Estimations in a ordered field?

Question

My Problem: I am stuck with a proof strategy on the following:

So i have got an ordered field $ (K,+,*,<) $ given. I also have

$x,y\in K$

and $0\le y < x$

I have to proof that, for every n $\in \mathbb{N}, n \ge 2$ there is:

$ny^{n-1} < \frac{x^n-y^n}{x-y} < nx^{n-1}$

What i got so far:

It can be easily proven that $ny^{n-1} < nx^{n-1}$. It won't help us much, i guess.

We also know that the fraction in the middle is >0 because nominator and denominator will always stay positive (because $0\le y < x$). This might be of some use.

This inequality just doesn't ring a bell. I know i have to estimate the fraction somehow. Also i wonder why the task is to prove it for a specific field, not just for $\mathbb{R}$ (What difference does it make?)

Any light you can shed on this is much appreciated.

Answer 1

Since $\;x-y>0\;$ , we have that

$$ny^{n-1}<\frac{x^n-y^n}{x-y}<nx^{n-1}\iff ny^{n-1}(x-y)<x^n-y^n<nx^{n-1}(x-y)$$

and for example (left inequality):

$$ny^{n-1}(x-y)<x^n-y^n\iff nxy^{n-1}-ny^n<x^n-y^n\iff y^n(1-n)<x^n-nxy^{n-1}$$

But

$$x^n-nxy^{n-1}>x^n-nx^n=x^n(1-n)>y^n(1-n)$$

using what you mention that "can be easily proven", and thus we're done.

Try to do the rightmost inequality in the same way.

Answer 2

Outline: Factor. We get $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots +y^{n-1}).$$ On the right, we have $x-y$ times a total of $n$ terms, each $\ge y^{n-1}$, and each $\le x^{n-1}$. If $n\ge 2$, there are at least $2$ terms on the right, so the inequalities are strict.

Answer 3

Let $K^r$ be the real closure of the ordered field $K$. For each integer $n$, the claim

$$0 \leq y < x \implies ny^{n-1} < \frac{x^n-y^n}{x-y} < nx^{n-1} $$

is a first-order statement in the language of ordered fields. Because the theory of real closed fields is complete, this means the inequality is true for the field $K^r$ if and only if it is true for the field $\mathbb{R}$.

And because $K$ is an ordered subfield of $K^r$, if it is true for $K^r$, then it must be true for $K$ as well.

Thus, if you can prove the statement for the real numbers by your favorite means (e.g. calculus?) for all $n \geq 2$, then for each particular $n \geq 2$ we can invoke the logical machinery to conclude it must be true for all ordered fields, and thus we infer it for all $n \geq 2$ for any ordered field.

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If you don't want to invoke the logical machinery to prove things automatically, you can still use the basic idea: first use calculus to prove it for $\mathbb{R}$, then digging into the $\epsilon-\delta$ definition of limits and stuff, you can often directly translate the calculus proof into pure algebra.