Baby Rudin question
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From: guyc on 23 Jun 2010 02:04 On Jun 23, 10:59 am, Jonathan Groves <JGro...(a)Kaplan.edu> wrote: > On 6/22/2010 at 8:20 pm, Porky Pig Jr wrote: > > > > > On Jun 22, 7:59 pm, guyc <guy.corrig...(a)gmail.com> > > wrote: > > > Let A be the set of all positive rationals p such > > that p^2>2. > > > > Let B be the set of all positive rationals p such > > that p^2<2. > > > > In order to show that A has no infimum, B no > > supremum, Rudin considers > > > the number q = p - (p^2 - )/(p +2) > > > > Note that Bryant in the same proof considers the > > number q = p/2 +1/p > > > > Very clever - but what is the motivation for these > > q? Where did they > > > come from? Lots of trial and error? A flash of > > inspiration? > > > If you get a hold on Lay, Analysis with an > > introduction to proof (in > > general it's more elementary than Baby Rudin), > > > he proves somewhat similar theorem: > > > Theorem 12.11: Let p be a prime number. Then there > > exists a positive > > real number x such that x^2 = p. > > > He uses the similar machinery and in a footnote > > carefully discusses > > the motivation of choosing those funny numbers. > > > PPJ. > > This is good advice for seeing why these funny numbers are chosen. > I'm sure that luck and insight were required for the ones seeking > such a proof to find this number q. I wonder if there are any other > commentaries in this proof out there; I'm sure there are, but I'm > not as of yet aware of them. > > The main idea behind this proof is this: A containing no > largest number means that, for every p in A, there is a q in A so that > q > p. To say that B has no smallest number means that, for every > p in B, there is a q in B so that q < p. The sets A and B described > are the reverse of what Rudin uses. So the proof is seeking an > example of a q in terms of p so that q meets these conditions: > If p is in A, then q is in A and q < p. If p is in B, then q is > in B and q > p. Since this q does, the statement is proven. > > An annoying feature of mathematical proofs is that such seemingly > unmotivated steps as these are allowed in proofs, and such steps > are not uncommon. Gauss was notorious for writing proofs in which > all tracks of his methods of discovery were hidden from his proofs. > As long as all the steps follow logically from the previous step(s), > the proof is valid--even if some of the steps are completely unclear > as to where they came from. Even the method of proof might not be > clear as to where it came from. For example, the method of proof of > the Urysohn Lemma in topology is the kind of proof that, when you > look at it for the first time, it leaves you wondering, "How the > heck did Urysohn know how to prove it that way?" A proof can be > found in Munkres' "Topology." Munkres introduces the proof as > a very clever one, one that few of us are capable of discovering > without "copious hints." Other ones are Gauss' proofs of the > quadratic reciprocity law. Some mathematicians try not to > present proofs in that way, but not all avoid doing so. > > Jonathan Groves Thanks for all these interesting and helpful responses. In the meantime I looked at Hardy, who has a nice concrete approach; STEP 1 We can construct a chain of rational numbers of which a is the first and b is the last in such a way as any two consecutive numbers differ by as little as we please, say d (property of the rationals). STEP 2 given sets A and B as in my first post, we can prove that an x can be found in A and a y in B such that (2-x^2) and (y^2-2) are as small as we please, say less than d. Substituting d/4 for d above, we can choose two numbers x and y such that (y-x)<d/4. We can suppose both x and y are each less than 2 without loss of generality, so; y+x<4, y^2 - x^2 = (y-x)(y+x) < 4(y-x) < d and since x^2<2 and y^2>2 it follows a fortiori that 2-x^2 and y^2-2 are each less than d. STEP3 It follows that there can be no largest member of A or smallest member of B. for, if x is any member of A then x^2<2. Suppose x = 2 - d. Then we can find a member x1 of A such that x1^2 differs from 2 by less than d, and so x1^2 > x^2 or x1>x. Thus there are larger members of A than x; and, since x is ANY member of A, it follows that no member of A can be larger than all the rest. ------ Wordy, but concrete and easy to follow. Guy
From: David C. Ullrich on 24 Jun 2010 13:05 On Tue, 22 Jun 2010 16:59:34 -0700 (PDT), guyc <guy.corrigall(a)gmail.com> wrote: >Let A be the set of all positive rationals p such that p^2>2. > >Let B be the set of all positive rationals p such that p^2<2. > >In order to show that A has no infimum, B no supremum, Rudin considers >the number q = p - (p^2 - )/(p +2) What? A certainly has an inf, and B has a sup! At least if we're allowing real numbers. A has no _rational_ inf, but that can't be what Rudin't proving here, since the detail below is irrelevant to that. Surely what Rudin says he's proving is that A has no smallest element and B has no largest element? The difference between "inf" and "smallest element" is hugely important. >Note that Bryant in the same proof considers the number q = p/2 +1/p > >Very clever - but what is the motivation for these q? Where did they >come from? Lots of trial and error? A flash of inspiration?
From: Jean-Claude Evard on 14 Jul 2010 21:05 > Let A be the set of all positive rationals p such > that p^2>2. > > Let B be the set of all positive rationals p such > that p^2<2. > > In order to show that A has no infimum, B no > supremum, Rudin considers > the number q = p - (p^2 - )/(p +2) > > Note that Bryant in the same proof considers the > number q = p/2 +1/p > > Very clever - but what is the motivation for these q? > Where did they > come from? Lots of trial and error? A flash of > inspiration? This is an important question, because to prepare an examination on this topic, it would be stupid to just memorize the formula for q and only be able to check that everything works with the formula, without knowing where the formula comes from. It would be wrong to assume that finding the expression for q was just a lucky guess. This expression can be obtained by standards methods in a logical way. It is useful to see these methods, and more importantly, it is useful to get ideas about how to proceed in similar cases. This question is very well chosen, because it is simple, it is quite relevant, it corresponds to a lot of similar cases, and it is not trivial. Tim Little has already given some ideas about how to find q, but I think that it is useful to give a complete answer; more so, because at the same time that we answer the question, we show some of the methods that authors of textbooks use to obtain seemingly mysterious formulas, and we also show some usual problems about publishing in mathematics, for example, the problem of deciding how much to explain every point. In addition, giving a complete answer to the question will lead us to a big surprise: It seems that we cannot answer without committing mathematical crimes. As I have only the first of the two books mentioned in the original posting, I answer only the question about this first book. Concerning the method mentioned from the second book, this method does not work when p^2 < 2, because it gives q^2 > 2. The first book mentioned in the original posting is the following: ------------------------------------------------ Principles of Mathematical Analysis Walter Rudin 3rd edition McGraw-Hill, 1976, 342 pages ISBN-10: 0-07-054235-X ------------------------------------------------ The part of the book of Walter Rudin corresponding to the first question of the original posting is located, in the introduction to Chapter 1, in Example 1.1, on page 2 of this book. In this chapter, Walter Rudin assumes that the reader is familiar with the field Q of rational numbers, but not with the construction of the field R of real numbers. He wants to motivate the reader for the need of an extension from the field Q to the field R, and for the need of precise definitions about the latter. For this, in the first part of Example 1.1, Walter Rudin proves that there are no rational numbers whose square is equal to 2. This is a first motivation for the extension from Q to R. This extension is constructed at the end of Chapter 1. In the second part of Example 1.1, Walter Rudin wants to prepare the reader to the concept of ordered sets with the least-upper-bound property, which is defined later, in Part 1.10, page 4. This does not look symmetric, but in Theorem 1.11, on page 5, Walter Rudin proves that, in a symmetric way, the least-upper-bound property is equivalent to the greatest-lower-bound property, and this leads to the existence of the infimum and supremum of a non-empty bounded subset of an ordered set possessing the above two equivalent symmetric properties. In the second part of Example 1.1, Walter Rudin wants to show that the field Q of rational numbers does not have any of these two properties. This will give a second motivation for the extension from Q to R. This second part of Example 1.1 is the subject of the first question of the original positing. As pointed out by Jonathan Groves implicitly and by David Ullrich explicitly, the original posting contains several mistakes in the copy of the part of the book of Walter Rudin corresponding to this question. Consequently, it is useful that I rewrite this part exactly. This part is the following: ------------------------------------------------------------------------- "Let A be the set of all positive rationals p such that p^2 < 2 and let B consist of all positive rationals p such that p^2 > 2. We shall 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, and for every p in B we can find a rational q in B such that q < p. To do this, we associate with each rational p > 0 the number q = p - (p^2 - 2) / (p + 2) = (2p + 2) / (p + 2)." ------------------------------------------------------------------------- The first question of the original positing is to find where this expression for q comes from. I think that as many authors of books of mathematics, Walter Rudin used the method of working backward to find this expression for q. This method consists to work from the end to the beginning, that is, from what we want to obtain to what we have. For example, to find an idea to prove that an assertion C implies an assertion G, the method consists to start from what we want, which is G, and perhaps see that we might get G from a certain assertion F, and F from E, and E from D, and D from C, and then we try to construct a proof of the form C implies D, which implies E, which implies F, which implies G, and if all of this works, it would prove that C implies G. There are two parts to prove an assertion with this backward method: Part 1: The inductive part: Find an idea for the proof by working backward. For this part, it is usual to not be rigorous, to take some risks, to cheat a little and to use some dangerous simplifications, if we feel that it may help. The objective is to find an idea that might work or not. All mistakes can be corrected later. It is better to get an idea with some mistakes, rather that being frozen and terrified to make any mistake, and remain without any idea about how to start in front of a blank sheet of paper. Part 2: The deductive part. Use the idea from Part 1 to try to construct a rigorous proof. Then keep working, eliminate all mistakes, all cheating, all dangerous simplifications, until the proof becomes 100% correct and rigorous. So, to answer the question of the original positing, and find an idea for the expression for q, we first work backward, and for this first part of the work, we can cheat a little. For example, it helps us intuitively to use s = [The positive square root of 2], in spite of the fact that this is mathematically illegal for a valid proof, since we are in the situation where we have only the rational numbers, and the irrational number s has not yet been constructed on page 2 of the book. To obtain an idea to construct q, it is also helpful to commit a second mathematical crime, and use the fact that the relation of order in the set R of real numbers that we want to define will be an extension of the relation of order in the set Q of rational numbers, with similar properties. In particular, for any positive real numbers a and b, we will have a < b if and only if a^2 < b^2. We already have this property in Q. We will also get it in R later. With this double cheating of using s, which is not yet defined, and using the relation of order in R, which is not yet defined either, everything becomes easier and more intuitive. For example, the set A defined above in a complicated way is simply the set of positive rational numbers that are smaller than the irrational number s, and B is simply the set of all rational numbers that are greater than the irrational number s. Because the set A is totally ordered, saying that A has no greatest element is equivalent to saying that for every element p of A, there exists an element q of A that is greater than p, otherwise, p would be the greatest element of A. Saying that q is in A implies intuitively that q < s. So, to show that A has no greatest element, we first consider an arbitrary element p of A, and then, we are looking for an element q of A such that, intuitively, p < q < s, where the relation of order q < s will be defined in R later in the book. Since p and q are in A, they are positive rational numbers by definition of A. To get q from p, we will increase the rational number p by a small rational number x, that will be small enough so that q = p + x will be smaller than s, that is to say, we are looking for a positive rational number x such that p + x < s, and since x will be positive, we will have p < p + x < s. At this point, we have to remember that, by definition, s is a square root of 2. To eliminate the irrational number s from the inequality p + x < s, we square both sides. Since all of the real numbers p, x, and s are positive, squaring does not reverse the order. This gives (p + x)^2 < s^2, that is, p^2 + 2px + x^2 < 2, that is, x^2 + 2px + p^2 < 2. Then, subtracting 2 on both sides of this inequality gives x^2 + 2px + p^2 - 2 < 0. Thus, a rational number x satisfies the inequality p + x < s if and only if it satisfies the inequality x^2 + 2px + p^2 - 2 < 0. By replacing the first inequality by the second, we eliminate the irrational number s. Now, we have a polynomial of second degree P(x) = x^2 + 2px + (p^2 - 2), and we are looking for a positive rational number x such that P(x) < 0. Since the leading coefficient of this polynomial is 1, and 1 is positive, we have that P(x) < 0 if and only if x lies inside the open interval (a, b) bounded by the two roots a and b of the polynomial P(x). By the special form of the quadratic formula when the term of degree 1 in x is 2 times a constant times x, these two roots are a = -p - sqrt[p^2 - (p^2 - 2)] and b = -p + sqrt[p^2 - (p^2 - 2)], where sqrt(x) denotes the positive square root of any non-negative real number x. By simplifying, we obtain that the two roots of the polynomial P(x) are a = -p - sqrt[2] and b = -p + sqrt[2] ,that is, a = -p - s and b = -p + s. At this point, we see that in Part 1, when we are looking for an idea about how to find q, not only we can cheat a little, but in addition, in our case, we are forced to cheat. We have to use the quadratic formula, which forces us to work with irrational square roots, in spite of the fact that on page 2 of the book, we are not supposed to know what an irrational number is. This may be the reason why Walter Rudin did not want to explain where the expression for q comes from: It is possible that he did not want to cheat, and confuse the students by mixing what we want to construct later, namely, the irrational number s, with what we are constructing now, which is the rational number q. As said above, we are looking for a positive rational number x that must lie in the open interval (a, b), that is, we are looking for a rational number x such that a < x < b, that is, such that -p - s < x < -p + s, that is, such that -(p + s) < x < s - p. Since x must be positive, this reduces to finding a rational number x such that 0 < x < s - p. At this point, we have a technical difficulty: As s is the positive square root of 2, the decimal expansion of s starts with s = 1.414,213,562,373,1... Since p is in the set A, we have 0 < p < s. Therefore, in the decimal expansion of p, the first digit that is not equal to the corresponding digit of s much be smaller than this corresponding digit of s. For example, we could have p = 1.414,213,52, which is rational, because its decimal expansion is finite, and in this example, all of the digits of p are the same as the corresponding digits of s, except the last one, so that p is very close from s. Then we would have p - s = 0.000,000,042,373,1... and we could chose x = 0.000,000,04, which would give what we want, which is 0 < x < p - s. This method of looking for the first digit that is different would always work, but it is not elegant. In general, when we have a difficulty with an inequality, we can try to use some algebraic transformation of the inequality. One of these transformations is the consider the reciprocals: Since x is positive, the inequality x < s - p is equivalent to 1 / (s - p) < 1 / x. Let us introduce the notation y = 1/x. Then the wanted inequality becomes 1 / (s - p) < y, and we have to find a rational number y satisfying this inequality. Since s is irrational, the denominator (s - p) of the fraction is irrational too. The standard way to rationalize this denominator is to multiply numerator and denominator of the fraction by the "conjugate" of s - p. For this, it is usual to write the rational number -p first, and the square root s second, which gives s - p = -p + s. Then the "conjugate" of -p + s is -p - s. Multiplying numerator and denominator of the fraction 1 / (s - p) by this "conjugate" (-p - s) gives 1 / (s - p) = (-p - s) / [(s - p)(-p - s)]. Multiplying numerator and denominator of the second fraction by -1 gives 1 / (s - p) = (p + s) / [(s - p)(p + s)], that is, 1 / (s - p) = (p + s) / (s^2 - p^2), that is, 1 / (s - p) = (p + s) / (2 - p^2). So, the wanted inequality 1 / (s - p) < y becomes (p + s) / (2 - p^2) < y. Now it is a piece of cake to find a rational number y satisfying this inequality. Indeed, we know that s < 2, because this is equivalent to s^2 < 2^2, that is, 2 < 4, which is true. Therefore, we have s < 2, consequently, p + s < p + 2. On the other hand, we have p < s, consequently, p^2 < s^2, that is, p^2 < 2, consequently, 0 < 2 - p^2, that is to say, (2 - p^2) is positive. Therefore, we can divide both sides of the inequality p + s < p + 2 by the positive rational number (2 - p^2), and we obtain (p + s) / (2 - p^2) < (p + 2) / (2 - p^2). Consequently, we can choose y = (p + 2) / (2 - p^2), and the above inequality becomes (p + s) / (2 - p^2) < y, which is what we wanted. It follows that x = 1/y = (2 - p^2) / (p + 2), and q = p + x = p + (2 - p^2) / (p + 2), that is q = p - (p^2 - 2) / (p + 2), which is exactly the expression of q given by Walter Rudin on page 2 of his book. As said at the beginning, this proves that the set A does not have a greatest element. Of course, we can use a similar method to prove that the set B does not have a smallest element. At the end of Example 1.1, Walter Rudin shows directly, in a very short way, that the above value of q works for both, the proof about the set A and the proof about the set B, so that we do not have to redo all of the above work for the set B. Additional remarks: Remark 1. In the first part of Example 1.1, Walter Rudin proves that there is no rational number whose square is equal to 2. This is equivalent to saying that the positive square root of 2 is irrational, but Walter Rudin does not want to say this, because the irrational numbers, in particular "the positive square root of 2", is what he wants to construct in Chapter 1, and we are not supposed to know what it is on page 2. Remark 2. Walter Rudin uses the complication to consider only positive rational numbers for the definitions of the two sets A and B. This avoids making mistakes, because with negative real numbers, the function f = [Take the square] defined by f(x) = x^2 for every real number x is decreasing, that is, order reversing, on the set of negative real numbers, and it is increasing, that is, order preserving, only on the set of positive real numbers. For example, we have (-3) < (-2), but (-3)^2 > (-2)^2, that is, 9 > 4. Remark 3. In the statement of the main theorem of Chapter 1, namely, Theorem 1.19, on page 8, about the existence of the field R of real numbers, it would be better to replace "R contains Q as a subfield" by "R contains Q as a sub-ordered-field", because it is a very important property that the relation of order in the set R is an extension of the relation of order in the set Q. We are not in a crazy situation where a rational number is positive with the order in Q and negative with the order in R. Remark 4. The expression of q from the second book mentioned in the original posting is clearly much simpler, as well as its explanation provided by Tim Little. However, it seems that to obtain it, we also have to commit some mathematical crimes, because it relies on Newton's method, which relies on Calculus, and so far, I have never seen a book of Calculus that uses only the rational numbers, without ever using an irrational number. Are there some publications of Analysis over the field Q of rational numbers that do not use R or Analysis over the field Q(i) of Gaussian numbers that do not use the field C of complex numbers?
From: Tim Little on 15 Jul 2010 02:01 On 2010-07-15, Jean-Claude Evard <jc.evard1(a)insightbb.com> wrote: > Remark 4. The expression of q from the second book mentioned in the > original posting is clearly much simpler, as well as its explanation > provided by Tim Little. However, it seems that to obtain it, we also > have to commit some mathematical crimes because it relies on > Newton's method, Heh, that is rather an overstatement. You are allowed to use any method you like to arrive at a result, up to and including hunches based on tea leaves, so long as you can prove that the result is correct. If you like, you could think of Newton's method as a "pre-formal" result that you then formally verify. > which relies on Calculus, and so far, I have never seen a book of > Calculus that uses only the rational numbers, without ever using an > irrational number. You do not actually need much of calculus to derive Newton's formula for square roots. You can define a derivative and prove that it exists for rational functions, prove that they obey the usual algebraic rules, etc in the usual way without any difficulty. Obviously you cannot prove that the method will converge toward a solution, as there is no such solution in rationals. You can prove that the sequence is Cauchy though, which is almost as useful. - Tim
From: master1729 on 15 Jul 2010 11:12
David Ullrich wrote : > On Tue, 22 Jun 2010 16:59:34 -0700 (PDT), guyc > <guy.corrigall(a)gmail.com> wrote: > > >Let A be the set of all positive rationals p such > that p^2>2. > > > >Let B be the set of all positive rationals p such > that p^2<2. remember david , A and B are sets of RATIONALS. > > > >In order to show that A has no infimum, B no > supremum, Rudin considers > >the number q = p - (p^2 - )/(p +2) > > What? A certainly has an inf, and B has a sup! ohhh david forgot it ! A has no inf and B has no sup , just as RUDIN PROVED. > > At least if we're allowing real numbers. AGAIN DAVID , remember the key word , RATIONAL. RRRRRAAAATTTTIONAAAALLL. > A has no > _rational_ inf, > but that can't be what Rudin't proving here, since > the > detail below is irrelevant to that. sigh. > > Surely what Rudin says he's proving is that A has no > smallest > element and B has no largest element? you have trouble with rudin hmm. > The difference > between "inf" and "smallest element" is hugely > important. thats new. plz tell us , david , what is the ( important ) difference between inf and smallest element. and plz tell us ; what is the inf of A in " Let A be the set of all positive rationals p such that p^2>2 " and what is the smallest element of A in " Let A be the set of all positive rationals p such that p^2>2 " dont forget RRAAATTIIIOOONNAAALLLLSSSS FRACTIONS if you understand that better. your really amazing. pulling a ' real ' solution out of a set with rationals. is that how you got your phd ? maybe we should abolish fractions as DeTurck asked , so that people like you cant be confused anymore. and also abolish sets of reals for that matter. otherwise you might confuse reals and rationals as you did here. > > >Note that Bryant in the same proof considers the > number q = p/2 +1/p > > > >Very clever - but what is the motivation for these > q? Where did they > >come from? Lots of trial and error? A flash of > inspiration? > inspiration ? good one lol regards the master tommy1729 |