1 The Transcendence of

Definitions

 A complex number is algebraic over if it is a root of a polynomial equation with rational coefficients.

Thus a is algebraic if there are rational numbers ₀,₁,...,_n not all 0, such that ₀aⁿ + ₁a^n-1 + ... + _n-1a + _n = 0.

A complex number is transcendental if it is not algebraic, so it is not the root of any polynomial equation with rational coefficients.

(The material in the rest of this paper is taken from notes issued by Ian Stewart as an adjunct to a series of lectures in 1970 at the University of Warwick)

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  is transcendental over

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!.

However, no naturally occurring real number (such as e or ) was proved transcendental until Hermite (1873) disposed of e. 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 , if a0 or 1 and b is irrational, prove a^b is transcendental.

This was solved independently in 1934 by the Russian, Gelfand, and a German, Schneider.

Before proving transcendence of 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.

Proof

Let I_n(x) = _-1 ^{+ 1 n} cos  dx

Integrating by parts we have

²I _n = 2n I_n-1 - 4n I_n-2

which implies that

²ⁿ⁺¹I _n = n! (*)

where P_n,Q_n are polynomials of degree < 2n + 1 in with integer coefficients.

Put = , and assume is rational, so that = , a,b

From we deduce that J_n = is an integer. On the other hand J_n 0 as n since b is fixed and I_n is bounded by

J_n is an integer, 0. Thus J_n = 0 for some n. But this integrand is continuous, and is > 0 in most of the range , so J_n0. Contradiction.

Proof

Assume  ² = , a,b .

Define f(x) = ,

G = bⁿ

(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 and G are integers. Now since f = 0

Getting more involved, now:

Proof

Suppose a_me^m + ... + a ₁e + a₀ = 0 (a_i ).

WLOG a₀0

Define f(x) =

where for the moment p is arbitrary and prime.

Define F = f + f^' + ... + f .

Now if 0 < x < m,

so that

We show that in the exceptional case the term is NOT divisible by p. Clearly f = ^p... ^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 . This is a contradiction.

Proof

If satisfies an algebraic equation with coefficents in , so does i (i = ). Let this equation be ₁ = 0, with roots i = ₁,...,_n. Now eⁱ + 1 = 0 so

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 ₁ + ₂,₁ + ₃,..._n-1 + _n. The s 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 s and are also rational. Thus the pairs are roots of the equation ₂ = 0 with rational coefficients. Similarly sums of 3 s are roots of ₃ = 0, etc. Then the equation

₁₂..._n = 0

is a polynomial equation over whose roots are all sums of s. Deleting zero roots from this, if any, we get

= 0

= cx^r + c ₁x^r-1 + ...c _r

and c_r0 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 ₁,..._r. The original equation becomes

e^₁ + ...e^_r + e⁰ + ...e⁰ = 0

ie

e^_i + k = 0

where k is an integer > 0 (0 since the term 1...1 exists)

Now define

f = c^sx^p-1

where s = rp - 1 and p will be determined later. Define

F = f + f^' + ... + f .

= -e^-xf as before.

Hence we have

e^-xF - F = - ₀^xe^-yf  dy.

Putting y = x we get

F - e^xF = -x ₀¹e^xf  d.

Let x range over the _i and sum. Since e^_i + k = 0 we get

_{j = 1}^rF + kF = - _{j = 1}^r_{j 0}¹e^_jf d.

CLAIM:

For large enough p the LHS is a non-zero integer.

_{j = 1}^rf = 0 by definition of f. Each derivative of order p or more has a factor p and a factor c^s, since we must differentiate ^p enough times to get 0. And f is a polynomial in _j of degree at most s. The sum is symmetric, and so is an integer provided each coefficient is divisible by c^s, which it is. (symmetric functions are polynomials in coefficients = polynomials in of degree < s). Thus we have

_{j = 1}^rf = pk_t t = p,...p + s.

Thus LHS = + kF . What is F?

f = 0 t = 0,...,p - 2.

f = c^sc_r^p

f = p (some integer) t = p,p + 1,....

So the LHS is an integer multiple of p + c^sc_r^pk. This is not divisible by p if p > k,c,c_r. So it is a non-zero integer. But the RHS 0 as p and we get the usual contradiction.