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.
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.)