Square of Number Always Exists
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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:
- $\forall x \in \mathbb K: x \times x \in \mathbb K$
$\blacksquare$