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

There is one thing in this proof I do not get. Why can he say that $k < y$?

enter image description here

share|cite|improve this question
Why don't you just prove it? You are given precisely what $k$ is. – Andrés E. Caicedo Jul 2 '14 at 21:57
up vote 5 down vote accepted

Note that $$ k = \frac{y^n - x}{ny^{n-1}} = \frac 1n y - \frac{x}{ny^{n-1}} < \frac 1n y \leq y $$

share|cite|improve this answer
Well, $n$ can be $1.$ – mfl Jul 2 '14 at 22:00
Doesn't matter, $x$ is positive. – André Nicolas Jul 2 '14 at 22:03
I refer only to the last strict inequality. – mfl Jul 2 '14 at 22:04
@mfl fixed it, thanks – Omnomnomnom Jul 2 '14 at 22:04

You know that $$k=\frac{y^n-x}{ny^{n-1}}.$$

Then you can write it as: $$=\frac{y^n}{ny^{n-1}}-\frac{x}{ny^{n-1}}$$

Now, using $x>0$, $y>0$, $n>0$, the second term must be positive, thus you have the inequalities: $$<\frac{y^n}{ny^{n-1}}\leq\frac{y}{n}\leq y.$$

Using this chain of equalities/indequalities you have $k< y$.

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.