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I am reading through a textbook on Analysis and have come across a question that I can't seem to make any headway with. A proof is outlined, but I can't make any sense out of it.

The problem is as follows: Let $n$ be a natural in $E^{1}$, and $p,a>0$ be elements of an ordered field $F$. Prove that if $p^{n}>a$, then $(\exists x \in F)|p>x>0$ and $x^{n}>a$.

This is the proof in the book: Let $x=p-d$ with $0<d<p$. Use the Bernoulli inequality to find $d$ such that $x^{n}=(p-d)^{n}>a$
This is where I run into trouble. The Bernoulli inequality states that $\frac{1}{p^n}(1-\frac{d}{p})^{n} \le\frac{1}{p^n}(1-\frac{nd}{p})$. This is fine, but the proof then makes the following step:
$\ldots\implies (1-\frac{d}{p})^{n} \ge (1-\frac{nd}{p})>\frac{a}{p^n}$
Here is my problem - I can't see why the Bernoulli part goes between the other two terms. If anyone could explain it would be much appreciated.

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1 Answer 1

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Part of the problem is that you have Bernoulli’s inequality backwards: it should be

$$\frac{1}{p^n}\left(1-\frac{d}{p}\right)^{n} \ge\frac{1}{p^n}\left(1-\frac{nd}{p}\right)\;.$$

Now multiply through by $p^n$ to get

$$\left(1-\frac{d}{p}\right)^{n} \ge1-\frac{nd}{p}\;.$$

For the other inequality, remember that $d$ isn’t a given: you’re actually choosing $d$ to make everything work. In particular, you’re going to choose $d$ to make the second inequality true. You know that $a<p^n$, so $\dfrac{a}{p^n}<1$, and therefore $1-\dfrac{a}{p^n}>0$. Choose $d$ small enough so that $$0<d\left(\frac{n}p\right)<1-\frac{a}{p^n}\;,$$

and you’ll have

$$\left(1-\frac{d}{p}\right)^{n} \ge 1-\frac{nd}{p}>\frac{a}{p^n}\;.$$

(Given $x,y>0$, you can always choose $z>0$ so that $xz<y$: $z=\dfrac{y}{2x}$ will do, for instance.)

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