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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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
,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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 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 Property,
√ 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 = b2 (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 only if 3x −1 > 0 and 2x3 −3x +1 > 0. The √ first
inequality is equivalent to x > 1/3. Since 2x3 −
√ 3x + 1 = (x − 1)(2x√
2
+ 2x − 1) implies that x = 1, (−1 ± 3)/2,
the second
√ 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 √ by (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 and b) ≤ A(a, b). On
√ G(a,√
the other hand, since 0 ≤ a ≤ b √ we have A(a, b) = (a + b)/2 ≤ 2b/2
√ 2= b and G(a, b) = ab ≥ √a2 = a. Finally,
√ √
A(a, b) = G(a, b) if and only if 2 ab = a + b if and only if ( a − b) = 0 if and only if a = b if and only if
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. Thus n > 499 ≈ 22.33, i.e., n ≥ 23.
1.2.9. a) mn−1 + pq −1 = mqq −1 n−1 + pq −1 nn−1 = (mq + pn)n−1 q −1 . But n−1 q −1 nq = 1 and uniqueness
of multiplicative inverses implies (nq)−1 = n−1 q −1 . Therefore, mn−1 + pq −1 = (mq + pn)(nq)−1 . Similarly,
mn−1 (pq −1 ) = mpn−1 q −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 6= 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−1 qn < pq −1 nq = np.
1.2.10. 0 ≤ (cb − ad)2 = c2 b2 − 2abcd + a2 d2 implies 2abcd ≤ c2 b2 + a2 d2 . Adding a2 b2 + c2 d2 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 6= ∅, 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.
2
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, d) False. Let A = B = [0, 1]. Then A − B = [−1, 1] so sup(A − B) = 1 6= 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 > x2
2 2
implies x < 3/2, inf E = 0, sup E = 3/2. c) Since p /q < 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 − rn | < 1/n.
√ √ √ √ √
√− 2 < b − 2. Choose r ∈ Q such
1.3.3. a < b implies a √ 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.
e ≤ m. If m and m
1.3.4. If m is a lower bound of E then so is any m e are both infima of E then m ≤ m
e and
e ≤ m, i.e., m = m.
m e
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 E, then
there is an x ∈ E such that 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 : 2n b ≤ 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 < 2n b, i.e., q < b.
On the other hand, m0 ∈ E implies 2n b ≤ m0 , so
m0 1 m0 − 1
a = b − (b − a) < n
− n = = q.
2 2 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 s1 ≥ s2 ≥ . . . .
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.
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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−k bk is the
product of two negative numbers, hence positive. If they’re both even, then an−k bk is the product of two positive
numbers, hence positive. Thus by the Binomial Formula,
n µ ¶
X n µ ¶
X
n n−k k n n−k k
(a + b)n = a b = an + nan−1 b + a b =: an + nan−1 b + C.
k k
k=0 k=2
Since C is a sum of positive numbers, the promised inequality follows at once.
d) True. By the Binomial Formula,
µ ¶n n µ ¶
X n µ ¶
X
1 1 a−2 n 1 (a − 2)n−k n (a − 2)n−k
= + = = .
2n a 2a k ak 2n−k an−k k an 2n−k
k=0 k=0
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.
Pn ¡ ¢ Pn ¡ ¢
1.4.2. a) 0 = (1 − 1)n = k=0 nk 1n−k (−1)k = k=0 nk (−1)k .
b) (a + b)n = an + nan−1 b + · · · + bn ≥ an + nan−1 b.
c) By b), (1 + 1/n)n ≥ 1n + n1n−1 (1/n) = 2.
Pn ¡ ¢ Pn ¡ ¢ Pn−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 ≤ 2n + n < 2n + 2n = 2n+1 .
c) Now n2 ≤ 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 < 2n + 2n = 2n+1 < 2n+1 + 1.
d) We claim that 3n2 + 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 n3 ≤ 3n holds by inspection for n = 1, 2, 3. Suppose it holds for some n ≥ 3. Then
(n + 1)3 = n3 + 3n2 + 3n + 1 ≤ 3n + 2 · 3n = 3n+1 .
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Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 1.4.4. a) The formula holds for n = 1. If it holds for n then
n+1
X n(n + 1) n (n + 1)(n + 2)
k= + n + 1 = (n + 1)( + 1) = .
2 2 2
k=1
b) The formula holds for n = 1. If it holds for n then
n+1
X n(n + 1)(2n + 1) n+1 (n + 1)(n + 2)(2n + 3)
k2 = + (n + 1)2 = (n(2n + 1) + 6(n + 1)) = .
6 6 6
k=1
c) The formula holds for n = 1. If it holds for n then
n+1
X a−1 1 a−1 1
= 1 − n + n+1 = 1 − n+1 .
ak a a a
k=1
d) The formula holds for n = 1. If it holds for n then
n+1
X n(4n2 − 1) 2n + 1
(2k − 1)2 = + (2n + 1)2 = (2n2 + 5n + 3)
3 3
k=1
2n + 1 (n + 1)(4n2 + 8n + 3)
= (2n + 3)(n + 1) =
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 .
√ √
By convention, n 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
obtain a = ( a) ≥ ( b) = b, a contradiction.
n
1.4.6. The result is true for n = 1. Suppose it’s true for some odd number ≥ 1, i.e., 22n−1 + 32n−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,
n µ ¶
X µ ¶
n n n n n(n − 1)(n − 2)
2 = (1 + 1) = > = .
k 3 6
k=0
√
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
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, √ √ √ √ √ √
b) If n + 3 + n ∈ Q then √ 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 n2 + 3n is a perfect
square when n = 1 but if n > 1 then
(n + 1)2 = n2 + 2n + 1 < n2 + 2n + n = n22 + 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 < n22 + 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 and a20 + b20 = c20 . Suppose that cn−1 − bn−1 = 1 and
a2n−1 + b2n−1= c2n−1 hold for some n ≥ 0. By definition, cn − bn = cn−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,
a2n + b2n = (an−1 + 2)2 + (2an−1 + bn−1 + 2)2
= a2n−1 + 4an−1 + 4 + (2an−1 + 2)2 + 2bn−1 (2an−1 + 2) + b2n−1
= c2n−1 + 2(an−1 + 2) + (2an−1 + 2)2 + 2(cn−1 − 1)(2an−1 + 2)
= c2n−1 + (2an−1 + 2)2 + 2cn−1 (2an−1 + 2)
= (2an−1 + cn−1 + 2)2 ≡ c2n .
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) 6= ∅.
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,
−1
f (x) = 1/ log x. By looking at the graph, we see that f (E) = (1, ∞).
γ) f is 1–1 on (π/2, 3π/2) because f ′ (x) = sec2 x > 0 there. The inverse is f −1 (x) = arctan(x − π). By looking
at the graph, we see that f (E) = (−∞, ∞).
= 2x + 2 < 0 for x < −6, f is 1–1 on [−∞, −6]. Since y = x2 + 2x − 5 is a quadratic in x, we
δ) Since f ′ (x) p
√
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) = −1 − 6 + x. By looking at the graph, we see that f (E) = [19, ∞).
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
f −1 (x) = x−2 2<x≤4
(x + 2)/3 x > 4,
6
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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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
,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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 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 Property,
√ 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 = b2 (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 only if 3x −1 > 0 and 2x3 −3x +1 > 0. The √ first
inequality is equivalent to x > 1/3. Since 2x3 −
√ 3x + 1 = (x − 1)(2x√
2
+ 2x − 1) implies that x = 1, (−1 ± 3)/2,
the second
√ 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 √ by (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 and b) ≤ A(a, b). On
√ G(a,√
the other hand, since 0 ≤ a ≤ b √ we have A(a, b) = (a + b)/2 ≤ 2b/2
√ 2= b and G(a, b) = ab ≥ √a2 = a. Finally,
√ √
A(a, b) = G(a, b) if and only if 2 ab = a + b if and only if ( a − b) = 0 if and only if a = b if and only if
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. Thus n > 499 ≈ 22.33, i.e., n ≥ 23.
1.2.9. a) mn−1 + pq −1 = mqq −1 n−1 + pq −1 nn−1 = (mq + pn)n−1 q −1 . But n−1 q −1 nq = 1 and uniqueness
of multiplicative inverses implies (nq)−1 = n−1 q −1 . Therefore, mn−1 + pq −1 = (mq + pn)(nq)−1 . Similarly,
mn−1 (pq −1 ) = mpn−1 q −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 6= 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−1 qn < pq −1 nq = np.
1.2.10. 0 ≤ (cb − ad)2 = c2 b2 − 2abcd + a2 d2 implies 2abcd ≤ c2 b2 + a2 d2 . Adding a2 b2 + c2 d2 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 6= ∅, 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.
2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, d) False. Let A = B = [0, 1]. Then A − B = [−1, 1] so sup(A − B) = 1 6= 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 > x2
2 2
implies x < 3/2, inf E = 0, sup E = 3/2. c) Since p /q < 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 − rn | < 1/n.
√ √ √ √ √
√− 2 < b − 2. Choose r ∈ Q such
1.3.3. a < b implies a √ 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.
e ≤ m. If m and m
1.3.4. If m is a lower bound of E then so is any m e are both infima of E then m ≤ m
e and
e ≤ m, i.e., m = m.
m e
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 E, then
there is an x ∈ E such that 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 : 2n b ≤ 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 < 2n b, i.e., q < b.
On the other hand, m0 ∈ E implies 2n b ≤ m0 , so
m0 1 m0 − 1
a = b − (b − a) < n
− n = = q.
2 2 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 s1 ≥ s2 ≥ . . . .
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
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,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−k bk is the
product of two negative numbers, hence positive. If they’re both even, then an−k bk is the product of two positive
numbers, hence positive. Thus by the Binomial Formula,
n µ ¶
X n µ ¶
X
n n−k k n n−k k
(a + b)n = a b = an + nan−1 b + a b =: an + nan−1 b + C.
k k
k=0 k=2
Since C is a sum of positive numbers, the promised inequality follows at once.
d) True. By the Binomial Formula,
µ ¶n n µ ¶
X n µ ¶
X
1 1 a−2 n 1 (a − 2)n−k n (a − 2)n−k
= + = = .
2n a 2a k ak 2n−k an−k k an 2n−k
k=0 k=0
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.
Pn ¡ ¢ Pn ¡ ¢
1.4.2. a) 0 = (1 − 1)n = k=0 nk 1n−k (−1)k = k=0 nk (−1)k .
b) (a + b)n = an + nan−1 b + · · · + bn ≥ an + nan−1 b.
c) By b), (1 + 1/n)n ≥ 1n + n1n−1 (1/n) = 2.
Pn ¡ ¢ Pn ¡ ¢ Pn−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 ≤ 2n + n < 2n + 2n = 2n+1 .
c) Now n2 ≤ 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 < 2n + 2n = 2n+1 < 2n+1 + 1.
d) We claim that 3n2 + 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 n3 ≤ 3n holds by inspection for n = 1, 2, 3. Suppose it holds for some n ≥ 3. Then
(n + 1)3 = n3 + 3n2 + 3n + 1 ≤ 3n + 2 · 3n = 3n+1 .
4
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
, 1.4.4. a) The formula holds for n = 1. If it holds for n then
n+1
X n(n + 1) n (n + 1)(n + 2)
k= + n + 1 = (n + 1)( + 1) = .
2 2 2
k=1
b) The formula holds for n = 1. If it holds for n then
n+1
X n(n + 1)(2n + 1) n+1 (n + 1)(n + 2)(2n + 3)
k2 = + (n + 1)2 = (n(2n + 1) + 6(n + 1)) = .
6 6 6
k=1
c) The formula holds for n = 1. If it holds for n then
n+1
X a−1 1 a−1 1
= 1 − n + n+1 = 1 − n+1 .
ak a a a
k=1
d) The formula holds for n = 1. If it holds for n then
n+1
X n(4n2 − 1) 2n + 1
(2k − 1)2 = + (2n + 1)2 = (2n2 + 5n + 3)
3 3
k=1
2n + 1 (n + 1)(4n2 + 8n + 3)
= (2n + 3)(n + 1) =
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 .
√ √
By convention, n 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
obtain a = ( a) ≥ ( b) = b, a contradiction.
n
1.4.6. The result is true for n = 1. Suppose it’s true for some odd number ≥ 1, i.e., 22n−1 + 32n−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,
n µ ¶
X µ ¶
n n n n n(n − 1)(n − 2)
2 = (1 + 1) = > = .
k 3 6
k=0
√
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
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, √ √ √ √ √ √
b) If n + 3 + n ∈ Q then √ 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 n2 + 3n is a perfect
square when n = 1 but if n > 1 then
(n + 1)2 = n2 + 2n + 1 < n2 + 2n + n = n22 + 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 < n22 + 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 and a20 + b20 = c20 . Suppose that cn−1 − bn−1 = 1 and
a2n−1 + b2n−1= c2n−1 hold for some n ≥ 0. By definition, cn − bn = cn−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,
a2n + b2n = (an−1 + 2)2 + (2an−1 + bn−1 + 2)2
= a2n−1 + 4an−1 + 4 + (2an−1 + 2)2 + 2bn−1 (2an−1 + 2) + b2n−1
= c2n−1 + 2(an−1 + 2) + (2an−1 + 2)2 + 2(cn−1 − 1)(2an−1 + 2)
= c2n−1 + (2an−1 + 2)2 + 2cn−1 (2an−1 + 2)
= (2an−1 + cn−1 + 2)2 ≡ c2n .
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) 6= ∅.
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,
−1
f (x) = 1/ log x. By looking at the graph, we see that f (E) = (1, ∞).
γ) f is 1–1 on (π/2, 3π/2) because f ′ (x) = sec2 x > 0 there. The inverse is f −1 (x) = arctan(x − π). By looking
at the graph, we see that f (E) = (−∞, ∞).
= 2x + 2 < 0 for x < −6, f is 1–1 on [−∞, −6]. Since y = x2 + 2x − 5 is a quadratic in x, we
δ) Since f ′ (x) p
√
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) = −1 − 6 + x. By looking at the graph, we see that f (E) = [19, ∞).
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
f −1 (x) = x−2 2<x≤4
(x + 2)/3 x > 4,
6
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