A.3. TRANSCENDENTAL NUMBERS 369
Defining fk (x) and Ik (s) as in Lemma A.3.4,
fk (x)≡v(m+1)pQp
k (x)xp−1
(p−1)!
and as before, let p be a very large prime number. It follows from Lemma A.3.4 that foreach k = 1, · · · ,n,
ck
mk
∑i=1
Ik (β (k)i) =
(Kk + ck
mk
∑i=1
eβ (k)i
)deg( fk)
∑j=0
f ( j)k (0)
−
(Kk
deg( fk)
∑j=0
f ( j)k (0)+ ck
mk
∑i=1
deg( fk)
∑j=0
f ( j)k (β (k)i)
)
This is exactly the same computation as in the beginning of that lemma except one addsand subtracts Kk ∑
deg( fk)j=0 f ( j)
k (0) rather than K ∑deg( fk)j=0 f ( j)
k (0) where the Kk are chosen suchthat their sum equals K and the term on the left converges to 0 as p→∞. By Lemma A.3.4,
ck
mk
∑i=1
Ik (β (k)i) =
(Kk + ck
mk
∑i=1
eβ (k)i
)(U p +Nk p)
−Kk (U p +Nk p)− ckN′k p
=
(Kk + ck
mk
∑i=1
eβ (k)i
)U p−KU p +Mk p
where Mk is some integer. Now add.
m
∑k=1
ck
mk
∑i=1
Ik (β (k)i) =U p
(K +
m
∑k=1
ck
mk
∑i=1
eβ (k)i
)−KmU p +Mp
If K +∑mk=1 ck ∑
mki=1 eβ (k)i = 0, then if p > max(K,m,U) you would have −KmU p +Mp
an integer so it cannot equal the left side which will be small if p is large. Therefore, ∗follows.
Next is an even more interesting Lemma which follows from the above corollary.
Lemma A.3.7 If b0,b1, · · · ,bn are non zero integers, and γ1, · · · ,γn are distinct alge-braic numbers, then
b0eγ0 +b1eγ1 + · · ·+bneγn ̸= 0
Proof: Assumeb0eγ0 +b1eγ1 + · · ·+bneγn = 0 (1.12)
Divide by eγ0 and letting K = b0,
K +b1eα(1)+ · · ·+bneα(n) = 0 (1.13)
where α (k) = γk− γ0. These are still distinct algebraic numbers. Therefore, α (k) is a rootof a polynomial
Qk (x) = vkxmk + · · ·+uk (1.14)