A SIMPLE PROOF THAT π IS IRRATIONAL
It is well known that π is an irrational number; it is not so well known
how to prove this fact, although these days a Google search using some
combination of ‘proof’, ‘pi’ and ‘irrational’ will lead to many web pages
describing the proof below, such as [2]. The original proof that π is irrational
is due to Johann Lambert, in 1761; this simple proof seems to be due to Ivan
Niven [1].
The proof proceeds by contradiction: we start by assuming that π is
rational, so that we can write π = pq with p and q integers, and we will show
that this assumption leads to a contradiction.
Consider the functions fn defined by
fn (x) =
1
1 n n
q x (π − x)n = xn (p − qx)n .
n!
n!
(1)
Here, n is a natural number, which is large but fixed; we will see just how
large n needs to be in a few lines.
The functions fn have some nice properties:
1. fn is a polynomial, of degree 2n, whose coefficients (apart from the
factor n!1 ) are integers;
2. fn (x) > 0 if 0 < x < π;
3. The maximum value of fn on [0, π] occurs at π2 :
Mn = max {fn (x)} = fn (π/2) =
x∈[0,π]
(This follows easily if we write (1) as fn (x) =
2n
1 n π
q
n!
2
1 n
q [x(π
n!
.
− x)]n .)
Recall that for any number a, limn→∞ (an /n!) = 0 ; from this and Property 3
above (with a = q(π/2)2 ), it follows that limn→∞ Mn = 0 .
We now define a number In by the integral
In =
Z
0
π
fn (x) sin x dx.
(2)
Since both fn (x) and sin x are positive for 0 < x < π, we see that In > 0.
On the other hand, we can estimate
In <
Z
0
π
Mn sin x dx = Mn
1
Z
0
π
sin x dx = 2Mn .
Thus, if n is large enough that Mn < 21 , we will have 0 < In < 1. (Note that
we could in principle work out how large n has to be, by using equation (3).)
The contradiction, and the proof, will be completed by showing that In is in
fact an integer for every n.
Since the integrand in In is a polynomial multiplied by sin x, it is natural to try to evaluate the integral using integration by parts (differentiating
fn (x) repeatedly, and anti-differentiating the trigonometric function). At this
point, we need another property of the polynomials fn :
4. Let fn(m) denote the m-th derivative of fn . Then fn(m) (0) and fn(m) (π)
are integers.
To see why Property 4 is true, notice that when we differentiate fn m times,
we will obtain a sum of terms, in each of which the factor xn will have been
differentiated some number k times, where 0 ≤ k ≤ m. If k < n, there is still
at least one factor of x remaining in the term, so the term vanishes when
x = 0. If k ≥ n, the formula
dn
(xn ) = n!
dxn
implies that the factor n!1 is cancelled, and the term will yield an integer when
x = 0. Since fn(m) (0) is a sum of integer-valued terms, it is itself an integer.
A similar argument takes care of fn(m) (π).
Now we integrate (2) by parts 2n + 1 times. We will obtain many, many
boundary terms, but each of these will involve some derivative of fn (x), multiplied by either sin x or cos x, and evaluated at 0 or π. Each of these boundary
terms is therefore an integer. The remaining integral involves fn(2n+1) , which
is identically zero, since fn has degree n. We conclude that In must be an
integer, comnpleting the contradiction and showing that π is irrational.
References
[1] I. Niven, A Simple Proof that π is Irrational, Bull. Amer. Math. Soc.
53(6), p.509 (1947).
[2] H. Richter, http://www.lrz-muenchen.de/~hr/numb/pi-irr.html
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