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
apply legal reaso
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 = x 2 + 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} .
, But this equals 3
{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 demonstrat e 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 |(n 1 — k1λ) — (n 2 — k2λ)| < ϵ or
|(n1 — n 2) — (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
√
z3 = v3e3iθ = (veiθ)3 = 2eiπ/4 .
, It follows that v = 2 1/6 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 2e iπ/4 replaced by 2ei17π/4,
we find the third cube root 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
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) = a 0 + a 1z + a 2z2 + · · · + akzk
be a polynomial with real coefficients aj. If α is a root of this polynomial
then
p(α) = a 0 + a 1α + a 2α 2 + · · · + a kα k = 0 .
Conjugating this equation gives
p(α) = a 0 + a 1α + a 2α 2 + · · · + a kα k = 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.
, 5 the set
15. The set of all complex numbers with rational real part contains
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 α 1 be a root of p. So
p(α) = 0. Now let us think about dividing p(z) by (z— α 1). By the
Euclidean algorithm,
p(z) = (z — α 1) · q 1(z) + v1(z) . (∗)
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrat e
knowledge of legal statutes, case law, and their ability to apply legal reaso
Here q 1 is the “quotient” and v 1 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 q 1 has degree
k — 1 ≥ 1 and we may apply the Fundamental Theorem of Algebra to
q 1 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 q 2.
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
apply legal reaso
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 = x 2 + 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} .
, But this equals 3
{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 demonstrat e 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 |(n 1 — k1λ) — (n 2 — k2λ)| < ϵ or
|(n1 — n 2) — (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
√
z3 = v3e3iθ = (veiθ)3 = 2eiπ/4 .
, It follows that v = 2 1/6 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 2e iπ/4 replaced by 2ei17π/4,
we find the third cube root 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
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) = a 0 + a 1z + a 2z2 + · · · + akzk
be a polynomial with real coefficients aj. If α is a root of this polynomial
then
p(α) = a 0 + a 1α + a 2α 2 + · · · + a kα k = 0 .
Conjugating this equation gives
p(α) = a 0 + a 1α + a 2α 2 + · · · + a kα k = 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.
, 5 the set
15. The set of all complex numbers with rational real part contains
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 α 1 be a root of p. So
p(α) = 0. Now let us think about dividing p(z) by (z— α 1). By the
Euclidean algorithm,
p(z) = (z — α 1) · q 1(z) + v1(z) . (∗)
Law ExamsLaw exams typically have a more structured format compared to business exams, often requiring students to demonstrat e
knowledge of legal statutes, case law, and their ability to apply legal reaso
Here q 1 is the “quotient” and v 1 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 q 1 has degree
k — 1 ≥ 1 and we may apply the Fundamental Theorem of Algebra to
q 1 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 q 2.
This process can be repeated k — 2 more times to produce a total of
k roots of the polynomial p.
