9.2. TRANSCENDENTAL NUMBERS 203
Proof: The first step is to verify 9.10 for f (x) as given in 9.6 for p large prime numbers.Let p be a large prime number. Then 9.10 follows right away from the definition of I
(β j
)and the definition of f (x) .
∣∣∣I(β j
)∣∣∣≤ ∫ 1
0
∣∣∣β j f(
tβ j
)eβ j−tβ j
∣∣∣dt ≤∫ 1
0
∣∣∣∣∣∣∣|v|(m−1)p
∣∣∣Q(tβ j
)∣∣∣p t p−1∣∣∣β j
∣∣∣p−1
(p−1)!
∣∣∣∣∣∣∣dt
which clearly converges to 0 using considerations involving convergent series which showthe integrand converges uniformly to 0. The degree of f (x) is n≡ pm+ p−1 where p willbe a sufficiently large prime number from now on.
From 9.4,
cm
∑i=1
I (β i) = cm
∑i=1
(eβ i
n
∑j=0
f ( j) (0)−n
∑j=0
f ( j) (β i)
)
=
(K + c
m
∑i=1
eβ i
)n
∑j=0
f ( j) (0)−
(K
n
∑j=0
f ( j) (0)+ cm
∑i=1
n
∑j=0
f ( j) (β i)
)(9.11)
Here K ∑nj=0 f ( j) (0) is added and subtracted. From Lemma 9.2.3,
vp(m+1)up +m1 (p) p+m2 (p) p = Kn
∑j=0
f ( j) (0)+ cm
∑i=1
n
∑j=0
f ( j) (β i)
Thus, if p is very large,
cm
∑i=1
I (β i) = small = Kvp(m+1)up +M (p) p+
(K + c
m
∑i=1
eβ i
)n
∑j=0
f ( j) (0)
Let p be prime and larger than max(K,v,u). If K + c∑mi=1 eβ i = 0, the above is impossible
because it would requiresmall = Kvp(m+1)up +M (p) p
Now the right side is a nonzero integer because p cannot divide Kvp(m+1)up so the rightside cannot equal something small. ■
Note that this shows π is irrational. If π = k/m where k,m are integers, then both iπand −iπ are roots of the polynomial with integer coefficients, m2x2 + k2 which wouldrequire, 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. It is like the above lemma except it involves
several polynomials. First is a lemma.
Lemma 9.2.5 Let vk,uk,mk be integers for k = 1,2...,m,uk,vk nonzero. Then for each kthere exists αk an integer such that α
mk+2k vmk+1
k uk is U for some non zero integer.
Proof: Let U ≡(
∏mj=1 v
m j+1j u j
)∏
mj=1(m j+2)
2
≡ αmk+2k vmk+1
k uk where αk is an integerchosen to make this so.■