24 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS We now consider the ring of polynomials C[z] over C. Exercise 1.9.25. Show that C[z] is an integral domain. Determine the elements in this domain which have multiplicative inverses. Definition 1.9.26. Let F be a field. We say that F is algebraically closed if every nonconstant polynomial in F [x] has a root in F . That is, F is algebraically closed if, for every nonconstant p(x) F [x], there is an element r F such that p(r) = 0. The most important example of an algebraically closed field is the com- plex numbers. There are a semi-infinite number of proofs of this theorem. We will present one of these as a project in Section 2.7.2 using the properties of continuous functions developed in Chapter 2. Exercise 1.9.27. Let F be a field and suppose that p(x) F [x]. Show that r is a root of p(x) if and only if (x r) is a factor of p(x). That is, we can write p(x) = (x r)q(x) for some q(x) F [x]. Definition 1.9.28. Let A be the collection of all complex roots of poly- nomials in Z[x]. The set A is called the set of algebraic numbers in C. The set AR = A R is called the set of real algebraic numbers. A real number which is not a real algebraic number is called transcendental. Example 1.9.29. Among the more famous algebraic numbers are i and −i. For real algebraic numbers, the most famous one is probably 2. The most famous transcendental numbers are π and e. Exercise 1.9.30. Show that A and AR are fields. Exercise 1.9.31. Show that the field A of algebraic numbers is count- able. Remark 1.9.32. It follows from the exercise above that the field AR of real algebraic numbers is countable and hence the set of transcendental numbers is uncountable. 1.10. Independent Projects 1.10.1. Another Construction of R. Definition 1.10.1. A subset α of Q is said to be a cut (or a Dedekind cut) if it satisfies the following: (a) the set α = and α = Q (b) if r α and if s Q satisfies s r, then s α (c) if r α, then there exists s Q with s r and s α. Let R denote the collection of all cuts. Definition 1.10.2. For α, β R, we define α + β = {r + s | r α and s β}. Let 0 = {r Q | r 0}.
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