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The Transcendence of π
The proof that π is transcendental is not well-known despite the fact that it
isn't too difficult for a university mathematics student to follow. The purpose of this
paper is to make the proof more widely available. A bonus is that the proof also shows
that e is transcendental as well.
The material in these notes are not mine; it is taken from a supplement issued by Ian Stewart as an
adjunct to a Rings and Fields course in 1970 at the University of Warwick.
Definition
A complex number is algebraic over Q if it is a
root of a polynomial equation with rational coefficients.
Thus a is algebraic if there are rational numbers α0,α1,…,αn not all 0, such that α0an+α1an−1+…+αn−1a+αn=0.
Definition
A complex number is transcendental if it is not algebraic, so it is not
the root of any polynomial equation with rational coefficients.
In proving that it is impossible to 'square the circle' by a ruler-and-compass
construction we have to appeal to the theorem:
The real number π is transcendental over Q
The purpose of this supplement is to indicate, for those who may be
interested, how this theorem may be proved.
It is possible to prove that there exist transcendental real numbers by using
infinite cardinals, as was first done by Cantor in 1874. Earlier Liouville
(1844) had actually constructed transcendentals, for example n=1∑∞10−n! is transcendental.1
However, no naturally occurring real number (such as e or π) was proved transcendental
until Hermite (1873) disposed of e. π held out until 1882 when Lindemann, using methods
related to those of Hermite, disposed of that. In 1900 David Hilbert proposed the problem:
If a,b are real numbers algebraic over Q, if a=0 or 1
and b is irrational, prove ab is transcendental.
This was solved independently in 1934 by the Russian, Gelfond, and a
German, Schneider.
Before proving transcendence of π we shall prove a number of similar
theorems, using simpler versions of the final method, as an aid to
comprehension. The tools needed are first-year analysis.2
Theorem
π is irrational
Proof.
Let In(x)=∫−1+1(1−x2)ncos(αx)dx
Integrating by parts we have
α2In=2n(2n−1)In−1−4n(n−1)In−2(n≥2)
which implies that
α2n+1In=n!(Pnsin(α)+Qncos(α))(∗)
where Pn,Qn are polynomials of degree <2n+1 in α with integer coefficients.
Remark. degPn=n,degQn=n−1
Put α=2π, and assume π is rational, so that π=ab,a,b∈Z
From (*) we deduce that Jn=n!b2n+1In is an integer. On the other hand
Jn→0 as n→∞ since b is fixed and In is bounded by
∫−1+1cos(2πx)dx
Jn is an integer, →0. Thus Jn=0 for some n. But this integrand is continuous, and is >0 in most
of the range (−1,+1), so Jn=0. Contradiction.
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Theorem
π2 is irrational (so π does not lie in any quadratic extension of Q)
(superscripts indicating differentiations). We see that the value of any
derivative of f at 0 or 1 is either 0 or an integer. Also G(0) and G(1) are integers.
Now
We claim that each f(i)(j) is an integer, divisible by p except when j=0 and i=p−1.
For only non-zero terms arise from terms where the factor (x−j)p has been differentiated
p times, and then p! cancels (p−1)! and leaves p, except in the exceptional case.
We show that in the exceptional case the term is NOT divisible by p. Clearly
f(p−1)(0)=(−1)p…(−m)p. We CHOOSE p larger than m, when this product cannot have a prime factor
p. Hence the right-hand side of the above equation is an integer =0.
But as p→∞ the left-hand side tends to 0, using the above
estimate for ∣f(x)∣. This is a contradiction.
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Theorem (Lindemann)
π is transcendental over Q
Proof.
If π satisfies an algebraic equation with coefficents in Q, so
does iπ (i=−1). Let this equation be θ1(x)=0, with roots
iπ=α1,…,αn. Now eiπ+1=0 so
(eα1+1)…(eαn+1)=0
We now construct an algebraic equation with integer coefficients whose roots
are the exponents of e in the expansion of the above product. For example,
the exponents in pairs are α1+α2,α1+α3,…,αn−1+αn.
The αs satisfy a polynomial equation over Q so their elementary symmetric functions
are rational. Hence the elementary symmetric functions of the sums of pairs
are symmetric functions of the αs and are also rational. Thus the pairs are roots of the equation θ2(x)=0 with rational coefficients. Similarly sums of 3 αs are roots of
θ3(x)=0, etc. Then the equation
θ1(x)θ2(x)…θn(x)=0
is a polynomial equation over Q whose roots are all sums of
αs. Deleting zero roots from this, if any, we get
θ(x)θ(x)=0=cxr+c1xr−1+…cr
and cr=0 since we have deleted zero roots. The roots of this equation
are the non-zero exponents of e in the product when expanded. Call these
β1,…βr. The original equation becomes
eβ1+…eβr+e0+…e0=0
ie
∑eβi+k=0
where k is an integer >0(=0 since the term 1…1 exists)
Now define
f(x)=csxp−1(p−1)![θ(x)]p
where s=rp−1 and p will be determined later.
Define
F(x)=f(x)+f′(x)+…+f(s+p)(x).
dxd[e−xF(x)]=−e−xf(x) as before.
Hence we have
e−xF(x)−F(0)=−∫0xe−yf(y)dy
Putting y=λx we get
F(x)−exF(0)=−x∫01e(1−λ)xf(λx)dλ.
Let x range over the βi and sum. Since ∑eβi+k=0 we get
j=1∑rf(t)(βj)=0(0<t<p) by definition of f.
Each derivative of order p or more has a factor p and a factor cs, since we must differentiate
[θ(x)]p enough times to get =0. And f(t)(βj) is a polynomial in
βj of degree at most s. The sum is symmetric, and so is an integer
provided each coefficient is divisible by cs, which it is. (symmetric
functions are polynomials in coefficients = polynomials in cci
of degree ≤s). Thus we have
So the LHS is an integer multiple of p+cscrpk. This is not
divisible by p if p>k,c,cr. So it is a non-zero integer. But the RHS
→0 as p→∞ and we get the usual contradiction.
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