B.3. TRANSCENDENTAL NUMBERS 397

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 2.3,

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. ■

B.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 B.3.1 Let the α i be distinct nonzero algebraic numbers and let the ai benonzero algebraic numbers. Then ∑

ni=1 aieα i ̸= 0.

I am following the interesting Wikepedia article on this subject. You can also look at thebook by Baker [5], Transcendental Number Theory, Cambridge University Press. There arealso many other treatments which you can find on the web including an interesting articleby Steinberg and Redheffer which appeared in about 1950.

The proof makes use of the following identity. For f (x) a polynomial,

I (s)≡∫ s

0es−x f (x)dx = es

deg( f )

∑j=0

f ( j) (0)−deg( f )

∑j=0

f ( j) (s) . (2.4)

where f ( j) denotes the jth derivative. It is like the convolution integral discussed earlierwith Laplace transforms. In this formula, s ∈ C and the integral is defined in the naturalway as ∫ 1

0s f (ts)es−tsdt (2.5)