Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
,Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
Solutions Manual
a supplement to
Real Analysis: Foundations
by Sergei Ovchinnikov
Springer 2021
ISBN 978-3-030-64700-1
Contents
0 Using the Manual 2
1 Rational Numbers 3
2 Real Numbers 7
3 Continuous Functions 12
4 Differentiation 17
5 Integration 22
6 Infinite Series 27
7 Appendix A: Natural Numbers and Integers 29
1
,Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
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
September 2021
2
, Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
1 Rational Numbers
1.1. Evidently, (mp)n = m(np) for all m, n, p ∈ Z. Hence, (mp, np) ~ (m, n).
We need p /= 0 to make sure that (mp, np) is a fraction.
n m
1.2. Suppose that = . 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),
m+n m n
ϕ(m + n) = = + = ϕ(m) + ϕ(n),
1 1 1
and m·n m n
ϕ(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.
/ 0, let b be an element of F such that a ·b = 1. By Property M3, we
For a =
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
a·c a c a
(e) By part (d), = · = .
b·c b c b
(f) We have
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
Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
,Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
Solutions Manual
a supplement to
Real Analysis: Foundations
by Sergei Ovchinnikov
Springer 2021
ISBN 978-3-030-64700-1
Contents
0 Using the Manual 2
1 Rational Numbers 3
2 Real Numbers 7
3 Continuous Functions 12
4 Differentiation 17
5 Integration 22
6 Infinite Series 27
7 Appendix A: Natural Numbers and Integers 29
1
,Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
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
September 2021
2
, Solutions Manual for Real Analysis Foundations by Sergei Ovchinnikov
1 Rational Numbers
1.1. Evidently, (mp)n = m(np) for all m, n, p ∈ Z. Hence, (mp, np) ~ (m, n).
We need p /= 0 to make sure that (mp, np) is a fraction.
n m
1.2. Suppose that = . 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),
m+n m n
ϕ(m + n) = = + = ϕ(m) + ϕ(n),
1 1 1
and m·n m n
ϕ(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.
/ 0, let b be an element of F such that a ·b = 1. By Property M3, we
For a =
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
a·c a c a
(e) By part (d), = · = .
b·c b c b
(f) We have
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
