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 $2$.