Rational Root Theorem

This article needs to be tidied. In particular: Arrange into sections: theorem, proof, sources. Not clear what is being proved. Source code needs a little tweaking. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

This article needs to be linked to other articles. In particular: including category You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

Let $P \in \Z \sqbrk X$.

Let $p, q \in \Z$ such that:

$p \perp q$

$a_n, a_0 \ne 0$

where $\perp$ denotes coprimality.

\(\ds \map P X\)

\(=\)

\(\ds a_n X^n + a_{n - 1} X^{n - 1} + \cdots + a_1 X + a_0\)

Definition of $P$

Assume that $\dfrac p q$ is a root of $P$,

To clear denominators, multiply both sides by $q^n$:

\(\ds \map P {\dfrac p q}\)

\(=\)

\(\ds a_n \paren {\dfrac p q}^n + a_{n - 1} \paren {\dfrac p q}^{n - 1} + \cdots + a_1 \paren {\dfrac p q} + a_0\)

\(\ds \)

\(=\)

\(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n\)

multiplying both sides by $q^n$

\(\ds 0\)

\(=\)

\(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n\)

as $\frac p q$ is a root of $P$

\(\ds -a_0 q^n\)

\(=\)

\(\ds a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1}\)

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:

$p \divides a_0$

Shift $a_n$ term to right hand side and factor out $q$:

\(\ds -a_0 p^n\)

\(=\)

\(\ds q \paren {a_{n - 1} p^{n - 1} + a_{n - 2} p q^{n - 2} + \cdots + a_0 q^{n - 1} }\)

There is believed to be a mistake here, possibly a typo. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mistake}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

It follows that:

$q \divides a_n$