372 APPENDIX A. CLASSIFICATION OF REAL NUMBERS

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) (1.19)

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 1.18 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 1.19 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 A.2.5,

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 ele-

mentary symmetri c 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 A.2.7 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 A.3.7.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 = 0which cannot happen if ln(a) is algebraic according to the above theorem. If a is algebraicand sin(a) ̸= 0, then sin(a) is transcendental because

12i

eia− 12i

e−ia +(−1)sin(a)e0 = 0

which cannot occur if sin(a) is algebraic. There are doubtless other examples of numberswhich are transcendental by this amazing theorem. For example, π is also transcendental.This is because 1+eiπ = 0. This couldn’t happen if π were algebraic because then so wouldbe iπ .

Of course this marvelous theorem is insufficient to classify an arbitrary real number,even many which are well specified like π + e.