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Covers All 14 Chapters
SOLUTIONS TO EXERCISES
, An Introduction to Analysis
Table of Contents
Chapter 1: The Real Number System
1.2 Ordered field axioms ............................................................... 1
1.3 The Completeness Axiom….................................................... 2
1.4 Mathematical Induction… ....................................................... 4
1.5 Inverse Functions and Images…............................................. 6
1.6 Countable and uncountable sets… ......................................... 8
Chapter 2: Sequences in R
2.1 Limits of Sequences… ............................................................. 10
2.2 Limit Theorems ....................................................................... 11
2.3 Bolzano-Weierstrass Theorem ............................................... 13
2.4 Cauchy Sequences… ............................................................... 15
2.5 Limits Supremum and Infimum ............................................ 16
Chapter 3: Functions on R
3.1 Two-Sided Limits… ................................................................ 19
3.2 One-Sided Limits and Limits at Infinity… ............................. 20
3.3 Continuity… ............................................................................ 22
3.4 Uniform Continuity…............................................................. 24
Chapter 4: Differentiability on R
4.1 The Derivative… ..................................................................... 27
4.2 Differentiability Theorem… ....................................................28
4.3 The Mean Value Theorem… .................................................. 30
4.4 Taylor’s Theorem and l’Hôpital’s Rule… ............................. 32
4.5 Inverse Function Theorems .................................................... 34
Chapter 5: Integrability on R
5.1 The Riemann Integral…........................................................... 37
5.2 Riemann Sums .......................................................................... 40
5.3 The Fundamental Theorem of Calculus… ............................. 43
5.4 Improper Riemann Integration… ........................................... 46
5.5 Functions of Bounded Variation… ......................................... 49
5.6 Convex Functions… ................................................................ 51
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Chapter 6: Infinite Series of Real Numbers
6.1 Introduction… ........................................................................... 53
6.2 Series with Nonnegative Terms… .......................................... 55
6.3 Absolute Convergence… ......................................................... 57
6.4 Alternating Series… ................................................................. 60
6.5 Estimation of Series… .............................................................. 62
6.6 Additional Tests… ................................................................... 63
Chapter 7: Infinite Series of Functions
7.1 Uniform Convergence of Sequences… ................................... 65
7.2 Uniform Convergence of Series… .......................................... 67
7.3 Power Series….......................................................................... 69
7.4 Analytic Functions… ...............................................................72
7.5 Applications… ......................................................................... 74
Chapter 8: Euclidean Spaces
8.1 Algebraic Structure… ............................................................. 76
8.2 Planes and Linear Transformations… ................................... 77
8.3 Topology of Rn .......................................................................................................................79
8.4 Interior, Closure, and Boundary… ........................................ 80
Chapter 9: Convergence in Rn
9.1 Limits of Sequences… .............................................................. 82
9.2 Heine-Borel Theorem............................................................... 83
9.3 Limits of Functions… ............................................................... 84
9.4 Continuous Functions… .......................................................... 86
9.5 Compact Sets… ........................................................................ 87
9.6 Applications… .......................................................................... 88
Chapter 10: Metric Spaces
10.1 Introduction… ............................................................................ 90
10.2 Limits of Functions… ................................................................ 91
10.3 Interior, Closure, and Boundary… .......................................... 92
10.4 Compact Sets… ......................................................................... 93
10.5 Connected Sets… ......................................................................94
10.6 Continuous Functions… ........................................................... 96
10.7 Stone-Weierstrass Theorem...................................................... 97
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
,Chapter 11: Differentiability on Rn
11.1 Partial Derivatives and Partial Integrals… ............................... 99
11.2 The Definition of Differentiability….......................................... 102
11.3 Derivatives, Differentials, and Tangent Planes… .................... 104
11.4 The Chain Rule… ........................................................................ 107
11.5 The Mean Value Theorem and Taylor’s Formula…................. 108
11.6 The Inverse Function Theorem .................................................. 111
11.7 Optimization… ............................................................................. 114
Chapter 12: Integration on Rn
12.1 Jordan Regions…...........................................................................117
12.2 Riemann Integration on Jordan Regions… ................................ 119
12.3 Iterated Integrals… ....................................................................... 122
12.4 Change of Variables… .................................................................. 125
12.5 Partitions of Unity… .....................................................................130
12.6 The Gamma Function and Volume ............................................ 131
Chapter 13: Fundamental Theorems of Vector Calculus
13.1 Curves… ........................................................................................ 135
13.2 Oriented Curves… ........................................................................ 137
13.3 Surfaces… ...................................................................................... 140
13.4 Oriented Surfaces… ...................................................................... 143
13.5 Theorems of Green and Gauss… ................................................. 147
13.6 Stokes’s Theorem ........................................................................... 150
Chapter 14: Fourier Series
14.1 Introduction… ............................................................................... 156
14.2 Summability of Fourier Series… .................................................. 157
14.3 Growth of Fourier Coefficients…................................................. 159
14.4 Convergence of Fourier Series… ................................................. 160
14.5 Uniqueness…................................................................................. 163
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SOLUTIONS TO EXERCISES
CHAPTER 1
1.2 Ordered field axioms.
1.2.0. a) False. Let a = 2/3, b = 1, c = −2, and d = −1.
b) False. Let a = −4, b = −1, and c = 2.
c) True. Since a ≤ b and b ≤ a + c, |a − b| = b − a ≤ a + c − a = c.
d) True. No a ∈ R satisfies a < b − ε for all ε > 0, so the inequality is vacuously satisfied. If you want
a more constructive proof, if b ≤ 0 then a < b − ε < 0 + 0 = 0. If b > 0, then for ε = b, a < b − ε = 0.
1.2.1. a) If a < b then a + c < b + c by the Additive Property. If a = b then a + c = b + c since + is a
function. Thus a + c ≤ b + c holds for all a ≤ b.
b) If c = 0 then ac = 0 = bc so we may suppose c > 0. If a < b then ac < bc by the Multiplicative Property. If
a = b then ac = bc since · is a function. Thus ac ≤ bc holds for all a ≤ b and c ≥ 0.
1.2.2. a) Suppose 0 ≤ a < b and 0 ≤ c < d. Multiplying the first inequality by c and the second by b, we have
0 ≤ ac ≤ bc and bc < bd. Hence by the Transitive P r o per
√ ty, ac < bd. √
√ √
b) Suppose 0 ≤ a < b. By (7), 0 ≤ a2 < b2. If a ≥ b then a = ( a)2 ≥ ( b)2 = b, a contradiction.
c) If 1/a ≤ 1/b, then the Multiplicative Property implies b = ab(1/a) ≤ ab(1/b) = a, a contradiction. If 1/b ≤ 0
then b = b 2(1/b) ≤ 0 a contradiction.
d) To show these statements may not hold when a < 0, let a = −2, b = −1, c = 2 and d = 5. Then a <
b and c < d but ac = −4 is not less than bd = −5, a2 = 4 is not less than b2 = 1, and 1/a = −1/2 is not
less than 1/b = −1.
1.2.3. a) By definition,
µ ¶
|a| + a |a| − a 2a
a+ − a− = − = =a
2 2 2
and
µ ¶
|a| + a |a| − a 2|a|
a+ + a− = + = = |a|.
2 2 2
b) By Definition 1.1, if a ≥ 0 then a+ = (a + a)/2 = a and if a < 0 then a+ = (−a + a)/2 = 0. Similarly,
a− = 0 if a ≥ 0 and a− = −a if a < 0.
1.2.4. a) |2x + 1| < 7 if and only if −7 < 2x + 1 < 7 if and only if − − 4 < x < 3.
b) |2 − x| < 2 if and only if −2 < 2 − x < 2 if and only if −4 < −x < 0 if and only if 0 < x < 4.
c) |x3 −3x+1| < x3 if and only if −x3 < x3 − 3x+1 < x3 if and only2 if 3x− 1 > 0 and 2x3 −3x+1 > 0. T h√e first
inequality is equivalent to x > 1/3. Since 2x3 −√ 3x + 1 = (x − 1)(2x + 2x − 1) implies that x = 1, (−1 3)/2,
±
√
the sec√ond inequality is equivalent to (−1 − 3)/2 < x < (−1 + 3)/2 or x > 1. Therefore, the solution is
(1/3, ( 3 − 1)/2) ∪ (1, ∞).
d) We cannot multiply by the denominator x − 1 unless we consider its sign.
Case 1. x − 1 > 0. Then x < x − 1 so 0 < −1, i.e., this case is empty.
Case 2. x − 1 < 0. Then by the Second Multiplicative Property, x > x − 1 so 0 > −1, i.e., every number
from this case works. Thus the solution is (−∞, 1).
e) Case 1. 4x2 − 1 > 0. Cross multiplying, we obtain 4x2 < 4x2 − 1, i.e., this case is empty.
Case 2. 4x2 − 1 < 0. Then by the Second Multiplicative Property, 4x2 > 4x2 − 1, i.e., 0 > −1.
Thus the solution is (−1/2, 1/2).
√ √
1.2.5. a) Suppose a > 2. Then a − 1 > 1 so 1 < a − 1 < a − 1 by (6). Therefore, 2 < b = 1 + a − 1 <
1 + (a − 1) = a. √ √
b) Suppose 2 < a < 3. Then 0 < a − 2 < 1 so 0 < a − 2 a − 2 < 1 √b y (6). Therefore, 0 < a < 2 a−√ 2 = b.
+
<
c) Suppose 0 < a√< 1. Then 0 > −a > −1, so 0 < 1 − a < 1. Hence 1 − a is real and by (6), 1 − a < 1 − a.
Therefore, b = 1 − 1 − a < 1 − (1 − a) = a. √ √
d) Suppose 3 < a < 5. Then 1 < a − 2 < 3 so 1 < a − 2 < a − 2 by (6). Therefore, 3 < 2 + a − 2 = b < a.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, √ √ √ √
1.2.6. a + b − 2 ab = ( a − b)2 ≥ 0 for all a, b ∈ [0, ∞). Thus 2 ab ≤ a + b an√d G( a ,√b ) ≤ A(a, b). On
the other hand, since 0 ≤ a ≤ b √w e have A(a, b) = (a + b)/2 ≤ 2b/√2 = b and G(a, b) = ab ≥ √a2 = a. Finally,
√ √
A(a, b) = G(a, b) if and only ab = a + b if and only if ( a b) 2 = 0 if and only if a b if and only if
if 2 − =
a = b.
1.2.7. a) Since |x + 2| ≤ |x| + 2, |x| ≤ 2 implies |x2 − 4| = |x + 2| |x − 2| ≤ 4|x − 2|.
b) Since |x + 3| ≤ |x| + 3, |x| ≤ 1 implies |x2 + 2x − 3| = |x + 3| |x − 1| ≤ 4|x − 1|.
c) Since |x − 2| ≤ |x| + 2, −3 ≤ x ≤ 2 implies |x2 + x − 6| = |x + 3| |x − 2| ≤ 6|x − 2|.
d) Since the minimum of x2 + x − 1 on (−1, 0) is −1.25, −1 < x < 0 implies |x3 − 2x + 1| = |x2 + x − 1| |x − 1| <
5|x − 1|/4.
1.2.8. a) Since (1− − n)/(1 n2) = 1/(1 + n), the inequality is equivalent to 1/(n + 1) < .01 =
1/100. Since 1 + ∈ n > 0 for all n N, it follows that n + 1 > 100, i.e., n > 99.
b) By factoring, we see that the inequality is equivalent to 1/(2n + 1) < 1/40, i.e., 2n + 1 > 40. Thus n > 39/2,
i.e., n ≥ 20.
√
c) The inequality is equivalent to n2 + 1 > 500. 499 ≈ 22.33, i.e., n ≥ 23.
Thus n >
1.2.9. a) mn−1 + pq−1 = mqq−1 n−1 + pq−1nn−1 = (mq + pn)n−1q−1. But n−1q−1 nq = 1 and
uniqueness of multiplicative inverses implies (nq)−1 = n−1q−1. Therefore, mn−1 + pq−1 = (mq +
pn)(nq)−1. Similarly, mn−1(pq−1) = mpn−1q−1 = mp(nq)−1. By what we just proved and (2),
m −m m −m 0
+ = = = 0.
n n n n
Therefore, by the uniqueness of additive inverses,
− (m/n)
− = ( m)/n. Similarly, (m/n)(n/m) = (mn)/(mn) =
mn(mn) −1 = 1, so (m/n) −1 = n/m by the uniqueness of multiplicative inverses.
b) Any subset of R which contains 0 and 1 will satisfy the Associative and Commutative
Properties, the Distributive Law, and have an additive identity 0 and a multiplicative identity 1. By
part a), Q satisfies the Closure Properties, has additive∈ inverses, and every nonzero q Q has a
multiplicative inverse. Therefore, Q satisfies Postulate 1.
c) If r ∈ Q, x∈ R\ Q but q := r +∈x Q, then x−= q∈ r Q, a contradiction. Similarly, ∈ if rx / Q
and r = ∈ 0, then x Q, a contradiction. However, the product of any irrational with 0 is a rational.
d) By the First Multiplicative Property, mn−1 < pq−1 if and only if mq = mn−1qn < pq−1nq = np.
1.2.10. 0 ≤ (cb − ad) 2 = c2b2 − 2abcd + a2d2 implies 2abcd ≤ c2b2 + a2d2. Adding a2b2 + c2d2 to both sides,
we conclude that (ab + cd) 2 ≤ (a2 + c2)(b2 + d2).
1.2.11. Let P := R+.
a) Let x ∈ R. By the Trichotomy Property, either x−> 0, x > 0, or x = 0. Thus P satisfies i). If x
> 0 and y > 0, then by the Additive Property, x + y > 0 and by the First Multiplicative Property, xy
> 0. Thus P satisfies ii).
b) To prove the Trichotomy Property, suppose a, b ∈ R. By i), either a − b ∈ P, b − a = −(a − b) ∈ P, or
a − b = 0. Thus either a > b, b > a, or a = b.
To prove the Transitive Property, suppose a < b and b < c. Then b − a, c − b ∈ P and it follows from ii) that
c − a = b − a + c − b ∈ P, i.e., c > a.
Since b − a = (b + c) − (a + c), it is clear that the Additive Property holds.
Finally, suppose a < b, i.e., b − a ∈ P. If c > 0 then c ∈ P and it follows from ii) that bc − ac = (b − a)c ∈
P, i.e., bc > ac. If c < 0 then −c ∈ P, so ac − bc = (b − a)(−c) ∈ P, i.e., ac > bc.
1.3 The Completeness Axiom.
1.3.0. a) True. If A∩ B = ∅ , then sup(A ∩B) := −∞ and there is nothing to prove. If ∩ A / B∅ = , then
use the Monotone Property.
b) True. If x ∈ A, then x ≤ sup A. Since ε > 0, we have εx ≤ ε sup A, so the latter is an upper bound
of B. It follows that sup B ≤ ε sup A. On the other hand, if x ∈ A, then εx ∈ B, so εx ≤ sup B, i.e., sup B/ε
is an upper bound for A. It follows that sup A ≤ sup B/ε.
c) True. If x ∈ A and y ∈ B, then x + y ≤ sup A + sup B, so sup(A + B) ≤ sup A + sup B. If this
inequality is strict, then sup(A + B) − sup B < sup A, and it follows from the Approximation Property
that there is an a0 ∈ A such that sup(A + B) −sup B < a0. This implies that sup(A + B) − a0 < sup B, so
by the Approximation Property again, there is a b0 ∈ B such that sup(A + B) − a0 < b0. We conclude
that sup(A + B) < a0 + b0, a contradiction.
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d) False. Let A = B = [0, 1]. Then A − B = [−1, 1] so sup(A − B) = 1 /= 0 = sup A − sup B.
1.3.1. a) Since x2 + 2x − 3 = 0 implies x = 1, −3, inf E = −3, sup E =√1. b) Since x2 − 2x + 3√ >x
2
implies x < 3/2, inf E = 0, sup E = 3/2. c) Since p2/q2 < 5 implies p/q < 5, inf E =
0, sup E = 5. d)
Since 1 + (−1)n/n = 1 − 1/n when n is odd and 1 + 1/n when n is even, inf E = 0 and sup E = 3/2.
e) Since 1/n + (−1)n = 1/n + 1 when n is even and 1/n − 1 when n is odd, inf E = −1 and sup E
= 3/2. f) Since 2 − (−1)n/n2 = 2 − 1/n2 when n is even and 2 + 1/n2 when n is odd, inf E = 7/4 and
sup E = 3.
1.3.2. Since a − 1/n < a + 1/n, choose rn ∈ Q such that a − 1/n < rn < a + 1/n, i.e., |a − r n| < 1/n.
√ √ √ √ √
1.3.3. a < b implies a√− 2 < b − 2. Choose r ∈ Q s √u c h that a − 2 < r < b − 2. Then a < r + 2 < b.
By Exercise 1.2.9c, r + 2 is irrational. Thus set ξ = r + 2.
1.3.4. If m is a lower bound of E then so is any m˜ ≤ m. If m and m̃ are both infima of E then m ≤ m̃ and
m̃≤ m, i.e., m = m̃.
1.3.5. Suppose that E is a bounded, nonempty subset of Z. Since −E is a bounded, nonempty subset
of Z, it has a supremum by the Completeness Axiom, and that supremum belongs to −E by Theorem
1.15. Hence by the Reflection Principle, inf E = − sup(−E) ∈ −(−E) = E.
1.3.6. a) Let ǫ > 0 and m = inf E. Since m + ǫ is not a lower bound of E, there is an a ∈ E such that m
+ ǫ > a. Thus m + ǫ > a ≥ m as required.
b) By Theorem 1.14, there is an a ∈ E such that sup(−E) − ǫ < −a ≤ sup(−E). Hence by the
Second Multiplicative Property and Theorem 1.20, inf E + ǫ = −(sup(−E) − ǫ) > a > − sup(−E) = inf
E.
1.3.7. a) Let x be an upper bound of E and∈x E. If M is any upper bound of E then≥M x. Hence
by definition, x is the supremum of E.
b) The correct statement is: If x is a lower bound of E and x ∈ E then x = inf E.
PROOF. −x is an upper bound of −E and −x ∈ −E so −x = sup(−E). Thus x = − sup(−E) = inf E.
c) If E is the set of points xn such that xn =—1 1/n for odd n and xn = 1/n for even n, then sup E = 1,
inf E = 0, but neither 0 nor 1 belong to E.
1.3.8. Since A⊆ E, any upper bound of E is an upper bound of A. Since A is nonempty, it
follows from the Completeness Axiom that A has a supremum. Similarly, B has a supremum.
≤
Moreover, by the Monotone Property, sup A, sup B sup E.
Set M := max{ sup A, sup}B and observe that M is an upper bound of both A and B. If M < sup
∈ is an x E such that
E, then there ≤ M < x sup ∈ E. But x ∈E implies
∈ x A or x B. Thus M is not
an upper bound for one of the sets A or B, a contradiction.
1.3.9. By induction, 2n > n. Hence by the Archimedean Principle, there is an n ∈ N such that 2n >
1/(b − a). Let E := {k ∈ N : 2nb ≤ k}. By the Archimedean Principle, E is nonempty. Hence let m0 be the
least element in E and set q = (m0 − 1)/2n. Since b > 0, m0 ≥ 1. Since m0 is least in E, it follows that m0
− 1 < 2nb, i.e., q < b. On the other hand, m0 ∈ E implies 2nb ≤ m0, so
m0 1 m0 − 1
a = b − (b − a) < − = = q.
2n 2n 2n
1.3.10. Since |xn| ≤ M , the set En = {xn, xn+1, . . . } is bounded for each n ∈ N. Thus sn := sup En
exists and is finite by the Completeness Axiom. Moreover, since En+1 ⊆ En, it follows from the
Monotone Property, sn ≥ sn+1 for each n ∈ N. Thus s 1 ≥ s 2 ≥ . . . .
By the Reflection Principle, it follows that t1 ≤ t2 ≤ · · · .
Or, if you prefer a more direct approach, σn := sup{−xn, −xn+1, . . . } satisfies σ1 ≥ σ2 ≥ Since tn = −σn
for n ∈ N, it follows from the Second Multiplicative Property that t1 ≤ t2 ≤ . . . .
1.3.11. Let E = {n ∈ Z : n ≤ a}. If a ≥ 0, then 0 ∈ E. If a < 0, then by the Archimedean Principle,
there is an m ∈ N such that m > −a, i.e., n := −m ∈ E. Thus E is nonempty. Since E is bounded above
(by a), it follows from the Completeness Axiom and Theorem 1.15 that n0 = sup E exists and belongs
to E.
Set k = n0 + 1. Since k > sup E, k cannot belong to E, i.e., a < k. On the other hand, since n0 ∈ E and
b − a > 1,
k = n0 + 1 ≤ a + 1 < a + (b − a) = b.
We conclude that a < k <
b.
3
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 1.4 Mathematical Induction.
1.4.0. a) False. If a =— b = 1 and n = 2, then (a + b) n = 0 is NOT greater than b2 = 1.
b) False. If a =— 3, b = 1, and n = 2, then (a + b)n = 4 is not less than or equal to bn = 1.
c) True. If n is even, then n
— k and k are either both odd or both even. If they’re both odd, then an−kbk is
the product of two negative numbers, hence positive. If they’re both even, then an−kbk is the product of
two positive numbers, henceµpositive.
¶ Thus by the Binomial µFormula,
¶
n n
k=0 k=2
Σ n Σ n
(a + b) n = k an−kbk = an + nan−1b + k an−kbk =: an + nan−1b + C.
Since C is a sum of positive numbers, the promised inequality follows at once.
d) True. By the Binomial Formula,
µ ¶n Σ µ ¶ Σ µ ¶
1 1 a −2 n
n 1 (a − 2)n−k n
n (a − 2)n−k
= + = = .
2n a 2a k=0
k ak 2n−ka n−k k=0
k an2n−k
1.4.1. a) By hypothesis, x1 > 2. Suppose xn > 2. Then by Exercise 1.2.5a, 2 < xn+1 < xn. Thus by
induction, 2 < xn+1 < xn for all n ∈ N.
b) By hypothesis, 2 < x1 < 3. Suppose 2 < xn < 3. Then by Exercise 1.2.5b, 0 < xn < xn+1.
Thus by induction, 0 < xn < xn+1 for all n ∈ N.
c) By hypothesis, 0 < x1 < 1. Suppose 0 < xn < 1. Then by Exercise 1.2.5c, 0 < xn+1 < xn. Thus by
induction this inequality holds for all n ∈ N.
d) By hypothesis, 3 < x1 < 5. Suppose 3 < xn < 5. Then by Exercise 1.2.5d, 3 < xn+1 < xn.
Thus by induction this inequality holds for all n ∈ N.
Σn ¡ ¢ Σn ¡ ¢
1.4.2. a) 0 = (1 − 1)n = k=0 nk 1n−k(−1)k = k=0 nk (−1)k.
b) (a + b) = a + na
n n n−1 b + · · · + b ≥ a + na b.
n n n−1
c) By b), (1 + 1/n)n ≥ 1n + n1n−1 (1/n) = 2.
Σn ¡ ¢ Σn ¡ ¢ Σ n−1
d) 2n = (1 + 1) n = k=0 nk so k=1 nk = 2n − 1. On the other hand k=0 2k = 2n − 1 by induction.
1.4.3. a) This inequality holds for n = 3. If it holds for some n ≥ 3 then
2(n + 1) + 1 = 2n + 1 + 2 < 2n + 2 < 2n + 2n = 2n+1.
b) The inequality holds for n = 1. If it holds for n then
n + 1 < 2n + 1 ≤ 2 n + n < 2n + 2n = 2n+1.
c) Now n 2 ≤ 2n + 1 holds for n = 1, 2, and 3. If it holds for some n ≥ 3 then by a),
(n + 1)2 = n2 + 2n + 1 < 2 n + 2n = 2n+1 < 2 n+1 + 1.
d) We claim that 3n 2 + 3n +≤1 2 ·3n for n = 3, 4, . . . . This inequality holds for n = 3. Suppose it
holds for some n. Then
3(n + 1) 2 + 3(n + 1) + 1 = 3n2 + 3n + 1 + 6n + 6 ≤ 2 · 3n + 6(n + 1).
Similarly, induction can be used to establish 6(n + 1) ≤ 4 · 3n for n ≥ 1. (It holds for n = 1, and if it holds for n
then 6(n + 2) = 6(n + 1) + 6 ≤ 4 · 3n + 6 < 4 · 3n + 8 · 3n = 4 · 3n+1.) Therefore,
3(n + 1) 2 + 3(n + 1) + 1 ≤ 2 · 3n + 6(n + 1) ≤ 2 · 3n + 4 · 3n = 2 · 3n+1.
Thus the claim holds for all n ≥ 3.
Now n 3 ≤ 3n holds by inspection for n = 1, 2, 3. Suppose it holds for some n ≥ 3. Then
(n + 1)3 = n 3 + 3n2 + 3n + 1 ≤ 3n + 2 · 3n = 3n+1.
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1.4.4. a) The formula holds for n = 1. If it holds for n then
Σn+1 n(n + 1) n (n + 1)(n + 2)
k= + n + 1 = (n + 1)( + 1) = .
k=1
2 2 2
b) The formula holds for n = 1. If it holds for n then
Σ
n+1
n(n + 1)(2n + 1) n+1 (n + 1)(n + 2)(2n + 3)
k2 = + (n + 1)2 = (n(2n + 1) + 6(n + 1)) = .
k=1 6 6 6
c) The formula holds for n = 1. If it holds for n then
Σ
n+1
a −1 1 a −1 1
= 1− + =1− .
k=1 ak an an+1 an+1
d) The formula holds for n = 1. If it holds for n then
n+1
Σ 2
n(4n − 1) 2n + 1
(2k − 1)2 = 3 + (2n + 1)2 = 2
3 (2n + 5n + 3)
k=1
2n + 1 (n + 1)(4n2 + 8n +
= (2n + 3)(n + 1)
3 = 3) 3
(n + 1)(4(n + 1)2— 1)
= .
3
1.4.5. 0 an < bn holds for n = 1. If it holds for n then by (7), 0 an+1 < bn+1.
≤ √n √ √ √ √ ≤
By convention,
√ b√≥ 0. If n a < n b is false, then n a ≥ n b ≥ 0. Taking the nth power of this inequality, we
n n n n
obtain a = ( a) ≥ ( b) = b, a contradiction.
1.4.6. The result is true for n = 1. Suppose it’s true for some odd number ≥ 1, i.e., 2 2n−1 + 3 2n−1 =
5ℓ for some ℓ, n ∈ N. Then
22n+1 + 32n+1 = 4 · 22n−1 + 9 · 32n−1 = 4 · 5ℓ + 5 · 32n−1
is evidently divisible by 5. Thus the result is true by induction.
1.4.7. We first prove that 2n! + 2 ≤ (n + 1)! for n = 2, 3, . . . . It’s true for n = 2. Suppose that it’s
true for some n ≥ 2. Then by the inductive hypothesis,
2(n + 1)! + 2 = 2(n + 1)n! + 2 = 2n! + 2 + 2n · n! ≤ (n + 1)! + 2n · n!.
But 2 < n + 1 so we continue the inequality above by
2(n + 1)! + 2 < (n + 1)! + n · (n + 1)! = (n + 2) · (n + 1)! = (n + 2)!
as required.
To prove that 2n≤ n! + 2, notice first that it’s true for n = 1. If it’s true for some
≥ n 1, then by the
inequality already proved,
2n+1 = 2 · 2n ≤ 2(n! + 2) = 2n! + 2 + 2 ≤ (n + 1)! + 2
as
required.
1.4.8. If n = 1 or n = 2, the result is trivial. If n ≥ 3, then by the Binomial Formula,
Σ µ ¶ µ ¶
n n n n(n − 1)(n − 2)
2n = (1 + 1)n = > = .
k=
6
0
k 3
√
1.4.9. a) If m = k 2, then m = k by definition. On the other hand, if m is not a perfect square,
√
then by Remark 1.28, m is irrational. In particular, it cannot be rational.
5
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, √ √ √ √ √ √
b) If n + 3 + n ∈ Q the√n n + 3 + 2 n + 3 n + n = ( n + 3 + n)2 ∈ Q. Since Q is closed under subtraction
and division, it follows that n2 + 3n ∈ Q. In particular, n2 + 3n = m2 for some m ∈ N. Now n 2 + 3n is a perfect
square when n = 1 but if n > 1 then
(n + 1)2 = n2 + 2n + 1 < n2 + 2n + n = n2 + 3n =< n2 + 4n + 4 = (n + 2)2.
Therefore, the original expression is rational if and only if n = 1.
c) By repeating the steps in b), we see that the original expression is rational if and only if n(n+7) = n2+7n = m2
for some m ∈ N. If n > 9 then
(n + 3)2 = n2 + 6n + 9 < n2 + 7n < n2 + 8n + 16 = (n + 4)2.
Thus the original expression cannot be rational when n > 9. On the other hand, it is easy to check that
n2 + 7n is not a perfect square for n = 1, 2, . . . , 8 but is a perfect square, namely 144 = 122, when n = 9.
Thus the original expression is rational if and only if n = 9.
1.4.10. The result holds for n = 0 since c0 − b0 = 1 and0 a 0+ b 0
2 2 = c2. Suppose that c
n−1 − b n−1 = 1 and
a2n−1 + bn−1
2 = c2 hold for some n ≥ 0. By definition, c − b = c
n−1 n n n−1 − bn−1 = 1, so by induction, this difference
is always 1. Moreover, by the Binomial Formula, the inductive hypothesis, and what we just proved,
a2 + b2 = (an−1 + 2)2 + (2an−1 + bn−1 + 2)2
n n
= a2 + 4an−1 + 4 + (2an−1 + 2)2 + 2bn−1(2an−1 + 2) + b2
n−1 n−1
= c + 2(an−1 + 2) + (2an−1 + 2)2 + 2(cn−1 − 1)(2an−1 + 2)
2
n−1
= c2 + (2an−1 + 2)2 + 2cn−1(2an−1 + 2)
n−1
= (2an−1 + cn−1 + 2)2 ≡ cn2 .
1.5 Inverse Functions and Images.
1.5.0. a) False. Since (sin x)′ = cos x is negative on [π/2, 3π/2], f is 1–1 there, but the domain of
arcsin
− x is [ π/2, π/2]. Thus here, f −1(x)
− = arcsin(π x).
b) True. By elementary set algebra and Theorem 1.37,
(f −1(A) ∩ f −1(B)) ∪ f −1(C) = f −1(A ∩ B) ∪ f −1(C) ⊃ f −1(A ∩ B) /= ∅.
c) False. If X = [0, 2], A = [0, 1] and B = {1}, then B \ A = ∅ but (A \ B)c = [0, 1)c = [1, 2].
d) False. Let f (x) = x + 1 for −1 ≤ x ≤ 0 and f (x) = 2x − 1 for 0 < x ≤ 1. Then f takes [−1, 1] onto [−1,
1] and f (0) = 1, but f −1(f (0)) = f −1(1) = {0, 1}.
1.5.1. α) f is 1–1 since f ′(x) = 3 > 0 for∈x R. If y = 3x− 7 then x = (y +7)/3. Therefore f −1(x) =
(x+7)/3. By looking at the graph, we see that f (E) = R.
β) f is 1–1 since f ′(x) = −e1/x/x2 > 0 for x ∈ (0, ∞). If y = e1/x then log y = 1/x, i.e., x = 1/ log y. Therefore,
f (x) = 1/ log x. By looking at the graph, we see that f (E) = (1, ∞).
−1
γ) f is 1–1 on (π/2, 3π/2) because f ′(x) = sec 2 x > 0 there. The inverse is f −1(x) = arctan(x − π). By
looking at the graph, we see that f (E) = (−∞, ∞).
δ) Since f ′(x) =
√ 2x + 2 < 0 for x < −6,√f is 1–1 on [−∞, −6]. Since y = x + 2x − 5 is a quadratic in x, we
2
have x = (−2 ± 4 + 4(5 +√y))/2 = −1 ± 6 + y. But x is negative on (−∞, −6], so we must use the negative
sign. Hence f −1(x) − = − 6 + x. By looking at the graph, we see that f (E) = [19, ∞).
1 ,
ε) By definition, , 3x + 2 x≤0
f (x) = x + 2 0 <x 2
≤
,
,
3x − 2 x > 2.
Thus f is strictly increasing, hence 1–1, and
,
, (x − 2)/3 x≤2
, x− 2
f −1(x) = , 2<x≤4
(x + 2)/3 x > 4,
Covers All 14 Chapters
SOLUTIONS TO EXERCISES
, An Introduction to Analysis
Table of Contents
Chapter 1: The Real Number System
1.2 Ordered field axioms ............................................................... 1
1.3 The Completeness Axiom….................................................... 2
1.4 Mathematical Induction… ....................................................... 4
1.5 Inverse Functions and Images…............................................. 6
1.6 Countable and uncountable sets… ......................................... 8
Chapter 2: Sequences in R
2.1 Limits of Sequences… ............................................................. 10
2.2 Limit Theorems ....................................................................... 11
2.3 Bolzano-Weierstrass Theorem ............................................... 13
2.4 Cauchy Sequences… ............................................................... 15
2.5 Limits Supremum and Infimum ............................................ 16
Chapter 3: Functions on R
3.1 Two-Sided Limits… ................................................................ 19
3.2 One-Sided Limits and Limits at Infinity… ............................. 20
3.3 Continuity… ............................................................................ 22
3.4 Uniform Continuity…............................................................. 24
Chapter 4: Differentiability on R
4.1 The Derivative… ..................................................................... 27
4.2 Differentiability Theorem… ....................................................28
4.3 The Mean Value Theorem… .................................................. 30
4.4 Taylor’s Theorem and l’Hôpital’s Rule… ............................. 32
4.5 Inverse Function Theorems .................................................... 34
Chapter 5: Integrability on R
5.1 The Riemann Integral…........................................................... 37
5.2 Riemann Sums .......................................................................... 40
5.3 The Fundamental Theorem of Calculus… ............................. 43
5.4 Improper Riemann Integration… ........................................... 46
5.5 Functions of Bounded Variation… ......................................... 49
5.6 Convex Functions… ................................................................ 51
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Chapter 6: Infinite Series of Real Numbers
6.1 Introduction… ........................................................................... 53
6.2 Series with Nonnegative Terms… .......................................... 55
6.3 Absolute Convergence… ......................................................... 57
6.4 Alternating Series… ................................................................. 60
6.5 Estimation of Series… .............................................................. 62
6.6 Additional Tests… ................................................................... 63
Chapter 7: Infinite Series of Functions
7.1 Uniform Convergence of Sequences… ................................... 65
7.2 Uniform Convergence of Series… .......................................... 67
7.3 Power Series….......................................................................... 69
7.4 Analytic Functions… ...............................................................72
7.5 Applications… ......................................................................... 74
Chapter 8: Euclidean Spaces
8.1 Algebraic Structure… ............................................................. 76
8.2 Planes and Linear Transformations… ................................... 77
8.3 Topology of Rn .......................................................................................................................79
8.4 Interior, Closure, and Boundary… ........................................ 80
Chapter 9: Convergence in Rn
9.1 Limits of Sequences… .............................................................. 82
9.2 Heine-Borel Theorem............................................................... 83
9.3 Limits of Functions… ............................................................... 84
9.4 Continuous Functions… .......................................................... 86
9.5 Compact Sets… ........................................................................ 87
9.6 Applications… .......................................................................... 88
Chapter 10: Metric Spaces
10.1 Introduction… ............................................................................ 90
10.2 Limits of Functions… ................................................................ 91
10.3 Interior, Closure, and Boundary… .......................................... 92
10.4 Compact Sets… ......................................................................... 93
10.5 Connected Sets… ......................................................................94
10.6 Continuous Functions… ........................................................... 96
10.7 Stone-Weierstrass Theorem...................................................... 97
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
,Chapter 11: Differentiability on Rn
11.1 Partial Derivatives and Partial Integrals… ............................... 99
11.2 The Definition of Differentiability….......................................... 102
11.3 Derivatives, Differentials, and Tangent Planes… .................... 104
11.4 The Chain Rule… ........................................................................ 107
11.5 The Mean Value Theorem and Taylor’s Formula…................. 108
11.6 The Inverse Function Theorem .................................................. 111
11.7 Optimization… ............................................................................. 114
Chapter 12: Integration on Rn
12.1 Jordan Regions…...........................................................................117
12.2 Riemann Integration on Jordan Regions… ................................ 119
12.3 Iterated Integrals… ....................................................................... 122
12.4 Change of Variables… .................................................................. 125
12.5 Partitions of Unity… .....................................................................130
12.6 The Gamma Function and Volume ............................................ 131
Chapter 13: Fundamental Theorems of Vector Calculus
13.1 Curves… ........................................................................................ 135
13.2 Oriented Curves… ........................................................................ 137
13.3 Surfaces… ...................................................................................... 140
13.4 Oriented Surfaces… ...................................................................... 143
13.5 Theorems of Green and Gauss… ................................................. 147
13.6 Stokes’s Theorem ........................................................................... 150
Chapter 14: Fourier Series
14.1 Introduction… ............................................................................... 156
14.2 Summability of Fourier Series… .................................................. 157
14.3 Growth of Fourier Coefficients…................................................. 159
14.4 Convergence of Fourier Series… ................................................. 160
14.5 Uniqueness…................................................................................. 163
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SOLUTIONS TO EXERCISES
CHAPTER 1
1.2 Ordered field axioms.
1.2.0. a) False. Let a = 2/3, b = 1, c = −2, and d = −1.
b) False. Let a = −4, b = −1, and c = 2.
c) True. Since a ≤ b and b ≤ a + c, |a − b| = b − a ≤ a + c − a = c.
d) True. No a ∈ R satisfies a < b − ε for all ε > 0, so the inequality is vacuously satisfied. If you want
a more constructive proof, if b ≤ 0 then a < b − ε < 0 + 0 = 0. If b > 0, then for ε = b, a < b − ε = 0.
1.2.1. a) If a < b then a + c < b + c by the Additive Property. If a = b then a + c = b + c since + is a
function. Thus a + c ≤ b + c holds for all a ≤ b.
b) If c = 0 then ac = 0 = bc so we may suppose c > 0. If a < b then ac < bc by the Multiplicative Property. If
a = b then ac = bc since · is a function. Thus ac ≤ bc holds for all a ≤ b and c ≥ 0.
1.2.2. a) Suppose 0 ≤ a < b and 0 ≤ c < d. Multiplying the first inequality by c and the second by b, we have
0 ≤ ac ≤ bc and bc < bd. Hence by the Transitive P r o per
√ ty, ac < bd. √
√ √
b) Suppose 0 ≤ a < b. By (7), 0 ≤ a2 < b2. If a ≥ b then a = ( a)2 ≥ ( b)2 = b, a contradiction.
c) If 1/a ≤ 1/b, then the Multiplicative Property implies b = ab(1/a) ≤ ab(1/b) = a, a contradiction. If 1/b ≤ 0
then b = b 2(1/b) ≤ 0 a contradiction.
d) To show these statements may not hold when a < 0, let a = −2, b = −1, c = 2 and d = 5. Then a <
b and c < d but ac = −4 is not less than bd = −5, a2 = 4 is not less than b2 = 1, and 1/a = −1/2 is not
less than 1/b = −1.
1.2.3. a) By definition,
µ ¶
|a| + a |a| − a 2a
a+ − a− = − = =a
2 2 2
and
µ ¶
|a| + a |a| − a 2|a|
a+ + a− = + = = |a|.
2 2 2
b) By Definition 1.1, if a ≥ 0 then a+ = (a + a)/2 = a and if a < 0 then a+ = (−a + a)/2 = 0. Similarly,
a− = 0 if a ≥ 0 and a− = −a if a < 0.
1.2.4. a) |2x + 1| < 7 if and only if −7 < 2x + 1 < 7 if and only if − − 4 < x < 3.
b) |2 − x| < 2 if and only if −2 < 2 − x < 2 if and only if −4 < −x < 0 if and only if 0 < x < 4.
c) |x3 −3x+1| < x3 if and only if −x3 < x3 − 3x+1 < x3 if and only2 if 3x− 1 > 0 and 2x3 −3x+1 > 0. T h√e first
inequality is equivalent to x > 1/3. Since 2x3 −√ 3x + 1 = (x − 1)(2x + 2x − 1) implies that x = 1, (−1 3)/2,
±
√
the sec√ond inequality is equivalent to (−1 − 3)/2 < x < (−1 + 3)/2 or x > 1. Therefore, the solution is
(1/3, ( 3 − 1)/2) ∪ (1, ∞).
d) We cannot multiply by the denominator x − 1 unless we consider its sign.
Case 1. x − 1 > 0. Then x < x − 1 so 0 < −1, i.e., this case is empty.
Case 2. x − 1 < 0. Then by the Second Multiplicative Property, x > x − 1 so 0 > −1, i.e., every number
from this case works. Thus the solution is (−∞, 1).
e) Case 1. 4x2 − 1 > 0. Cross multiplying, we obtain 4x2 < 4x2 − 1, i.e., this case is empty.
Case 2. 4x2 − 1 < 0. Then by the Second Multiplicative Property, 4x2 > 4x2 − 1, i.e., 0 > −1.
Thus the solution is (−1/2, 1/2).
√ √
1.2.5. a) Suppose a > 2. Then a − 1 > 1 so 1 < a − 1 < a − 1 by (6). Therefore, 2 < b = 1 + a − 1 <
1 + (a − 1) = a. √ √
b) Suppose 2 < a < 3. Then 0 < a − 2 < 1 so 0 < a − 2 a − 2 < 1 √b y (6). Therefore, 0 < a < 2 a−√ 2 = b.
+
<
c) Suppose 0 < a√< 1. Then 0 > −a > −1, so 0 < 1 − a < 1. Hence 1 − a is real and by (6), 1 − a < 1 − a.
Therefore, b = 1 − 1 − a < 1 − (1 − a) = a. √ √
d) Suppose 3 < a < 5. Then 1 < a − 2 < 3 so 1 < a − 2 < a − 2 by (6). Therefore, 3 < 2 + a − 2 = b < a.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, √ √ √ √
1.2.6. a + b − 2 ab = ( a − b)2 ≥ 0 for all a, b ∈ [0, ∞). Thus 2 ab ≤ a + b an√d G( a ,√b ) ≤ A(a, b). On
the other hand, since 0 ≤ a ≤ b √w e have A(a, b) = (a + b)/2 ≤ 2b/√2 = b and G(a, b) = ab ≥ √a2 = a. Finally,
√ √
A(a, b) = G(a, b) if and only ab = a + b if and only if ( a b) 2 = 0 if and only if a b if and only if
if 2 − =
a = b.
1.2.7. a) Since |x + 2| ≤ |x| + 2, |x| ≤ 2 implies |x2 − 4| = |x + 2| |x − 2| ≤ 4|x − 2|.
b) Since |x + 3| ≤ |x| + 3, |x| ≤ 1 implies |x2 + 2x − 3| = |x + 3| |x − 1| ≤ 4|x − 1|.
c) Since |x − 2| ≤ |x| + 2, −3 ≤ x ≤ 2 implies |x2 + x − 6| = |x + 3| |x − 2| ≤ 6|x − 2|.
d) Since the minimum of x2 + x − 1 on (−1, 0) is −1.25, −1 < x < 0 implies |x3 − 2x + 1| = |x2 + x − 1| |x − 1| <
5|x − 1|/4.
1.2.8. a) Since (1− − n)/(1 n2) = 1/(1 + n), the inequality is equivalent to 1/(n + 1) < .01 =
1/100. Since 1 + ∈ n > 0 for all n N, it follows that n + 1 > 100, i.e., n > 99.
b) By factoring, we see that the inequality is equivalent to 1/(2n + 1) < 1/40, i.e., 2n + 1 > 40. Thus n > 39/2,
i.e., n ≥ 20.
√
c) The inequality is equivalent to n2 + 1 > 500. 499 ≈ 22.33, i.e., n ≥ 23.
Thus n >
1.2.9. a) mn−1 + pq−1 = mqq−1 n−1 + pq−1nn−1 = (mq + pn)n−1q−1. But n−1q−1 nq = 1 and
uniqueness of multiplicative inverses implies (nq)−1 = n−1q−1. Therefore, mn−1 + pq−1 = (mq +
pn)(nq)−1. Similarly, mn−1(pq−1) = mpn−1q−1 = mp(nq)−1. By what we just proved and (2),
m −m m −m 0
+ = = = 0.
n n n n
Therefore, by the uniqueness of additive inverses,
− (m/n)
− = ( m)/n. Similarly, (m/n)(n/m) = (mn)/(mn) =
mn(mn) −1 = 1, so (m/n) −1 = n/m by the uniqueness of multiplicative inverses.
b) Any subset of R which contains 0 and 1 will satisfy the Associative and Commutative
Properties, the Distributive Law, and have an additive identity 0 and a multiplicative identity 1. By
part a), Q satisfies the Closure Properties, has additive∈ inverses, and every nonzero q Q has a
multiplicative inverse. Therefore, Q satisfies Postulate 1.
c) If r ∈ Q, x∈ R\ Q but q := r +∈x Q, then x−= q∈ r Q, a contradiction. Similarly, ∈ if rx / Q
and r = ∈ 0, then x Q, a contradiction. However, the product of any irrational with 0 is a rational.
d) By the First Multiplicative Property, mn−1 < pq−1 if and only if mq = mn−1qn < pq−1nq = np.
1.2.10. 0 ≤ (cb − ad) 2 = c2b2 − 2abcd + a2d2 implies 2abcd ≤ c2b2 + a2d2. Adding a2b2 + c2d2 to both sides,
we conclude that (ab + cd) 2 ≤ (a2 + c2)(b2 + d2).
1.2.11. Let P := R+.
a) Let x ∈ R. By the Trichotomy Property, either x−> 0, x > 0, or x = 0. Thus P satisfies i). If x
> 0 and y > 0, then by the Additive Property, x + y > 0 and by the First Multiplicative Property, xy
> 0. Thus P satisfies ii).
b) To prove the Trichotomy Property, suppose a, b ∈ R. By i), either a − b ∈ P, b − a = −(a − b) ∈ P, or
a − b = 0. Thus either a > b, b > a, or a = b.
To prove the Transitive Property, suppose a < b and b < c. Then b − a, c − b ∈ P and it follows from ii) that
c − a = b − a + c − b ∈ P, i.e., c > a.
Since b − a = (b + c) − (a + c), it is clear that the Additive Property holds.
Finally, suppose a < b, i.e., b − a ∈ P. If c > 0 then c ∈ P and it follows from ii) that bc − ac = (b − a)c ∈
P, i.e., bc > ac. If c < 0 then −c ∈ P, so ac − bc = (b − a)(−c) ∈ P, i.e., ac > bc.
1.3 The Completeness Axiom.
1.3.0. a) True. If A∩ B = ∅ , then sup(A ∩B) := −∞ and there is nothing to prove. If ∩ A / B∅ = , then
use the Monotone Property.
b) True. If x ∈ A, then x ≤ sup A. Since ε > 0, we have εx ≤ ε sup A, so the latter is an upper bound
of B. It follows that sup B ≤ ε sup A. On the other hand, if x ∈ A, then εx ∈ B, so εx ≤ sup B, i.e., sup B/ε
is an upper bound for A. It follows that sup A ≤ sup B/ε.
c) True. If x ∈ A and y ∈ B, then x + y ≤ sup A + sup B, so sup(A + B) ≤ sup A + sup B. If this
inequality is strict, then sup(A + B) − sup B < sup A, and it follows from the Approximation Property
that there is an a0 ∈ A such that sup(A + B) −sup B < a0. This implies that sup(A + B) − a0 < sup B, so
by the Approximation Property again, there is a b0 ∈ B such that sup(A + B) − a0 < b0. We conclude
that sup(A + B) < a0 + b0, a contradiction.
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d) False. Let A = B = [0, 1]. Then A − B = [−1, 1] so sup(A − B) = 1 /= 0 = sup A − sup B.
1.3.1. a) Since x2 + 2x − 3 = 0 implies x = 1, −3, inf E = −3, sup E =√1. b) Since x2 − 2x + 3√ >x
2
implies x < 3/2, inf E = 0, sup E = 3/2. c) Since p2/q2 < 5 implies p/q < 5, inf E =
0, sup E = 5. d)
Since 1 + (−1)n/n = 1 − 1/n when n is odd and 1 + 1/n when n is even, inf E = 0 and sup E = 3/2.
e) Since 1/n + (−1)n = 1/n + 1 when n is even and 1/n − 1 when n is odd, inf E = −1 and sup E
= 3/2. f) Since 2 − (−1)n/n2 = 2 − 1/n2 when n is even and 2 + 1/n2 when n is odd, inf E = 7/4 and
sup E = 3.
1.3.2. Since a − 1/n < a + 1/n, choose rn ∈ Q such that a − 1/n < rn < a + 1/n, i.e., |a − r n| < 1/n.
√ √ √ √ √
1.3.3. a < b implies a√− 2 < b − 2. Choose r ∈ Q s √u c h that a − 2 < r < b − 2. Then a < r + 2 < b.
By Exercise 1.2.9c, r + 2 is irrational. Thus set ξ = r + 2.
1.3.4. If m is a lower bound of E then so is any m˜ ≤ m. If m and m̃ are both infima of E then m ≤ m̃ and
m̃≤ m, i.e., m = m̃.
1.3.5. Suppose that E is a bounded, nonempty subset of Z. Since −E is a bounded, nonempty subset
of Z, it has a supremum by the Completeness Axiom, and that supremum belongs to −E by Theorem
1.15. Hence by the Reflection Principle, inf E = − sup(−E) ∈ −(−E) = E.
1.3.6. a) Let ǫ > 0 and m = inf E. Since m + ǫ is not a lower bound of E, there is an a ∈ E such that m
+ ǫ > a. Thus m + ǫ > a ≥ m as required.
b) By Theorem 1.14, there is an a ∈ E such that sup(−E) − ǫ < −a ≤ sup(−E). Hence by the
Second Multiplicative Property and Theorem 1.20, inf E + ǫ = −(sup(−E) − ǫ) > a > − sup(−E) = inf
E.
1.3.7. a) Let x be an upper bound of E and∈x E. If M is any upper bound of E then≥M x. Hence
by definition, x is the supremum of E.
b) The correct statement is: If x is a lower bound of E and x ∈ E then x = inf E.
PROOF. −x is an upper bound of −E and −x ∈ −E so −x = sup(−E). Thus x = − sup(−E) = inf E.
c) If E is the set of points xn such that xn =—1 1/n for odd n and xn = 1/n for even n, then sup E = 1,
inf E = 0, but neither 0 nor 1 belong to E.
1.3.8. Since A⊆ E, any upper bound of E is an upper bound of A. Since A is nonempty, it
follows from the Completeness Axiom that A has a supremum. Similarly, B has a supremum.
≤
Moreover, by the Monotone Property, sup A, sup B sup E.
Set M := max{ sup A, sup}B and observe that M is an upper bound of both A and B. If M < sup
∈ is an x E such that
E, then there ≤ M < x sup ∈ E. But x ∈E implies
∈ x A or x B. Thus M is not
an upper bound for one of the sets A or B, a contradiction.
1.3.9. By induction, 2n > n. Hence by the Archimedean Principle, there is an n ∈ N such that 2n >
1/(b − a). Let E := {k ∈ N : 2nb ≤ k}. By the Archimedean Principle, E is nonempty. Hence let m0 be the
least element in E and set q = (m0 − 1)/2n. Since b > 0, m0 ≥ 1. Since m0 is least in E, it follows that m0
− 1 < 2nb, i.e., q < b. On the other hand, m0 ∈ E implies 2nb ≤ m0, so
m0 1 m0 − 1
a = b − (b − a) < − = = q.
2n 2n 2n
1.3.10. Since |xn| ≤ M , the set En = {xn, xn+1, . . . } is bounded for each n ∈ N. Thus sn := sup En
exists and is finite by the Completeness Axiom. Moreover, since En+1 ⊆ En, it follows from the
Monotone Property, sn ≥ sn+1 for each n ∈ N. Thus s 1 ≥ s 2 ≥ . . . .
By the Reflection Principle, it follows that t1 ≤ t2 ≤ · · · .
Or, if you prefer a more direct approach, σn := sup{−xn, −xn+1, . . . } satisfies σ1 ≥ σ2 ≥ Since tn = −σn
for n ∈ N, it follows from the Second Multiplicative Property that t1 ≤ t2 ≤ . . . .
1.3.11. Let E = {n ∈ Z : n ≤ a}. If a ≥ 0, then 0 ∈ E. If a < 0, then by the Archimedean Principle,
there is an m ∈ N such that m > −a, i.e., n := −m ∈ E. Thus E is nonempty. Since E is bounded above
(by a), it follows from the Completeness Axiom and Theorem 1.15 that n0 = sup E exists and belongs
to E.
Set k = n0 + 1. Since k > sup E, k cannot belong to E, i.e., a < k. On the other hand, since n0 ∈ E and
b − a > 1,
k = n0 + 1 ≤ a + 1 < a + (b − a) = b.
We conclude that a < k <
b.
3
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 1.4 Mathematical Induction.
1.4.0. a) False. If a =— b = 1 and n = 2, then (a + b) n = 0 is NOT greater than b2 = 1.
b) False. If a =— 3, b = 1, and n = 2, then (a + b)n = 4 is not less than or equal to bn = 1.
c) True. If n is even, then n
— k and k are either both odd or both even. If they’re both odd, then an−kbk is
the product of two negative numbers, hence positive. If they’re both even, then an−kbk is the product of
two positive numbers, henceµpositive.
¶ Thus by the Binomial µFormula,
¶
n n
k=0 k=2
Σ n Σ n
(a + b) n = k an−kbk = an + nan−1b + k an−kbk =: an + nan−1b + C.
Since C is a sum of positive numbers, the promised inequality follows at once.
d) True. By the Binomial Formula,
µ ¶n Σ µ ¶ Σ µ ¶
1 1 a −2 n
n 1 (a − 2)n−k n
n (a − 2)n−k
= + = = .
2n a 2a k=0
k ak 2n−ka n−k k=0
k an2n−k
1.4.1. a) By hypothesis, x1 > 2. Suppose xn > 2. Then by Exercise 1.2.5a, 2 < xn+1 < xn. Thus by
induction, 2 < xn+1 < xn for all n ∈ N.
b) By hypothesis, 2 < x1 < 3. Suppose 2 < xn < 3. Then by Exercise 1.2.5b, 0 < xn < xn+1.
Thus by induction, 0 < xn < xn+1 for all n ∈ N.
c) By hypothesis, 0 < x1 < 1. Suppose 0 < xn < 1. Then by Exercise 1.2.5c, 0 < xn+1 < xn. Thus by
induction this inequality holds for all n ∈ N.
d) By hypothesis, 3 < x1 < 5. Suppose 3 < xn < 5. Then by Exercise 1.2.5d, 3 < xn+1 < xn.
Thus by induction this inequality holds for all n ∈ N.
Σn ¡ ¢ Σn ¡ ¢
1.4.2. a) 0 = (1 − 1)n = k=0 nk 1n−k(−1)k = k=0 nk (−1)k.
b) (a + b) = a + na
n n n−1 b + · · · + b ≥ a + na b.
n n n−1
c) By b), (1 + 1/n)n ≥ 1n + n1n−1 (1/n) = 2.
Σn ¡ ¢ Σn ¡ ¢ Σ n−1
d) 2n = (1 + 1) n = k=0 nk so k=1 nk = 2n − 1. On the other hand k=0 2k = 2n − 1 by induction.
1.4.3. a) This inequality holds for n = 3. If it holds for some n ≥ 3 then
2(n + 1) + 1 = 2n + 1 + 2 < 2n + 2 < 2n + 2n = 2n+1.
b) The inequality holds for n = 1. If it holds for n then
n + 1 < 2n + 1 ≤ 2 n + n < 2n + 2n = 2n+1.
c) Now n 2 ≤ 2n + 1 holds for n = 1, 2, and 3. If it holds for some n ≥ 3 then by a),
(n + 1)2 = n2 + 2n + 1 < 2 n + 2n = 2n+1 < 2 n+1 + 1.
d) We claim that 3n 2 + 3n +≤1 2 ·3n for n = 3, 4, . . . . This inequality holds for n = 3. Suppose it
holds for some n. Then
3(n + 1) 2 + 3(n + 1) + 1 = 3n2 + 3n + 1 + 6n + 6 ≤ 2 · 3n + 6(n + 1).
Similarly, induction can be used to establish 6(n + 1) ≤ 4 · 3n for n ≥ 1. (It holds for n = 1, and if it holds for n
then 6(n + 2) = 6(n + 1) + 6 ≤ 4 · 3n + 6 < 4 · 3n + 8 · 3n = 4 · 3n+1.) Therefore,
3(n + 1) 2 + 3(n + 1) + 1 ≤ 2 · 3n + 6(n + 1) ≤ 2 · 3n + 4 · 3n = 2 · 3n+1.
Thus the claim holds for all n ≥ 3.
Now n 3 ≤ 3n holds by inspection for n = 1, 2, 3. Suppose it holds for some n ≥ 3. Then
(n + 1)3 = n 3 + 3n2 + 3n + 1 ≤ 3n + 2 · 3n = 3n+1.
, @LECTSOLUTIONSSTUVIA
1.4.4. a) The formula holds for n = 1. If it holds for n then
Σn+1 n(n + 1) n (n + 1)(n + 2)
k= + n + 1 = (n + 1)( + 1) = .
k=1
2 2 2
b) The formula holds for n = 1. If it holds for n then
Σ
n+1
n(n + 1)(2n + 1) n+1 (n + 1)(n + 2)(2n + 3)
k2 = + (n + 1)2 = (n(2n + 1) + 6(n + 1)) = .
k=1 6 6 6
c) The formula holds for n = 1. If it holds for n then
Σ
n+1
a −1 1 a −1 1
= 1− + =1− .
k=1 ak an an+1 an+1
d) The formula holds for n = 1. If it holds for n then
n+1
Σ 2
n(4n − 1) 2n + 1
(2k − 1)2 = 3 + (2n + 1)2 = 2
3 (2n + 5n + 3)
k=1
2n + 1 (n + 1)(4n2 + 8n +
= (2n + 3)(n + 1)
3 = 3) 3
(n + 1)(4(n + 1)2— 1)
= .
3
1.4.5. 0 an < bn holds for n = 1. If it holds for n then by (7), 0 an+1 < bn+1.
≤ √n √ √ √ √ ≤
By convention,
√ b√≥ 0. If n a < n b is false, then n a ≥ n b ≥ 0. Taking the nth power of this inequality, we
n n n n
obtain a = ( a) ≥ ( b) = b, a contradiction.
1.4.6. The result is true for n = 1. Suppose it’s true for some odd number ≥ 1, i.e., 2 2n−1 + 3 2n−1 =
5ℓ for some ℓ, n ∈ N. Then
22n+1 + 32n+1 = 4 · 22n−1 + 9 · 32n−1 = 4 · 5ℓ + 5 · 32n−1
is evidently divisible by 5. Thus the result is true by induction.
1.4.7. We first prove that 2n! + 2 ≤ (n + 1)! for n = 2, 3, . . . . It’s true for n = 2. Suppose that it’s
true for some n ≥ 2. Then by the inductive hypothesis,
2(n + 1)! + 2 = 2(n + 1)n! + 2 = 2n! + 2 + 2n · n! ≤ (n + 1)! + 2n · n!.
But 2 < n + 1 so we continue the inequality above by
2(n + 1)! + 2 < (n + 1)! + n · (n + 1)! = (n + 2) · (n + 1)! = (n + 2)!
as required.
To prove that 2n≤ n! + 2, notice first that it’s true for n = 1. If it’s true for some
≥ n 1, then by the
inequality already proved,
2n+1 = 2 · 2n ≤ 2(n! + 2) = 2n! + 2 + 2 ≤ (n + 1)! + 2
as
required.
1.4.8. If n = 1 or n = 2, the result is trivial. If n ≥ 3, then by the Binomial Formula,
Σ µ ¶ µ ¶
n n n n(n − 1)(n − 2)
2n = (1 + 1)n = > = .
k=
6
0
k 3
√
1.4.9. a) If m = k 2, then m = k by definition. On the other hand, if m is not a perfect square,
√
then by Remark 1.28, m is irrational. In particular, it cannot be rational.
5
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, √ √ √ √ √ √
b) If n + 3 + n ∈ Q the√n n + 3 + 2 n + 3 n + n = ( n + 3 + n)2 ∈ Q. Since Q is closed under subtraction
and division, it follows that n2 + 3n ∈ Q. In particular, n2 + 3n = m2 for some m ∈ N. Now n 2 + 3n is a perfect
square when n = 1 but if n > 1 then
(n + 1)2 = n2 + 2n + 1 < n2 + 2n + n = n2 + 3n =< n2 + 4n + 4 = (n + 2)2.
Therefore, the original expression is rational if and only if n = 1.
c) By repeating the steps in b), we see that the original expression is rational if and only if n(n+7) = n2+7n = m2
for some m ∈ N. If n > 9 then
(n + 3)2 = n2 + 6n + 9 < n2 + 7n < n2 + 8n + 16 = (n + 4)2.
Thus the original expression cannot be rational when n > 9. On the other hand, it is easy to check that
n2 + 7n is not a perfect square for n = 1, 2, . . . , 8 but is a perfect square, namely 144 = 122, when n = 9.
Thus the original expression is rational if and only if n = 9.
1.4.10. The result holds for n = 0 since c0 − b0 = 1 and0 a 0+ b 0
2 2 = c2. Suppose that c
n−1 − b n−1 = 1 and
a2n−1 + bn−1
2 = c2 hold for some n ≥ 0. By definition, c − b = c
n−1 n n n−1 − bn−1 = 1, so by induction, this difference
is always 1. Moreover, by the Binomial Formula, the inductive hypothesis, and what we just proved,
a2 + b2 = (an−1 + 2)2 + (2an−1 + bn−1 + 2)2
n n
= a2 + 4an−1 + 4 + (2an−1 + 2)2 + 2bn−1(2an−1 + 2) + b2
n−1 n−1
= c + 2(an−1 + 2) + (2an−1 + 2)2 + 2(cn−1 − 1)(2an−1 + 2)
2
n−1
= c2 + (2an−1 + 2)2 + 2cn−1(2an−1 + 2)
n−1
= (2an−1 + cn−1 + 2)2 ≡ cn2 .
1.5 Inverse Functions and Images.
1.5.0. a) False. Since (sin x)′ = cos x is negative on [π/2, 3π/2], f is 1–1 there, but the domain of
arcsin
− x is [ π/2, π/2]. Thus here, f −1(x)
− = arcsin(π x).
b) True. By elementary set algebra and Theorem 1.37,
(f −1(A) ∩ f −1(B)) ∪ f −1(C) = f −1(A ∩ B) ∪ f −1(C) ⊃ f −1(A ∩ B) /= ∅.
c) False. If X = [0, 2], A = [0, 1] and B = {1}, then B \ A = ∅ but (A \ B)c = [0, 1)c = [1, 2].
d) False. Let f (x) = x + 1 for −1 ≤ x ≤ 0 and f (x) = 2x − 1 for 0 < x ≤ 1. Then f takes [−1, 1] onto [−1,
1] and f (0) = 1, but f −1(f (0)) = f −1(1) = {0, 1}.
1.5.1. α) f is 1–1 since f ′(x) = 3 > 0 for∈x R. If y = 3x− 7 then x = (y +7)/3. Therefore f −1(x) =
(x+7)/3. By looking at the graph, we see that f (E) = R.
β) f is 1–1 since f ′(x) = −e1/x/x2 > 0 for x ∈ (0, ∞). If y = e1/x then log y = 1/x, i.e., x = 1/ log y. Therefore,
f (x) = 1/ log x. By looking at the graph, we see that f (E) = (1, ∞).
−1
γ) f is 1–1 on (π/2, 3π/2) because f ′(x) = sec 2 x > 0 there. The inverse is f −1(x) = arctan(x − π). By
looking at the graph, we see that f (E) = (−∞, ∞).
δ) Since f ′(x) =
√ 2x + 2 < 0 for x < −6,√f is 1–1 on [−∞, −6]. Since y = x + 2x − 5 is a quadratic in x, we
2
have x = (−2 ± 4 + 4(5 +√y))/2 = −1 ± 6 + y. But x is negative on (−∞, −6], so we must use the negative
sign. Hence f −1(x) − = − 6 + x. By looking at the graph, we see that f (E) = [19, ∞).
1 ,
ε) By definition, , 3x + 2 x≤0
f (x) = x + 2 0 <x 2
≤
,
,
3x − 2 x > 2.
Thus f is strictly increasing, hence 1–1, and
,
, (x − 2)/3 x≤2
, x− 2
f −1(x) = , 2<x≤4
(x + 2)/3 x > 4,
