Integer Power Function is Bijective iff Index is Odd

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $f_n: \R \to \R$ be the real function defined as:

$\map {f_n} x = x^n$

Then $f_n$ is a bijection if and only if $n$ is odd.

Proof

Even Index

Suppose $n$ is even.

Let $x \ne 0$.

Then $1^n = \paren {-1}^n$ by Power of Ring Negative, so $f_n$ is not injective.

Also, by Even Power is Non-Negative, $f_n$ is not surjective.

By definition, a bijection is both injective and surjective.

It follows that for even $n$, $f_n$ is not bijective.

$\Box$

Odd Index

Now suppose $n$ is odd.

From Odd Power Function is Strictly Increasing, $f_n$ is injective.

From Odd Power Function is Surjective, $f^n$ is surjective.

So when $n$ is odd, $f_n$ is both injective and surjective, and so by definition bijective.

Hence the result.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \beta$