Proof that p is irrational

Suppose p is rational.  Then p² is rational, so p²=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/p, or paⁿ/n! < 1.

Choose an integer N ³ M.

It follows that pa^N/N! < 1.

Define a polynomial

f(x) = (x^N(1-x)^N)/N!

Expanding the polynomial,

f(x) = (1/N!)(c_Nx^N + c_N+1x^N+1 + c_N+2x^N+2 + ... + c_2Nx^2N), where each c_N is an integer.

For integers k such that N £ k £ 2N, the k^th derivative f^(k)(x) is given by

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 (x^N(1-x)^N)/N! = ((1-x)^Nx^N)/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!c_k/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,

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(px) - pF(x) cos(px)

It follows that

g'(x) = F''(x) sin(px) + pF'(x) cos(px) - pF'(x) cos(px) + p²F(x) sin(px)

g'(x) = [ F''(x) + p²F(x) ] sin(px)

Now

F(x) = b^N [ p^2N f(x) - p^2N-2 f''(x) + p^2N-4 f⁽⁴⁾(x) - ... + (-1)^j f^(2N)(x) ], and so

F''(x) = b^N [ p^2N f''(x) - p^2N-2 f⁽⁴⁾(x) + p^2N-4 f⁽⁶⁾(x) - ... + (-1)^j p² f^(2N+2)(x) ].  Also,

p²F(x) = b^N [ p^2N+2 f(x) - p^2N f''(x) + p^2N-2 f⁽⁴⁾(x) - ... + (-1)^j p² 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) + p² F(x) = b^N p⁽²ⁿ⁺²⁾ f(x).  So

g'(x) = b^N p^(2N+2) f(x) sin(px)

g'(x) = (a^N/p^(2N)) p^(2N+2) f(x) sin(px)

g'(x) = p² a^N f(x) sin(px)

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(p) - pF(1) cos(p) = pF(1), and
g(0) = F'(0) sin(0) - pF(0) cos(0) = -pF(0).

So g(1)-g(0) = pF(1) + pF(0) = p [F(1) + F(0)], and thus

p [F(1) + F(0)] = p² a^N f(c) sin(pc)

F(1) + F(0) = p a^N f(c) sin(pc)

Since 0 < c < 1, 0 < sin(pc) < 1, and also 0 < 1-c < 1.

Since f(c) = (c^N (1-c)^N) / N! < 1/N!,
and remember that we selected N in the first place so that pa^N/N! < 1,
so it follows that

0 <  p a^N f(c) < p a^N/N! < 1

It now follows that 0 <  p a^N f(c) sin(pc) < 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.

Other Irrationality Proofs

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.

Internet References

http://www.meikleriggs1.free-online.co.uk/pi/index.htm 

Mathworld: Irrational Number

Related Pages in this website

Number Theory

The Gauss Series for π

Perfect Squares -- proof that sqrt(n) is irrational, as long as n isn't a perfect square.

The webmaster and author of the Math Help site is Graeme McRae.
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