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So I've been working on Rudin's 1.1.3 for a while and I think I'm close to a figuring it out.

In the set $A$ there are positive rationals $p$ such that $p^2<2$ and we can find a rational $q$ in $A$ such that $p<q$. So we can also say $q^2<2$.

I want to say that $q=p+x$ where $x$ is very small so that I can say $p^2<q^2$ because $q^2=(p+x)^2=p^2+2px+x^2$ and that's bigger than just $p^2$

Now $p^2<q^2<2$ and I can move things around so that I get $0<q<p+(2-p^2)/(q+p)$ which I can then rearrange into what I have below.

So, unless I'm doing things wrong, I can get $p<q<p-(p^2-2)/(p+q)$

The latter part of that relation is somewhat close to what Rudin has in 1.1.3 but I still have inequalities and my denominator still has a $q$.

=Update=

So reading a helpful reply, I've simplified by attempts:

We know that $q^2<2$ and if $q=p+x$ then we have $(p+x)^2<2$. We can then the $x$ on one side and get this; $x(2p+x)<2-p^2$

But still not sure completely clear what the next methodological step is.

=Update 2=

For some reason I still can't make the connection from $2p+x = p + p +x$ and the $p +2 $. I understand that the latter expression is probably chosen due to elegance in how it will nicely simplify eventually.

For this last stage, I understand that

$0<p^2<q^2=(p+x)^2<2$ and because $(p+x)^2<2$ then either $p$ or $x$ is positive and less than $sqrt(2)$. I suppose we can think of it as $p+x<sqrt(2)$ even though we shouldn't be using sqrt(2). Like, sure, $p+x<sqrt(2)<2$ but then how do we get rid of that inequality when we're replacing that $p+x$ with $2$.

I'm pretty sure I understand the theory of what Rudin's doing and your replies, but the actual computation with the inequalities is messing me up.

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  • $\begingroup$ Are you talking about Equation (3) in Section 1.1 of Rudin's Principles of Mathematical Analysis? $\endgroup$ – user49640 Jun 14 '17 at 5:28
  • $\begingroup$ Yes, on page 2. Sorry, should have mentioned the book since he has multiple ones. $\endgroup$ – Ally Jun 14 '17 at 5:31
  • $\begingroup$ You already know that there are infinite many $q$ that can be selected, So there are two focused questions you might ask:<br> Can someone help me show that Rudin's formula for $q$ works? <br> How was Rudin motivated to come up with his formula for $q$? $\endgroup$ – CopyPasteIt Jun 14 '17 at 13:06
  • $\begingroup$ Guess you spent enough time on this. If you would like to see a proof that Rudin's $q$ works, see math.stackexchange.com/a/2217492/432081 Bunch of further info and proofs there. $\endgroup$ – CopyPasteIt Jun 14 '17 at 22:24
  • $\begingroup$ You can't get a rational number that is "infinitely close" to $\sqrt{2}$. That statement is not meaningful. You can get a rational number as close as you like to it, but depending on how close you say you want it, you have to pick a different number. Also, it's not that you "know that $p + x < 2$, it's that you're looking at what the consequences are if this is true, and then at the end you're picking $x$ so that it is true. Then you say that for a number to be in the set $A$, it has to be less that $2 - p^2$. That's not true. $\endgroup$ – user49640 Jun 18 '17 at 22:39
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I don't think it's necessarily easy to go from what you're doing to what he does.

You are basically looking for some small, positive number $x$ that satisfies the inequality $$x(2p + x) < 2 - p^2.$$

At this point, there are many ways to pick $x$. For example, you could pick $x$ to be any number that is both less than $p$ and less than $(2-p^2)/3p$.

But if you want to get Rudin's result, you first impose the condition $x < 2 - p$. Now it is sufficient for the inequality $$x(p+2) \leq 2 - p^2$$ to be satisfied.

So it's enough to take $$x = \frac{2-p^2}{p + 2}$$ and to check that this value of $x$ satisfies $x < 2-p$. This is straightforward to check.

What I think Rudin is really doing is this. Draw the parabola $y = x^2 - 2$. You already know a point $(p,p^2 - 2)$ on the parabola, and you want to find a better approximation than $p$ to $\sqrt{2}$, the new approximation being on the same side of $\sqrt{2}$ as $p$ is.

To do this you draw a line of positive (rational) slope $m$ through the point $(p,p^2 - 2)$ you know, and take the intersection $(q,0)$ with the $x$-axis. The fact that $m > 0$ guarantees that $q$ and $\sqrt{2}$ are on the same side of $p$. To be certain that you don't overshoot $\sqrt{2}$, all you need to do is make sure that the slope $m$ of the line is larger than that of the secant line from $(p,p^2 - 2)$ to $(\sqrt{2},0)$, which is $p + \sqrt{2}$. For convenience, Rudin takes $m = p + 2$, but he could have taken $p + 3/2$ or a lot of other things.

Addendum

When you reach $x(2p + x) < 2 - p^2$, you have a lot of freedom in how to pick $x$. In this problem, all we need to do is find one positive value of $x$ that works. The condition is equivalent to $$x < \frac{2-p^2}{2p + x}.$$ Because $x$ appears on the right side, it wouldn't make sense to simply say "Pick $x$ to be less than $(2-p^2)/(2p + x)$." Nonetheless, since $x$ can be picked as small as we like (clearly, if $(p + x)^2 < 2$, this will remain true if $x$ is replaced with a smaller positive number), we can simplify things by assuming that $x$ is less than some convenient value. In particular, $2p + x$ will be much simpler if we replace $x$ with $p$, since $x$ will no longer appear in the expression.

Consider the following chain of inequalities, which may or may not be true for a given value of $x$. $$x(2p + x) < x(3p) \leq 2 - p^2.$$ The first inequality will be true if $x < p$. The second will be true if $x \leq (2-p^2)/3p$. So any choice of $x$ that is less than the minimum of $p$ and $(2 - p^2)/3p$ will necessarily be a solution of the original inequality.

Addendum 2

In response to a comment, I will add details about the part of the answer that explains Rudin's expression for $q$.

This is very much like what I explained above for $3p$, but with $p + 2$ instead. Essentially, in order for the inequality $x(2p + x) < 2 - p^2$ to hold, it is sufficient for the following two inequalities to hold: $$x(2p + x) < x(p + 2) \leq 2 - p^2.$$ I have not written $p + 2$ here for any reason other than that it's the particular expression Rudin chose. Other expressions would also work, including $p + 3/2$, or more generally $p + a$ where $a$ is any rational number larger than $\sqrt{2}$.

When does the double inequality hold? The first part holds when $2p + x < p + 2$ or, equivalently, when $x < 2 - p$. The second part holds when $x \leq (2 - p^2)/(p + 2)$. As I mentioned in my answer, the second inequality actually implies the first, because we certainly have $(2-p^2)/(p + 2) < 2 - p$. Therefore it is enough to take $x = (2 - p^2)/(p + 2)$, or any smaller rational number.

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  • $\begingroup$ Thanks for the help. That first inequality of yours is great because I see that I was going down a too complicated path. I went a step back and simplified my question. I'm not sure what prompts the condition other than something like if q^2 < 2 then q <2 and so p+x < 2. But even then, I'm also sure where your second inequality comes from. $\endgroup$ – Ally Jun 14 '17 at 7:45
  • $\begingroup$ I've added a clarification of what I said about how you can select $x$. $\endgroup$ – user49640 Jun 14 '17 at 8:03
  • $\begingroup$ Yes, it's getting more clear. I still have to think about it more but this will be good to consider in the meantime. Thanks. $\endgroup$ – Ally Jun 14 '17 at 8:21
  • $\begingroup$ I like the geometric description of what Rudin might be doing. Makes me want to develop the theory of linear quadratic systems over the rational numbers. $\endgroup$ – CopyPasteIt Jun 14 '17 at 14:47
  • $\begingroup$ You know, I think I get it now because of similar asked questions. I've got to where I have "x (2p + x) < 2 - p^2" and I know we have to d something about the terms in the brackets so we can have x only on the left side. Apparently we have "2p + x" becoming "p+2" but that's because we somehow know that "p + x < 2". Without knowing the answer, how do you get from "2p+x" to "p+2"? I understand the "p+p+x" part but what's the intuition about the "p+x<2" to make the magic happen? the 2 seems arbitrary to me, unless Rudin wanted a number which could perhaps relate to both A and B simultaneously? $\endgroup$ – Ally Jun 17 '17 at 10:09

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