From: "Raymond E. Griffith" Subject: Re: pi does not repeat Date: Tue, 24 Oct 2000 13:05:57 -0400 Newsgroups: alt.math.recreational,sci.math Summary: [missing] "Marc Fleury" wrote in message news:39f5b070.70406633@news.rdc1.on.home.com... > "CALLA" wrote: > > >You are talking in circles friend. I said nothing about pi being rational > >although for a repeating sequence it does imply that. > > Actually, > > >> >But pi is irrational , so our assumption that pi repeats must be false. > > This part is where the circular argument comes in. Everyone knows that > irrational numbers don't repeat perfectly, so the rest of the argument > becomes pointless if you use this fact. Just say "irrational numbers > don't repeat, pi is irrational, therefore pi doesn't repeat." > > The real problem is proving that pi is irrational. > > -- Marc. We can do this, but it will take a bit of work. In order, I think you would need to prove the following: First: the square root of 2 is an irrational number. (Proof by ancients.) Second: the sum of a rational number and an irrational number is an irrational number. (Proof by construction) Third: the square root of an irrational number is an irrational number. Eg. Suppose the square root of an irrational number x is rational, say a/b, where a and b are relatively prime integers. so, sqrt(x) = a/b, then x = a^2/b^2, where a^2 and b^2 are also relatively prime integers. So x is rational. Contradiction. Fourth: construct pi as a cascading sequence of the square roots of irrational numbers. It goes like this: From trigonometry we are able to construct the half-angle formulas of sine and cosine. sin(t/2) = sqrt((1-cos t)/2) cos(t/2) = sqrt((1+cos t)/2) Now consider that pi is half the distance around a circle of radius 1 (this would have to have already been shown, btw). For an angle of t = pi/4, sin(t) = sqrt(2)/2 and cos(t) = sqrt(2). Note that pi/4 *is* a fourth of pi (yes, I know it is not necessary to say it, but it *is* easy to overlook). Now you can use sin(pi/4) as an approximation of the curve of the circle to that angle (although it is *very rough*). In which case you have an approximation to pi of 4*sqrt(2)/2. As I said -- *very rough* (btw, 2sqrt(2) is an irrational number...). The calculated value is 2.82842712.... What can you do now? We need a better approximation. So we need to cut our angle in half and work with t/2 = pi/8. We can now use 8 * sin(pi/8) as an approximation of pi. It will be a bit better. sin(t/2) = sqrt((1-cos(t)/2) = sqrt((1-sqrt(2)/2)/2) = sqrt(2 - sqrt(2))/2 The new calculated value is 3.06146745.... To get sin(pi/16), we need cos(pi/8). You can do the math, but it works out to sqrt(2 + sqrt(2))/2 ... You will notice that the constructed pattern for sin(pi/(2^n)) = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + ... ...)))/2, where there are n-1 2's in the numerator, a single 2 in the denominator. The first sign is subtration, the rest are all addition. pi is approximately equal to 2^n * sin(pi/(2^n)). This sequence converges to pi as n gets increasingly large. And this number has all of the characteristics of an irrational number, no matter how far out you go. Each iteration involves the addition and/or subtraction of an irrational number from a rational number (so irrational), and taking the square root of an irrational number. This makes pi irrational. Neat, eh? Raymond ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: pi does not repeat Date: 24 Oct 2000 16:18:32 -0400 Newsgroups: alt.math.recreational,sci.math In article <39f5b070.70406633@news.rdc1.on.home.com>, Marc Fleury wrote: [...] >The real problem is proving that pi is irrational. An elementary proof, using only first year Calculus (and patience), is finding its way to this medium for about fourth time. Its originator would be hard to trace. Fixed spacing is assumed. The following proof was suggested in book IV of the series "Elements of Mathematics", called "Functions of one real variable" Chapter III, Sec. 2.4, Exercise 8. Author of the book: Nicolas Bourbaki For any positive number q and for every nonnegative integer n, consider the integrals n / pi q | n A(n) = -- | (x (pi - x)) sin x dx n! / 0 These numbers are positive, and they converge to 0 as n goes to infinity because their sum is finite. (The sum is pi / | qx(pi-x) | e sin x dx / 0 since the series of integrands converges uniformly on [0, pi].) We calculate directly for n = 0, 1, ... (integrating by parts where needed) 2 A(0) = 2, A(1) = 4q, A(n+2) = (4n+6)qA(n+1) - (q*pi) A(n) Now we are ready to argue by contradiction: If q were a positive integer such that q*pi would also be an integer, then all A(n) would be positive integers (by induction from the recursion formula), but then they would not converge to 0, being always >=1. Hence no such integer q exists, and pi is irrational. Cheers, ZVK(Slavek).