Irrationality of PiDate: 01/09/98 at 05:14:57 From: Dennis Chitwood Subject: C/d = a rational number if actually measured I think that perhaps the motion of making a measurement involving length is distorted, very minutely, but enough to force the rationality of the measure of the circumference of a circle divided by its diameter. I have lots of ideas that support my claim, I think... I hope I'm not as nuts as I must be to think I have the answer. Help. Date: 01/09/98 at 12:04:49 From: Doctor Rob Subject: Re: C/d = a rational number if actually measured If you think that Pi = a/b, then you will believe that Pi has either a terminating or an eventually periodic decimal expansion. This is false. Suppose you thought that Pi = a/b, for some specific a and b. If you draw a circle with diameter b, you should get circumference a. If the measurements are accurate enough, you will be able to see that the circumference is actually larger or smaller than a, and not equal to it. Here's a proof of the irrationality of Pi from Robert Simms: ______________________________________________________________________ Theorem: pi is irrational Proof: Suppose pi = p / q, where p and q are integers. Consider the functions f_n(x) defined on [0, pi] by f_n(x) = q^n x^n (pi - x)^n / n! = x^n (p - q x)^n / n! Clearly f_n(0) = f_n(pi) = 0 for all n. Let f_n[m](x) denote the m-th derivative of f_n(x). Note that f_n[m](0) = - f_n[m](pi) = 0 for m <= n or for m > 2n; otherwise some integer max f_n(x) = f_n(pi/2) = q^n (pi/2)^(2n) / n! By repeatedly applying integration by parts, the definite integrals of the functions f_n(x) sin x can be seen to have integer values. But f_n(x) sin x are strictly positive, except for the two points 0 and pi, and these functions are bounded above by 1 / pi for all sufficiently large n. Thus for a large value of n, the definite integral of f_n sin x is some value strictly between 0 and 1, a contradiction. ______________________________________________________________________ You will need to understand calculus to understand that proof, however. All proofs of this fact that I know use calculus in some form or other. The value of Pi can be computed by computing the perimeter of regular polygons with 2^n sides inscribed and circumscribed about a circle of radius 1. The inscribed polygon will have smaller perimeter than the circumference of the circle, and the circumscribed polygon will have larger perimeter than the circumference of the circle. Start with n = 2, so the polygons are squares. That will tell you that 4*Sqrt[2] < 2*Pi < 4*2, so 2.828427125 < Pi < 4. Now use n = 3, so there are octagons. This will tell you that 8*Sqrt[2-Sqrt[2]] < 2*Pi < 8*(Sqrt[2]-1) so 3.061467458921 < Pi < 3.31370849898. This is enough to show that Pi is not a rational number with denominator of 4 or less. Now use n = 4, so there are 16-gons. Then 16*Sqrt[2-Sqrt[2+Sqrt[2]]] < 2*Pi < 16*[(2+Sqrt[2])*Sqrt[2-Sqrt[2]]-1-Sqrt[2] so 3.121445152 < Pi < 3.182597878. This is enough to show that Pi is not a rational number with denominator of 5 or less. As you use larger and larger values of n, you get a better and better approximation to Pi. Eventually you will be able to prove that Pi is not equal to any fraction with denominator smaller than or equal to any particular number you might guess. See also the following URL: http://mathforum.org/dr.math/faq/faq.pi.html -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 11/16/2004 at 16:32:11 From: Mohamed Subject: pi How can you prove that pi is not a rational number? Date: 11/16/2004 at 16:46:47 From: Doctor Vogler Subject: Re: pi Hi Mohamed, Thanks for writing to Dr Math. A search of the internet, such as pi proof irrational on google.com, yields many hits to various places on the internet where you can find a proof, such as Pi is Irrational http://www.lrz-muenchen.de/~hr/numb/pi-irr.html and Pi is Irrational http://pi314.at/math/irrational.html and Proof that Pi is Irrational http://www.mathpages.com/home/kmath313.htm If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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