|
|
|
Suppose π is rational. Then π² is rational, so π²=a/b, where a and b are integers (b not zero).
Since the limit as n approaches infinity of an/n! = 0, it follows
that
there exists an M > 0 such that if n ³ M, an/n!
< 1/π, or πan/n! < 1.
Choose an integer N ³ M.
It follows that πaN/N! < 1.
Define a polynomial
f(x) = (xN(1-x)N)/N!
Expanding the polynomial,
f(x) = (1/N!)(cNxN + cN+1xN+1 + cN+2xN+22 + ... + c2Nx2N), where each cN is an integer.
For integers k such that N ≤ k ≤ 2N, the kth derivative f(k)(x) is given by
f(k)(x) = (1/N!)
2N ∑ n=k (n!/(n-k)!) cNxN-k
Now I will show by induction that f(k)(x) = (-1)(k)f(k)(1-x) for all integers, k.
It is true when k=0; f(0)(x) = (-1)(0)f(0)(1-x), because (xN(1-x)N)/N! = ((1-x)NxN)/N!
Suppose f(k)(x) = (-1)(k)f(k)(1-x)
f(k+1)(x) = d(f(k)(x))/dx
f(k+1)(x) = d[ (-1)(k) f(k)(1-x) ]/dx
f(k+1)(x) = (-1)(k) d[ f(k)(1-x) ]/dx
f(k+1)(x) = (-1)(k) (-1)f(k+1)(1-x)
f(k+1)(x) = (-1)(k+1) f(k+1)(1-x)
For integer values of k such that 0 ≤ k ≤ N, each term in the expansion of f(k)(0)
is 0,
and f(k)(1) = (-1)f(k)(0) = 0
For integer values of k such that N ≤ k ≤ 2N, the only term that does not
contain a positive integer power of x is the first (when n=k), so
f(k)(0) = k!ck/N!, which is an integer, since k ³
N. f(k)(1) is also an integer because f(k)(1) =
(-1)f(k)(0).
Now define another function,
| F(x) = (bN) |
|
(-1)j π2N-2j f(2j)(x) |
| F(x) = (bN) |
|
(-1)j (a/b)(N-j) f(2j)(x) |
| F(x) = (bN) |
|
(-1)j a(N-j)b(j-N) f(2j)(x) |
| F(x) = |
|
(-1)j a(N-j)bj f(2j)(x) |
It follows that F(0) and F(1) are both integers, and thus F(0)+F(1) is an integer. Now define
g(x) = F'(x) sin(πx) - πF(x) cos(πx)
It follows that
g'(x) = F''(x) sin(πx) + πF'(x) cos(πx) - πF'(x) cos(πx) + π²F(x) sin(πx)
g'(x) = [ F''(x) + π²F(x) ] sin(πx)
Now
F(x) = bN [ π2N f(x) - π2N-2 f''(x) + π2N-4 f(4)(x) - ... + (-1)j f(2N)(x) ], and so
F''(x) = bN [ π2N f''(x) - π2N-2 f(4)(x) + π2N-4 f(6)(x) - ... + (-1)j π2 f(2N+2)(x) ]. Also,
π²F(x) = bN [ π2N+2 f(x) - π2N f''(x) + π2N-2 f(4)(x) - ... + (-1)j π2 f(2N)(x) ]
Notice that f(x) is a polynomial of degree 2N, so f(2N+2)(x) = 0 for all x. Thus
F''(x) + π2 F(x) = bN π(2n+2) f(x). So
g'(x) = bN π(2N+2) f(x) sin(πx)
g'(x) = (aN/π(2N)) π(2N+2) f(x) sin(πx)
g'(x) = π2 aN f(x) sin(πx)
Since g(x) is continuous on [0,1], and g' exists on (0,1), by the
Mean Value Theorem, there exists a c in (0,1) such that
g(1)-g(0)=g'(c).
Now g(1) = F'(1) sin(π) - πF(1) cos(π) = πF(1), and
g(0) = F'(0) sin(0) - πF(0) cos(0) = -πF(0).
So g(1)-g(0) = πF(1) + πF(0) = π [F(1) + F(0)], and thus
π [F(1) + F(0)] = π2 aN f(c) sin(πc)
F(1) + F(0) = π aN f(c) sin(πc)
Since 0 < c < 1, 0 < sin(πc) < 1, and also 0 < 1-c < 1.
Since f(c) = (cN (1-c)N) / N! < 1/N!,
and remember that we selected N in the first place so that πaN/N! <
1,
so it follows that
0 < π aN f(c) < π aN/N! < 1
It now follows that 0 < π aN f(c) sin(πc) < 1, and so 0 < F(1) + F(0) < 1 is an integer. But this is a contradiction because F(1)+F(0) is an integer, and there can be no integer between 0 and 1.
Proof that cos a is irrational, when a is rational and not zero
Rumor has it that this proof that pi is irrational can be adapted to one that shows cos a irrational whenever a is rational and not zero. Mathworld: Irrational Number says cos a is irrational for every rational number a not equal to 0, and credits Niven 1956, and Stevens 1999 for proving this.
http://www.meikleriggs1.free-online.co.uk/pi/index.htm
Mathworld: Irrational Number
Perfect Squares -- proof that sqrt(n) is irrational, as long as n isn't a perfect square.
The webmaster and author of this Math Help site is Graeme McRae.