Square of Real Number is Non-Negative

Theorem
Let $x \in \R$.

Then:
 * $0 \le x^2$

Proof
There are two cases to consider:
 * $(1): \quad x = 0$
 * $(2): \quad x \ne 0$

Let $x = 0$.

Then:
 * $x^2 = 0$

and thus
 * $0 \le x^2$

Let $x \ne 0$.

From Square of Non-Zero Real Number is Strictly Positive it follows that:
 * $0 < x^2$

and so by definition:
 * $0 \le x^2$