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I have been looking at intro to real analysis. I am using the text book "Principals of Mathematical Analysis, third edition" by Walter Rudin. I have some questions about things I found confusing and wanted to clarify.

1) The chapter begins with the discussion of the equation $p^2=2$, and the proof of how $p$ is not rational. I can follow this proof, but then it says, Let $A$ be the set of all positive rationals p such that $P^2<2$ and let $B$ consist of all positive rationals such that $p^2>2$. " We shall now show that A Contains no largest number and $B$ contains no smallest.

More explicitly for every $p$ in $A$ we can find a rational $q\in A$ such that $p<q$ for every $p\in B$ we can find a rational $q\in B$ such that $q<p$ This is where I get confused. Now it says to do this we associate with each rational $p>0$ the number $$q=p-( p^2-2/{2}{p+2})=\frac{2p+2}{p+2}\tag 3$$ Then $$q^2-2=\frac2{p^2-2}{p+2}^2\tag 4$$

If $p\in A$ then $p^2-2<0$ and $(3)$ shows that $q>p$ and $(4)$ shows that $q^2<2$. Thus $q\in A$.

If $p\in B$ then $p^2-2>0$ and $(3)$ shows that $0<q<p$ and $(4)$ shows that $q^2>2$ thus $q\in B$.

Can anyone help explain this last part? How and why do we associate $p$ with that different mix of numbers..

My second question is in regard to the Archiemedian Property. I also follow the proof right up until the end.. a) If $x\in\mathbb R$, $y\in\mathbb R$ and $x>0$, then there is a positive integer $n$ such that $nx>y$.

proof; Let A be the set of all $nx$, where $n$ runs through the positive integers if (a) were false then y would be an upper bound of A. But then A has least upper bound in R. Put $\alpha=\sup A$ since $x>0$, $\alpha x< \alpha$ and $\alpha-x$ is not an upper bound of $A$. Hence $\alpha-x < mx$ for some positive integer $m$. But then $\color{red}{ \alpha< (m+l)x \in A}$ which is impossible since $\alpha$ is an upper bound of $A$.

I can follow along right about until " But then $\color{red}{\alpha< (m+1) x \in A}$ which is impossible since it is an upper bound. What I am confused about is as follows: yes $\alpha$ is the least upper bound, so how does this mean it cant be smaller than anything in the set? I thought by least upper bound it would actually have to be smaller then anything in the set because it is the least upper bound.

Thanks so much for any help.

voldemort
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Quality
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  • See math notation guide. Also, please separate questions separately –  Aug 12 '14 at 23:22
  • Your re-edition has no sens ! what mean 2/2p in the formula (3) ? And you can correct as well the red formula ! – idm Aug 12 '14 at 23:35
  • I have a second edition Rudin, not the third, but I do not find anything in it that is at all like the OP's equations (3) and (4). The equations that do appear in this portion of the second edition's discussion of $p^2=2$ are unnumbered and quite different (for one thing, they are correct). – Barry Cipra Aug 13 '14 at 00:07
  • @BarryCipra Thanks so much your input really helped. Is there a reason you feel the need to act so superior to just someone asking a question? Why dont you look at the online editions of the third edition and see for yourself. – Quality Aug 13 '14 at 00:10

1 Answers1

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For $2)$ I guess you mean $\alpha < (m+1)x$. But $(m+1)x$ is an element in the set $A$! Now, $\alpha$ is an upper bound for $A$ means that for all $a \in A$, $a \leq \alpha$. Hence the contradiction.

$1)$ is easy enough. I would recommend that you think it through. Maybe write a "value" for $p$ and see what the formula for $q$ tells you- and then it would be clear.

For example, if $p^2-2<0$ then $q=p-$(negative number) by $3$ in your question, and so $q>p$.. Just carry out the computations- there's nothing very deep happening here

voldemort
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