Chapter The Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Section 3.3 The Completeness Axiom Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide There is one additional axiom that distinguishes completeness axiom from It is called the Before presenting this axiom, let’s look at why it’s needed When we graph the function f (x) = x2 – 2, it appears to cross the horizontal axis at a point between and 2.But does it really? How can we be sure that there is a “number” x on f (x) the axis such that x – = 0? f (x) = x – It turns out that if the x-axis consists only of rational numbers, then no such number exists –2 That is, there is no rational number whose square is 2 x –1 In fact, we can easily prove the more general result p that is irrational (not rational) for any prime number p Recall that an integer p > is prime iff its only divisors are and p Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Theorem 3.3.1 Let p be a prime number Then p Proof: We suppose that p If pis not a rational number is rational and obtain a contradiction p is rational, then we can write = m / n, where m and n are integers Then with no common factors other than p n2 = m2, so p must be a factor of m2 Now the prime factored forms for m and m have exactly the same prime factors, so p is a factor of m so that n 2   = k   p That is, m = k  p for some integer k But then p 2  n = k   p  , Thus p is a factor of n , and as above we conclude that p is also a factor of n Hence p is a factor of both m and n, contradicting the fact that they have no common factors other than ♦ In Section 2.4 we learned that there are uncountably many irrational numbers Thus, if we were to restrict our analysis to rational numbers, our “number line” would have uncountably many “holes” in it It is these holes in the number line that the completeness axiom fills Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide In order to state the completeness axiom for , we need some preliminary definitions Upper Bounds and Suprema Definition 3.3.2 Let S be a subset of If there exists a real number m such that m ≥ s for all s ∈ S, then m is called an upper bound of S, and we say that S is bounded above If m ≤ s for all s ∈ S, then m is a lower bound of S and S is bounded below The set S is said to be bounded if it is bounded above and bounded below If an upper bound m of S is a member of S, then m is called the maximum (or largest element) of S, and we write m = max S Similarly, if a lower bound of S is a member of S, then it is called the minimum (or least element) of S, denoted by S While a set may have many upper and lower bounds, if it has a maximum or a minimum, then those values are unique (Exercise 6) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Example 3.3.3 (a) Let S = {2, 4, 6, 8} max • • • • lower bounds 10 upper bounds Then S is bounded above by 8, 9, 8.5, π , and any other real number greater than or equal to 8.Since ∈ S, we have max S = Similarly, S has many lower bounds, including 2, which is the largest of the lower bounds and the minimum of S It is easy to see that any finite set is bounded and always has a maximum and a minimum Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Example 3.3.3 (b) The interval [0, ∞) is not bounded above no maximum [ –2 no upper bounds lower bounds It is bounded below by any nonpositive number, and of these lower bounds, is the largest Since 0 ∈ [0, ∞), is the minimum of [0, ∞) (c) The interval (0, 4] has a maximum of 4, and this is the smallest of the upper bounds max no ( lower bounds ] upper bounds It is bounded below by any nonpositive number, and of these lower bounds, is the largest Since ∉ (0,1], the set has no minimum Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Example 3.3.3 (d) The empty set ∅ is bounded above by any m ∈ Note that the condition m ≥ s for all s ∈ ∅ is equivalent to the implication “if s ∈ ∅, then m ≥ s.” This implication is true since the antecedent is false Likewise, ∅ is bounded below by any real m Definition 3.3.5 Let S be a nonempty subset of If S is bounded above, then the least upper bound of S is called its supremum and is denoted by sup S Thus m = sup S iff (a) m ≥ s, for all s ∈ S, (b) if That is, m is an upper bound of S m′ < m, then there exists s′ ∈ S such that s′ > m′ Nothing smaller than m is an upper bound of S If S is bounded below, then the greatest lower bound of S is called its infimum and is denoted by inf S Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide { Let T = q ∈ } :0 ≤ q ≤ Does T have a supremum? sup T = If we think of T as a subset of the real numbers, then 2But is not a rational number So T does not have a supremum in When considering subsets of least upper bound , it has been true that each set bounded above has had a This supremum may be a member of the set, as in the interval [0,1], or it may be outside the set, as in the interval [0,1), exists as a real number It depends on the context but in both cases the supremum This fundamental difference between and is the basis for our final axiom of the real numbers, the completeness axiom: Every nonempty subset S of that is bounded above has a least upper bound That is, sup S exists and is a real number It follows readily from this that every nonempty subset S of that is bounded below has a greatest lower bound So, inf S exists and is a real number Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide Theorem 3.3.7 Given nonempty subsets A and B of , let C denote the set C = {x + y : x ∈ A and y ∈ B} If A and B have suprema, then C has a supremum and sup C = sup A + sup B Proof: Let Thus If z ∈ C, then z = x + y for some x ∈ A and y ∈ B sup A = a and sup B = b z = x + y ≤ a + b, so a + By the completeness axiom, b is an upper bound of C C has a least upper bound, say sup C = c least upper bound of C, we have To see that a + b ≤ c, choose any ε c≤ We must show that c Since = a + cb.is the a + b > Since a = sup A, of A, and there must exist x ∈ A such that x > a – ε exists y ∈ B such that But, Similarly, since b = sup B, there Combining these inequalities, we have y > b – ε That is, a + b < x + y ∈ C and c = sup C, so c > a + b – 2ε Thus by Theorem 3.2.8, a – ε is not an upper bound a + b ≤ c x + y > a + b – 2ε c + 2ε for every ε > Finally, since c ≤ a + b and c ≥ a + b, we conclude that c  = a + b ♦ B A y x • a – ε • A+B • • a b – ε Copyright © 2013, 2005, 2001 Pearson Education, Inc • • • b a + b – 2ε x+y • • c Section 3.3, Slide 10 Theorem 3.3.8 Suppose that D is a nonempty set and that x, y ∈ D, f (x) ≤ g Furthermore,  ( y), then f : D →  and g : D →  f (D) is bounded above and g (D) is bounded below f sup  (D) ≤ inf g  (D) Proof: Given any z ∈ D, we have f (x) ≤ g Thus f (D) is bounded  (z), for all x ∈ D It follows that the least upper bound of above by g (z) That is, sup If for every f (D) is no larger than g Since this last inequality holds for all z ∈ D, g f (D) ≤ g (z) bounded below by sup Thus the greatest lower bound of g f (D) than sup f (D) That is, inf g  (D) ≥ sup • g sup f (D) • lower bound of g(D) Copyright © 2013, 2005, 2001 Pearson Education, Inc  (D) is no smaller z f least upper bound of f (D)  (D) is f (D) ♦ D f (D)  (z) inf g(D) g(z) • • g (D) upper bound of f (D) greatest lower bound of g(D) Section 3.3, Slide 11 Theorem 3.3.9 The Archimedean Property of The set of natural numbers is unbounded above in Proof: If were bounded above, then by the completeness axiom it would have a least upper bound, say sup Thus there exists an n in Since m is a least upper bound, m – is not an upper bound of = m such that n > m – this contradicts m being an upper bound of But then n +and > m, since n + ∈ , ♦ The Archimedean Property is widely used in analysis and there are several equivalent forms (which are also sometimes referred to as the Archimedean Property) We state them and establish their equivalence in the following theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide 12 Theorem 3.3.10 Each of the following is equivalent to the Archimedean property (a) For each z∈ , there exists an n ∈ such that n > z (b) For each x > and for each y ∈ (c) For each x > 0, there exists an n ∈ , there exists an n ∈ such that nx > y such that < 1/n < x Proof: We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby establishing their equivalence Theorem 3.3.9  (a) If (a) were not true, then for some z0 ∈ we would have n ≤ z0 for all n ∈ But then z0 would be an upper bound of , contradicting Theorem 3.3.9 Thus the Archimedean property implies (a) (a)  (b) Given x > and y ∈ , let z = y /x Then there exists n ∈ such that n > y/x, so that n x > y Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide 13 Theorem 3.3.10 Each of the following is equivalent to the Archimedean property (a) For each , there exists an n ∈ z∈ such that n > z (b) For each x > and for each y ∈ (c) For each x > 0, there exists an n ∈ , there exists an n ∈ such that nx > y such that < 1/n < x Proof: We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby establishing their equivalence (b)  (c) Given Since n ∈ Then n x > 0, take y = in part (b) ,  x > 1, so that 1/n < x n > and also 1/n > (c)  Theorem 3.3.9 Suppose that That is, n ≤ m for all n ∈ were bounded above by some real number, say m This contradicts (c) with Let k = m + x = 1/k Copyright © 2013, 2005, 2001 Pearson Education, Inc Then n ≤ k – < k and 1/n > 1/k for all n Thus (c) implies the Archimedean property ♦ Section 3.3, Slide 14 In Theorem 3.3.1 we showed that pis not rational when p is prime We are now in a position to prove there is a positive real number whose square is p, thus illustrating that we actually have filled in the “holes” in Theorem 3.3.12 Let p be a prime number Then there exists a positive real number x such that x = p Proof: Let S = {r ∈ : r > and r < p} Furthermore, if r ∈ S, then Since r < p < p , so r < p p > 1, ∈ S and S is nonempty Thus S is bounded above by p, Let x = sup S and by the completeness axiom, sup S exists as a real number It is clear that x > 0, and we claim that x  = p To prove this, we shall show that 2 neither x < p nor x > p is consistent with our choice of x Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide 15 Theorem 3.3.12 Let p be a prime number Then there exists a positive real number x such that x = p Proof: Let S = {r ∈ Suppose that x < : r > and r < p} and x = sup S • x • ( x+ ) 2 n p • p We want to find something positive, say 1/n, that we can add on to x and still have its square be less than p This will contradict x as an upper bound of S We have ( p − x2 < 2x + n n∈ such that  p – x )/(2x + 1) > 0, so that Theorem 3.3.10(c) implies the existence of some The text shows where this estimate comes from Then 1 2x 1 1  2 x + = x + + = x + x +  ÷  ÷ n n n2 n n  ≤ x + ( x + 1) < x + ( p − x ) = p n It follows that x + 1/n ∈ S, which contradicts our choice of x as an upper bound of S Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide 16 Theorem 3.3.12 Let p be a prime number Then there exists a positive real number x such that x = p Proof: Let S = {r ∈ Now suppose that x > : r > and r < p} and x = sup S • • ( x− ) p m p • x We want to find something positive, say 1/m, that we can subtract from x and still have its square be greater than p This will contradict x as the least upper bound of S We have (x –  p)/(2x) > 0, so there exists an m ∈ such that Then x2 − p < 2x m 1 2x 2x  2 2 x − = x − + > x − > x − ( x − p ) = p  ÷ m m m m   This implies that x – 1/m > r, for all r ∈ S, so x – 1/m is an upper bound of S Since x – 1/m < x, this contradicts our choice of x as the least upper bound of S 2 Finally, since neither x < p nor x > p is a possibility, we conclude by the trichotomy law that in fact x = p ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.3, Slide 17 Theorem 3.3.13 The density of in If x and y are real numbers with x < y, then there exists a rational number r such that x < r < y Proof: We begin by supposing that x > there exists an n ∈ not difficult to show Using the Archimedean property 3.3.10(a), such that n > 1/( That is, n  y – x) (Exercise 9) that there exists m ∈ such that m – ≤ n  x + 1 < n y Since n  x > 0, it is  x < m But then m ≤ n  x + 1 < n y, so that n x < m < n y It follows that the rational number r = m/n satisfies x < r < y Finally, if x ≤ 0, choose an integer k such that k > |  x | Then apply the argument above to the positive numbers x + k and y + k If q is a rational satisfying x + k < q < y + k, then the rational r = q – k satisfies x < r < y ♦ It follows easily from this that the irrational numbers are also dense in the real numbers (Theorem 3.3.15) That is, if x and y are real numbers with x < y, then there exists an irrational number w such that Copyright © 2013, 2005, 2001 Pearson Education, Inc x < w < y Section 3.3, Slide 18 ... such that x = p Proof: Let S = {r ∈ Suppose that x < : r > and r < p} and x = sup S • x • ( x+ ) 2 n p • p We want to find something positive, say 1/n, that we can add on to x and still have its... above by 8, 9, 8.5, π , and any other real number greater than or equal to 8.Since ∈ S, we have max S = Similarly, S has many lower bounds, including 2, which is the largest of the lower bounds and... – is not an upper bound of = m such that n > m – this contradicts m being an upper bound of But then n +and > m, since n + ∈ , ♦ The Archimedean Property is widely used in analysis and there