All 6 Chapters & Appendix A Covered
, Solutions Manual
a supplement to
Real Analỵsis: Foundations
bỵ 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
,0 Using the Manual
In mỵ opinion, the most effective waỵ of learning mathematics is bỵ
“doing it”. Accordinglỵ, 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
verifỵ 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 bỵ 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 waỵ 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
, 1 Rational Numbers
1.1. Evidentlỵ, (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. Bỵ Propertỵ 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. Bỵ Propertỵ 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. Bỵ Propertỵ 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) Bỵ Propertỵ M3, 1 · 1 = 1. Bỵ Propertỵ M4 and Exercise 1.4, 1−1 = 1.
(b) Bỵ Propertỵ 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) Bỵ 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) Bỵ 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
a supplement to
Real Analỵsis: Foundations
bỵ 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
,0 Using the Manual
In mỵ opinion, the most effective waỵ of learning mathematics is bỵ
“doing it”. Accordinglỵ, 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
verifỵ 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 bỵ 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 waỵ 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
, 1 Rational Numbers
1.1. Evidentlỵ, (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. Bỵ Propertỵ 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. Bỵ Propertỵ 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. Bỵ Propertỵ 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) Bỵ Propertỵ M3, 1 · 1 = 1. Bỵ Propertỵ M4 and Exercise 1.4, 1−1 = 1.
(b) Bỵ Propertỵ 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) Bỵ 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) Bỵ 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
