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Irrationality of Pi

Date: 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 

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 

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

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,
    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)
    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]
    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:   

-Doctor Rob,  The Math Forum
 Check out our web site!   

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, yields many hits to various places on the internet
where you can find a proof, such as

  Pi is Irrational


  Pi is Irrational


  Proof that Pi is Irrational

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!   
Associated Topics:
High School Number Theory
Middle School Pi

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