Solution Manual For
Real Analysis and Foundations Fourth Edition by Steven G. Krantz
Chapter 1-12Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to
apply legal reaso
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have distinct areas of focus
but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore the nature of exams in business, law,
and mathematics, their format, and how they evaluate students’ comprehension and application
Chapter 1
Number Systems
1.1 The Real Numbers
1. The set (0, 1] contains its least upper bound 1 but not its greatest lower
bound 0. The set [0, 1) contains its greatest lower bound 0 but not its
least upper bound 1.
3. We know that α ≥ a for every element a ∈ A. Thus —α ≤ —a for
every element a ∈ A hence —α ≤ b for every b ∈ B. If b' > —α is a
lower bound for B then —b ' < α is an upper bound for A, and that is
impossible. Hence —α is the greatest lower bound for B.
Likewise, suppose that β is a greatest lower bound for A. Define
B = {—a : a ∈ A}. We know that β ≤ a for every element a ∈ A.
Thus —β ≥ —a for every element a ∈ A hence —β ≥ b for every b ∈ B.
If b' < —β is an upper bound for B then —b' > β is a lower bound for
A, and that is impossible. Hence —β is the least upper bound for B.
5. We shall treat the least upper bound. Let α be the least upper bound
for the set S. Suppose that α' is another least upper bound. It α' > α
then α' cannot be the least upper bound. If α' < α then α cannot be
the least upper bound. So α' must equal α.
7. Let x and y be real numbers. We know that
(x + y)2 = x2 + 2xy + y2 ≤ |x|2 + 2|x||y| + |y|2 .
Taking square roots of both sides yields
|x + y| ≤ |x| + |y| .
9. We treat commutativity. According to the definition in the text, we
add two cuts C and D by
C + D = {c + d : c ∈ C, d ∈ D} .
,IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have distinct areas of focus
but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore the nature of exams in business, law,
and mathematics, their format, and how they evaluate students’ comprehension and application
But this equals
{d + c : c ∈ C, d ∈ D}
and that equals D + C.
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reaso
11. Consider the set of all numbers of the form
j
√
k 2
for j, k relatively prime natural numbers and j < k. Then certainly
each of these numbers lies between 0 and 1 and each is irrational.
Furthermore, there are countably many of them.
* 13. Notice that if n— kλ = m—lλ then (n —m) = (k — l)λ. It would follow
that λ is rational unless n = m and k = l. So the numbers n— kλ are
all distinct.
Now let ϵ > 0 and choose an positive integer N so large that
λ/N < ϵ. Consider ϕ(1), ϕ(2), . . . , ϕ(N). These numbers are all
distinct, and lie in the interval [0, λ]. So two of them are distance not
more than λ/N < ϵ apart. Thus |(n1 — k1λ) — (n2 — k2λ)| < ϵ or
|(n1 — n2) — (k1 — k2)λ| < ϵ. Let us abbreviate this as |m — pλ| < ϵ.
It follows then that the numbers
(m — pλ), (2m — 2pλ), (3m — 3pλ), . . .
are less than ϵ apart and fill up the interval [0, λ]. That is the definition
of density.
1.2 The Complex Numbers
1. We calculate that
z z·z |z| 2
z· = = = 1.
|z|2 |z|2 |z|2
So z/|z|2 is the multiplicative inverse of z.
3. Write
√
1+i= 2eiπ/4 .
, We seek a complex number z = veiθ such that
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have
distinct areas of focus but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore
the nature of exams in business, law, and mathematics, their format, and how they evaluate students’ comprehension and application
√
z3 = v3e3iθ = (veiθ)3 = 2eiπ/4 .
It follows that v = 21/6 and θ = π/12. So we have found the cube root
c1 = 21/6eiπ/12 .
√ √
Now we may repeat this process with 2eiπ/4 replaced by 2ei9π/4.
We find the second cube root
c2 = 21/6ei9π/12 .
√ √
Repeating the process a third time with 2eiπ/4 replaced by 2ei17π/4,
we find the third cube rootLaw ExamsLaw exams typically have a more structured format compared to business
exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reaso
c3 = 21/6ei17π/12 .
5. We see that
φ(x + x') = (x + x') + i0 = (x + i0) + (x' + i0) = φ(x) + φ(x') .
Also
φ(x · x') = (x · x') + i0 = (x + i0) · (x' + i0) = φ(x) · φ(x') .
7. Let p(z) = a0 + a1z + a2z2 + · · · + akzk
be a polynomial with real coefficients aj. If α is a root of this polynomial
then
p(α) = a0 + a1α + a2α2 + · · · + akαk = 0 .
Conjugating this equation gives
p(α) = a0 + a1α + a2α2 + · · · + akαk = 0 .
Hence α is a root of the polynomial p. We see then that roots of p
occur in conjugate pairs.
9. The function ϕ(x) = x + i0 from R to C is one-to-one. Therefore
card(R) ≤ card(C) .
Since the reals are uncountable, we may conclude that the complex
, numbers are uncountable.
11. The defining condition measures the sum of the distance of z to 1 + i0
plus the distance of z to —1 + i0. If z is not on the x-axis then |z —
1| + |z + 1| > 2 (by the triangle inequality). If z is on the x axis but
less than —1 or greater than 1 then |z — 1| + |z + 1| > 2. So the only z
that satisfy |z — 1| + |z + 1| > 2 are those elements of the x-axis that
are between —1 and 1 inclusive.
15. The set of all complex numbers with rational real part contains the set
of all complex numbers of the form 0 + yi, where y is any real number.
This latter set is plainly uncountable, so the set of complex number
with rational real part is also uncountable.
17. The set S = { z ∈ C : | z| = 1} can be identified with F = { eiθ : 0 ≤
θ < 2π }. The set F can be identified with the interval [0, 2π), and that
interval is certainly an uncountable set. Hence S is uncountable.
19. Let p be a polynomial of degree k ≥ 1 and let α1 be a root of p. So
p(α) = 0. Now let us think about dividing p(z) by (z— α1). By the
Euclidean algorithm,
p(z) = (z — α1) · q1(z) + v1(z) . (∗)
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate
knowledge of legal statutes, case law, and their ability to apply legal reaso
Here q1 is the “quotient” and v1 is the “remainder.” The quotient
will have degree k — 1 and the remainder will have degree less than
the degree of z — α1. In other words, the remainder will have degree
0—which means that it is constant. Plug the value z = α1 into the
equation (∗). We obtain
0 = 0 + v1 .
Hence the remainder, the constant v1, is 0.
If k = 1 then the process stops here. If k > 1 then q1 has degree
k — 1 ≥ 1 and we may apply the Fundamental Theorem of Algebra to
q1 to find a root α2. Repeating the argument above, we divide (z — α2)
into q1 using the Euclidean algorithm. We find that it divides in evenly,
producing a new quotient q2.
This process can be repeated k — 2 more times to produce a total of
k roots of the polynomial p.
Real Analysis and Foundations Fourth Edition by Steven G. Krantz
Chapter 1-12Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to
apply legal reaso
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have distinct areas of focus
but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore the nature of exams in business, law,
and mathematics, their format, and how they evaluate students’ comprehension and application
Chapter 1
Number Systems
1.1 The Real Numbers
1. The set (0, 1] contains its least upper bound 1 but not its greatest lower
bound 0. The set [0, 1) contains its greatest lower bound 0 but not its
least upper bound 1.
3. We know that α ≥ a for every element a ∈ A. Thus —α ≤ —a for
every element a ∈ A hence —α ≤ b for every b ∈ B. If b' > —α is a
lower bound for B then —b ' < α is an upper bound for A, and that is
impossible. Hence —α is the greatest lower bound for B.
Likewise, suppose that β is a greatest lower bound for A. Define
B = {—a : a ∈ A}. We know that β ≤ a for every element a ∈ A.
Thus —β ≥ —a for every element a ∈ A hence —β ≥ b for every b ∈ B.
If b' < —β is an upper bound for B then —b' > β is a lower bound for
A, and that is impossible. Hence —β is the least upper bound for B.
5. We shall treat the least upper bound. Let α be the least upper bound
for the set S. Suppose that α' is another least upper bound. It α' > α
then α' cannot be the least upper bound. If α' < α then α cannot be
the least upper bound. So α' must equal α.
7. Let x and y be real numbers. We know that
(x + y)2 = x2 + 2xy + y2 ≤ |x|2 + 2|x||y| + |y|2 .
Taking square roots of both sides yields
|x + y| ≤ |x| + |y| .
9. We treat commutativity. According to the definition in the text, we
add two cuts C and D by
C + D = {c + d : c ∈ C, d ∈ D} .
,IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have distinct areas of focus
but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore the nature of exams in business, law,
and mathematics, their format, and how they evaluate students’ comprehension and application
But this equals
{d + c : c ∈ C, d ∈ D}
and that equals D + C.
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reaso
11. Consider the set of all numbers of the form
j
√
k 2
for j, k relatively prime natural numbers and j < k. Then certainly
each of these numbers lies between 0 and 1 and each is irrational.
Furthermore, there are countably many of them.
* 13. Notice that if n— kλ = m—lλ then (n —m) = (k — l)λ. It would follow
that λ is rational unless n = m and k = l. So the numbers n— kλ are
all distinct.
Now let ϵ > 0 and choose an positive integer N so large that
λ/N < ϵ. Consider ϕ(1), ϕ(2), . . . , ϕ(N). These numbers are all
distinct, and lie in the interval [0, λ]. So two of them are distance not
more than λ/N < ϵ apart. Thus |(n1 — k1λ) — (n2 — k2λ)| < ϵ or
|(n1 — n2) — (k1 — k2)λ| < ϵ. Let us abbreviate this as |m — pλ| < ϵ.
It follows then that the numbers
(m — pλ), (2m — 2pλ), (3m — 3pλ), . . .
are less than ϵ apart and fill up the interval [0, λ]. That is the definition
of density.
1.2 The Complex Numbers
1. We calculate that
z z·z |z| 2
z· = = = 1.
|z|2 |z|2 |z|2
So z/|z|2 is the multiplicative inverse of z.
3. Write
√
1+i= 2eiπ/4 .
, We seek a complex number z = veiθ such that
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines not only have
distinct areas of focus but also require unique approaches to assessment, with each exam testing different cognitive abilities, analytical skills, and subject-specific knowledge. In this essay, we will explore
the nature of exams in business, law, and mathematics, their format, and how they evaluate students’ comprehension and application
√
z3 = v3e3iθ = (veiθ)3 = 2eiπ/4 .
It follows that v = 21/6 and θ = π/12. So we have found the cube root
c1 = 21/6eiπ/12 .
√ √
Now we may repeat this process with 2eiπ/4 replaced by 2ei9π/4.
We find the second cube root
c2 = 21/6ei9π/12 .
√ √
Repeating the process a third time with 2eiπ/4 replaced by 2ei17π/4,
we find the third cube rootLaw ExamsLaw exams typically have a more structured format compared to business
exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reaso
c3 = 21/6ei17π/12 .
5. We see that
φ(x + x') = (x + x') + i0 = (x + i0) + (x' + i0) = φ(x) + φ(x') .
Also
φ(x · x') = (x · x') + i0 = (x + i0) · (x' + i0) = φ(x) · φ(x') .
7. Let p(z) = a0 + a1z + a2z2 + · · · + akzk
be a polynomial with real coefficients aj. If α is a root of this polynomial
then
p(α) = a0 + a1α + a2α2 + · · · + akαk = 0 .
Conjugating this equation gives
p(α) = a0 + a1α + a2α2 + · · · + akαk = 0 .
Hence α is a root of the polynomial p. We see then that roots of p
occur in conjugate pairs.
9. The function ϕ(x) = x + i0 from R to C is one-to-one. Therefore
card(R) ≤ card(C) .
Since the reals are uncountable, we may conclude that the complex
, numbers are uncountable.
11. The defining condition measures the sum of the distance of z to 1 + i0
plus the distance of z to —1 + i0. If z is not on the x-axis then |z —
1| + |z + 1| > 2 (by the triangle inequality). If z is on the x axis but
less than —1 or greater than 1 then |z — 1| + |z + 1| > 2. So the only z
that satisfy |z — 1| + |z + 1| > 2 are those elements of the x-axis that
are between —1 and 1 inclusive.
15. The set of all complex numbers with rational real part contains the set
of all complex numbers of the form 0 + yi, where y is any real number.
This latter set is plainly uncountable, so the set of complex number
with rational real part is also uncountable.
17. The set S = { z ∈ C : | z| = 1} can be identified with F = { eiθ : 0 ≤
θ < 2π }. The set F can be identified with the interval [0, 2π), and that
interval is certainly an uncountable set. Hence S is uncountable.
19. Let p be a polynomial of degree k ≥ 1 and let α1 be a root of p. So
p(α) = 0. Now let us think about dividing p(z) by (z— α1). By the
Euclidean algorithm,
p(z) = (z — α1) · q1(z) + v1(z) . (∗)
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrate
knowledge of legal statutes, case law, and their ability to apply legal reaso
Here q1 is the “quotient” and v1 is the “remainder.” The quotient
will have degree k — 1 and the remainder will have degree less than
the degree of z — α1. In other words, the remainder will have degree
0—which means that it is constant. Plug the value z = α1 into the
equation (∗). We obtain
0 = 0 + v1 .
Hence the remainder, the constant v1, is 0.
If k = 1 then the process stops here. If k > 1 then q1 has degree
k — 1 ≥ 1 and we may apply the Fundamental Theorem of Algebra to
q1 to find a root α2. Repeating the argument above, we divide (z — α2)
into q1 using the Euclidean algorithm. We find that it divides in evenly,
producing a new quotient q2.
This process can be repeated k — 2 more times to produce a total of
k roots of the polynomial p.
