9.2. TRANSCENDENTAL NUMBERS 207
Then consider the big product
∏(i1,··· ,in)
ik∈{1,··· ,mk}
(a(1)i1 eα(1)+a(2)i2 eα(2)+ · · ·+a(n)in eα(n)
)(9.18)
the product taken over all ordered lists (i1, · · · , in) . Since one of the factors in this productequals 0, this product equals
0 = b1eβ (1)+b2eβ (2)+ · · ·+bNeβ (N) (9.19)
where the β ( j) are the distinct exponents which result and the bk result from combiningterms corresponding to a single β (k). The β (i) are clearly algebraic because they are thesum of the α (i). I want to show that the bk are actually rational numbers. Since theproduct in 9.18 is taken for all ordered lists as described above, it follows that for a givenk,if a(k)i is switched with a(k) j , that is, two of the roots of vkxmk + · · ·+uk are switched,then the product is unchanged and so 9.19 is also unchanged. Thus each bl is a symmetricpolynomial in the a(k) j , j = 1, · · · ,mk for each k. Consider then a particular bk.It follows
bk = ∑( j1,··· , jmn )
A j1,··· , jmn a(n) j11 · · ·a(n)
jmnmn
and this is symmetric in the{
a(n)1 , · · · ,a(n)mn
}(note n is distinguished) the coefficients
A j1,··· , jmn being in the commutative ring Q [A(1) , · · · ,A(n−1)] where A(p) denotes
a(k)1 , · · · ,a(k)mp
and so from Theorem 9.1.4,
bk = ∑( j1,··· , jmn )
B j1,··· , jmn p j11
(a(n)1 · · ·a(n)mn
)· · · p jmn
mn
(a(n)1 · · ·a(n)mn
)where the B j1,··· , jmn are symmetric in
{a(k) j
}mk
j=1for each k ≤ n− 1. and the pl
k are el-
ementary symmetric polynomials. Now doing to B j1,··· , jmn what was just done to bk andcontinuing this way, it follows bk is a finite sum of rational numbers times powers of el-ementary polynomials in the various
{a(k) j
}mk
j=1for k ≤ n. By Theorem 9.1.6 this is a
rational number. Thus bk is a rational number as desired. Multiplying by the product of allthe denominators, it follows there exist integers ci such that
0 = c1eβ (1)+ c2eβ (2)+ · · ·+ cNeβ (N)
which contradicts Lemma 9.2.7. ■This theorem is sufficient to show e is transcendental. If it were algebraic, then
ee−1 +(−1)e0 ̸= 0
but this is not the case. If a ̸= 1 is algebraic, then ln(a) is transcendental. To see this, notethat
1eln(a)+(−1)ae0 = 0