Chapter The Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide Section 3.2 Ordered Fields Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide The real numbers are an example of an ordered field We have two operations, + and ⋅ , called addition and multiplication, such that the following properties apply: A1 For all x, y ∈ A2 For all x, y ∈ , x+y∈ and if x = w and y = z, then x + y = w + z , x + y = y + x A3 For all x, y, z ∈ Commutative Property , x + ( y + z) Associative Property = (x + y) + z A4 There is a unique real number such that x + = x for all x ∈ A5 For each x ∈ M1 For all x, y ∈ M2 For all x, y ∈ , x⋅y∈ , x ⋅ ( y ⋅ z) Additive Identity there is a unique real number − x such that Multiplication Commutative Property Associative Property = (x ⋅ y) ⋅ z M4 There is a unique real number such that ≠ and x ⋅ = x for all x ∈ M5 For each DL For all x, y, z ∈ Add Inverse x + (− x) = and if x = w and y = z, then x ⋅ y = w ⋅ z , x ⋅ y = y ⋅ x M3 For all x, y, z ∈ Addition , x ⋅ ( y + z) x∈ with x ≠ 0, there is a unique real number 1/ x = (x ⋅ y) + (x ⋅ z) Mult Identity such that x ⋅ (1/ x) = Distributive Law Mult Inverse These 11 axioms are called the field axioms Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide In addition to the field axioms, the real numbers also satisfy four order axioms These axioms indentify the properties of the relation “ b means b < a trichotomy law O1 For all x, y ∈ O2 For all x, y, z ∈ O3 For all x, y, z ∈ , exactly one of the relations x = y, x > y, or x < y holds , if x < , if x < O4 For all x, y, z ∈ transitive property y and y < z, then x < z y then x + z < y + z , if x < y and z > 0, then x ⋅ z < y ⋅ z Our first theorem shows how the axioms may be used to derive some familiar algebraic properties Theorem 3.2.1 Let x, y, and z be real numbers (a) If x + z = y + z, then x = y (c) – = (e) x ⋅ y = iff x = or y = (b) x ⋅ = (d) (–1) ⋅ x = – x (f) x < y iff – y < – x (g) If x < y and z < 0, then x ⋅ z > y ⋅ z We illustrate the proofs by doing parts (a) and (d) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide Theorem 3.2.1 Let x, y and z be real numbers (a) If x + z = y + z, then x = y Proof: If x + z = y + z, then (x + z) + (– z) = ( y + z) + (– z) by A5 (add inverse) and A1 (addition) x + [z + (– z)] = y + [z + (– z)] x+0 = y+0 x = y Copyright © 2013, 2005, 2001 Pearson Education, Inc by A3 (assoc property) by A5 (add inverse) by A4 (add identity) Section 3.2, Slide Theorem 3.2.1 Let x, y and z be real numbers (d) For any real x, (–1) ⋅ x = – x Question: What is –1? Answer: –1 is the number which when added to gives Question: What is – x? Answer: – x is the number which when added to x gives Question: So how we show that (–1) ⋅ x = – x? Answer: We show that (–1) ⋅ x satisfies definition of – x Namely, when (–1) ⋅ x is added to x, the result is Proof: We must show that x + (–1) ⋅ x = x + (–1) ⋅ x We have Thus (–1) ⋅ x by M2 (commutative) = x + x ⋅ (–1) = x ⋅ (1) + x ⋅ (–1) by M4 (mult identity) = x ⋅ [1 + (–1)] by DL (distributive law) = x⋅0 by A5 (add inverse) = by part (b) = – x by the uniqueness of – x in A5 ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide Any mathematical system that satisfies these fifteen axioms is called an ordered field The real numbers The rational numbers But there are other examples as well are an example of an ordered field are another example Example 3.2.6 For a more unusual example of an ordered field, let That is, be the set of all rational functions A typical element of is the set of all quotients of polynomials an x n + L + a1x + a0 , k bk x + L + b1x + b0 looks like where the coefficients are real numbers and bk ≠ Using the usual rules for adding, subtracting, multiplying, and dividing polynomials, it is not difficult to verify that is a field quotient such as above is positive iff We can define an order on by saying that a an and bk have the same sign; that is, an ⋅ bk > 3x + x − > 0, For example, since (3)(7) > 7x + x5 + 3x − < 0, 2But 6x − 7x Copyright © 2013, 2005, 2001 Pearson Education, Inc since (4)(–7) < Section 3.2, Slide If p /q and f /g are rational functions, then we say that That is, p f > q g iff p f > q g iff p f − > q g pg − fq > qg Practice 3.2.7* Which is larger, x2 or ? x2 We have x2 21 − x − x + 21 − = = < x2 7 x2 x2 x2 So, < x The verification that “> ” satisfies the order axioms is Exercise 11 It turns out that the ordered field has a number of interesting properties, as we shall see later There is one more algebraic property of the real numbers to which we give special attention because of its frequent use in proofs in analysis, and because it may not familiar Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide Theorem 3.2.8 Let x, y ∈ such that x ≤ y + ε for every ε > Then x ≤ y Proof: We shall establish the contrapositive By axiom O1 (the trichotomy law), the negation of that x > y and we must show that there exists an ε Question: If Thus we suppose x ≤ y is x > y > such that x > y + ε x > y, what positive ε can we add on to y so that x > y + ε ? ε We could take ε y Let ε = (x – y) /2 x Since x > y, ε > 0 y +ε = y + the distance from x to y Furthermore, x− y x+ y x+x = < = x, 2 Copyright © 2013, 2005, 2001 Pearson Education, Inc equal to half as required ♦ Section 3.2, Slide Recall the definition of absolute value from Section 1.4 Definition 3.2.9 If x ∈ , then the absolute value of x, denoted by | x |, is defined by  x, if x ≥ 0, | x| = − x, if x < The basic properties of absolute value are summarized in the following theorem Theorem 3.2.10 Let x, y ∈ and let a > Then (a) | x | ≥ 0, (b) | x | ≤ a iff − a ≤ x ≤ a, (c) | x y | = | x | ⋅ | y |, (d) | x + y | ≤ | x | + | y | We will prove parts (b) and (d) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide 10 Theorem 3.2.10 (b) Let x, y ∈ and let a > Then | x | ≤ a iff − a ≤ x ≤ a Proof: Since x = | x | or x = − | x |, it follows that − | x | ≤ x ≤ | x | If | x | ≤ a, then we have − a ≤ − | x | ≤ x ≤ | x | ≤ a Conversely, suppose that − If x ≥ 0, then | x | = x ≤ a a ≤ x ≤ a And if x  Then | x + y | ≤ | x | + | y | Proof: As in part (b), we have − | x | ≤ x ≤ | x | and − | y | ≤ y ≤ | y | Adding the inequalities together, we obtain which implies that | x + y | ≤| Copyright © 2013, 2005, 2001 Pearson Education, Inc − (| x | + | y |) ≤ x + y ≤ | x | + | y |,  x | + | y | by part (b) ♦ Section 3.2, Slide 11 Part (d) of Theorem 3.2.10 is referred to as the triangle inequality: | x + y | ≤ | x | + | y | Its name comes from its being used with vectors in the plane, where | x | represents the length of vector x x+ y y x It says that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.2, Slide 12 ... polynomials, it is not difficult to verify that is a field quotient such as above is positive iff We can define an order on by saying that a an and bk have the same sign; that is, an ⋅ bk > 3x + x − > 0,... Thus (–1) ⋅ x by M2 (commutative) = x + x ⋅ (–1) = x ⋅ (1) + x ⋅ (–1) by M4 (mult identity) = x ⋅ [1 + (–1)] by DL (distributive law) = x⋅0 by A5 (add inverse) = by part (b) = – x by the uniqueness... 3.2.10 is referred to as the triangle inequality: | x + y | ≤ | x | + | y | Its name comes from its being used with vectors in the plane, where | x | represents the length of vector x x+ y y x It