Baby Rudin question
in [Math]
|
Prev: Absulote Logic Is Equator Of Self-contradiction
Next: BP loves Waxman-Obama cap&trade (circa Kyoto Protocol, or Waxman's '91 bill)
From: guyc on 22 Jun 2010 19:59 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?
From: porky_pig_jr on 22 Jun 2010 20:20 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.
From: Tim Little on 22 Jun 2010 20:39 On 2010-06-22, guyc <guy.corrigall(a)gmail.com> wrote: > 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? The idea is to find a function that for a rational p, yields another rational closer to sqrt(2). Any formula that achieves this aim will suffice, though it is preferable if the proving the property is easy. The function f(p) = p - (p^2 - 2) / (p + 2) is fairly explicitly constructed: start with p, use p^2-2 to move in the correct direction, but divide by something (in this case p+2) to ensure that "large" values of p don't overshoot. It just happens that p+2 is simple and works. A denominator of [(355/113) p^2 + 42] would also work, but is practically more difficult to work with. The function f(p) = p/2 + 1/p is easy to derive if you have seen Newton's method (aka Babylonian method) for finding square roots, known for at least 2000 years. To find sqrt(n), start with a guess p and replace it with the average of p and n/p. In the special case n=2, you get Bryant's formula. - Tim
From: Jonathan Groves on 22 Jun 2010 16:58 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
From: Jonathan Groves on 22 Jun 2010 16:59
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 |