# 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.

Thanks in Advance !

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

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## marked as duplicate by lhf, kahen, Douglas S. Stones, Norbert, ArkamisOct 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

That is not from a proof of the existence of the real numbers. That is stated much later in the chapter and the proof is in an appendix if memory serves me right.

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.

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Exactly, that blemish is what is troubling me. Is there any other approach to this? or any other proof? –  Raghavendra Kumar Oct 30 '12 at 8:49