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

$0 \le x^2$

$\Box$

Let $x \ne 0$.

From Square of Non-Zero Real Number is Strictly Positive it follows that:

$0 < x^2$

and so by definition:

$0 \le x^2$

$\blacksquare$