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Unformatted text preview: Instructor’s Solutions Manual
to Accompany Introduction to Analysis, 39
By
William Wade
@2004 SOLUTIONS TO EXERCISES CHAPTER 1
1.1 Ordered ﬁeld axioms.
1. a) By deﬁnition, +_ _=a+a_ a.—~a _2_(i_
a a 2 (2 ‘2‘“ and , ,—, 2,
a++a—= lal2+a +<lal2 0) =%.l=lal. b) By Deﬁnition 1.1, ifa 2 0 then (1+ = ((z+a)/2 = a. and ifa < 0 then (1+ = (—a.+a)/2 = 0. Similarly, a‘ = 0
ifaZOanda‘=—aifa.<0. 2. a) [ac—2 <5 ifandonlyif—5<$—2<5 ifandonlyif—3<.r<7. b) 1—.r <4 ifandonlyif—4< 1—rc<4 ifandonlyif—5< —m<31fandonlyif—3<rc<5. c) lmz—m—ll < m2 ifand only if—an2 <rL'2—z— 1 < .732 ifand only ifrc+1 > Oand 2xz—m—1>0. The
ﬁrst inequality is equivalent to .r > —1. Since 2.12 — :1: — 1 has roots 1, —1/2, the second inequality is equivalent to
m > 1 or is < —1/2. Therefore, the solution is (—1, —1/2) U (1,00). d) [$2 +11 < 2 ifand only if —2 < .172 +.r < 2 if and only ifJ:2 +ar+2 > 0 and $2 +rc—2 < 0. The ﬁrst
inequality always holds Since .12 + rt: + 2 has no real roots. Since :52 + :L' — 2 has roots 1, —2, the second inequality
is equivalent to —2 < .7: < 1. 3. a) If a < b then a + c < b + c by the Additive Property. If a = b then a + c = b + 0 since + is a function.
Thus a + c S b + 0 holds for all a S b. b) If c = 0 then ac = O = be so we may suppose c > 0. If a < b then ac < be by the Multiplicative Property. If
a = b then ac = be since  is a function. Thus ac S be holds for all a. S b and c 2 0. 4. a) Suppose 0 S a < b and 0 S c < d. Multiplying the ﬁrst inequality by c and the second by b, we have
0 S ac S be and be < bd. Hence by the Transitive Property, ac < bd. b) Suppose 0 S a < b. By (7), 0 S a2 < b2. If.\/5 2 \/b then a. = 2 = b, a contradiction. c) If l/a S l/b, then the Multiplicative Property implies b = ab(1/a) S ab(1/b) = a, a contradiction. If 1/b S 0
then b = b2(1/b) S 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 l/a = ——1/2 is not less than
l/b = —1. 5. a) Suppose 0 < a <1. Then 0 > —a > —1,so 0 <1—a <1. Hence V1 — a is real and by (6), l—a < \/1 — (1.
Therefore, b = 1 —‘ V1 — a < 1 — (1 — a.) = a. b) Suppose a > 2. Then a—l > 1 so 1 < x/a. — 1 < a—l by Therefore, 2 < b = 1+\/a — 1 < 1+(a—1): a. c) a. + b — wa— = (f— «(3)2 2 0 for all a,b e [0,oo). Thus 2% g a + b and C(a, b) g A(a,b). On the
other hand, since 0 S a S b we have A(a, b) = (a + b)/2 S 2b/2 = b and C(a, b) = x/E 2 x/a.—2 = a, Finally,
A(a,b) = G(a.,b) if and only if 2% = a + b if and only if — x/b)2 = 0 if and only if \/a_ = \/b if and only if
a. = b. 6. a) 77171—1 +pq”1 = 71qu‘1n_1 +pq‘1nn‘1 = (mg +pn)n_1q‘1. But n‘lq‘lnq = 1 and uniqueness of multiplicative inverses implies (7.111)‘1 = n—lrfl. Therefore, mn‘1 +19q‘1 = (mg +pn)(nq)_1. Similarly, vim—1(pq‘1) = 7npn‘1q“1 = 7n.p(nq)_1. By what we just proved and (2),
m —m _ m — m __ 0 _ 0
n n T n _ n _ ' Therefore, by the uniqueness of additive inverses, —(7n./n) = (—m)/n. Similarly, (711/17,)(71/771.) = (771.n)/(mn) =
mn(mn)‘1 = 1, so ("L/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 E Q has a multiplicative inverse. Therefore, Q
satisﬁes Postulate 1. c) If 7' E Q, .17 E R\ Q but q := 1‘ + .7: E Q, then :5 = q — r E Q, a contradiction. Similarly, if rm 6 Q and 7' 75 0,
then :5 E Q, a contradiction. However, the product of any irrational with 0 is a rational. d) By the First Multiplicative Property, mn‘l < pq”1 if and only if mq = mn‘lqn < pq‘1nq = np. 7. S 1 implies 0 S .1:+1 S 2. Thus [3:2 — 1] = m+1 lm— 1 5 2.r— 1]. b) —15m$ 2implies 1 git+2 34. Thus m2+m—2 = II— 1I+2I S4lm—1l. c) S 1 implies —3 S m — 2 S —1. Thus [.722 — m — 2 = + 1 — 2] S 3rc + 1]. (1)].r— ll < 1 implies 0 < .77 < 2 implies $2 +rc+2 < 8. Thus rc3 +17 — 2] = lz—1H$2 +rc+2 < 8l.1;— 1].. 8. a) Since (1 —n)/(1—n2) = 1/(1 +17), the inequality is equivalent to 1/(n+1) < .01 = 1/100. Since 1+n > 0
for all n 6 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., 271+ 1 > 40. Thus n > 39/2,
i.e., n 2 20. ' c) The inequality is equivalent to n2 + 1 > 500. Thus 71. > W R: 22.33, i.e., n _>_ 23. 9. 0 S (azbl — a1b2)2 = ugh? — 2a1b1a2b2 + agbg implies 2alb1a2b2 S agbf + afbg. Adding afbf + a.ng to both
sides, we conclude that (albl + (12112)2 S (a? + ag)(b¥ + 10. a) [my— abl = [fry—mb—l—xb— ab g Iy— b + — 0.] lb] < [TIEl Ible. Moreover, g Irc— a] + Ial < c + a.
Thus — ab] < (e + lal)6 + be = + b)e + 62. a b) Similarly, rc2y — a2bl = x2y — (123/ + a2y — azb] 5 lg! — a + a. + a2y — b < (6 + Ibl)€(6 + 2]a) + lalze =
€(a2 + 2labl) + E2(lbl + 2M) + 63 11. a) Let :5 E R. By the Trichotomy Property, either :0 > 0, ~rc > 0, or .7: = 0. Thus P satisﬁes i). If x >_ 0
and y > 0, then by the Additive Property, .7: + y > 0 and by the First Multiplicative Property, my > 0. ThusP
satisﬁes ii). b) To prove the Trichotomy Property, suppose a,b E R. By i), either a — b E P, b — a = —(a — b) E P, er
a—b=0. Thus eithera>b,b>a,ora=b. ' To prove the Transitive Property, suppose a < b and b < 0. Then b — a,c — b E P and it follows from ii) that
c—a=b—a+c—bEP,i.e.,c>a. '5‘ Since b — a. = (b+ c) — (a + c), it is clear that the Additive Property holds. , 3 Finally, suppose a < b, i.e., b — a E P. If c > 0 then c E P and it follows from ii) that be — ac = (b — a.)c ESP,
i.e., bc > ac. If e < 0 then —c E P, so ac — bc 2 (b — a)(—c) E P, i.e., ac > be. " 1.2 The WellOrdering Principle. ,
1. a) The formula holds for n = 1. If it holds for n then _ " n+1 2 Zk=M+n+1=(n+1)(ﬂ+1)=W. 7,3
2 2 2 :12:
16:1 '
b) The formula holds for n = 1. If it holds for n. then
n+1 y. 7
, . ,  1 2 2 3
Z“ = ———"("+1)6(2"+1) +(n+1)2 = "31(n(2n+1)+6(nx+1)) = ———~("+ X"; X ”+ ).
k=l .
c) The formula holds for n = 1. If it holds for n then ‘
n+1 _ 1...
a a" a" a"
k=1
(1) The formula holds for n. = 1. If it holds for n. then
n+1 2 _ 2 1
2(2): — 1)2 = "—(4—"3—1) + (271+ 1)2 = (2n2 +5n+3)
k=1
1 4 2 8 3
= 2";1(2n+3)(n+1) : w (n+1)(4(n + 1)2 — 1)
*3 . 2. a) 2n = (1+1)" = 2:001) 1))(0. + b)" = a." + na,"‘1b +   ~ + b” 2 a” + nan—1b.
C.) By b), (1 +1/n)" Z 1" + n1"‘1(1/n) = 2.
3. By the Binomial Theorem, 7]. (.7: + h)" — 9:" 1 i _k k k. ,
——~—. :  {L'n = Z .Tn— fill—1
h h. k=1 It [(1 [6:1 4. a) By hypothesis, 0 < .771 < 1. Suppose 0 < 70,, < 1. Then by Exercise 5a in 1.1, 0 < $714.1 < mn < 1. Thus
by induction this inequality holds for all n E N. b) By hypothesis, m1 2 2. Suppose 1‘" 2 2. If .717, = 2 then rck = 2 for all k 2 72.. If .12" > 2 then by Exercise 5b
in 1.1, 2 < mu“ < in". Thus by induction, 2 S 1771+] S It?” 5 $1 for all n E N. 5. Suppose 0 < In < 2. Since 1:2 — rc — 2 < 0 for 0 < .7: < 2, we have by (8) that 0 < 1:" < 2 + 17,, = rcn+1 <
V2 + 2 = 2. Thus by induction, 0 < .7?” < mn+1 < 2 for all n E N. 6. a) The inequality holds for n = 1. If it holds for n then n+1<2"+1s2"+n<2"+2"=2"+1. b) We ﬁrst prove by induction that 27?. + 1 < 2" for n = 3,4,. . . . This inequality holds for n = 3. If it holds for
some n 2 3 then 2(n+1)+1=2n+1+2<2"+2<2"+2"=2"+1. Now 17,2 S 2" + 1 holds for n = 1, 2, and 3. If it holds for some n 2 3 then
(n+1)2 =n2+2n+1 <2"+2" =2"+1 < 2n+1+1. c) We claim that 311.2 + 3n + 1 S 2  3" 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+652~3"+6(n+1). Now 6(n+1) S 43" for n 21. (It holds for n = 1, and if it holds for n then 6(n+2) = 6(n+1)+6 g 43" +6 <
4  3" + 8 ' 3" = 4  3"+1.) Therefore, 3(n+1)2+3(n+1)+1323"+6(n+1)s2~3"+43”=23"+1. Thus the claim holds for all n 2 3.
Now 113 S 3" holds by inspection for n = 1, 2,3. Suppose it holds for some n 2 3. Then (n+1)3=n3+3n2+3n+1S3"+2~3"=3"+1. 7. 0 S a" < b" holds for n = 1. If it holds for n then by (7), 0 S a"+1 < bn‘H. By convention, {/5 2 0. If < {/13 is false, then {/5 Z {/17 2 0. Taking the nth power of this inequality, we
obtain a = “75)” 2 = b, a contradiction. 8. a) If \/n+ 3 + ﬂ 6 Q then 71. + 3 + 2y/n +3\/7—1 + n = (y/n + 3 + ﬂy E Q. Since Q is closed under
subtraction and division, it follows that \/n2 +377. E Q. In particular, 712 + 3n = m2 for some m. E N. Now
11.2 + 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.
b) By repeating the steps in a), we see that the original expression is rational if and only if n(n+7) = n2+7n = m2
for some m E N. If n > 9 then (n+3)2=n2+Gn.+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 71.2 + 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. 9. The result is true for n = 1. Suppose it’s true for some odd number 2 1, i.e., 22’“1 + 32".1 = 58 for some
3,17. 6 N. Then
22n+1 + 321141 = 4 ' 2211—1 + 9 ' 32n—1 : 4 _ + 5 _ 32n—1 is evidently divisible by 5. Thus the result is true by induction. 10. The result holds for n = 0 since c0 — ()0 = l and ag + ()3 = 03. Suppose that (:n_1 — bn_1 = 1 and
(13,4 +b2_l = c3,_1 hold for some n 2 0. By deﬁnition, cn — bn = cn_1 — b,,_1 = 1, so by induction, this difference
is always 1. Moreover, by the Binomial Formula, the inductive hypothesis, and what we just proved, 0,2, + b3, = (an_1+ 2)2 +(2an_1+ I;,,_1 + 2)2
= a3,_1+ 4a,,_1 + 4 + (2a,,_1 + 2)2 + 2b,,_1(2a,,_1 + 2) + b3,_,
= ci_1 + 2(an_1 + 2) + (2an_1 + 2)2 + 2(cn_1 — 1)(20.n_1 + 2)
= c3,_, + (2a,,_1 + 2)2 + 2cn_1(2an_1 + 2)
= (2a,,_1+ cn_1 + 2)2 E 03,. 1.3 The Completeness Axiom. 1. a) infE = 1, supE = 8. b) infE = (3 —— \/2_9)/2, supE = (3 + c) infE = a, supE = b. (l) p/q E E
if and only if 0 < p/q < Thus infE = 0, supE = 0) Since E = {0,2}, infE = 0 and supE = 2. f)
Since 1/n — (—1)" = 1/n — 1 when n is even and 1/n + 1 when n is odd, infE = —1 and supE = 2. g) Since
1+ (—1)"/n =1— l/n when n is odd and 1+1/n when n is even, infE = 0 and supE = 3/2. 2. By Theorem 1.28, it sufﬁces to prove the result for sup E. First, we show that supE must be an integer.
Indeed, if k: < supE < k +1 for some k E Z then by the Approximation Property there is an n E E C Z such that
k < n < k + 1. In particular, 0 < n — k < 1, i.e., by Remark 1.1, n — [C does not belong to Z. This contradicts the
fact that Z is closed under subtraction. Next, we suppose to the contrary that n := supE ¢ E. Then k < n for all k E E. But by the Approximation
Property, there is a k E E C Z such that k > n — 1. Therefore, 71 — 1 < k < n, i.e., 0 < n — k < 1. In particular,
we obtain the contradiction that n — k cannot be an integer. 3. a < 1) implies a— x/2 < b— Choose 7* E Q such that a— \/2< r < b— Then a < r+\/2 <1). By
Exercise (5c in 1.1, r 4— \/2 is irrational. Thus set 8' = 7‘ + 4. Since a, — 1/n < a + l/n, choose rn E Q such that a — 1/n < r" < a +1/n, i.e., la — rnf<1/n. 5. a) Let 6 > 0 and m = inf E. Since 7". + 6 is not a lower bound of E, there is an a E E such that m. + F > 0..
Thus 77'). + e > a 2 m as required. b) By Theorem 1.20, there is an a E E such that sup(—E) — e < —a S sup(—E). Hence by the Second
Multiplicative Property and Theorem 1.28, infE + e = —(sup(—E) — 6) > a > — sup(—E) = inf E. 6. a) If m is a lower bound of E then so is any 771. S m. .If m. and 777. are both inﬁma of E then m _<_ 771, and
771, S m, i.e., 7n. = 777.. b) If E is nonempty and bounded below, then —E is nonempty and bounded above. Hence by the Completeness
Axiom and Theorem 1.28, infE = —sup(—E) exists and is ﬁnite. 7. a) Let .T be an upper bound of E and .1‘ E E. If Ill is any upper bound of B then AI _>_ .z‘. Hence by deﬁnition,
.1: is the suprernum of E. b) The correct statement is: If as is a lower bound of E and a: E E then m = inf E. PROOF. —$ is an upper bound of —E and —.T E —E so —a: = sup(—E). Thus a: = —sup(—E) = inf E. e) If E is the set of points In such that .77., = 1 — 1/n for odd n and :1?7, = 1/n for even n, then supE = 1,
infE = 0, but neither 0 nor 1 belong to E. 8. Since [377,] S Al, the set En = {mn,mn+1,...} is bounded for each n E N. Thus 5,, := sup En exists and
is finite by the Completeness Axiom. Moreover, by the Monotone Property, 5,, 2 sn+1 for each n E N. Thus
81 2 $2 2 . . . . Similarly, 0,, := sup{—.Tn, —.’l‘n+1, . . satisﬁes 01 2 02 Z Since tn = —(7,, for n E N, it follows from the
Second Multiplicative Property that t1 3 t2 3 9. By induction, 10" > n. Hence by the Archimedean Principle, there is an n E N such that 10" > 1/(b —— a).
Let E := E N : 10"b S By the Archimedean Principle, E is nonempty. Hence let me be the least element
in E and set q = (mo — 1)/10”. Since b > 0, mo 2 1. Since mo is least in E, it follows that me — 1 < 10"b, i.e.,
q < b. On the other hand, mo 6 E implies 10”b 3 me, so mo 1 7710—1
,=b— 11—, ————=
a ( “<10” 10" 10” =q' 4 10. Since A Q 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,supB S sup E. Set A! :2 max{supA,sup B} and observe that Ill is an upper bound of both A and B. If [VI < sup E, then there is an .r E E such that Al < 1' S sup E. But at E E implies .1: E A or m E B. Thus Ill is not an upper bound
for one of the sets A or B, a contradiction. 1.4 Functions, Countability, and the Algebra of Sets. 1. a) f is 1~1 since f’(.r) = 3 75 0 for :5 E E. If y = 3x — 7 then .1': = (y + 7)/3. Therefore f—1(rr) = (1' + b) f is 1—1 since = —el/I/z2 7E 0 for :5 E E. If y = 81/1 then logy = l/m, i.e., .1: = 1/logy. Therefore,
.rle) = l/logz. c) f is 1~1 on (—7r/2,7r/2) because f’(.r) 2 sec2 :5 is nonzero there. The inverse is f‘1(.1t) = arctan (1) Since f’(rr) = 2.77 + 3 aé 0 for .1: > —3/2, f is 1~1 on [—3/2,oo). Since y = m2 + 3. — 6 is a quadratic in m,
we have m = (—3 :t But .7: is eventually positive on {—3/2, 00), so we must use the positive sign.
Hence f_1(.T) = (—3 + \/33 + 4m)/2. C) By definition, 3m+2 x30
f(m)= 1+2 0<mS2
3m—2 .r>2. Thus f is strictly increasing, hence 1—1, and (m—2)/3 $52
f‘1(m)= rc—2 2<x§4
(.7:+2)/3 :I:>4. f) Since f’(.77) = (1 — + 1)2 is never zero on (—1, 1), f is 1»1 on [—l,1]. By the quadratic formula,
y = implies m = (1 i V1 ~ 4y2)/2y. Since 2: E [——1, 1] we must take the minus sign. Hence (1—v1—4rc2)/2:r maéO
0 —1
f { .1: = 0. 2. By deﬁnition, there is an n E N and a 1—1 function 45 which takes Z := {1, 2, . . . ,n} onto A. Let 1/)(.r) z: for IL‘ 6 Z. Since f and ¢ are 1—1, 1/)(rc) = 1/1(y) implies = 43(3)) implies .7: = y. Moreover, since f
and <15 are onto, given I) E B there is an a E A such that f(a) = b, and an is E Z such that = (1, hence
2/)(m) E = f(a.) = b. Thus 1/1 is 1—1 from Z onto B. By deﬁnition, then, B is ﬁnite. 3. Let = 2n —— 1. Then f is 1~1 from N onto the set of odd integers. 4. a) f decreases and f(—3) ‘= 16, f(1) = —4. Therefore, f(E) = (—4, 16). Since = —3 implies m = 4/5
and = 1 implies a: = 0, we also have f“1(E) = (0, 4/5). b) The graph of f is a parabola whose absolute minimum is 0 at x = 0 and whose maximum on [—1,4] is 16 at
.1: = 4. Therefore, = [0,16]. Since f takes £2 to 4 and 0 to 0, f"1(E) = [—2, 2]. c) The graph of f is a parabola whose absolute minimum is —1/4 at .’L‘ = —1/2. Since f(—2) = 2 and
f(1) = 2, it follows that f(E) = {—1/4, 2]. Since .12 +m = 1 implies m = (—1:l: x/5)/2, we also have f“1(E) =
((—l — V5)/2),(1+ x/EVQl d) The graph of m2 + a: + 1 is a parabola whose minimum is 3/4 at .1: = —1/2. Since log increases on (0, oo),
f(1/2) = log(7/4), and f(5) = log(31), it follows that = (log(7/4),log(31)]. Since 1/2 = log(.’c2 + I + 1)
implies .7: = (—1i\/4\/e— and 5 = log(.’c2 + .z‘ + 1) implies .T. = (—1 :t V465 —— 3)/2, we also have f“1(E)=[—1—\/4e5 — 3)/2, (—1 _ 4,/5— 3)/2) o ((—1 + 4,/é— 3)/2, —1 + «435 — 3)/2]. 0) Since sinm is periodic with maximum 1 and minimum —1, f = [—1,1]. Since sinm is nonnegative when
2k7r S rc 3 (21: + 1)7r for some'k E Z, it follows that f'1(E) = U [21m (2k + 1m. keZ [ 5. The minimum of .7: — 1 on [0, 1] is —1 and the maximum of .77 + 1 on [0,1] is 2. Thus Umelmﬂm — 1, re +1] =
—l, 2 . b) The maximum ofm— 1 on [0, 1] is 0 and the minimum ofm+1 on [0, 1] is 1. Thus ﬂxqmﬂm — 1,1+ 1] = [0, 1]. c) The maximum of 1/k for k E N is 1. Thus UkeN[0, = [0,1]. (1) The minimum of 1/k for k E N is 0 and 0 E [0, for all k E N. Thus ﬂk€N[0, 1/k] = 6. a) implies b). By deﬁnition, f(A \ B) 2 f(A) \ f(B) holds whether f is 1'1 or not. To prove the reverse
inequality, suppose f is 1—1 and y E f(A\B). Then 3/ = f(a) for some a E A \ B. Since f is 1—1, a. = f‘1({y}).
Thus y 76 f(b) for any b E B. In particular, 1} E f(A) \ f(B). b) implies c). By deﬁnition, A Q f‘1(f(A)) holds whether f is 171 or not. Conversely, suppose m E f'1(f(A)).
Then E f(A) so f(m) = f(a) for some a E A. If :5 ¢ A, then it follows from b) that f(A) = f(A \ =
f(A) \ i.e., ¢ f(A), a contradiction. (3) implies d). By Theorem 1.43, f(A I") B) Q f(A) ﬂ Conversely, suppose y E f(A) ﬂ Then
1; = f(a.) = for some a E A and b E B. If y ¢ f(A n B) then (1. ¢ B and b ¢ A. Consequently,
f‘1(f({a})) 2 {(1,1)} 3 {a}, which contradicts c). (1) implies a). If f is not 1~1 then there exist (1,1) 6 X such that a. ;L b and y := f(u.) = f(b). Hence by d),
{y} = f({a.}) n = 0, a contradiction. 7. Suppose m belongs to the left side of (16), i.e., .1: E X and .77 ¢ ﬂaeAEa. By definition, .r E X and .1: ¢ Ea for
some 01 E A. Therefore, a: E E; for some a e A, i.e., :5 belongs to the right side of (16). These steps are reversible. 8. By deﬁnition, .7: E f“1(UQeAEQ) if and only if 6 EC, for some a E A if and only if ."c E erAf_1(Ea). b) By definition, :5 E f_1(r‘lae,4Ea) if and only if E E, for all a E A if and only if .1: E ﬂaeAf‘1(Ea). c) To Show f(f—1(E)) = E, let .1: E E. Since E Q f(X), choose a. E X such that x = f(a.). By deﬁnition,
a E f_1(E) so .1: E f(a.) E f(f_1(E)). Conversely, if .1: E f(f—1(E)), then m = f(a) for some a. E f_1(E). By
deﬁnition, this means .7: = f(a) and f(a) E E. In particular, .1: E E. To show E Q f‘1(f(E)), let in e E. Then E f(E), so by deﬁnition, in E f‘1(f(E)). 9. a) This was done in Exercise 2 above. 1)) By Deﬁnition 1.42, it is clear that f takes A onto f(A). Suppose f_1(.r) = f‘1(y) for some 13,7; 6 f(A).
Since f is 171 from A onto f(A), it follows from Theorem 1.31 that .7: = f(f‘1(rc)) = f(f_1(y)) = y. Thus f‘1 is
1~1 on f(A). . c) If f is 1~1 (respectively, onto), then it follows from part a) that g o f is 171 (respectively, onto). Conversely, if gof is 1‘1 (respectively, onto), then by pats a) and b), f E 9‘1 ogof is 1— 1 (respectively, onto). 10. a) We prove this result by induction on 71.. Suppose n = 1. Since d) : {1} —> {1}, it must satisfy (15(1) = 1. In particular, in this case (15 is both 1771 and onto
and there is nothing to prove. Suppose that the result holds for some integer n 2 1 and let ¢ : {1,2, . . . ,n +1} —> {1,2,...,n + 1}. Set
[:0 = d>(n + 1) and define 1/1 by
Z Z < kc W): { e—i €>ko. The 2/) is 1‘1 from {1,2,...,ko — Lko +1,...,n+ 1} onto {1,2,...,n}. Suppose 45 is 1'1 on {1,2,. . . ,7; +1}. Then 4) is 1~1 on {1,2, . . . ,n}, hence 1/1 o ()5 is 1‘1 from {1,2,...,n} into
{1, 2, . . . , n}. It follows from the inductive hypothesis that 2/10 qS takes {1, 2, . . . ,77} onto {1, 2, . . . ,n}. By Exercise
9, (1) takes {1,2, . ..,n} onto {1,2, . . .,ko — 1,160 + 1,. . . ,n +1}. Since (25(n + 1) 2 k0, we conclude that (15 takes
{1,2,...,n+1} onto {1,2,...,n+1}. Conversely, if ¢ takes {1, 2, . . . , n+1} onto {1,2, . . .,n+1}, then ¢ takes {1, 2, . . .,n} onto {1, 2, . . .,k0 — 1, k0 +
1, . . . ,n + 1}, so 1/) o ([5 takes {1, 2,. . . , 77.} onto {1, 2, . . . , It follows from the inductive hypothesis that 1/1 0 d) is
1—1 on {1, 2,. . . ,n}. Hence by Exercise 9 and construction, d) is 171011 {1, 2, . . . ,n +1}. b) We may suppose that E is nonempty. Hence by hypothesis, there is an n E N and a 17~1 function 45 from E
onto {1,2, . . . , 71}. Moreover, by Exercise 9b), the function (f4 is 1—1 from {1, 2, . . .,n} onto E. Consider the function ¢‘1 0 f o ()5. Clearly, it takes {1, 2, . . . ,n} into {1, 2, . . . , 71}. Hence by part a), ()5‘1 o f o (f)
is 11 if and only if it is onto. In particular, it follows from Exercise 9c) that f is 1—1 if and only if f is onto. 11. a) Let q = If k = 0 then nq = 1 is a root of the polynomial m — 1. If k > 0 then 72" is a root of the
polynomial mj — n". If k < 0 then nq is a root of the polynomial n‘kmj —— 1. Thus 11" is algebraic. b) By Theorem 1.38, there are countany many polynomials with integer coefﬁcients. Each polynomial of degree
n has at most n roots. Hence the class of algebraic numbers of degree n is a countable union of ﬁnite sets, hence countable. c) Since any number is either algebraic or transcendental,
the set of transcendental numbers. By b) by the argument of Remark 1.39. R is the union of the set of algebraic numbers and
, the former set is countable. Therefore, the latter must be uncountable ...
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This note was uploaded on 05/05/2008 for the course MATH 415 taught by Professor Dont know during the Spring '08 term at Iowa State.
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