1  The Transcendence of p

Definitions

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

Thus a is algebraic if there are rational numbers a₀,a₁,¼,a_n not all 0, such that a₀aⁿ+a₁a^n-1+¼+a_n-1a+a_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 p is transcendental over \BbbQ

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 å_{n = 1}^¥10^-n!.

However, no naturally occurring real number (such as e or p) was proved transcendental until Hermite (1873) disposed of e. p 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 \BbbQ, if a ¹ 0 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 p 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.

p is irrational

Proof

Let I_n(x) = ò_-1⁺¹( 1-x²) ⁿcos( ax)  dx

Integrating by parts we have

       a²I_n = 2n( 2n-1) I_n-1-4n( n-1)I_n-2              ( n ³ 2)

which implies that

       a²ⁿ⁺¹I_n = n!( P_nsin( a)+Q_ncos( a) )        (*)

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

degP_n = n, degQ_n = n-1

Put a = [(p)/2], and assume p is rational, so that p = \dfracba,    a,b Î \BbbZ

From ( *) we deduce that J_n = \dfracb²ⁿ⁺¹I_nn! is an integer. On the other hand J_n® 0 as n®¥ since b is fixed and I_n is bounded by

ó õ +1 -1  cos æ ç è px 2 ö ÷ ø  dx

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 (-1,+1) , so J_n ¹ 0. Contradiction.

p² is irrational - so p does not lie in any quadratic extension of \BbbQ

Proof

Assume  p² = \dfracab,        a,b Î \BbbZ.

Define       f(x) = \dfracxⁿ( 1-x) ⁿn!,

              G( x) = bⁿ[ p²ⁿf( x) -p^2n-2f^¢¢( x) +¼+( -1) ⁿp⁰f^{( 2n)} ( x) ]

(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 since f⁽²ⁿ⁺²⁾ ( x) = 0

d dx [ G¢( x) sin( px) -pG( x) cos( px) ] = [ G¢¢( x) +p2G( x) ] sin( px) = bnp2n+2f( x) sin( px) = p2ansin( px) f( x)

so that

p ó õ 1 0  ansin( px) f( x)  dx = [ \dfracG¢( x) sin( px) p-G( x) cos( px) ] 01 = 0+G( 0) +G( 1) = integer.

But again the integral is non-zero and ® 0 as n®¥. Thus again we have a contradiction.

Getting more involved, now:

e is transcendental over \BbbQ (Hermite)

Proof

Suppose a_me^m+¼+a₁e+a₀ = 0       (a_i Î \BbbZ).

WLOG a₀ ¹ 0

Define       f(x) = \dfracx^p-1( x-1) ^p( x-2)^p¼( x-m) ^p( p-1) !

where for the moment p is arbitrary and prime.

Define       F( x) = f( x) +f^¢( x)+¼+f^{( mp+p-1)} ( x) .

Now if 0 < x < m,

| f( x) | £ \dfracmp-1mmp(p-1) ! = \dfracmmp+p-1( p-1) !

Also       [(d)/(dx)]( e^-xF( x) ) = e^-x[F^¢( x) -F( x) ] = -e^-xf(x)

so that

aj ó õ j 0  e-xf( x)  dx = aj[ -e-xF(x) ] 0j = ajF( 0) -aje-jF( j) .

Multiplying by e^j and summing over j = 0,1,¼m we get

m å j = 0  ajej ó õ j 0  e-xf( x)  dx = F( 0) .0- m å j = 0  ajF( j) = -  m å j = 0  mp+p-1 å i = 0  ajf( i)( j) .

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) ^p¼( -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 ¹ 0. But as p® ¥ the left-hand side tends to 0, using the above estimate for | f( x) | . This is a contradiction.

p is transcendental over \BbbQ (Lindemann)

Proof

If p satisfies an algebraic equation with coefficents in \BbbQ, so does ip (i = Ö[(-1)]). Let this equation be q₁(x) = 0, with roots ip = a₁,¼,a_n. Now e^ip+1 = 0 so

( ea1+1) ¼( ean+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 a₁+a₂,a₁+a₃,¼a_n-1+a_n. The a 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 a s and are also rational. Thus the pairs are roots of the equation q₂( x) = 0 with rational coefficients. Similarly sums of 3 a s are roots of q₃( x) = 0, etc. Then the equation

       q₁( x) q₂( x) ¼q_n( x) = 0

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

       q( x) = 0

       q( x) = cx^r+c₁x^r-1+¼c_r

and c_r ¹ 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 b₁,¼b_r. The original equation becomes

       e^b₁+¼e^b_r+e⁰+¼e⁰ = 0

ie

       åe^b_i+k = 0

where k is an integer > 0        ( ¹ 0 since the term 1¼1 exists)

Now define

       f( x) = c^sx^p-1\dfrac[ q( x)] ^p( p-1) !

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

       F( x) = f( x) +f^¢( x)+¼+f^{( s+p)} ( x) .

       [(d)/(dx)][ e^-xF( x) ] = -e^-xf(x) as before.

Hence we have

       e^-xF( x) -F( 0) = -ò₀^xe^-yf(y)  dy.

Putting y = lx we get

       F( x) -e^xF( 0) = -xò₀¹e^{(1-l) x}f( lx)  dl.

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

       å_{j = 1}^rF( b_j) +kF( 0) = -å_{j = 1}^rb_jò₀¹e^{( 1-l)b_j}f( lb_j)  dl.

CLAIM:

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

å_{j = 1}^rf^{( t)} ( b_j) = 0      ( 0 < t < p) by definition of f. Each derivative of order p or more has a factor p and a factor c^s, since we must differentiate [ q( x) ] ^p enough times to get ¹ 0. And f^{( t)} ( b_j) is a polynomial in b_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 \dfracc_ic of degree £ s). Thus we have

       å_{j = 1}^rf^{( t)} ( b_j) = pk_t       t = p,¼p+s.

Thus LHS = ( \funcinteger) +kF( 0) .       What is F( 0) ?

       f^{( t)} ( 0) = 0       t = 0,¼,p-2.

       f^{( p-1)} ( 0) = c^sc_r^p       (c_r ¹ 0)

       f^{( t)} ( 0) = p (some integer)       t = p,p+1,¼.

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.