# Doubt - Proof of completeness of number system [duplicate]

Possible Duplicate:
Choice of $q$ in Baby Rudin’s Example 1.1

I am going through Walter S. Rudin - Introduction to Mathematical Analysis,3rd Edition.

In his proof of the existence of real numbers, shown below, I am unable to understand equation-3. Specifically, How did he get the formulation for 'q' in terms of 'p' that seems to help the proof in such a nice way.

[EDIT] - The proof is in Page 2 of the book.

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## marked as duplicate by lhf, kahen, Douglas S. Stones, Norbert, EmilyOct 30 '12 at 15:46

You'll get used to Rudin's slick proofs. Often you just need to experiment a lot in order to get your proofs this short. – wj32 Oct 30 '12 at 7:46
Haha. Yes. I have read a few books on calculus. But I was not satisfied with the material in them. Especially the proofs. Calculus always seems to be missing something. I wanted to cover that gap with this book. Hope I have made the right choice. My aim is to get a thorough understanding of the mathematics behind most modern engineering and physics. Vector Calculus, Differential Equations, etc. – Raghavendra Kumar Oct 30 '12 at 8:33
The name is Principles of Mathematical Analysis. :) – user43081 Oct 30 '12 at 10:21

Note that this is "just" a proof that $\mathbb Q$ is not Dedekind complete by exhibiting something kind of like a Dedekind cut $(A,B)$ (except only with positive rationals) where $A$ doesn't have a supremum (in $\mathbb Q$) and $B$ doesn't have an infimum (in $\mathbb Q$).
While the proof of this fact is perfectly correct and clear, it does suffer from the slight blemish that his formula for $q$ seems to be pulled out of thin air.
In response to your comment: It's a standard exercise in real analysis to prove that when $a>0$ then the recursively defined sequence $\bigl(\tfrac12(x_n + \frac{a}{x_n})\bigr)_{n\geq1}$ converges monotonically to $\sqrt a$ for any choice of $x_1>\sqrt a$ (this is the Babylonian method for computing square roots). Using this, you could instead define $q = \tfrac12(p + \frac2p)$ for $p>\sqrt2$.
Rudin's idea is based on something similar: make $q$ a better approximation to $\sqrt2$ than $p$ without accidentally getting to the other side of $\sqrt2$. That's what subtracting $\frac{p^2-2}{p+2}$ from $p$ does. It's just enough to get closer, but not enough to overshoot.