1   The Transcendence of p





Definitions

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 a 0,a 1,... ,a n not all 0, such that a 0an+a 1an-1+... +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 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 ån=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 Q, if a¹ 0 or 1 and b is irrational, prove ab 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.

Theorem 1   p is irrational

Proof

Let In(x)=ò-1+1( 1-x2) ncos ( a x) dx

Integrating by parts we have

     a 2In=2n( 2n-1) In-1-4n( n-1) In-2          ( n³ 2)

which implies that

     a 2n+1In=n!( Pnsin ( a ) +Qncos ( a ) )      (* )

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



Remark 2   deg Pn=n, deg Qn=n-1

Put a =p /2, and assume p is rational, so that p =b/a,   a,bÎ Z

From ( * ) we deduce that Jn=b2n+1In/n! is an integer. On the other hand Jn® 0 as n® ¥ since b is fixed and In is bounded by
ó
õ
+1


-1
cos æ
ç
ç
è
p x
2
ö
÷
÷
ø		 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 Jn¹ 0. Contradiction.



Theorem 3   p 2 is irrational - so p does not lie in any quadratic extension of Q

Proof

Assume p 2=a/b,     a,bÎ Z.

Define     f(x)=xn( 1-x) n/n!,

          G( x) =bn[ p 2nf( x) -p 2n-2f' ' ( x) +... +( -1) np 0f( 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
é
ë		 G
'
 
( x ) sin ( p x ) -p G ( x ) cos ( p x ) ù
û
=
é
ë		 G
' '
 
( x ) +p 2G ( x ) ù
û		 sin ( p x )
  =
bnp 2n+2f ( x ) sin ( p x )
  =
p 2ansin ( p x ) f ( x )
so that

p ó
õ
1


0
ansin ( p x ) f ( x ) dx
=
é
ê
ê
ë
G
'
 
( x ) sin ( p x )
p
-G ( x ) cos ( p x ) ù
ú
ú
û
1



0
  =
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:

Theorem 4   e is transcendental over Q (Hermite)

Proof

Suppose amem+... +a1e+a0=0     (aiÎ Z).

WLOG a0¹ 0

Define    f(x)=xp-1( x-1) p( x-2) p... ( x-m) p/( p-1) !

where for the moment p is arbitrary and prime.

Define    F( x) =f( x) +f' ( x) +... +f( mp+p-1) ( x) .

Now if 0<x<m,
| f ( x ) |
£
mp-1mmp
( p-1 ) !
  =
mmp+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 ) ]
j
 
0
  =
ajF ( 0 ) -aje-jF ( j ) .
Multiplying by ej and summing over j=0,1,... m 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) 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 ¹ 0. But as p® ¥ the left-hand side tends to 0, using the above estimate for | f( x) | . This is a contradiction.



Theorem 5   p is transcendental over Q (Lindemann)

Proof

If p satisfies an algebraic equation with coefficents in Q, so does ip (i=Ö-1). Let this equation be q 1( x) =0, with roots ip =a 1,... ,a n. Now eip +1=0 so
æ
è		 e
a 1
 
+1 ö
ø		 ... æ
è		 e
a n
 
+1 ö
ø		 =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 a 1+a 2,a 1+a 3,... a 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 q 2( x) =0 with rational coefficients. Similarly sums of 3 a s are roots of q 3( x) =0, etc. Then the equation

    q 1( x) q 2( x) ... q n( x) =0

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

    q ( x) =0

    q ( x) =cxr+c1xr-1+... cr

and cr¹ 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 b 1,... b r. The original equation becomes

    eb 1+... eb r+e0+... e0=0

ie

    å eb i+k=0

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

Now define

    f( x) =csxp-1[ q ( x) ] p/( p-1) !

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

    F( x) =f( x) +f' ( x) +... +f( s+p) ( x) .

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

Hence we have

    e-xF( x) -F( 0) =-ò0xe-yf( y) dy.

Putting y=l x we get

    F( x) -exF( 0) =-xò01e( 1-l ) xf( l x) dl .

Let x range over the b i and sum. Since å eb i+k=0 we get

    åj=1rF( b j) +kF( 0) =-åj=1rb jò01e( 1-l ) b jf( l b j) dl .

CLAIM:

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

åj=1rf( 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 cs, 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 cs, which it is. (symmetric functions are polynomials in coefficients = polynomials in ci/c of degree £ s). Thus we have

    åj=1rf( t) ( b j) =pkt     t=p,... p+s.

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

    f( t) ( 0) =0     t=0,... ,p-2.

    f( p-1) ( 0) =cscrp     ( cr¹ 0)

    f( t) ( 0) =p (some integer)    t=p,p+1,... .

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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