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I was going through walter Rudin's first example when trying to show $A = \{p \in Q_+ : 0<p^2<2 \}$ has no smallest element (not $Q_+$ are positive rationals). In the first part of the proof he has the statement:

$$ q = p - \frac{p^2 - 2}{p+2} = \frac{2p+2}{p+2} $$

and the goal of it is to show $p < q$. For me its obvious just from the first statement that $q > p$ because $p \in A \implies p^2 < 2 \iff p^2 - 2 < 0 $. So it means $ - \frac{p^2 - 2}{p+2}$ has to be positive and therefore $q$ is a little bigger than $p$ since it starts off at $p$ and only increases.

However, if I only had the second statement I would have no idea if $q > p$. Is there a way to see that the second statement is larger than $p$ also without appealing to the first statement? Or what is the purpose of the second statement on Rudin's example?

Charlie Parker
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  • Is there a square missing in your definition of $A$? – Servaes Dec 04 '16 at 00:43
  • @Servaes yes. oops – Charlie Parker Dec 04 '16 at 00:43
  • Your statement isn't true. Consider $p = \sqrt{3}$, $\frac{2\sqrt{3} + 2}{\sqrt{3}+2} = 2\sqrt{3}-2 = 2(\sqrt{3} - 1) < \sqrt{3}$ – Eli Sadoff Dec 04 '16 at 00:44
  • @EliSadoff $p = \sqrt 3$ is not in $A$. Only rationals are allowed thats why I had $Q_+$. Also, $p^2 = (\sqrt 3)^2 = 3$, so even if reals were allowed your p is not in $A$. – Charlie Parker Dec 04 '16 at 00:45
  • @Servaes why did u delete ur answer? – Charlie Parker Dec 04 '16 at 00:46
  • Because it didn't adress your question; I did not read it thoroughly. – Servaes Dec 04 '16 at 00:48
  • @Pinocchio I missed the $p^2$. I used $\sqrt{3}$ because despite not being in the condition itself, it was an easy number to use computationally that has infinitesimal close rational approximations allowing it to work. – Eli Sadoff Dec 04 '16 at 00:50
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    @EliSadoff sorry you might have seen my question before I corrected it (it was missing the $p^2$). Thanks for the help though, it was fun to think about your objection. :) – Charlie Parker Dec 04 '16 at 00:52
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    @Pinocchio No worries. I must have. I put it into mathematica to check because I thought that there's no way that is true for $0 \lt p \lt 2$. – Eli Sadoff Dec 04 '16 at 00:53

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Since $p^2<2$, we have $p^2+2p<2+2p$, so $p(p+2)<2p+2$, whence $p<\dfrac{2p+2}{p+2}$. (Note that $p+2>0$.)

Ted Shifrin
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  • your argument is definitively correct. Though, Im not sure if I would have thought of adding $2p$ to both sides. Seemed to come out of nowhere. It doesn't help that Rudin never explicitly says how either equality shows $q>p$ but I guess he could have had your proof in mind (which I don't really find as an "easily" way to see that its what we wanted). – Charlie Parker Dec 04 '16 at 00:57
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    @Pinocchio: As usual, one works backwards and then rewrites the proof forwards. If you're not used to that, play with it. :) Make sure all the steps are reversible, though. – Ted Shifrin Dec 04 '16 at 01:19
  • I didn't realize thats what you did. I see, that way you started of with a true statement in the forward direction and the conclusion was found at the end. I guess I was wondering what the point of even writing $ \frac{ 2p + 2}{ p+2}$ was, since it doesn't reveal much by its own. But looking at it, now I realize its a lot more clear that $q \in Q$ since the expression doesn't involve a square at all, which is definitively nice (though, not necessary since $p \in Q \implies p^2 \in Q$. The thing is that otherwise that second statement looks rather random to me). – Charlie Parker Dec 04 '16 at 01:23
  • There's also a good discussion here about how $q$ could be constructed by working backwards (and committing mathematical crimes along the way). – Yibo Yang May 23 '18 at 01:16