Editing Rational root theorem (section)

===Elementary proof===

Let <math>P(x) \ =\ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0</math> with <math>a_0, \ldots, a_n \in \mathbb{Z}, a_0,a_n \neq 0.</math> 

Suppose {{math|1=''P''(''p''/''q'') = 0}} for some [[coprime]] {{math|''p'', ''q'' ∈ '''ℤ'''}}:
<math display="block">P\left(\tfrac{p}{q}\right) = a_n\left(\tfrac{p}{q}\right)^n + a_{n-1}\left(\tfrac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\tfrac{p}{q}\right) + a_0 = 0.</math>

To clear denominators, multiply both sides by {{math|''q''<sup>''n''</sup>}}: 
<math display="block">a_n p^n + a_{n-1} p^{n-1}q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0.</math>

Shifting the {{math|''a''<sub>0</sub>}} term to the right side and factoring out {{mvar|p}} on the left side produces:
<math display="block">p \left (a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1} \right ) = -a_0q^n.</math>

Thus, {{mvar|p}} divides {{math|''a''<sub>0</sub>''q<sup>n</sup>''}}. But {{mvar|p}} is coprime to {{mvar|q}} and therefore to {{math|''q<sup>n</sup>''}}, so by [[Euclid's lemma]] {{mvar|p}} must divide the remaining factor {{math|''a''<sub>0</sub>}}.

On the other hand, shifting the {{math|''a''<sub>''n''</sub>}} term to the right side and factoring out {{mvar|q}} on the left side produces:
<math display="block">q \left (a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1} \right ) = -a_np^n.</math>

Reasoning as before, it follows that {{mvar|q}} divides {{math|''a<sub>n</sub>''}}.<ref>{{cite book |first1=D. |last1=Arnold |first2=G. |last2=Arnold |title=Four unit mathematics |publisher=Edward Arnold |year=1993 |isbn=0-340-54335-3 |pages=120–121 }}</ref>