Solutions Manual a supplement to Real Analysis: Foundations by Sergei Ovchinnikov Springer 2021 ISBN 978-3-030-64700-1
Contents 0 Using the Manual
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1 Rational Numbers
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2 Real Numbers
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3 Continuous Functions
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4 Differentiation
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5 Integration
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6 Infinite Series
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7 Appendix A: Natural Numbers and Integers
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0
Using the Manual
In my opinion, the most effective way of learning mathematics is by “doing it”. Accordingly, I urge the student not to read the solutions in advance, but rather to make a concerted effort to find a solution to the problem in the exercise before consulting the Solution Manual to verify correctness. If the student’s solution differs from the one given in the Manual, a comparison might reveal an unjustified assumption that had been made by the student or a misapplication of a theorem. Meanwhile, the instructor can use the solutions to create balanced assignments and research projects. Solutions in the Manual are labeled in the same way as exercises in the book. For instance, item 1.10 in the Manual is a solution to the problem in Exercise 1.10. Sergei Ovchinnikov sergei@sfsu.edu September 2021
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1
Rational Numbers
1.1. Evidently, (mp)n = m(np) for all m, n, p ∈ Z. Hence, (mp, np) ∼ (m, n). We need p 6= 0 to make sure that (mp, np) is a fraction. n m = . Then (n, 1) ∼ (m, 1), that is, n · 1 = m · 1. Hence, 1 1 n = m, so ϕ is one-to-one. Furthermore (cf. (1,1) on p. 4 in the book),
1.2. Suppose that
ϕ(m + n) =
m n m+n = + = ϕ(m) + ϕ(n), 1 1 1
and ϕ(m · n) =
m n m·n = · = ϕ(m) · ϕ(n), 1 1 1
for all m, n ∈ Z. 1.3. Straightforward verification of the properties defining a field. 1.4. Suppose that a + b = 0 in F. Then, −a + a + b = −a. By Property A4, b = −a. Hence, −a is a unique additive inverse of a. For a 6= 0, let b be an element of F such that a · b = 1. By Property M3, we have a−1 = 1 · a−1 = a · b · a−1 = b · a · a−1 = b · 1 = b. Hence, a−1 is a unique multiplicative inverse of a.
1.5. By Property D, 0 + 0 = 0 implies a · 0 + a · 0 = a · 0. Hence, a · 0 = −a · 0 + a · 0 + a · 0 = −a · 0 + a · 0 = 0. 1.6. (a) By Property M3, 1 · 1 = 1. By Property M4 and Exercise 1.4, 1−1 = 1. (b) By Property M1, a−1 ·b−1 ·a·b = a−1 ·a·b−1·b = 1. Hence, (a·b)−1 = a−1 ·b−1 . (c) If c · b = a, then c = c · b · b−1 = a · b−1. If c = a · b−1, then c · b = a · b−1 · b = a. (d) By part (b) above, a·c a c = (a · c) · (b · d)−1 = a · c · b−1 · d−1 = a · b−1 · c · d−1 = · . b·d b d (e) By part (d), (f) We have
a·c a c a = · = . b·c b c b
a·d+c·b = (a · d + c · b)(b · d)−1 = (a · d + c · b)(b−1 · d−1 ) b·d = a · d · b−1 · d−1 + c · b · b−1 · d−1 a c = a · b−1 + c · d−1 = + . b d 3
a c = , then (a · b−1 ) · b · d = (c · d−1 ) · b · d, so a · d = b · c. If a · d = b · c, b d a c then (a · d) · b−1 · d−1 = (b · c) · b−1 · d−1 , so = a · b−1 = c · d−1 = . b d (h) Because, clearly, (b−1 )−1 = b, a −1 b = (a · b−1 )−1 = a−1 · b = . b a (g) If
a·d . It is easy to (i) Correction. The right hand side of the equality must be b·c c −1 a c a d a·d d verify that = . Therefore, by part (d) above, ÷ = · = . d c b d b c b·c
1.7. Note that Z[x] is an integral domain (cf. Exercise A.14)) and repeat the steps in the proof of Theorem 1.1. 1.8. Repeat the steps in the proof of Theorem 1.4.
1.9. Suppose that 1 < 0 and let a 6= 0 be a positive element of F. Then a · a−1 = 1 < 0. By part (b) of Definition 1.7, a = a · a−1 · a < a · 0 = 0, a contradiction. 1.10. Correction. The function τ in the proof of Theorem 1.8 is defined only for nonnegative integers, not for all n ∈ Z. For negative integers, we set τ (−n) = −τ (n) for n ∈ N. Thus Exercise 1.10 is, in fact, a part of the definition of the function τ . 1.11. By Definition 1.7(b), ab−1 < cd−1 is equivalent to ad < cb, because b > 0 and d > 0. 1.12. By mathematical induction. The case n = 3 follows immediately from the transitivity property of the linear order < (cf. Definition A.7). By the same property, the case of n = k + 1 follows from the case of n = k. 1.13. First, suppose that x > 0. By the Archimedean Property, there is n ∈ N such that x < n, so the set {k ∈ N : x < k} is nonempty. By the Well-Ordering Principle (cf. Theorem A.6), there is m ∈ N such that m − 1 ≤ x < m. For a non-positive x, note that the claim is trivial if x ∈ Z \ N. If x ∈ / Z \ N, apply the previous argument to −x. 1.14. It is shown in Exercise 1.8 that Z(x) is a field. An order on Z(x) is defined in Example 1.3. To show that Z(x) endowed with this order is an ordered field, repeat the steps in the proofs of Theorems 1.6 and 1.7. 1.15. Suppose to the contrary that D = p2 /q 2 for some p, q ∈ N. We may assume that q is the least natural number satisfying this equation (cf. the proof of Theorem 1.10.). Because D is not a perfect square, there is t ∈ N such that t2 < D < (t + 1)2 . 4
(Why?) It follows that tq < p < (t + 1)q. Let q 0 = p − tq and p0 = Dq − tp. It is easy to verify that 0 < q 0 < q and p0 > 0. We have p02 − Dq 02 = (Dq − tp)2 − D(p − tq)2
= D2 q 2 − 2Dqtp + t2 p2 − Dp2 + 2Dptq − Dt2 q 2
= (D2 q 2 − Dt2 q 2 ) − (Dp2 − t2 p2 ) = Dq 2 (D − t2 ) − p2 (D − t2 ) = (Dq 2 − p2 )(D − t2 ) = 0,
because D = p2 /q 2 . This contradicts the minimality of q. 1.16. Let a = x − y, b = y − z, so x − z = a + b. We need to show that |a + b| ≤ |a| + |b|. By Exercise 1.17(e), −|a| ≤ a ≤ |a| and −|b| ≤ b ≤ |b|. Hence, −(|a| + |b|) ≤ a + b ≤ |a| + |b|. The desired result follows from Exercise 1.17(f). 1.17. We prove only part (h) using the triangular inequality from the proof of Theorem 1.12 and other parts of this exercise. We have |a| = |(a − b) + b| ≤ |a − b| + |b|. Therefore, |a| − |b| ≤ |a − b|. On the other hand, |b| = |a + (b − a)| ≤ |a| + |b − a| = |a| + |a − b|. Hence, |a| − |b| ≥ −|a − b|. The result follows from part (e). 1.18. If a ≤ b ≤ c, then |a − b| = b − a and |b − c| = c − b. Hence, |a − b| + |b − c| = (b − a) + (c − b) = c − a = |a − c|. In the opposite direction, the proof is by contradiction. Suppose that b < a, so also b < c. Then |a − b| + |b − c| = (a − b) + (c − b) = a + c − 2b > a + c − 2a = c − a = |a − c|, a contradiction. Similarly, we obtain a contradiction by assuming that b > c. Therefore, a ≤ b ≤ c. 1.19. Without loss of generality, we may assume that x > y. We have b − a = y − a + x − y + b − x ≥ x − y, because y − a ≥ 0 and b − x ≥ 0. Hence the result. 1.20. Correction. The inequality |ak | > ε in Exercise 1.20 must be replaced with |ak − a| > ε. 5
To form the negation of the statement in Definition 1.12, we must replace the quantifier “for every positive ε” by “there exists ε > 0”, the quantifier “there exists” by “for every”, and the quantifier “for all” by “for some”. 1.21. Standard Calculus exercises. 1.22. Inasmuch as (an ) is Cauchy, for ε > 0 there is N ∈ N such that |am − an | < ε,
for all m, n > N .
By Exercise 1.17(h), |am | − |an | ≤ |am − an | < ε,
for all m, n > N .
Therefore, (|an |) is Cauchy. 1.23. Apply Theorem 1.20 for a constant sequence (bn ) = (k, k, . . .). 1.24. First note that the sequence (an ) is bounded below by a positive rational number. Indeed, because lim a2n = p > 0, there is only a finite number of terms of (a2n ) that are smaller than p/2. Hence, (a2n ) is bounded below by a positive rational number (cf. Exercise 1.25). It follows that there is a positive r ∈ Q such that an > r for all n ∈ N. By Theorem 1.16, (a2n ) is Cauchy, that is, for ε > 0 there is N ∈ N such that |a2m − a2n | < 2r · ε for all m, n > N . We have |am − an | =
2r · ε |a2m − a2n | < = ε, am + an 2r
for all m, n > N .
Therefore, (an ) is a Cauchy sequence. 1.25. The proof is by induction on the cardinality of the set A. The claim is trivial if A consists of one or two elements. Suppose that it holds for every n-element set and let A be a set containing n + 1 elements. For an element x of A, let A0 = A \ {x} and a = min A0 , b = max A0 . If x < a, we set min A = x, max A = b, if x > b, we set max A = x, min A = a. Finally, if a ≤ x ≤ b, we set min A = a, max A = b. It is not difficult to verify that in all these cases, min A ≤ c ≤ max A for every c ∈ A. 1.26. Recall that a tail of the sequence (an ) is a subsequence (am , am+1 , . . .), m ∈ N. For ε > 0 there is N ∈ N such that |an −a| < ε for all n > N . It follows that |an − a| < ε for all n > max{m, N }. Hence, the subsequence converges to the same limit. 1.27. We define the mapping f recursively. By the Well-Ordering Principle (cf. Theorem A.6), the set S has a least element. We define f(1) to be this element. Suppose the values f(1), f(2), . . . , f(n) are defined and satisfy the required inequalities. By the same Principle, the nonempty set S\{f(1), . . . , f(n)} has a least element. We define f(n + 1) to be this element. Clearly, thus defined function f satisfies the required inequalities. 1.28. Straightforward algebra. 6