# All unnatural square roots are irrational

Theorem: If $n$ is a natural number^{} and $\sqrt[2]{n}$ is not whole, then $\sqrt[2]{n}$ must be irrational.

Proof Ad absurdum: Assume there exists a natural number $n$ that $\sqrt[2]{n}$ is not whole, but is rational.

Therefore $\sqrt[2]{n}$ can be notated as an irreducible fraction: $\frac{m}{d}$

Now break the numerator and denominator into their prime factors^{}:

$\sqrt[2]{n}=\frac{m}{d}=\frac{{m}_{1}\times {m}_{2}\times \mathrm{\dots}\times {m}_{k}}{{d}_{1}\times \mathrm{\dots}\times {d}_{l}}$

Because the fraction is irreducible, none of the factors can cancel each other out.

For any $i$ and $j$, ${m}_{i}\ne {d}_{j}$.

Now look at $n$:

$n=\frac{m_{1}{}^{2}\times m_{2}{}^{2}\times \mathrm{\dots}\times m_{k}{}^{2}}{d_{1}{}^{2}\times \mathrm{\dots}\times d_{l}{}^{2}}$

Because $n$ is a natural number, all the denominator factors are supposed to cancel out,

but this is impossible because for any $i$ and $j$, ${m}_{i}\ne {d}_{j}$.

Therefore $\sqrt[2]{n}$ must be irrational.

Unfortunately this means that a (non-integer) fraction can never become whole by simply squaring, cubing, etc.

I call this unsatisfying fact my ”Greenfield Lemma”.

Title | All unnatural square roots are irrational |
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Canonical name | AllUnnaturalSquareRootsAreIrrational |

Date of creation | 2013-03-22 17:37:03 |

Last modified on | 2013-03-22 17:37:03 |

Owner | ubershmekel (18723) |

Last modified by | ubershmekel (18723) |

Numerical id | 12 |

Author | ubershmekel (18723) |

Entry type | Theorem |

Classification | msc 11J82 |

Classification | msc 11J72 |

Synonym | The square of a fraction is always a fraction |

Related topic | RationalBriggsianLogarithmsOfIntegers |