Most numbers are like this. Here the algebraic numbers are those which are roots of a polynomial
equation having rational numbers as coefficients. By the fundamental theorem of algebra, all these
numbers are in ℂ. There are only countably many of these algebraic numbers, (Problem 41 on
Page 610). Therefore, most numbers are transcendental. Nevertheless, it is very hard to prove that
this or that number is transcendental. Probably the most famous theorem about this is the
Lindemannn Weierstrass theorem.
Theorem F.3.1Let the α_{i}be distinct nonzero algebraic numbers and let the a_{i}be nonzeroalgebraic numbers. Then
n
∑ aieαi ⁄= 0
i=1
I am following the interesting Wikepedia article on this subject. You can also look at the book
by Baker [4], Transcendental Number Theory, Cambridge University Press. There are also many
other treatments which you can find on the web including an interesting article by
Steinberg and Redheffer which appeared in about 1950 part of which I am following
here.
The proof makes use of the following identity. For f
(x)
a polynomial,
∫ s deg(f) deg(f)
I(s) ≡ es− xf (x)dx = es ∑ f(j)(0)− ∑ f(j)(s). (6.3)
0 j=0 j=0
(6.3)
where f^{}
(j)
denotes the j^{th} derivative. In this formula, s ∈ ℂ and the integral is defined in the
natural way as
∫ 1
sf (ts)es−tsdt (6.4)
0
(6.4)
The identity follows from integration by parts.
∫ ∫
1 s− ts s 1 −ts
0 sf (ts)e dt = se 0 f (ts)e dt
[ e− ts ∫ 1e−ts ]
= ses − ---f (ts)|10 + ----sf′(st)dt
[ s 0 s∫ 1 ]
= ses − e−-sf (s)+ 1f (0) + e− tsf′(st)dt
s s 0
s ∫ 1 s−ts ′
= e f (0)− f (s)+ se f (st)dt
∫0s
≡ esf (0)− f (s)+ es−xf′(x )dx
0
Continuing this way establishes the identity since at the right end looks just like what we started
with except with a derivative on the f.
Lemma F.3.2If K and c are nonzero integers, and β_{1},
⋅⋅⋅
,β_{m}are the roots of a singlepolynomial with integer coefficients,
Q (x) = vxm + ⋅⋅⋅+ u
where v,u≠0, then
( )
K + c eβ1 + ⋅⋅⋅+eβm ⁄= 0.
Letting
v(m+1)pQp-(x)xp−1
f (x ) = (p− 1)!
and I
(s)
be defined in terms of f
(x)
as above, it follows,
∑m
pli→m∞ I (βi) = 0
i=1
and
∑n
f (j)(0) = vp(m+1)up +m1 (p)p
j=0
m n
∑ ∑ (j)
f (βi) = m2 (p)p
i=1 j=0
where m_{i}
(p)
is some integer.
Proof: Let p be a large prime number. Then consider the polynomial f
which clearly converges to 0. This proves the claim.
The next thing to consider is the term on the end in 6.5,
n m n
K ∑ f(j)(0)+ c∑ ∑ f (j)(βi) (6.6)
j=0 i=1j=0
(6.6)
The idea is to show that for large enough p it is always a nonzero integer. When this is done, it
can’t happen that K + c∑_{i=1}^{m}e^{βi} = 0 because if this were so, you would have a very small
number equal to an integer.
v(m+1)p(vxm-+-⋅⋅⋅+-u)pxp−1-
f (x) = (p− 1)!
Then f^{j}
(0)
= 0 unless j ≥ p − 1 because otherwise, that x^{p−1} term will result in some x^{r},r > 0
and everything is zero when you plug in x = 0. Now say j = p − 1. Then it is clear
that
f(p−1)(0) = upv(m+1 )p
So what if j > p − 1? Then by Liebniz formula,
( )
f j(x) = j v(m+1)p---d----[(vxm +⋅⋅⋅+ u)]p + Stuff
p − 1 dxj−(p−1)
where the Stuff equals 0 when x = 0. Thus f^{j}
(0)
= pm_{j} where m_{j} is some integer depending on
the integer coefficients of the polynomial Q
v(m+1)p((x-−-β1)(x-−-β2)⋅⋅⋅(x-−-βm))pxp−1
f (x) = (p− 1)!
it follows that for j < p, f^{}
(j)
(βi)
= 0. This is because for such derivatives, each term will have
that product of the
(x − βi)
in it.
To get something non zero, the nonzero terms must involve at least p derivatives of the
expression
((x− β1)(x− β2)⋅⋅⋅(x − βm))p
since otherwise, when evaluated at any β_{k} the result would be 0.
Now say j ≥ p. Then by Liebniz formula, f^{j}
(x)
is of the form
( )
v(m+1)p∑j j d p d p−1
(p−-1)! r dxr (((x − β1)(x − β2)⋅⋅⋅(x − βm))) dxj−rx
r=0 ( )
vp(m+1)−2p+1 ∑j j d p d p−1
= ---(p-−-1)!--- r dxr (((vx− vβ1)(vx − vβ2) ⋅⋅⋅(vx− vβm )) )dxj−r (vx )
r=0
Note that for r too small, the term will be zero when evaluated at any of the β_{i}. You only get
something nonzero if r ≥ p and so there will be a p! produced which will cancel with the
(p − 1)
!
to yield an extra p.
Now if you do the computations using the product rule and then replace x with β_{i} and sum
these over all vβ_{i}, you will get a symmetric polynomial in the quantities
{vβ1,⋅⋅⋅,vβm }
and by
Theorem F.1.4 this is an integer. To see this is symmetric note that switching vβ_{a},vβ_{b}
in
d
dxr ((vx− vβ1)(vx − vβ2)⋅⋅⋅(vx− vβm ))
does not change anything. The other term is just v^{p−1}
(p − 1)
(p− 2)
⋅⋅⋅
(p− j + r)
x^{p−j+r−1} or
zero if j − r > p − 1. It follows that when adding these over i,
Kvp (m+1 )up + m (p)p + L(p)p ≡ Kvp(m+1)up + M (p)p
for some integer M
(p)
. Summarizing, it follows
◜---⁄=◞0◟---◝
∑m ( ∑m )∑n
c I (βi) = K + c eβi f (j)(0)+ Kvp(m+1)up + M (p)p
i=1 i=1 j=0
where the left side is very small whenever p is large enough. Let p be larger than max
(K,v,u)
.
Since p is prime, it follows that it cannot divide Kv^{p(m+1)
}u^{p} and so the last two terms must sum
to a nonzero integer and so the equation 6.5 cannot hold unless
∑m
K + c eβi ⁄= 0 ■
i=1
Note that this shows π is irrational. If π = k∕m where k,m are integers, then both iπ and −iπ
are roots of the polynomial with integer coefficients,
m2x2 + k2
which would require, from what was just shown that
0 ⁄= 2+ eiπ + e− iπ
which is not the case since the sum on the right equals 0.
The following corollary follows from this.
Corollary F.3.3Let K and c_{i}for i = 1,
⋅⋅⋅
,n be nonzero integers. For each k between 1 and nlet
{β (k)i}
_{i=1}^{mk}be the roots of a polynomial with integer coefficients,
m
Qk(x) ≡ vkx k + ⋅⋅⋅+ uk
where v_{k},u_{k}≠0. Then
( ) ( ) ( )
m∑1 β(1) ∑m2 β(2) ∑mn β(n)
K +c1( e j) + c2( e j) + ⋅⋅⋅+ cn ( e j) ⁄= 0.
j=1 j=1 j=1
Proof: Defining f_{k}
(x )
and I_{k}
(s)
as in Lemma F.3.2, it follows from Lemma F.3.2 that for
each k = 1,
⋅⋅⋅
,n,
( )
∑mk ∑mk deg∑(fk) (j)
ck Ik (β (k )i) = Kk + ck eβ(k)i fk (0)
i=1 i=1 j=0
de∑g(fk) m∑k de∑g(fk)
− Kk f(jk)(0)− ck fk(j)(β(k)i)
j=0 i=1 j=0
This is exactly the same computation as in the beginning of that lemma except one adds and
subtracts K_{k}∑_{j=0}^{deg (fk)
}f_{
k}^{(j)
}
(0)
rather than K∑_{
j=0}^{deg (fk)
}f_{
k}^{(j)
}
(0)
where the K_{
k} are chosen
such that their sum equals K. By Lemma F.3.2,
m ( m )
∑ k ∑ k β(k)i ( (mk+1)p p )
cki=1Ik(β(k)i) = Kk + cki=1e vk uk +Nkp
( (mk+1)p p ) ′
− Kk vk uk + Nkp − ckNkp
and so
m ( m )
∑ k ∑ k β(k)i ( (mk+1)p p )
cki=1Ik(β(k)i) = Kk + cki=1e vk uk +Nkp
− Kkv (mk+1)pup + Mkp
k k
for some integer M_{k}. By multiplying each Q_{k}
(x)
by a suitable constant depending on k, it can be
assumed without loss of generality that all the v_{k}^{mk+1}u_{
k} are equal to a constant integer U. Then
the above equals
( )
m∑k m∑k β(k) p
ck Ik(β(k)i) = Kk + ck e i (U + Nkp)
i=1 i=1
p
− KkU + Mkp
Adding these for all k gives
∑n m∑k ( ∑n ∑mk )
ck Ik(β(k)i) = Up K + ck eβ(k)i − KU p + M p
k=1 i=1 k=1 i=1