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”.