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{{Proof
| title=Proof that the root of any integer is either an integer or irrational
| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap|''p'' / ''q''}}, where {{nowrap|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap|gcd(''p'', ''q'') {{=}} 1}}. Then 
{{nowrap|''p''<sup>n</sup> / ''q''<sup>n</sup> {{=}} ''m''}}, and {{nowrap|''p''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap|p {{=}} mr}}, so we now have {{nowrap|''m''<sup>n</sup>''r''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. Dividing both sides by ''m'' gives{{nowrap|''m''<sup>{{nowrap|n − 1}}</sup>''r''<sup>n</sup> {{=}} ''q''<sup>n</sup>}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime. {{qed}}
}}

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Proof that the root of any integer is either an integer or irrational

Assume [math]\displaystyle{ \sqrt[n]{m} }[/math] is p / q, where p, q ∈ ℤ⁺ and gcd(p, q) = 1. Then pⁿ / qⁿ = m, and pⁿ = mqⁿ. This means p is divisible by m. Therefore, there exists some integer r such that p = mr, so we now have mⁿrⁿ = mqⁿ. Dividing both sides by m gives m^n − 1rⁿ = qⁿ. This means that q must also be divisible by m, which is a contradiction, since p and q were assumed to be relatively prime. [math]\displaystyle{ \square }[/math]

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See also

• Template:Theorem