# How to prove it? (apostol)

Discussion in 'Physics & Math' started by alyosha, Jun 14, 2006.

1. ### alyoshaRegistered Senior Member

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I just purchased Tom Apostol's infamous calculus book. In the section on infimum and supremum, he notes, but does not prove, that sets such as all real z that satisfy z>x have no smallest member. Perhaps this is completely self evident (and it is) but many of his own proofs and those in the exercises involve proving things that are self evident to our intuition. The entire point, I suppose, is to verify that the axioms we have agreed to build our system on can at least be used to show things that are "self evident" to indeed be true. I have spent some time thinking about how this would be proved from the other axioms and theorems given in the book, but I am fairly new to the rigorous method. The opinions of those on this forum would be helpful. I imagine that any such proof would involve showing that a smallest member in such a set would make for a simple contradiction in inequalities, but, again, this concept seems so self evident that it is difficult to rid oneself of the stumbling block that is intuition.

3. ### alyoshaRegistered Senior Member

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By the way, if it is anything like his proof of the existence of unique nonnegative square roots for positive numbers using the concept of least upper bounds, then I believe all hope may be lost for me. That particular proof "makes sense", but just how one would come up with it......is beyond me at the moment. Maybe there is something more obvious than it seems about its construction, but in the condensed form it's given on the page it looks like the workings of a miracle.

5. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Hi alyosha,

Welcome to sciforums. In the case at hand I think the method of proof is suggested quite naturally. Let me explain. Intuitively, why does the set {z | z > x} not have a smallest member? Let y be our candidate for the least element. Obviously y cannot be equal to x. With this in mind, can you build another number from y and x which is smaller than y and bigger than x? What does this tell you?

7. ### przyksquishyValued Senior Member

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It's pretty simple, really. If x is your infimum, all you have to show is that for any z > x, there's a number between x and z (a y such that z > y > x). If this is true in general, it means there cannot be a minimum. Can you think of an example?

8. ### shmoeRegistred UserRegistered Senior Member

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If you're still having trouble after the hints above, try considering the special case where x=0, that is the set of real numbers z where z>0. You might find this easier to work with, and it's just a translation away from the full problem.

That's alright, it sounds like you can follow the steps and that's a good start. Most advanced math will look like arcane scribblings when you first encounter it. After pushing the symbols around a bit, you will become more and more comfortable and will start to develop an intuition for these things, and proofs like this will become simpler and simpler.

9. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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All the guys here know more math than me, but have not been explicit, but as I seldom get to answer a math question clearly and explicitly:

Starting with PM's candiadate y as "the smallest," construct as "step (1)" [(y-x)/2 + x] and call it "y2." Step (2) Note that y > y2 > x. Now rename y2 as y and return to step 1. There is no end to (exit from) this program loop.

I am not a mathematician, so perhaps that is not rigorous enough, but it convences me.

Last edited by a moderator: Jun 14, 2006
10. ### alyoshaRegistered Senior Member

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121
It is interesting that it was suggested that I prove that for x < y there must be some z that x < z < y. The fact is, that my original question appeared when I was trying to prove this exact statement (as an exercise).

Here's how that particular proof panned out:

Consider the set S = {t| t > x}

This set is nonempty (not only because of y but because for a given real there exists a positive integer larger than it) and is bounded below by x. Thus S has some infimum, say b. Now either b > x , b < x, or b = x.

b < x cannot be true because then b would not be the "greatest" lower bound.
If b > x, then by the definition of S, b is an element of S, and by the definition of an infimum that is an element of a set, it is the minimum element of that set. But, according to Apostol (and the proofs you have suggested), such a set can have no minimum element.

Then b = x.

By another theorem, S must have that property that given a postive number h, for some t in S

t < inf S +h

If this were not so, then all t > = inf S +h, and by definition inf S + h would become the new inf S.

We have shown that x = inf S. Also, because x < y, the number (y-x) must be positive. Let us return to the inequality with x = inf S and h = (y-x).

Then for some t in S,

t < x + (y-x) = y

and thus for some t in S,

t < y

But by the definition of S, t must also satisfy t > x. Then we have shown that there exists some t such that

x < t < y.

In my humble opinion, I thougt at the time that this proof was pretty airtight with the exception of the "assumption" that x was the infimum of S. I am new to this, is this proof too little or too much?

Last edited: Jun 14, 2006
11. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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PS that convenced Zeno also - what more proof do you need.

12. ### alyoshaRegistered Senior Member

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Billy's suggestion is clever and with it my only concern would be to demonstrate clearly that this y2 always y > y2 > x. It could also be used, I suppose, to demonstrate the idea of a z for x < y such that x < z < y. This demonstration seems like it could "clearly" be done with the following reasoning:

(y-x)/2 is positive and necessarily x < x + c where c is a given positive number.

(y-x)/2 + x < y is verified by solving until you have 0 < (y-x)/2, which was asserted earlier. My only qualms with this is that for the moment I feel uncomfortable venturing outside of the exact tools given the chapters I've read so far. I want to at least make an attempt at rigor. Perhaps, though, this is "rigorous" enough and I have only been making things too difficult.

13. ### funkstarratsknufValued Senior Member

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Consider the case where x=0.

Assume that y > 0 is the minimal element of {z | z > 0} i.e. the positive reals. Now consider y/2. Certainly, 0 < y/2 < y. But then y/2 is an element of {z | z > 0} that is strictly less than y, so we have a contradiction with the assumption that y was the minimal element. Hence, the positive reals have no least element.

Translate to the general case and you're done.

14. ### alyoshaRegistered Senior Member

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121

Yes, he gives ten axioms: the field axioms, the ordering axioms, and the least upper bound axiom, and builds from there.

What is the main difference between this and a "real" real analysis book, say, walter rudin's "principles of mathematical analysis"? I've heard that apostol's (and spivak's) books would more accurately be called introductions to analysis, than mere calculus books.

Also, is there any kind of "flowchart" for learning different topics in math, showing which topics are prerequisites for others? If I wanted to work my way from this book to something like topology or differential geometry, what else would I need to know? I've looked in the table of contents of real analysis books and they cover some topology before almost anything else, but I've heard from other places that topology books assume some familiarity with analysis....this.....is a little confusing. How much set theory and modern algebra is necessary for other topics?

15. ### draqonBannedBanned

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bratan, matematika ubiyot tebya, ey net koncza.

16. ### §outh§taris feeling causticRegistered Senior Member

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hello alyosha,

i'm a bit ahead of you so look here if you are having problems with some of the.. problem sets: http://www.sciforums.com/showthread.php?t=51312

If you like we can work through them together. I took an extended break after going through the introduction so I could focus on electrodynamics.

17. ### §outh§taris feeling causticRegistered Senior Member

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4,832
Z, the trichotomy law is a theorem not an axiom

18. ### alyoshaRegistered Senior Member

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121
Thank you for the help, South Star. It was your thread, in fact, that led me to these forums, which seem to be very legit.

Again, I have some questions about the progression of my interest in math. If I were to read Apostol's book and then Walter Rudin's "The principles of mathematical analysis", would I be prepared for something like Manfredo Do Carmo's books on Differential Geometry? found here:

http://www.amazon.com/gp/product/0132125897/102-9309023-0181723?v=glance&n=283155

I was initially drawn to math through physics, and eventually I became interested in the study of math in its own right. I dislike how physicists sometimes might ignore a rigorous approach. To me it is important to know what assumptions you are making about reality before you apply something purely theoretical (like math) to it. I was skimming through an introduction to quantum mechanics when I came across the mistake physicists originally made: attempting to calculate the energy of light spectra with the integral, which of course presupposes that the phenomenon is a continuous one. Little things like this, to me, highlight the importance of the deepest possible understanding of the higher mathematics when applying them to the physical world.

How did everyone else here get involved with mathematics, and how did your education in this subject progress?

19. ### alyoshaRegistered Senior Member

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121
P.S. How does Spivak's infamous "Calculus" book compare with Apostol's? When would be the right time to attempt something like Spivak's "Calculus on Manifolds"?

20. ### §outh§taris feeling causticRegistered Senior Member

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I haven't progressed past Apostol but I would strongly recommend mastering his stuff before moving on to Rudin. I have Rudin but I haven't dared read it yet because I want to have solid fundamentals. I think after going through Apostol Vol 1 and 2 you should be ready for Rudin though because, if I remember correctly, that's the regular US sequence.

As an aside, physics problems can only be solved by a good intuition and NOT merely mechanically knowing facts. That said the book I'm using for electrodynamics (by Griffiths) is already asking me to prove certain vector properties (I just started) and I expect I will also have to go through other derivations. You might also be interested in Morris Kline's calculus book from what you have said. It's certainly tough and thorough but not as comprehensive as Apostol (doesn't start from set theory but rather from limits, for example).

I also forget the difference between Apostol and Spivak. They are both certainly through and comprehensive but I personally prefer Apostol's formal prose to Spivaks, which is a little more loose. Why, Spivak sometimes provides as examples proofs to problems Apostol assigns and also explains them well. On page 29 of Apostol's book he presents the most horrific looking proof (to novices like ourselves) of why every nonnegative real # has a unique nonnegative real square root. Hitherto, and afterwards, his proofs are elegant but there I'm not sure what happened.

Myself I am only at the level of plug and chug calculus and I'm using this summer to treat myself to a solid understanding of the subject. I'm also learning electrodynamics because Apostol can become a little exasperating because none of the proofs he offers as exercises are solved in the back and so when you are stuck you are stuck... If I remember Apostol (or perhaps Spivak) provides examples from astrophysics in his book. Unfortunately these books cost a few pennies and so I won't be buying Spivak.

It sounds like you are very interested in rigour and so it should be a pleasure learning with you. As for these forums seeming legit, I am seeing otherwise day after day after day...

Oh. And I was looking through my library and came across Apostol's Introduction to Analytic Number Theory which he describes in the introduction as suitable even for 'sophisticated high school students' with no knowledge of calculus. I hadn't even previously looked at it. I'll keep at calculus though and maybe later on look at that. I'm hoping to get into some math competitions and flex my brain. The elements of real analysis, by Robert Bartle, is also another book seemingly on the level of Rudin. It's too bad I'm not at the level to soak up these books just yet.

Feel free to post any Apostol questions here and we'll see if we can help you. I'll do the same. Starting later today or definitely tomorrow. I'm dead tired right now.

Last edited: Jun 17, 2006
21. ### alyoshaRegistered Senior Member

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121
Actually Kline's book was the first exposure I ever had to calculus. I got all the way to the end, to triple integrals, when I just lost focus because I did not feel that I had an extremely solid idea of what the integral was. His book, and many others, makes the miracle jump from solving for the limit of a series of rectangles underneath a curve, to the fundamental theorem of calculus. Obviously "it works", but just telling me the definite integral and the direct limit method are the same isn't enough. Klines book would probably be a good supplement to a proof-theorem style book, because it gives one a birds eye view of the material. It seems that in apostol he actually proves that the definite integral is capable of giving area, and this is exactly what I was looking for. I'm not interested in solving arbitrary poblems for a standardized test, I just want to know exactly how it all works and then move on.

22. ### §outh§taris feeling causticRegistered Senior Member

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alyosha how did you prove the existence of the greatest integer in x?

23. ### alyoshaRegistered Senior Member

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If x is an arbitrary real, prove there is exactly one integer n satisfying

n <= x < n + 1

The first thing was to show that the set of integers less than or equal to x was nonempty, or in other words not bounded below by x. I did this in the same manner as apostol proved that the set of positive integers couldn't be bounded above. We first should agree that an integer is any number that can be obtained by starting with the number 1 and repeatedly either adding or subtracting 1. He mentions this offhandedly in the beginning of the chapter.

Consider the set of all integers, Z, and suppose Z is bounded below by arbitrary real x, so that for all integers i in Z,

Z = { i | i >= x}

now by theorem I.29 there exists some integer n > x so Z is nonempty and has a greatest lower bound (GLB) that we will call b, such that

b <= i

For all i in Z.

Consider b + 1, which cannot be the GLB because necessarily b+1 > b. Then there must be some integer n such that n < b + 1, if there weren't then b + 1 would be the GLB. However, for this n, we find that n - 1 < b. We already agreed that n was an integer and subtracting one from an integer necessarily results in another integer. Then we have n - 1, an integer smaller than b, the supposed GLB of Z, and the contradiction shows that Z, the set of all integers, cannot be bound from below. (this proof is enough for the previous problem in the book proving the existence of integers m and n for every arbitrary real such that m < x < n)

So, finally, we can say that given the set S of all integers i less than or equal to x, must be nonempty, that is,

S = { i | i <= x} is nonempty and bounded above by x.

Then S has a least upper bound (LUB), say, n, such that for all i in S,

n >= i

Now either n > x, n < x, or n=x.

n > x is not possible because then x would be an upper bound less than the LUB.

So n <= x, and in either case n is the largest member of S.

Now, x-n < 1, because if x-n >= 1 then x >= n+1 , and n+1 would be an integer larger than n in S, contradicting the notion of n being the LUB.

Then necessarily x-n < 1 and consequently x < n+1.

So we have shown the existence of some integer n such that

n <= x < n+1

Now it's time for uniqueness.

Suppose there were some other integer a in S satisfying

a <= x < a+1

Either a < n, a > n, or a=n.

a > n is impossible because n is the LUB.

If a < n, then necessarily n-a >= 1 (because they are both integers)

But then n >= a +1, and because we have already shown that n <= x,

x >= n >= a +1, and

x >= a +1.

But this contradicts the assertion that x < a +1 , so we have shown that both

a > n and a < n are both impossible, and necessarily

a = n completing the proof of the uniqueness.

My biggest problem with this proof is that I may have taken some liberties with the properties of integers.

Last edited: Jun 23, 2006