Square of Number Always Exists

Theorem

Let $x$ be a number.

Then its square $x^2$ is guaranteed to exist.

Proof

Whatever flavour of number under discussion, the algebraic structure $\struct {\mathbb K, +, \times}$ in which this number sits is at least a semiring.

The binary operation that is multiplication is therefore closed on that algebraic structure.

Therefore:

$\forall x \in \mathbb K: x \times x \in \mathbb K$

$\blacksquare$