What's wrong with my proof that $f(x) = x^2, x^4, \ldots$ are bijective?

Question

Let $n$ be an odd positive integer. Prove $f : \mathbb{R} \rightarrow > \mathbb{R}$ defined by $f(x) = x^n$ is bijective.

My attempt:

First we prove injectivity. Suppose we have $f(x) = f(y)$ so that $x^n = y^n$. Raising to the $(1/n)^{\text{th}}$ power, we get $x = y$, from which injectivity follows.

Now we prove surjectivity. We want to find $x$ so that $f(x) = y$. Taking $x = y^{1/n}$ works.

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My question:

1) Is this proof correct for odd $n$?

2) Why does it fail for $n$ even?

Answer 1

It is incorrect.

Your proof only holds for odd $n$!

Indeed, we have $f(1)=f(-1)$ for $n$ even.

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Your proof for $n$ is even breaks down because we have

$$(x^n)^{1/n} = |x|$$

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Here are hints how you can formally show that $x \mapsto x^n$ is bijective for $n$ odd (without assuming $n$-th roots exist!)

Injective: Because the function is strictly increasing (prove this using derivatives).

Surjective: By continuity and the mean value theorem.

Answer 2

Your proof is not correct because it assumes that there is a function $x\mapsto\sqrt[n]x$ which is an inverse of $x\mapsto x^n$. But you are supposed to prove that that inverse exists!

Answer 3

For $n$ even, $f(-1)=f(1)$ and for $n$ odd, $f(-1)=-f(1)$. So for $n$ even, the function cannot be one-to-one. For $n$ odd, it is.

Answer 4

The map $x\mapsto x^{1/n}$ only exists when $n$ is odd. As is, your proof is wrong because you have not shown the map exists. You have not even shown it exists for odd $n$, and you're lucky it does.

As an alternative, try showing $x\mapsto x^n$ is monotonously increasing and continuous to show it is a bijection (when $n$ is odd). It should be obvious why this does not work when $n$ is even.

Answer 5

I suppose you mean "fails for $n$ even". Your proof is fine for $n$ odd.

The problem depends on your perspective, I will express it in two ways:

1. "There is no $n^{\text{th}}$ root is $n$ is even."

2. "The standard $\sqrt[n]{x}$ is not an inverse function."

   I'll focus on the case $n=2$, but the argument is completely general.

Write $f(x) = x^{2}$ and $g(x) = \sqrt{x}$. The question is, what is $fg$ and what is $gf$. To answer this, we need to know what $g$ really means. Suppose you 'define' $$g(x^{2}) = x\quad \text{for every $x$}$$ The important thing to realise is that this is not a function! Why? Because the function has to be well defined by its input - if we take the input to be $1$, then we have simultaneously: $$g(1^{2}) = 1$$ $$g((-1)^{2}) = -1$$ which would be a contradiction. The usual definition of square rooting says that we take the positive square root, so that unambiguously, $g(1)=1$.

The problem now is that even though $f(g(x))$ equals $x$ for every $x \ge 0$, we do not always have $g(f(x)) = x$: for example, $g(f(-1)) = g(1) = 1$.

So, saying "taking $n^{\text{th}}$ roots" does not make sense for even $n$.