Even Power is Non-Negative

Theorem
Let $x \in \R$ be a real number.

Let $n \in \Z$ be an even integer.

Then $x^n \ge 0$.

That is, all even powers are positive.

Proof for n = 2
First we prove the result for $n=2$.


 * Suppose $x = 0$.

Then $x^2 = x \times x = 0 \times 0 = 0 \ge 0$.


 * Suppose $x > 0$.

Then $x^2 = x \times x > 0$ as the set of real numbers, being a field, are also a ring, and $\ge$ is an ordering compatible with the ring structure of $\R$.


 * Suppose $x < 0$.

Let $y = -x$. Then $y \ge 0$ and so $y^2 > 0$ from above.

So by Negative Product, $x \times x = \left({-y}\right) \times \left({-y}\right) = y^2 > 0$.

Proof for n > 2
Let $n \in \Z$ be an even integer greater than $2$.

Then $n = 2k$ for some $k \in \Z$.

Thus $z^n = z^{2k} = \left({x^k}\right)^2$.

The result now follows directly from the result for $n$ equal to $2$.