B.3. TRANSCENDENTAL NUMBERS 401
=
(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)
)(2.11)
Here K ∑nj=0 f ( j) (0) is added and subtracted. From Lemma B.3.3,
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 B.3.5 Let vk,uk,mk be integers for k = 1,2...,m,uk,vk nonzero. Then for eachk there 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.■
Corollary B.3.6 Let K and ci for i = 1, · · · ,n be nonzero integers. For each k between1 and n let {β (k)i}
mki=1 be the roots of a polynomial with integer coefficients,
Qk (x)≡ vkxmk + · · ·+uk
where vk,uk ̸= 0. Then
K + c1
(m1
∑j=1
eβ (1) j
)+ c2
(m2
∑j=1
eβ (2) j
)+ · · ·+ cn
(mn
∑j=1
eβ (n) j
)̸= 0. (∗)
Proof: Let Kk be nonzero integers which add to K. It is certainly possible to obtain thissince the Kk are allowed to change sign. They only need to be nonzero. Also let αk be as inthe above lemma such that α
mk+2k vmk+1
k uk = U some integer. Thus, replacing each Qk (x)