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The only thing I can come up with is this:

$$x<y$$

$$\Longleftrightarrow x\cdot x^{n-1}<y\cdot x^{n-1}$$

Surely $y\cdot x^{n-1}<y\cdot y^{n-1}$ because $x<y$

Then $x\cdot x^{n-1}<y\cdot x^{n-1}<y\cdot y^{n-1}$

$$x^n<y\cdot x^{n-1}<y^n,\text{ so }x^n<y^n$$

I feel this is wrong because I didn't use the fact that $n$ is odd anywhere, but it's all I could come up with.

Michael Hardy
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Juanma Eloy
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    The derivation you have does not seem to be correct. For example, you cannot claim $x^{n-1} < y^{n-1}$: what if $x=-100$, $y=1$ and $n=3$? It might be easier if you consider the possible sign combinations: i) $x < 0 \le y$, ii) $x < y \le 0$, iii) $0\le x < y$. – megas Dec 30 '14 at 23:13
  • When $n$ is odd, $x$ and $x^n$ have the same sign. You can't say that $x < y \implies x^n < y^n$ for even $n$ because in the case where $|x| > |y|$, it wouldn't hold. That makes your third step ("Surely ...") incorrect. Go with @m.a. 's suggestion. When you have negative $x$, rewrite it as $(-1)|x|$. – user123 Dec 30 '14 at 23:17
  • Note that if $n<0$ you could have a problem as while $2<3$ it isn't the case that $\frac{1}{2}<\frac{1}{3}$ as I recall. – JB King Dec 30 '14 at 23:43
  • Oh, also note that your third step is incorrect because it is self-referencing. You're assuming that $x < y \implies x^n < y^n$, which is what you're asked to prove for odd $n$. – user123 Dec 30 '14 at 23:43

3 Answers3

7

As was pointed out in the comments, the 3rd line of your attempt is wrong.

It's easy to prove this by examining the signs of $x$ and $y$:

  • If $x\le0\le y$ (one of them sharp), then $x^n\le 0\le y^n$ (one of them sharp) and we are done.
  • If $0\le x<y$, then $0\le\frac xy<1\implies (\frac xy)^n<1\implies x^n<y^n$.
  • If $x<y\le 0$, then $1>\frac yx\ge 0\implies 1>(\frac yx)^n\implies x^n<y^n$, it gets reversed since $x^n<0$.

Alternatively, you are trying to prove $f(t) = t^n$ is an increasing function if $n$ is odd. We have derivates for this: $$({t^n})'=nt^{n-1}=n(t^{(n-1)/2})^2\ge0$$ with equality only when $t=0$. Therefore the function is increasing on $\mathbb R\setminus\{0\}$ and hence necessarily on the whole $\mathbb R$.

user2345215
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  • The use of $(n-1)/2$ in the exponent is clever, +1 – Zubin Mukerjee Dec 30 '14 at 23:43
  • @ZubinMukerjee Thanks. It's way more obvious if you write $n=2k+1$, then the derivative is $(t^{2k+1})'=(2k+1) t^{2k}$, but I didn't want to introduce any new variables. – user2345215 Dec 30 '14 at 23:57
  • Based on your inequalities, we could have $x=0$ in the second bullet, at which point we've divided by $0$. However, we can substitute the strict inequality $0 < x < y$, since the case $x=0$ is handled by the first bullet. –  Dec 31 '14 at 00:20
  • @Strants Thanks, I missed that. – user2345215 Dec 31 '14 at 00:52
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Suppose for the sake of contradiction that there is a positive integer $k$ along with numbers $x,y$ with $x<y$ such that $x^{2k+1}-y^{2k+1}\geq 0$. It is well known (and a good inductive exercise as well) that $$x^{2k+1}-y^{2k+1} = (x-y)(x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k})$$ By assumption $x-y$ is a negative quantity, so we know $$(x-y)(x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k})\geq 0 \\ \implies x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k} \leq 0 $$ by division of $x-y$ to both sides. It should now be clear that if $x\geq 0$ and $y >0$ then the inequality is incorrect. So we must investigate the case of $x<0$ and $y \leq 0$. If $y=0$ the inequality will fail because everything cancels except for the term $x^{2k}>0$. So we now know $x<y<0$ is the last case we must look at. However, if both $x,y<0$ then $x^{2k}>0$, and $x^{2k-1}y>0$ because $x^{2k-1}$ and $y$ are negative. And $x^{2k-2}y^2>0$ because $x^{2k-2}$ and $y^2$ are positive. This argument holds for each term in the sum $$x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k}$$ because $x^{2k-i}y^{i}>0$ for all $i \in \{0,1,2,\dots, 2k\}$. So the sum has to be greater than zero, yielding your last contradiction.

graydad
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The point of restricting $n$ to odd integers is that if $x<0$ and $n$ is odd then $x^n<0$ (this is not true for even $n$ since $x>0$ $\implies$ $x^n>0$ for even $n$). As a result, your problem becomes much simpler when considering different cases based on the sign of $x$ and $y$. Let's consider the case where $x<0<y$. Let $n$ be an odd integer. Since $x<0$ and $n$ is odd then $x^n<0$. Since $y>0$ then clearly $y^n>0$. Hence, $x^n<y^n$ and this completes the proof for the case $x<0<y$.

I leave the cases of $0<x<y$ and $x<y<0$ to you (hint: use induction on $n$).

sardoj
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