Integer Power Function is Bijective iff Index is Odd
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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: $\S 12 \beta$