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Unformatted text preview: Advanced Calculus, Supplement and Solutions Horst R. Thieme, Arizona State University, Fall 2007. updated December 4, 2007 2 Chapter 1 The real numbers 1.1 Ordered Fields [1, Sec.11] We use the following symbols: N set of natural numbers (without 0), N = { 1 , 2 ,... } Z set of integers, Z = { ..., 2 , 1 , , 1 , 2 ,... } Q set of rational numbers R set of real numbers C set of complex numbers R is a field, i.e. R satisfies the following axioms. 1.1.1 Field axioms. There are functions + and defined on R 2 = R R , with values in R , called addition and multiplication which satisfy the following properties: (Ac) For all x,y R , x + y = y + x . [Additive commutative law] (Aa) For all x,y,z R , x + ( y + z ) = ( x + y ) = z . [Additive associative law] 3 4 CHAPTER 1. THE REAL NUMBERS (Aid) There exists a number 0 R such that x + 0 = x for all x R . [Additive identity] (Ain) For each x R there is a number x R such that x + ( x ) = 0. [Additive inverse] (Mc) For all x,y R , x y = y x . [Multiplicative commutative law] (Ma) For all x,y,z R , x ( y z ) = ( x y ) z . [Multiplicative associative law] (Mid) There exists a number 1 R such that 1 6 = 0 and x 1 = x for all x R . [Multiplicative identity] (Min) For each x R , x 6 = 0, there is a number 1 /x such that x (1 /x ) = 1. [Multiplicative inverse] (Dis) For all x,y,z R , x ( y + z ) = x y + x x . [Distributive law] 1.1 Remark. (a) The additive identity 0 and the multiplicative identity 1 are uniquely determined. (b) The additive and multiplicative inverses are uniquely determined. For the multiplicative inverse we also write 1 /x = 1 x = x 1 . (c) We require 1 6 = 0 for the two identities because, otherwise, R could be the trivial field { } with 0 acting as multiplicative identity as well. (d) Q and C are fields as well. (e) The set { , 1 } becomes a field by the following definitions: 0 + 0 = 0 , 0 + 1 = 1 = 1 + 0 , 1 0 = 0 = 0 , 1 1 = 1 as one expects, but 1 + 1 = 0. 1.1. ORDERED FIELDS [1, SEC.11] 5 (f) Often x y is written as xy . (g) We write x y for x + ( y ) and x/y for x (1 /y ). Proof. (a) Suppose there is a number 0 such that x + 0 = x for all x R . By the additive commutative law, 0 + x = x for all x R . Then 0 = 0 + 0 = 0. A similar argument shows the uniqueness of 1. (b) We explore the uniqueness of multiplicative inverses. Let x R , x 6 = 0. Suppose that there exists some y R such that x y = 1. By the multiplicative commutative law, y x = 1. By the commutative identity and associativity laws, y = y 1 = y x 1 x = ( y x ) 1 x = 1 /x....
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 Spring '07
 thieme
 Calculus, Real Numbers, Natural Numbers

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