1})=-a_0q^n$

Let $w=a_np^n-1 + a_{n-1}p^{n-2}q+a_{n-3}p^{n-2}q^2+...+a_1q^{n-1}\in\Z$

Thus, $w=-\frac{a_0q^n}{p}$

As $p$ is coprime to $q$ and $w\in\Z.$, thus $\frac{a_0}{p}\in\Z$.

Again,
$a_np^n + a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^2+...+a_1pq^{n-1}+a_0q^n=0$
$\Rightarrow q(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+...+a_1pq^{n-2}+a_0q^{n-1})=-a_np^n$

Let $(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+...+a_1pq^{n-2}+a_0q^{n-1})=v\in\Z$

Thus, $qv=-a_np^n$ $\Rightarrow v=-\frac{a_np^n}{q}$

As $q$ is coprime to $p$ and $v\in\Z.$, thus $\frac{a_n}{p}\in\Z$.

$\therefore$ For
$P(x) = a_{n}x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ...+ a_0$, if
$P(\frac{p}{q}) = 0$, (where $a_i\in\Z$ and $a_0, a_n \neq 0$) then
$\frac{(a_0)}{p}\in\Z$ and $\frac{a_n}{q}\in\Z$. ***\[Proved\]***

