# All unnatural square roots are irrational

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:

$\sqrt[2]{n}=\frac{m}{d}=\frac{m_{1}\times m_{2}\times\dots\times m_{k}}{d_{1}% \times\dots\times d_{l}}$

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

For any $i$ and $j$, $m_{i}\neq d_{j}$.

Now look at $n$:

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

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

but this is impossible because for any $i$ and $j$, $m_{i}\neq d_{j}$.

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 AllUnnaturalSquareRootsAreIrrational 2013-03-22 17:37:03 2013-03-22 17:37:03 ubershmekel (18723) ubershmekel (18723) 12 ubershmekel (18723) Theorem msc 11J82 msc 11J72 The square of a fraction is always a fraction RationalBriggsianLogarithmsOfIntegers