9.2. TRANSCENDENTAL NUMBERS 205

Lemma 9.2.7 If b0,b1, · · · ,bn are non zero integers, and γ1, · · · ,γn are distinct algebraicnumbers, then

b0eγ0 +b1eγ1 + · · ·+bneγn ̸= 0

Proof: Assumeb0eγ0 +b1eγ1 + · · ·+bneγn = 0 (9.12)

Divide by eγ0 and letting K = b0,

K +b1eα(1)+ · · ·+bneα(n) = 0 (9.13)

where α (k) = γk− γ0. These are still distinct algebraic numbers. Therefore, α (k) is a rootof a polynomial

Qk (x) = vkxmk + · · ·+uk (9.14)

having integer coefficients, vk,uk ̸= 0. Recall algebraic numbers were defined as roots ofpolynomial equations having rational coefficients. Just multiply by the denominators to getone with integer coefficients. Let the roots of this polynomial equation be{

α (k)1 , · · · ,α (k)mk

}and suppose they are listed in such a way that α (k)1 = α (k). Thus, by Theorem 9.1.6every symmetric polynomial in these roots is rational.

Letting ik be an integer in {1, · · · ,mk} it follows from the assumption 9.12 that

∏(i1,··· ,in)

ik∈{1,··· ,mk}

(K +b1eα(1)i1 +b2eα(2)i2 + · · ·+bneα(n)in

)= 0 (9.15)

This is because one of the factors is the one occurring in 9.13 when ik = 1 for every k. Theproduct is taken over all distinct ordered lists (i1, · · · , in) where ik is as indicated. Expandthis possibly huge product. This will yield something like the following.

K′+ c1

(eβ (1)1 + · · ·+ eβ (1)µ(1)

)+ c2

(eβ (2)1 + · · ·+ eβ (2)µ(2)

)+ · · ·+

cN

(eβ (N)1 + · · ·+ eβ (N)µ(N)

)= 0 (9.16)

These integers c j come from products of the bi and K. You group these exponentials ac-cording to which ci they multiply. The β (i) j are the distinct exponents which result, eachbeing a sum of some of the α (r)ir . Since the product included all roots for each Qk (x),interchanging their order does not change the distinct exponents β (i) j which result. Theymight occur in a different order however, but you would still have the same distinct ex-ponents associated with each cs as shown in the sum. Thus any symmetric polynomialin the β (s)1 ,β (s)2 , · · · ,β (s)

µ(s) is also a symmetric polynomial in the roots of Qk (x) ,α (k)1 ,α (k)2 , · · · ,α (k)mk

for each k.Doesn’t this contradict Corollary 9.2.6? This is not yet clear because we don’t know

that the β (i)1 , ...,β (i)µ(i) are roots of a polynomial having rational coefficients. For a

given r,β (r)1 , · · · ,β (r)µ(r) are roots of the polynomial

(x−β (r)1)(x−β (r)2) · · ·(

x−β (r)µ(r)

)(9.17)