# Is there a “one-line” proof of $x<y\Rightarrow x^n<y^n$ (for $n$ an odd natural number)?

Given the usual field and ordering axioms for the real numbers, it isn't difficult to prove that $x<y$, without any further restrictions on the signs of $x$ and $y$, implies $x^n<y^n$, with $n$ an odd natural number. However, all the proofs I've seen and come up with myself seem to rely on distinguishing between the different possible combinations of signs of $x$ and $y$. I find such "case-by-case" proofs, especially of elementary facts, rather unappealing.

Now, my question is: does there exist an elegant one-line proof of the fact that $x<y\Rightarrow x^n<y^n$ (for $n$ an odd natural number) which doesn't take into account the signs of $x$ and $y$?

• Are you looking for a deduction from the axioms? – Marco Disce Feb 23 '16 at 21:58
• @Marco Disce. All valid proofs of this are deductions from the axioms, one way or another. I'm looking for a fairly elementary proof, vague as that concept may be. – Kim Fierens Feb 23 '16 at 22:03
• This is a good question, since $n$ has to be odd I don't think you will get a one liner. – Brennan.Tobias Feb 24 '16 at 0:30
• $0 <x <y => x^{n+1}=x^nx <y^nx <y^n*y=y^{n+1}$ via induction for all (odd and even) natural numbers. – fleablood Feb 24 '16 at 1:48

The "One Liner": $$x^n - y^n = \frac{(X-Y)^n - (X+Y)^n}{2^{n/2}} = \sum_{k=0}^n \binom{n}{k} \frac{\left[ X^{n-k}Y^k \left( (-1)^k - 1 \right) \right]}{2^{n/2}} = N Y = P (x - y)$$ where $N$ is a negative and $P$ is a positive number (both dependent on the particular $x$, $y$).

Explanation: \begin{align} x^n - y^n &= \frac{1}{2^{n/2}} \left[ (X-Y)^n - (X+Y)^n \right] \\ &= \frac{1}{2^{n/2}} \sum_{k=0}^n \binom{n}{k} \left[ X^{n-k}(-1)^k Y^k - X^{n-k} Y^k \right] \\ &= \frac{1}{2^{n/2}} \sum_{k=0}^n \binom{n}{k} \left[ X^{n-k} Y^k ((-1)^k - 1) \right] \quad (*) \\ &= N Y \\ &= \frac{N}{\sqrt{2}} (y - x) \\ &= -\frac{N}{\sqrt{2}}(x - y) \\ &= P (x - y) \end{align} This one uses a coordinate transformation $(x,y) \to (X,Y)$, a rotation by $-45^\circ$, to move all points with $x < y$ into the upper half plane such that the expression simplifies.

Here are plots (3D, isolines) for $x^3 - y^3$ before and after the rotation:

$\quad$ $\quad \begin{matrix}\to \\ \\ \\ \\ \\ \end{matrix} \quad$ $\quad$

We see that $N$ consists mainly of the even powers of the binomial expansion, and then we transform back.

The necessary fact that $n$ is odd is used at $(*)$ to note that the differences $n-k$ are even, so $X$ shows up with even powers, while $Y$ stays with odd powers, so we can pull $Y$ out and are left with even powers of $X$ and $Y$.

• I don't mean to be critical but you could make any proof a "one-liner" by making the characters sufficiently small. The key word is "elegant". Binomial expansions are not elegant. – MathematicsStudent1122 Feb 23 '16 at 23:46
• I massaged it a bit. The only elegant idea here is the coordinate transformation. I thought about just putting the four images out as one liner, but I was not sure how many folks would get it. – mvw Feb 23 '16 at 23:56
• @MathematicsStudent1122: you have a strange notion of elegance in mathematics if you find the binomial theorem inelegant. You could argue that this particular use of the binomial theorem is inelegant. – Rob Arthan Feb 24 '16 at 0:07
• I happen to think this proof is nice and beautiful. Exactly the kind of thing I've been looking for but couldn't find myself. Well done! – Kim Fierens Feb 24 '16 at 1:03

$f(t)=t^{2k+1}$ is a strictly monotonic function, as $f'(t)>0$ (except at $t=0$).

• Yes, that certainly fits the description "one-line" proof. – Kim Fierens Feb 23 '16 at 21:57

\begin{align}y^n-x^n=(y-x)\sum_{k=0}^{n-1}x^ky^{n-1-k}=y^{n-1}(y-x)\underset{\text{>1+x/y>0 if} \ \lvert x/y\rvert<1}{\underbrace{\sum_{k=0}^{n-1}(x/y)^k}}&\ge y^{n-1}(y-x)\underset{>0\ \text{if}\ \lvert x/y\rvert>1}{\underbrace{\sum_{k=0}^{(n-1)/2}\lvert x/y\rvert^{2k}(1-\lvert x/y\rvert^{-1})}}\end{align}