MAT371

# MAT371 - Advanced Calculus Supplement and Solutions Horst R...

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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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MAT371 - Advanced Calculus Supplement and Solutions Horst R...

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