## Proofs That PI is Irrational

```The first proof of the irrationality of PI was found by Lambert in
1770 and published by Legendre in his "Elements de Geometrie".  A
simpler proof, essentially due to Mary Cartwright, goes like this:

For any integer n and real number r we can define a quantity A[n] by
the definite integral

/ 1
A[n]  =  |    (1 - x^2)^n cos(rx) dx
/
x=-1

If we integrate this by parts we find that the quantities A[n] for
n=2,3,4,...etc satisfy the recurrence relation

2n(2n-1) A[n-1]  -  4n(n-1) A[n-2]
A[n]  =   ----------------------------------
r^2

We can obviously express A[n-1] and A[n-2] in terms of lower members
of the recurrence, and so on, all the way back to A[0] and A[1].  The
result is
n!
A[n]  =  -------- [ P(r) sin(r) - Q(r) cos(r) ]
r^(2n+1)

where P(r) and Q(r) are polynomials in r of degree less than 2n-1 with
integer coefficients.  (See below if you're interested.)

Now let's assume that PI=a/b, where a and b are integers, and let's
set r=PI/2.  Of course, this means that r=a/(2b).  If we substitute
this value into the preceding equation, and remember that sin(PI/2)=1
and cos(PI/2)=0, we have

(a/2b)^(2n+1) A[n]  =  n! P(a/2b)

Multiplying both sides by (2b)^(2n+1) and dividing by n! gives

a^(2n+1) A[n]
-------------    =  (2b)^(2n+1) P(a/2b)
n!

Remember that P(a/2b) is a polynomial in a/2b with integer coefficients,
and its degree is less than 2n+1.  Therefore, when we multiply through
by the factor (2b)^(2n+1) we clear out all the 2b's in the denominators,
so the quantity on the right hand side of the preceding equation is
clearly an integer.  It follows that the left side is also an integer.

Now, recall that A[n] was defined as the integral of

(1 - x^2)^n cos(rx) dx

from x=-1 to +1.  Clearly the leading factor of this integrand,
(1 - x^2)^n, is always between 0 and 1, so an upper bound on the
value of A[n] is given by

/ 1
A[n]    <     |    cos(rx) dx
-     /
x=-1

This integral has some constant value C (doesn't matter what it is),
which is the upper bound on the value of A[n] for ANY n.  But now we
see the contradiction, because we showed previously that the quantity

a^(2n+1) A[n]
-------------
n!

is always an integer (assuming PI is rational).  But since A[n] is less
than C, it follows that the integer given by the above expression is
less than
a^(2n+1) C
----------
n!

But that's impossible, because n! increases faster than a^(2n+1),
so there is some value of n beyond which this ratio will be less
than 1.  This proves the quantity can't be an integer, so we have a
contradiction.  We are forced to conclude that PI is not a rational
number.  QED.

By the way, the actual expressions for the first few A[n] and the
polynomials P[r] and Q[r] are

2   /        \
A[0]  =  --- (  sin(r)  )
r   \        /

4   /                         \
A[1]  =  --- (  [1] sin(r) - [r] cos(r)  )
r^3  \                         /

16   /                              \
A[2]  =  ---- ( [3 - r^2] sin(r) - [3r] cos(r) )
r^5   \                              /

96   /                                       \
A[3]  =  ---- ( [15 - 6r^2] sin(r) - [15r - r^3] cos(r) )
r^7   \                                       /

768   /                                                  \
A[4]  =  ---- ( [105 - 45r^2 + r^4] sin(r) - [105r - 10r^3] cos(r) )
r^9   \                                                  /

and so on.  If we let double curly brackets {{*}} denote the "two-step
factorial function, e.g., {{5}} = 5*3*1, {{8}} = 8*6*4*2, and
{{-1}}={{0}}=1, then the general expression for A[n] is

2^(n+1) n!  /                              \
A[n]  =  ---------- (  P_n[r] sin(r) - Q_n[r] cos(r) )
a^(2n+1)   \                              /

where the coefficient of r^2k in the polynomial P_n is

/ n-k \  {{2n-2k-1}}
(-1)^k   (       ) -----------
\  k  /   {{2k-1}}

For example, to determine the coefficient of r^2 in A[4] we have
k=1 and n=4, which gives

/ 3 \  {{5}}
(-1)^1  (     ) -----    =   - 45
\ 1 /  {{1}}

Of course, the coefficients of odd powers of r in P_n are all
zero, as are the coefficients of all even powers of r in Q_n.  The
coefficient of r^(2k+1) in Q_n is

/ n-k-1 \  {{2n-2k-1}}
(-1)^k  (         ) -----------
\   k   /   {{2k+1}}

For example, the coefficient of r^3 in A[4] is found by setting
k=1 and n=4 to give

/ 2 \  {{5}}
(-1)^1  (     ) -----   =   -10
\ 1 /  {{3}}

A further simplification of the proof was devised by Ivan Niven in
1949.  Assume pi = a/b for integers a,b, and define the polynomial

n         n
x  (a - bx)
f(x) = ------------
n!

For some given positive integer n. This function is symmetrical in x and
pi - x, so, letting f^[k](x) denote the kth derivative of f[x], we have
f^[k](0) = +-f^[k](pi) for all k. Also, since every term in this polynomial
contains at least n powers of x, it follows that f^[k](0) = 0 for all k
less than n, and furthermore for k greater than or equal to n we know
f^[k](0) is an integer, because for any exponent m greater than n the n
differentiations apply a product of m!/(m-n)!, and so the coefficient of
that term is the integer binomial coefficient m!/[(m-n)!n!]. Now we define

F(x) = f^[0](x) - f^[2](x) + f^[4](x) - f^[6](x) + ...

Since each of the values f^[j](0) are non-negative integers, it follows
that F(0), and hence F(pi), are also integers. Also, notice that
F(x) + F"(x) = f(x). Now, since

d
--[F' sin(x) - F cos(x)]  =  (F" + F)sin(x) = f(x)sin(x)
dx

it follows that

pi
/                                           |pi
| f(x)sin(x)dx  =  F'(x)sin(x) - F(x)cos(x) |    =  F(0) + F(pi)
/                                           |0
0

which is an integer. But now we see a contradiction, because the
integrand is strictly positive but with magnitude less than (a pi)^n/n!,
which (for any value of a) is less than 1 for sufficiently large n.
Since there is no positive integer less than 1, this shows that we
have a contradiction, and hence our assumption was wrong, so pi does
not equal the ratio of any two integers.
```

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