364 APPENDIX A. CLASSIFICATION OF REAL NUMBERS
where the pi are elementary symmetric polynomials defined as the coefficients of p̂(x) =∏
nj=1 (x− x j) with pk (x1, ...,xn) of degree k since it is the coefficient of xn−k. Earlier we
had them ± these coefficients. Thus
f (anα1, · · · ,anαn) = ∑k1+···+kn=d
ak1···kn pk11 (anα1, · · · ,anαn) · · · pkn
n (anα1, · · · ,anαn)
Now the given polynomial in ∗, p(x) is of the form
an
n
∏j=1
(x−α j)≡ an
(n
∑k=0
pk (α1, · · · ,αn)xn−k
)
= anxn +an−1xn−1 + · · ·+a1x+a0
Thus, equating coefficients, an pk (α1, · · · ,αn) = an−k. Multiply both sides by ak−1n . Thus
pk (anα1, · · · ,anαn) = ak−1n an−k
an integer. Therefore,
f (anα1, · · · ,anαn) = ∑k1+···+kn=d
ak1···kn pk11 (anα1, · · · ,anαn) · · · pkn
n (anα1, · · · ,anαn)
and each pk (anα1, · · · ,anαn) is an integer. Thus f (anα1, · · · ,anαn) is indeed an integer.From this, it is obvious that f (α1, · · · ,αn) is rational. Indeed, from 1.3,
f (α1, · · · ,αn) = ∑k1+···+kn=d
ak1···kn pk11 (α1, · · · ,αn) · · · pkn
n (α1, · · · ,αn)
Now multiply both sides by aMn , an integer where M is chosen large enough that
aMn f (α1, · · · ,αn)
= ∑k1+···+kn=d
ah(k1,...,kn)n ak1···kn pk1
1 (anα1, · · · ,anαn) · · · pknn (anα1, · · · ,anαn)
where h(k1, ...,kn) is some nonnegative integer. Then the right side is an integer. Thusf (α1, · · · ,αn) is rational. If the f had rational coefficients, then m f would have inte-ger coefficients for a suitable m and so m f (α1, · · · ,αn) would be rational which yieldsf (α1, · · · ,αn) is rational.
A.3 Transcendental NumbersMost numbers are like this, transcendental. Here the algebraic numbers are those whichare roots of a polynomial equation having rational numbers as coefficients, equivalentlyinteger coefficients. By the fundamental theorem of algebra, all these numbers are in Cand they constitute a countable collection of numbers in C. Therefore, most numbers in Care transcendental. Nevertheless, it is very hard to prove that a particular number is tran-scendental. Probably the most famous theorem about this is the Lindermannn Weierstrasstheorem, 1884.
Theorem A.3.1 Let the α i be distinct nonzero algebraic numbers and let the ai benonzero algebraic numbers. Then ∑
ni=1 aieα i ̸= 0.