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I was reading Principles of Mathematical Analysis - Walter Rudin. In the start (pg-11), it is shown that the equation $p^2 = 2$ is not satisfied by any rational $p$ i.e. $\sqrt{2}$ is irrational.

After this the book says:

"We now examine this situation a little more closely. Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $В$ consist of all positive rationals $p$ such that $p^2 > 2$. We shall show that $A$ contains no largest number and $В$ contains no smallest.

More explicitly, for every $p \in A$ we can find a rational $q \in A$ such that $p < q$, and for every $p \in В$ we can find a rational $q \in В$ such that $q < p$.

To do this, we associate with each rational $p > 0$ the number $$ \begin{align} q &= p - \frac{(p^2 - 2)}{p + 2} &(2)\\ &=\frac{2p + 2}{p + 2} &(3)\\ &\Rightarrow q^2 - 2 = \frac{2(p^2 - 2)}{(p + 2)^2} &(4) \end{align} $$

If $p \in A$ then $p^2 — 2 < 0$, (3) shows that $q > p$, and (4) shows that $q^2 < 2$. Thus $q \in A$. If $p \in В$ then $p^2 — 2 > 0$, (3) shows that $0 < q < p$, and (4) shows that $q^2 > 2$. Thus $q \in B$."

I am unable to follow this part, specially how we construct the equations (3) and (4). Can somebody explain?

Ritu
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  • I tried to clean up and LaTeX the question; please check that I did not inadvertently introduce an error. Thanks. – Avraham Jul 03 '14 at 14:01
  • Please do not approve edits that undo the formatting of others. – Chris Brooks Jul 03 '14 at 14:28
  • seems that this is the same question or really close: http://math.stackexchange.com/questions/1932371/how-to-prove-that-the-set-a-left-pp2-2-p-in-bbb-q-right/2042690#2042690 – Charlie Parker Dec 04 '16 at 17:03

2 Answers2

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This is a "good old-fashioned" textbook which tells you the answer and proves it is true but never gives you a clue as to where the answer came from in the first place. It might have been something like this.$\def\eps{\varepsilon}$

We have a positive rational $p$ such that $p^2<2$; we want a rational bigger than $p$, such that its square is still less than $2$. So we want, let's say, $$(p+\eps)^2<2$$ with $\eps$ a positive rational. Expanding and rearranging, $$\eps^2+2p\eps<2-p^2\ .$$ Now we could solve this to find $\eps$, but it would be rather messy. More importantly, the formula would contain a square root, so it would be very hard to be sure that $\eps$ was rational. So we have to be a bit more careful. Write the above as $$\eps(2p+\eps)<2-p^2\ ;\tag{$*$}$$ we can get rid of the second $\eps$ by noting that we will certainly need $p+\eps<2$. So if $$\eps(p+2)\le2-p^2\ ,$$ then $(*)$ will be true, because its LHS is less than this LHS. Thus we can take $$\eps=\frac{2-p^2}{p+2}$$ and hence $$p+\eps=p+\frac{2-p^2}{p+2}\ ,$$ which is Rudin's $q$.

David
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  • Thanks... it was quite helpful :) – Ritu Jul 04 '14 at 03:45
  • how did you change from your (*) to the second line? I don't understand how the less than equals came from :( – Charlie Parker Jan 08 '15 at 05:45
  • @David just to clarify, I do understand that $p+\epsilon < 2$, I don't understand how that helps or completes your proof. Sorry if its really obvious to you. – Charlie Parker Jan 08 '15 at 06:29
  • @Pinocchio You mean from $(*)$ to the previous equation? All you have to do is multiply out the left hand side. – David Jan 08 '15 at 12:12
  • @David I mean how $\epsilon(2p+\epsilon) < 2 - p^2$ goes to $\epsilon (p + 2) \leq 2 - p^2$. I know you did something by splitting the $2p + \epsilon$ to $p + (p+ \epsilon)$ and then using $p+\epsilon < 2$, its just not clear what exactly you did and how the inequality turned from strictly less than to $\leq$. Thnx for your time. – Charlie Parker Jan 08 '15 at 16:09
  • It doesn't go that way, it goes the other way because in my answer we are working backwards in order to find a suitable $\eps$. If $p+\eps<2$ and$$\eps(p+2)\le2-p^2$$then$$\eps(2p+\eps)=\eps(p+p+\eps)<\eps(p+2)\le2-p^2\ .$$ – David Jan 08 '15 at 20:18
  • Sorry for the late question in the game, but how do you indeed know that $p+\epsilon < 2$? What is the intuition or rigorous proof for that part? – Charlie Parker Dec 24 '16 at 21:29
  • @CharlieParker The logic goes the other way, see my comment from 8/1/15 at 20:18. We are going to choose $\varepsilon$ to make sure that $p+\varepsilon<2$. – David Jan 08 '17 at 06:40
  • @David sorry for being repetitive, but it feels that just because we want that to be true it doesn't mean it is or maybe it is for some reason not obvious to me right now. – Charlie Parker Jan 10 '17 at 20:51
  • Please note that the logic goes the other way. This is how we might have worked out the problem. For the proof you have to start at the end and work backwards, because the logic goes the other way. – David Jan 10 '17 at 22:52
  • @David I saw your 8/1/15 comment. I understand the whole statement only this is bugging me: How making $p+\epsilon<2$ cause the inequality to be $\le$? – banned Aug 13 '20 at 11:11
  • @Mouse the $\le$ doesn't imply that $p+\epsilon=2,$ merely that it is possible given the derivation. Indeed, $\epsilon(p+2)=\epsilon(2p+\epsilon)-\epsilon(p+\epsilon-2),$ for which the $\epsilon(p+\epsilon-2)<0$, since $p+\epsilon<2,$ and we have therein $\epsilon(p+2)>\epsilon(2p+\epsilon)$, and $\epsilon(2p+\epsilon)<2-p^2.$ So, the $=$ part of $\le$ is possible but not necessary, and answer OP simply meant that from this inequality, we can be certain that $(*)$ is true. – mpnm Jan 06 '21 at 18:52
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Equation 2 is the definition of $q$. Rudin (or somebody) had the inspiration that this $q$ would work. Adding some steps:$$\begin{align} q &= p - \frac{(p^2 - 2)}{p + 2} &(2)\\ &=\frac {p(p+2)}{p+2}-\frac{(p^2 - 2)}{p + 2}\\ &=\frac{p^2+2p-p^2+2}{p+2}\\ &=\frac{2p + 2}{p + 2} &(3)\\ \Rightarrow q^2 - 2 &=\frac{4p^2+8p+4}{p^2+4p+4}-\frac{2p^2+8p+8}{p^2+4p+4}\\ &=\frac{2(p^2 - 2)}{(p + 2)^2} &(4) \end{align}$$

Ross Millikan
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