A proof that π is irrational
Gilles Cazelais
The first rigorous proof that π is irrational is from Johann Heinrich Lambert in 1761. He proved that if
x 6= 0 is rational, then tan x must be irrational. Since tan π4 = 1 is rational, then π must be irrational.
The simpler proof given here is due to Ivan Niven in 1947. It only assumes a knowledge of basic Calculus.
We will prove that π 2 is irrational. This stronger result implies that π is irrational. The proof uses the
function
xn (1 − x)n
f (x) =
.
n!
Lemma. For any integer n ≥ 1,
2n
(i) f (x) is a polynomial of the form f (x) =
(ii) For 0 < x < 1, we have 0 < f (x) <
1 X
ci xi and all the coefficients ci are integers.
n! i=n
1
.
n!
(iii) The derivatives f (k) (0) and f (k) (1) are integers for all k ≥ 0.
Proof. By expanding the binomial (1 − x)n and multiplying each term by xn , we get the polynomial
n n+1
n n+2
n
n
n
n n
x (1 − x) = x −
x
+
x
− · · · + (−1)
x2n
1
2
n
where all coefficients are integers. It then follows that part (i) holds. To see that part (ii) holds, observe
that for any 0 < x < 1, we have
0 < xn < 1
and 0 < (1 − x)n < 1
=⇒
0<
Let’s now show that part (iii) holds. From part (i) it is clear that
f (k) (0) = 0,
For n ≤ k ≤ 2n, we have f (k) (0) =
k!
n! ck
if k < n or k > 2n.
which is an integer. Since
f (x) = f (1 − x),
we have
f (k) (x) = (−1)k f (k) (1 − x).
Therefore, f (k) (1) = (−1)k f (k) (0) is also an integer for all k.
1
xn (1 − x)n
< 1.
n!
Theorem. π 2 is irrational.
Proof. Assume that π 2 is rational, i.e., π 2 = a/b for two positive integers a and b. Let
F (x) = bn π 2n f (x) − π 2n−2 f (2) (x) + π 2n−4 f (4) (x) − · · · + (−1)n f (2n) (x) .
(1)
Observe that for all 0 ≤ k ≤ n,
bn π 2n−2k = bn (π 2 )n−k = bn
a n−k
b
= an−k bk
is an integer. Since f (k) (0) and f (k) (1) are integers, we see that F (0) and F (1) are integers. Differentiating
F twice gives
(2)
F 00 (x) = bn π 2n f (2) (x) − π 2n−2 f (4) (x) + π 2n−4 f (6) (x) − · · · + (−1)n f (2n+2) (x) .
Observe that f (2n+2) (x) = 0. From (1) and (2), we get
F 00 (x) + π 2 F (x) = bn π 2n+2 f (x) = π 2 an f (x).
By differentiation, we get
d
0 ((((
0 ((((
πF(
(x) cos πx + π 2 F (x) sin πx
πF(
(x) cos πx + F 00 (x) sin πx − (
(F 0 (x) sin πx − πF (x) cos πx) = (
dx
= F 00 (x) + π 2 F (x) sin πx
= π 2 an f (x) sin πx,
from (3).
From the Fundamental Theorem of Calculus, we get
Z 1
1
π 2 an
f (x) sin πx dx = (F 0 (x) sin πx − πF (x) cos πx)
0
0
= F 0 (1) sin π − πF (1) cos π − F 0 (0) sin 0 + πF (0) cos 0
= π (F (1) + F (0)) .
Thus,
πan
Z
1
f (x) sin πx dx = F (1) + F (0)
0
is an integer. Since 0 < f (x) < 1/n! for 0 < x < 1, then
0 < f (x) sin πx <
1
,
n!
for 0 < x < 1.
Therefore, for any integer n ≥ 1 we have
n
Z
0 < πa
1
f (x) sin πx dx <
0
πan
.
n!
Since for any number a, we have
an
=0
n→∞ n!
n
we can choose n large enough so that πa
n! < 1. This gives us
Z 1
0 < πan
f (x) sin πx dx < 1
lim
0
R1
which is a contradiction since πan 0 f (x) sin πx dx is an integer. Therefore π 2 is irrational.
2
(3)