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:

$\sqrt[2]{n} = \frac{m}{d} = \frac{m_1 \times m_2 \times \dots \times m_{k}}{d_1 \times \dots \times d_{l}}$

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".