Pi is Irrational
By Jennifer, Luke, Dickson, and Quan
I.
Definition of Pi
II.
Proof of Lemma 2.5.1
III.
Proof that π is irrational
IV.
Ivan Niven’s Original Proof
Definition of π

Pi is the Greek letter used in the formula to find the circumference, or perimeter of a
circle.


Pi is the ratio of the circle’s circumference to its diameter π=C/d.
Pi is also the ratio of the circle’s area to the area of a square whose side is equal to the
radius π=A/r2
Lemma 2.5.1
Define a function 𝑓𝑛 (𝑥) =
properties:

1
0 < fn(x) < 𝑛!

and
𝑥 𝑛 1−𝑥 𝑛
𝑛!
. Then this function has the following
for 0 < x < 1
are both integers
Proof: Using the binomial theorem, we see that when the numerator of the function is multiplied
out, the lowest power of x will be n, and the highest power is 2n. Therefore, the function can be
written as
fn(x) =
where all coefficients are integers. It is clear from this expression that
for k > 2n.
Also, looking at the sum more carefully, we see that
...
But this implies that
is an integer for any k.
Moreover, since
we also have
Therefore
Also note that |
is an integer for any k.
|<
for n large enough ( n > 2a) and any a.
= 0 for k < n and
Niven's proof
The proof uses the characterization of π as the smallest positive zero of the sine function. As in
many proofs of irrationality, the argument proceeds by reductio ad absurdum.
Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may
be taken without loss of generality to be positive. Given any positive integer n, we define the
polynomial function
and denote by
the alternating sum of f and its first n even derivatives.
Claim 1: F(0) = F(π)
Proof: Since
the chain rule and mathematical induction imply
for all the derivatives, in particular
for j = 1, 2, ...,n and Claim 1 follows from the definition of F.
Claim 2: F(0) is an integer.
Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we
get the representation
Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n,
we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,
Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial
coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and
every derivative of f at 0 is an integer and so is F(0).
Claim 3:
Proof: Since f (2n + 2) is the zero polynomial, we have
The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x,
hence the product rule implies
By the fundamental theorem of calculus
Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization
of π as a zero of the sine function), Claim 3 follows from Claim 1.
Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of
the sine function), Claims 2 and 3 show that F(0) is a positive integer. Since
and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have
which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for
the positive integer F(0).
QED
http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
Niven’s Original proof at: http://www.ams.org/bull/1947-53-06/S0002-9904-1947-08821-2/S0002-99041947-08821-2.pdf