Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.