## Help with Principles of Mathematical Analysis

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s0792554
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### Help with Principles of Mathematical Analysis

Hi,

I recently started working my way through Rudin's Principles of Mathematical Analysis (3rd Edition) in my own time, and I'm having a few problems in the first chapter with some of the finer points of Rudin's proof of the density of the rationals in the real numbers. I don't currently have anyone at hand offline to help me with this, so I would be very grateful for some help.

I'll just briefly outline the form of the theorem and (most of) the proof, since it's so short.

Basically, we're given real numbers x,y with y > x, and we want to show we can construct a rational number between x and y, (using the Archimedean property and the L.U.B property).

The proof starts with the Archimedean property being used to produce a positive integer n such that n(y - x) > 1.

The next step involves using the Archimedean property again to find lower and upper bounds for nx, i.e positive integers m1 and m2 are introduced such that m1 > nx and m2 > -nx, leading to the inequality

-m2 < nx < m1

The next sentence states 'Hence there is an integer m (with -m2 <= m <= m1) such that m - 1 <= nx < m.'
(The remainder of the proof just combines the first and last inequalities and divides through by n to get x < m/n < y).

My problem is that I don't see how producing the bounds on nx justifies the conclusion that nx is between (or coincides with the smallest of) two consecutive integers. Perhaps the idea is that given an upper bound like m1 we can simply subtract 1 a finite number of times (since we'll eventually reach -m2) until we either reach nx or something smaller than it, whereupon we'll have the pair m -1, m such that m - 1 <= nx < m? That's my best guess, but I'm not really convinced it's right. It just seems as though a little bit is missing from that part of the proof. I understand this is par for the course with PoMA, and that's fine, except that I'm unsure how to fill in the gap.

I don't feel as though I've explained my misunderstanding very well. It's just that given the content of the proof, I don't see why the desired m isn't produced straight away by noting that ny - nx > 1 implies there's an integer between ny and nx, since it seems to rest on the assumption that consecutive integers are separated by 1 unit, which as far as I can tell is the same assumption employed later to get m by a more circuitous route. Obviously I'm missing something, since if the stuff with the extra bounds was redundant it wouldn't be in there.

Dark Avorian
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### Re: Help with Principles of Mathematical Analysis

I think it's using the well-ordering principle for integers there
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Tirian
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### Re: Help with Principles of Mathematical Analysis

I agree. Rudin seems to think it beneath mention that any non-empty subset of natural numbers contains a least element.

There is an elephant in the room when dealing with the construction of the real numbers, which is that it's a pain in the butt to deal with both positive and negative numbers at the same time. IMHO, people with good sense do all this work on the positive reals only and then extend it to the entire real line using the same trick that we use to get the integers from the naturals. Without that, you're often reduced to a proof by cases when dealing with both positive and negative numbers. Alas, Rudin here must have thought that proof by cases is inelegant but then starts throwing in this m₂ variable that covers the case of x being negative but it just makes everything much more confusing.

Qaanol
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### Re: Help with Principles of Mathematical Analysis

It appears Rudin is finding upper and lower bounds so that he can employ the well-ordering of finite sets of integers. Well-ordering of finite sets is fairly trivial.
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Tirian
Posts: 1891
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### Re: Help with Principles of Mathematical Analysis

Yeah, but nobody would have questions if he had done a proof by cases.

Let x and y be two distinct real numbers, and assume wlog that x < y.

Case I: x and y are both positive. Let n be the smallest natural number such that n(y-x) > 1. Let m be the smallest natural number such that m+1 > ny. Then m/n is a rational number between x and y.

Case II: x is negative, y is positive. Then 0 is a rational number between x and y.

Case III: x and y are both negative. Use Case I to find m/n between -y and -x, so that -m/n is between x and y.

Easy peasy. Rudin's proof is finding upper and lower bounds because he is trying to finish this proof in a single bite instead of three. Not surprisingly, that makes it unreasonably hard to swallow.

z4lis
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### Re: Help with Principles of Mathematical Analysis

Just a bit of advice. There are lots of people that will disagree with me on this, but...

Rudin is an abhorrently bad book to learn analysis out of for the first time. Absolutely terrible. It's wonderful if you already have exposure to analysis and are going through it a second time to flesh out what you already know, and it's also probably good if you have a fantastic lecturer. But having skimmed through the book a bit I have absolutely no idea how any human being has ever self-taught themselves analysis from that book. You need someone to explain the ideas and the "whys" the first time you learn about a topic. A list of theorems and their proofs is simply not sufficient.

On the other hand, I really can't offer you any advice on what else to use. I think the best route would be to actually get 2-3 analysis books and read them concurrently. One of them should be Rudin's. The point is that you can hopefully piece together what's *really* going on if you use several sources.

But please take my advice with a grain of salt. It's just what I feel about the book, and you might absolutely love it and do great. I hope you do!

EDIT: Take a look through some of these posts for an idea of a companion to Rudin. http://math.stackexchange.com/search?q=rudin
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polymer
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### Re: Help with Principles of Mathematical Analysis

Don't abuse yourself, but if Rudin is making sense and you are having fun, then I vote to stick with it. It's not an easy book, but I'm suspicious long explanations in most books only help because they force you to think about an idea longer. You could think about a page of Rudin for hours and nobody would think less of you.

Rudin was the first book where math made sense to me - I can't speak ill of it. But it wasn't my first Analysis book either.

s0792554
Posts: 3
Joined: Mon Feb 18, 2013 9:40 pm UTC

### Re: Help with Principles of Mathematical Analysis

Thanks for all the replies everyone.

I overlooked the well-ordering principle at the time I posted this question, but it makes the use of those bounds I was wondering about a lot clearer. It's unfortunate that standard properties of the natural numbers, integers, and rationals are assumed without any discussion, since otherwise PoMA seems not to assume much prior knowledge of the reader. I did find some good notes online that start with the construction of the natural numbers, and go from there to the integers and rationals.

Tirian:

Thanks for your example of the proof by cases approach, I agree that it seems a lot clearer. I do have one (probably stupid) question about the first case, which is what happens if y is integral? Then it would seem that the smallest natural number m s.t m+1 > ny is ny itself, so that the ratio m/n isn't strictly between x and y.

z4lis:

Your advice is appreciated, and I think I might actually be somewhat inclined to agree with you on Rudin as a first time analysis book, but as I should perhaps have explained in my first post, I do have a little bit of previous experience with analysis, although not too much. I'm in my early twenties and was until quite recently in the middle of studying for a degree in mathematics. Unfortunately, personal issues pretty much rendered me incapable of continuing, and I had to withdraw, although I hope to find myself back in academia in the not too distant future. So I've taken a few real analysis courses, and a lot of the material in the book will be at least somewhat familiar.

My aim is, as you stated, to flesh out some things I already know, gain some familiarity with some other things I didn't follow at all on first viewing, and to develop some mathematical maturity. Consulting multiple textbooks is a good idea, but I don't have access to actual physical copies currently, and I don't like using digital copies of large documents.

Also, thanks for the link to the stackexchange thread, though I think I've seen it already. A lot of good recommendations there.

I think things are going fairly well with PoMA, I'm enjoying myself (most of the time!) and I don't mind that my progress is slow.

Anyway, I've made a little bit of progress since my last question, and there's another theorem later in Chapter 1 I was wondering about.

Specifically, I was wondering how I might prove the Schwarz inequality in a more natural way than Rudin does in 1.35. The proof is not hard to follow, but it involves a first step that clearly has some sort of clever trick behind it that isn't ever explained. Rudin pulls out a particular sum that doesn't obviously relate to the S.I, and performs some fairly simple manipulations until the Schwarz inequality falls out at the end. I'm not too concerned with where Rudin got the sum at the beginning of the proof from at this time. I mostly want to know if my own attempt at a proof was along the right lines.

If you don't have the book to hand, the form of the Schwarz inequality that appears here is |a1b1 + ... + anbn|^2 <= (|a1|^2 + ... + |an|^2)(|b1|^2 + ... + |bn|^2), where a1, ..., an, b1, ..., bn are all complex numbers. Actually, the bk where they appear on the LHS are all conjugates of complex numbers, but on the RHS the conjugate sign is not there. Since the absolute value of a complex number and its conjugate are the same, whether the conjugate signs are there or not shouldn't affect the form of the inequality, I think.

I begin by using the triangle inequality repeatedly (which was proved in 1.33 (e)) to bound |a1b1 + ... anbn|^2 above by (|a1b1| + ... |anbn|)^2. To be rigorous, it seems that I should prove that the triangle inequality can be applied with n summands as opposed to the 2 we see in 1.33 (e), which could easily be done inductively, although I admit I have yet to do this. The form Rudin uses of the Schwarz inequality has the complex conjugates of each of the bk, but
I then multiply out the brackets to get n terms of form |akbk|^2 and (1/2)(n^2 - n) terms of form 2 |akbk||albl| for all cases in which k not equal to l.
I want to reduce these terms to something of form (|ak|^2)(|bl|^2) so I can factor to get the R.H.S.
This is done by noting that (a - b)^2 => a^2 + b^2 >= 2ab for a,b real, so of course this can be applied to the real absolute values we have here. Given something of form 2 |akbk||albl| (k n.eq l) it can be written equivalently as 2|akbl||albk| by some earlier established rules for manipulating abs. values of complex numbers. So then |akbl|^2 + |albk|^2 >= 2|akbl||albk| = 2|akbk||albl| and so I can get a better upper bound by replacing the 2|akbk||albl| terms with |akbl|^2 + |albk|^2.
This gives me |a1bk|^2, ..., |anak|^2 for k=1,..,n, so I can factor (using R's field properties) to get |a1|^2 (|b1|^2 + ... + |bn|^2) + ... + |an|^2(|b1|^2 + ... + |bn|^2), which with one more factorisation can be seen to be the desired R.H.S.

I'm sorry if that's hard to follow. My formatting probably leaves a lot to be desired.

Dark Avorian
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### Re: Help with Principles of Mathematical Analysis

About the Schwarz inequality, a more natural proof perhaps comes from the notion of an inner product. Particularly, for any pair of n-tuples of complex numbers we can define an inner product as follows

(a_1, ..., a_n) * (b_1, ..., b_n) = sum{i=1 to n} a_i conj(b_i)

Where conj() is the complex conjugate. This is a complex inner product, but you should be able to verify that the inner product of a tuple with itself is the sum of the squares of its components. It further is linear with respect to the first combination, and if we reverse the orders of the tuples we just get the conjugate. (Look up complex inner products for more info on this). We can prove a general form of the Schwarz inequality for any inner product satisfying rules (See Cauchy Schwarz Inequality). Also, this explains why there are complex conjugates on one side and not the other. Actually, there are conplex conjugates as |z|^2 = z * conj(z).
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Tirian
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### Re: Help with Principles of Mathematical Analysis

s0792554 wrote:Tirian:

Thanks for your example of the proof by cases approach, I agree that it seems a lot clearer. I do have one (probably stupid) question about the first case, which is what happens if y is integral? Then it would seem that the smallest natural number m s.t m+1 > ny is ny itself, so that the ratio m/n isn't strictly between x and y.

There's no such thing as a stupid question that you've obviously put thought into. Especially since you seem to be right. Einstein said that you should make things as simple as they can be but not simpler than that, and I seem to have gone a little too far just because I didn't want to search Unicode for a lax inequality symbol. I think I meant that smallest natural m such that m+1 ⩾ ny -- or m ⩾ ny -1 is clearer yet I suppose. Let's double check by finding a rational number between 3 and 4. The smallest natural n such that n(4-3) > 1 is n=2, and the smallest natural m such that m ⩾ 2(4)-1 is m=7, and 7/2 is indeed a rational number between 3 and 4.

I should be clear that the algebra and bounds of my argument are the same as Rudin's, it's just a difference of rhetorical skeleton. If you try doing a few different cases in Rudin's way, you'll find that at least one of his m bounds is 0 depending on the relative signs of x and y. In fact, I think we're going to produce the same rational number unless it's something like finding a rational number between -2.5 and -7.5 (where I would pick -7 and I think Rudin would pick -3).

s0792554
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### Re: Help with Principles of Mathematical Analysis

[TLDR]

I've been (slowly) working my way through the appendix of Chapter 1, which is concerned with the construction of the real numbers from the rationals using Dedekind cuts.

I think I need some clarification of a point that was only looked at in passing detail in the appendix.

I'm trying to show that the set of positive Dedekind cuts (those which have 0* as their proper subset) satisfies the multiplicative axioms of a field, where the definition of the multiplication of two positive cuts A,B is that AB is the set of all rational numbers less than or equal to any product ab, where a is in A, b is in B. This is going to be extended to deal with non-positive cuts a little bit later on, but I haven't yet progressed to that point. The text itself doesn't concern itself with the proofs of the field axioms for multiplication, instead averring that the proofs are almost identical to those for the additive axioms, which are provided in an earlier section. For the most part, this seems to be the case.

I've proved to my satisfaction that the product of two positive cuts is itself a (positive) cut, and I've established that this multiplication is associative, commutative, and has as its identity the rational cut 1*. I believe distributivity shouldn't be a problem, although I haven't attempted it yet. I'm having problems establishing that every positive cut has a (positive) multiplicative inverse, however.

We are not given the form of the multiplicative inverse of a positive cut A, but by analogy with the additive inverses discussed earlier, it should be the set of all positive rational numbers q s.t 1/q - r is not in A, for some r > 0, as well as 0 and all the negative rationals. Proving that A^(-1) as defined is a (positive) cut, and that A.A^(-1) is contained in 1* seems straightforward, but proving that 1* is contained in A.A^(-1) is more difficult.

By analogy with the proof that 0* is contained in A + (-A), I was able to come up with a (sketch) proof:

To show 1* is contained in A.A^(-1), we show that if v is in 1*, it is also in A.A^(-1).
We can note that if v <= 0, then v belongs to A.A^(-1), since A.A^(-1) is a positive cut, being the product of two positive cuts, and hence contains 0* (and 0).
So we need only consider the remaining case 0 < v < 1.
Let w = sqrt(1/v), which, from the main body of the first chapter, we know to be greater than 1, given that 0 < v < 1.
Given this w > 1, there should be an integer k s.t w^k is in A and w^(k+1) is not in A.
Then given that w^(k+1) is not in A, and w > 1, w^(k+2) > w^(k+1) => w^(k+2) is not in A and we can also find a positive r (the distance between w^(k+2) and w^(k+1)) such that w^(k+2) - r is not in A.
Hence 1/(w^(k+2)) is in A^(-1) by definition of A^(-1).
Then we note that w^k * 1/(w^(k+2)) = 1/(w^2) = 1/(1/v) = v and we have that v is the product of an element from A and an element from A^(-1) and hence v is in A.A^(-1) as required.
[Since I've already shown A.A^(-1) is contained in 1*, this shows that given positive A, we have positive A^(-1) s.t A.A^(-1) = 1*]

This seems valid, except that I wasn't able to justify the statement "Given w > 1, there's an integer k s.t w^k is in A and w^(k+1) is not in A". In the earlier proof that 0* is in A + (-A), the 'equivalent' statement is that given a w > 0, there's an integer n such that nw is in A, and (n+1)w is not in A, which is justified by the Archimedean property for the rationals. Unfortunately, I haven't been able to see how to prove the newer statement, although it seems quite intuitive. I've been trying to do this for a few days now, but haven't been able to come up with anything (well, I've been trying to prove that given w, v in Q, with w > 1, there's a natural number n such that w^n > v, which will lead to the statement I made in my proof, and v <= w is trivial) or even get anywhere with any approach.

I tried to reduce the exponential form to the more familiar Archimedean form, but I don't seem to be able to get anywhere with it. Maybe if I used logarithms, but since they haven't been introduced (they aren't mentioned at all until the exercises of Chap 1), I don't want to do that. I also tried to consider the rationals in the form 'p/q', but this seems no easier for integers than it is for rational numbers.

Since this wasn't referred to at all in the section of the appendix I mentioned, I'm wondering if I've missed something obvious or taken completely the wrong approach to the whole question. I would be very grateful for a small nudge in the right direction w.r.t this issue, since I'm not hopeful about getting any further on my own at the moment.

[TLDR]

In summary, how would I go about proving that given two rationals w,v, with w > 1, there exists a natural number n such that w^n > v? Hints preferred over a complete proof.

Tirian
Posts: 1891
Joined: Fri Feb 15, 2008 6:03 pm UTC

### Re: Help with Principles of Mathematical Analysis

Off the top of my head, I think it would help to show that w^n > wn for all sufficiently large n.

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