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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?

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    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.2012-09-03

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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 $x=y$.

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    @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.2012-09-04
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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.

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    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.2012-09-04
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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.

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    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 case2012-09-04