Square of Real Number is Non-Negative

Theorem
Let $x \in \R$.

Then:
 * $0 \le x^2$

Proof
From the Trichotomy Law for Real Numbers, there are three possibilities:
 * $x < 0$
 * $x = 0$
 * $x > 0$

Let $x = 0$.

Then:
 * $x^2 = 0$

and thus
 * $0 \le x^2$

Let $x > 0$.

Then:

Thus $x^2 > 0$ and so:
 * $0 \le x^2$

Let $x < 0$.

Then:

Thus $0 < x^2$ and so:
 * $0 \le x^2$