# Proof in Rudin, real numbers

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

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Why don't you just prove it? You are given precisely what $k$ is. –  Andres Caicedo Jul 2 at 21:57

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

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

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