Math 249A Fall 2010: Transcendental Number Theory
Math 249A Fall 2010: Transcendental Number Theory A course by Kannan Soundararajan LATEXed by Ian Petrow September 19, 2011 Contents 1 Introduction; Transcendence of e and π α is algebraic if there exists p 2 Z[x], p 6= 0 with p(α) = 0, otherwise α is called transcendental . Cantor: Algebraic numbers are countable, so transcendental numbers exist, and are a measure 1 set in [0; 1], but it is hard to prove transcendence for any particular number. p p 2 Examples of (proported) transcendental numbers: e, π, γ, eπ, 2 , ζ(3), ζ(5) ::: p p 2 Know: e, π, eπ, 2 are transcendental. We don't even know if γ and ζ(5), ζ(7);::: are irrational or rational, and we know that ζ(3) is irrational, but not whether or not it is transcendental! Lioville showed that the number 1 X 10−n! n=1 is transcendental, and this was one of the first numbers proven to be transcen- dental. Theorem 1 (Lioville). If 0 6= p 2 Z[x] is of degree n, and α is a root of p, α 62 , then Q a C(α) α − ≥ : q qn Proof. Assume without loss of generality that α < a=q, and that p is irreducible. Then by the mean value theorem, p(α) − p(a=q) = (α − a=q)p0(ξ) 1 for some point ξ 2 (α; a=q). But p(α) = 0 of course, and p(a=q) is a rational number with denominator qn. Thus 1=qn ≤ jα − a=qj sup jp0(x)j: x2(α−1,α+1) This simple theorem immediately shows that Lioville's number is transcen- dental because it is approximated by a rational number far too well to be al- gebraic.
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