# Proving the odd integer power function is injective over the reals

I am taking Abstract Algebra and in the first assignment we were asked to determine which values of $n$ make the following function injective.

$f: \mathbb{R}\rightarrow \mathbb{R}$

$x \mapsto x^n$ $| n \in \mathbb{N^+}$

Obviously the case where $n$ is even is quite easy to disprove. For the odd case I had a more difficult time. I understand that I could use the fact that the odd functions are continuous and because their derivative is positive everywhere (besides 0 which is dealt with separately), the function is increasing so $a > b \implies f(a) > f(b)$ which would prove it was injective. However, I do not think this is the proper way to go about it, as we have not and will not cover continuity and those sort of things. Not to mention this method seems out of place with the other problems in the homework which all deal with equivalence relations and general set theory questions.

So I just want to know, is there a simpler way of going about proving the odd power function is injective that does not use much Real Analysis as much?

-
I think the set theory tag is not in it's place. So you have it when the map is $1-1$, when is there a (real) solution for $a=x^n$ for $a\in\mathbb{R}$ ? – Belgi Sep 3 '12 at 23:42
Have you considered induction on $n$? For $n=1$, $a>b\implies f(a)=f(b)$ is trivial, and for $n>1$ you could say $x^n = x^2 \cdot x^{n-2}$ and use the fact that $x^2$ is positive except at 0. – MJD Sep 3 '12 at 23:43
Yes, I considered induction. The problem was that whenever I felt like the math was getting too involved I got the feeling that it was not the solution that was in mind when the problem was written. – Karl Sep 3 '12 at 23:58

Take any $x,y$. If $x,y$ have different signs, so do their odd powers, so they are distinct. So we can assume that they have the same sign, wlog both are positive. Then $$x^{2n+1}-y^{2n+1}=(x-y)\left(\sum_{j=0}^{2n}x^jy^{2n-j}\right)$$ And the second parenthesis is a positive expression, so the expression is zero iff $x=y$.
This was the first route that I considered. I dealt with the case that $n=3$, but I then had to show that $x^2+xy+y^2 was always positive. Showing this is easy with differentiating or using the quadratic formula, but I stopped myself because I felt the solution should be simpler. But looking at the comments I don't think an even simpler method is really possible. – Karl Sep 4 '12 at 0:00 By the way, is there a very straight forward method for showing the second term has no real roots that does not go into calculus or any "higher" math? – Karl Sep 4 '12 at 0:10 @Karl: a product and sum of positive terms is positive. No need to differentiate or anything for that... – tomasz Sep 4 '12 at 0:10 @ tomasz: Oops, skipped the line where you dealt with the negative case. Thanks, this seems right. I guess I'll have to put in a small lemma to show the factoring is correct as well though. – Karl Sep 4 '12 at 0:14 @Karl: I don't think I'd go out on a limb much by saying that the proof works in pretty much the same form in an arbitrary ordered domain. So it's as purely algebraic as it gets. – tomasz Sep 4 '12 at 0:51 Here's an algebraic proof. Suppose$a^{2m+1} = b^{2m+1}$. Then$(a/b)^{2m+1} = 1$, which reduces the problem to showing that$1$is the only real number$z$with$z^{2m+1} = 1$. The rest can be done with basic abstract algebra. First, a well known (and easy to prove) lemma on the number of roots to a polynomial over a field shows that there are at most$2m+1$roots to$x^{2m+1} -1 = 0$in$\mathbb{C}$. We can enumerate exactly$2m+1$distinct$(2m+1)$st roots of unity (as complex powers of$e$) and verify that only one of them is real, namely$z = 1$. Thus$z^{2m+1}$is injective over the reals. - I suppose you'd have to check that the polynomial is injective at$z=0$as a special case. – Carl Mummert Sep 3 '12 at 23:58 This seems perfectly acceptable, but I dunno if the teacher had this in mind when the problem set was compiled, it seems to use more information than we are "supposed" to know. But thanks, I think this method and MJD's are the best solutions if I can't find anything else. – Karl Sep 4 '12 at 0:07 If it's very early in the semester, my guess is they are looking for something like the answer by tomasz. You can tell that the large sum is positive because it is a sum of positive terms; positivity of sums and products of positive numbers follows from the axioms for an ordered field. The main benefit of this proof in particular is that it makes no use of the ordering, positivity, etc. – Carl Mummert Sep 4 '12 at 0:10 One way to know using elementary calculus is this. The case$x\mapsto x$is trivially injective. Suppose$n$is a positive integer. Then $$\left(x^{2n + 1}\right)' = (2n + 1)x^{2n} > 0 \qquad {x \not= 0}.$$ By calculus, the function$x\mapsto x^{2n+1}$is strictly increasing and is therefore 1-1. - OP already said that it is true that it is increasing, but wanted to avoid using this argument, as well as continuity, so what makes you think he will like an argument that uses the derivative? – tomasz Sep 3 '12 at 23:48 Here is an algebraic route using complex numbers. The roots of unity for$x^{2n + 1} -1$are all symmetric about the real axis. In fact, it's not hard to see that all but one of these roots ($x = 1$) lies off the$x$axis, since$-1$is not a root. Hence$x^{2n+1} - 1$is the product of an irreducible polynomial and$x - 1\$. This can easily be bootstrapped to the general case – ncmathsadist Sep 4 '12 at 0:02