Even Powers are Positive

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Theorem

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

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

Then $x^n \ge 0$.

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

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

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 $\R$.

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

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

$\blacksquare$

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 $2$.

$\blacksquare$