Solutions. Chapter A. A.1 Solutions to Chapter 1. Solutions to Exercises 1.1

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1 Chapter A Solutions A. Solutions to Chapter Solutions to Exercises... Suppose that f : A B is invertible. This means there is a g : B A with g(f(a)) a for all a and f(g(b)) b for all b B. If f(a) f(a ), then a g(f(a)) g(f(a )) a. Hence f is injective. If b B, then a g(b) satisfies f(a) b. Hence f is surjective. We conclude that f is bijective. Conversely, if f is bijective, for each b B, let g(b) denote the unique element a A satisfying f(a) b. Then by construction, f(g(b)) b. Moreover, because a is unique element of A mapping to f(a), we also have a g(f(a)). Thus g is the inverse of f. Hence f is invertible...2 Suppose that g : B A and g 2 : B A are inverses of f. Then f(g (b)) b f(g 2 (b)) for all b B. Because f is injective, this implies g (b) g 2 (b) for all b B, hence, g g For the first equation in De Morgan s law, choose a in the right side. Then a lies in every set A c, A F. Hence a does not lie in any of the sets A, A F. Hence a is not in {A : A F}, thus a is in the left side. If a is in the left side, then a is not in {A : A F}. Hence a is not in A for all A F. Hence a is in A c for all A F. Thus a is in {A c : A F}. This establishes the first equation. To obtain the second equation in De Morgan s law, replace F in the first equation by F {A c : A F}...4 {x} {t : t x} x. On the other hand, { x} is a singleton, so can equal x only when x is a singleton...5 If x b, there are two cases, x a and x a. In the first case, x a b, and in the second, x a b but not in a, so x a b a. O. Hijab, Introduction to Calculus and Classical Analysis, Undergraduate Texts in Mathematics, 267 DOI.7/ , Springer Science+Business Media, LLC 2

2 268 A Solutions..6 The elements of (a, b) are {a} and {a, b}, so their union is (a, b) {a, b}, and their intersection is (a, b) {a}. Hence (a, b) a b, (a, b) a b, (a, b) a and (a, b) a...7 If a S(x) x {x}, then either a x or a x. In the first case, a x, because x is hierarchical. In the second, a x x S(x). Thus S(x) is hierarchical...8 If a c and b d, then (a, b) {{a}, {a, b}} {{c}, {c, d}} (c, d). Conversely, suppose (a, b) (c, d). By Exercise..6, a (a, b) (c, d) c and a b (a, b) (c, d) c d and a b (a, b) (c, d) c d. By Exercise..5, b ((a b) a) (a b) ((c d) c) (c d) d. Thus a c and b d. Solutions to Exercises.2.2. a a + (a a) (a + a) a (a + a) a a( + ) a a a a a. This is not the only way..2.2 The number satisfies a a a for all a. If also satisfies b b b for all b, then choosing a and b yields. Hence. Now, suppose that a has two negatives, one called a and one called b. Then a+b, so a a+ a+(a+b) ( a+a)+b +b b. Hence b a, so a has a unique negative. If a has two reciprocals, one called /a and one called b, ab, so /a (/a) (/a)(ab) [(/a)a]b b b..2.3 Because a + ( a), a is the (unique) negative of a which means ( a) a. Also a + ( )a a + ( )a [ + ( )]a a, so ( )a is the negative of a or a ( )a..2.4 By the ordering property, a >, and b > implies ab >. If a < and b >, then a >. Hence ( a)b >, so ab [ ( a)]b ( )( a)b (( a)b) <. Thus negative times positive is negative. If a < and b <,

3 A. Solutions to Chapter 269 then b >. Hence a( b) <. Hence ab a( ( b)) a( )( b) ( )a( b) (a( b)) >. Thus negative times negative is positive. Also > whether > or < ( is part of the ordering property)..2.5 a < b implies b a > which implies (b+c) (a+c) > or b+c > a+c. Also c > implies c(b a) > or bc ac > or bc > ac. If a < b and b < c, then b a and c b are positive. Hence c a (c b)+(b a) is positive or c > a. Multiplying a < b by a > and by b > yields aa < ab and ab < bb. Hence aa < bb..2.6 If a b we know, from above, that aa bb. Conversely, if aa bb then we cannot have a > b because applying.2.5 with the roles of a, b reversed yields aa > bb, a contradiction. hence a b iff aa bb..2.7 A. By definition, inf A x for all x A. Hence inf A x for all x A. Hence inf A y for all y A. Hence inf A sup( A). Conversely, sup( A) y for all y A, or sup( A) x for all x A, or sup( A) x for all x A. Hence sup( A) inf A which implies sup( A) inf A. Because we already know that sup( A) inf A, we conclude that sup( A) inf A. Now, replace A by A in the last equation. Because ( A) A, we obtain supa inf( A) or inf( A) supa. B. Now, supa x for all x A, so (supa) + a x + a for all x A. Hence (supa) + a y for y A + a, so (supa) + a sup(a + a). In this last inequality, replace a by a and A by A + a to obtain [sup(a + a)] a sup(a+a a), or sup(a+a) (sup A)+a. Combining the two inequalities yields sup(a + a) (supa) + a. Replacing A and a by A and a yields inf(a + a) (inf A) + a. C. Now, supa x for all x A. Because c >, c supa cx for all x A. Hence c supa y for all y ca. Hence c supa sup(ca). Now, in this last inequality, replace c by /c and A by ca. We obtain (/c) sup(ca) sup A or sup(ca) c sup A. Combining the two inequalities yields sup(ca) csupa. Replacing A by A in this last equation yields inf(ca) c inf A. Solutions to Exercises.3.3. Let S be the set of naturals n for which there are no naturals between n and n+. From the text, we know that S. Assume n S. Then we claim n + S. Indeed, suppose that m N satisfies n + < m < n + 2. Then m, so m N (see.3) satisfies n < m < n +, contradicting n S. Hence n + S, so S is inductive. Because S N, we conclude that S N.

4 27 A Solutions.3.2 Fix n N, and let S {x R : nx N}. Then S because n n. If x S, then nx N, so n(x + ) nx + n N (because N + N N), so x + S. Hence S is inductive. We conclude that S N or nm N for all m N..3.3 Let S be the set of all naturals n such that the following holds. If m > n and m N, then m n N. From.3, we know that S. Assume n S. We claim n + S. Indeed, suppose that m > n + and m N. Then m > n. Because n S, we conclude that (m ) n N or m (n + ) N. Hence by definition of S, n + S. Thus S is inductive, so S N. Thus m > n implies m n N. Because, for m, n N, m > n, m n, or m < n, we conclude that m n Z, whenever m, n N. If n N and m N, then n N and m n m + ( n) N Z. If m N and n N, then m N and m n (n + ( m)) N Z. If n and m are both in N, then m and n are in N. Hence m n (( m) ( n)) Z. If either m or n equals zero, then m n Z. This shows that Z is closed under subtraction..3.4 If n is even and odd, then n + is even. Hence (n + ) n is even, say 2k with k N. But k implies 2k 2, which contradicts < 2. If n 2k is even and m N, then nm 2(km) is even. If n 2k and m 2j are odd, then nm 2(2kj k j + ) is odd..3.5 For all n, we establish the claim: if m N and there is a bijection between {,, n} and {,, m}, then n m. For n, the claim is clearly true. Now, assume the claim is true for a particular n, and suppose that we have a bijection f between {,, n + } and {,, m} for some m N. Then by restricting f to {,, n}, we obtain a bijection g between {,, n} and {,, k, k +,, m}, where k f(n + ). Now, define h(i) i if i k, and h(i) i if k + i m. Then h is a bijection between {,, k, k +,, m} and {,, m }. Hence h g is a bijection between {,, n} and {,, m }. By the inductive hypothesis, this forces m n or m n +. Hence the claim is true for n +. Hence the claim is true, by induction, for all n. Now, suppose that A is a set with n elements and with m elements. Then there are bijections f : A {,, n} and g : A {,, m}. Hence g f is a bijection from {,, n} to {,, m}. Hence m n. This shows that the number of elements of a set is well-defined. For the last part, suppose that A and B have n and m elements, respectively, and are disjoint. Let f : A {,, n} and g : B {,, m} be bijections, and let h(i) i+n, i m. Then h g : B {n +,, n + m} is a bijection. Now, define k : A B {,, n + m} by setting k(x) f(x) if x A, k(x) h g(x) if x B. Then k is a bijection, establishing the number of elements of A B is n + m..3.6 Let A R be finite. By induction on the number of elements, we show that max A exists. If A {a}, then a maxa. So maxa exists. Now,

5 A. Solutions to Chapter 27 assume that every subset with n elements has a max. If A is a set with n + elements and a A, then B A \ {a} has n elements. Hence max B exists. There are two cases: If a maxb, then maxb max A. Hence max A exists. If a > maxb, then a maxa. Hence maxa exists. Because, in either case max A exists when #A n +, by induction, maxa exists for all finite subsets A. Because A is finite whenever A is finite, mina exists by the reflection property..3.7 Let c sups. Because c is not an upper bound, choose n S with c < n c. If c S, then c maxs and we are done. Otherwise, c S, and c < n < c. Now, choose m S with c < n < m < c concluding that m n (m c) (n c) lies between and, a contradiction. Thus c maxs..3.8 If we write x a/b with a Z and b N, then b A so A is nonempty. Because A N, D min A exists. Because D A, we have N Dx Z, hence x N/D. Also if q N divides both N and D, then x (N/q)/(D/q) so D/q A. Because D min A, this implies q hence N and D have no common factor. Conversely, if x N/D with N and D having no common factor, then reversing the argument shows that D min A. Now x + n (N + nd)/d, so if q divides D and N + nd, then q divides N and D, so q. Thus D and N + nd have no common factor which shows D min{k N : k(x + n) Z} or D D(x + n)..3.9 If yq x, then q x/y. So {q N : yq x} is nonempty and bounded above, hence has a max, call it q. Let r x yq. Then r, or r R +. If r y, then x y(q+) r y, so q+ S, contradicting the definition of q. Hence r < y..3. Because N R is an inductive subset of R R, f N R. Now, A because (, a) f, and x A implies (x, y) f for some y implies (x +, g(x, y)) f implies x + A. Hence A is an inductive subset of R. We conclude that A N. Because f N R, we obtain A N. To show that f is a function, we need to show that, for each n, there is a unique y R with (n, y) f. Suppose that this is not so; then by Theorem.3.2, let n be the smallest natural such that (n, y) is in f for at least two distinct y s. We use this n to discard a pair (n, y) from f obtaining a strictly smaller inductive subset f. To this end, if n, let f f {(, y)}, where y a. If n >, let f f {(n, y)} where y g(n, f(n )). Because there are at least two y s corresponding to n and n is the least natural with this property, f(n ) is uniquely determined, and there is at least one pair (n, y) f with y g(n, af(n )). Because it is clear f is inductive, this contradicts the fact that f was the smallest. Hence f is a function. Moreover, by construction, f() a, and f(n + ) g(n, f(n)) for all n. If h : N R also satisfies h() a and h(n + ) g(n, h(n)), n, then B {n N : f(n) h(n)} is inductive, implying f h.

6 272 A Solutions.3. By construction, we know that a n+ a n a for all n. Let S be the set of m N, such that a n+m a n a m for all n N. Then S. If m S, then a n+m a n a m, so a n+(m+) a (n+m)+ a n+m a a n a m a a n a m+. Hence m + S. Thus S is inductive. Hence S N. This shows that a n+m a n a m for all n, m N. If n, then a n+m a n a m is clear, whereas n < implies a n+m a n+m a n a n a n+m n a n a m a n. This shows that a n+m a n a m for n Z and m N. Repeating this last argument with m, instead of n, we obtain a n+m a n a m for all n, m Z. We also establish the second part by induction on m with n Z fixed: If m, the equation (a n ) m a nm is clear. So assume it is true for m. Then (a n ) m+ (a n ) m (a n ) a nm a n a nm+n a n(m+). Hence it is true for m+, hence by induction, for all m. For m, it is clearly true, whereas for m <, (a n ) m /(a n ) m /a nm a nm..3.2 By +2+ +n, we mean the function f : N R satisfying f() and f(n + ) f(n) + (n + ) (Exercise.3.). Let h(n) n(n + )/2, n. Then h(), and h(n) + (n + ) n(n + ) 2 + (n + ) Thus by uniqueness, f(n) h(n) for all n. (n + )(n + 2) 2 h(n + )..3.3 Because p 2, < 2 n p n and n < 2 n p n for all n. If p k m p j q with k < j, then m p j k q pp j k q is divisible by p. On the other hand, if k > j, then q is divisible by p. Hence p k m p j q with m, q not divisible by p implies k j. This establishes the uniqueness of the number of factors k. For existence, if n is not divisible by p, we take k and m n. If n is divisible by p, then n n/p is a natural < p n. If n is not divisible by p, we take k and m n. If n is divisible by p, then n 2 n /p is a natural < p n 2. If n 2 is not divisible by p, we take k 2 and m n 2. If n 2 is divisible by p, we continue this procedure by dividing n 2 by p. Continuing in this manner, we obtain n, n 2, naturals with n j < p n j. Because this procedure ends in n steps at most, there is some k natural or for which m n/p k is not divisible by p and n/p k is divisible by p..3.4 We want to show that S N. If this is not so then N \ S would be nonempty, hence would have a least element n. Thus k S for all naturals k < n. By the given property of S, we conclude that n S, contradicting n / S. Hence N \ S is empty or S N..3.5 Because p S a, pa N. Because N is closed under multiplication, any multiple m of p satisfies ma N, hence m S a. Conversely, suppose m S a and write m pq + r with q N, r N {}, and r < p (Exercise.3.9). If r, multiplying by a yields ra ma q(pa) N because ma N and pa N. Hence r S a is less than p contradicting p min S a. Thus r, or p divides m.

7 A. Solutions to Chapter With a n/p, p S a because pa n N, and m S a because ma nm/p N. By the Exercise.3.5, mins a divides p. Because p is prime, min S a, or mins a p. In the first case, a a n/p N, hence p divides n; in the second case, by the previous exercise, p divides m..3.7 We use induction according to Exercise.3.2. For n, the statement is true. Suppose that the statement is true for all naturals less than n. Then either n is a prime or n is composite, n jk with j > and k >. By the inductive hypothesis, j and k are products of primes, hence so is n. Hence in either case, n is a product of primes. To show that the decomposition for n is unique except for the order, suppose that n p p r q q s. By the previous exercise, because p divides the left, hence the right side, p divides one of the q j s. Because the q j s are prime, we conclude that p equals q j for some j. Hence n n/p < n can be expressed as the product p 2 p r and the product q q j q j+ q s. By the inductive hypothesis, these ps and qs must be identical except for the order. Hence the result is true for n. By induction, the result is true for all naturals..3.8 If the algorithm ends, then r n for some n. Solving backward, we see that r n Q, r n 2 Q, and so on. Hence x r Q. Conversely, if x Q, then all the remainders r n are rationals. Now (see Exercise.3.8) write r N(r)/D(r). Then D(r) D(r + n) for n Z. Moreover, because r n <, N(r n ) < D(r n ). Then ( ) N (r n ) D D (q n+ + r n+ ) D (r n+ ) > N (r n+ ). r n Thus N(r ) > N(r ) > N(r 2 ) > is strictly decreasing. Hence N(r n ) for some n, so r n Let B be the set of n N satisfying A n is bounded above. Because A f ({}) is a single natural, we have B. Assume n B; then A n is bounded above, say by M. Because A n+ A n f ({n + }), A n+ is then bounded above by the greater of M and f ({n+}). Hence n+ B. Thus B is inductive hence B N..3.2 Because 2!, the statement is true for n. Assume it is true for n. Then 2 (n+) 22 n 2n! (n + )n! (n + )!..3.2 Because n! is the product of, 2,, n, and n n is the product of n, n,, n, it is obvious that n! n n. To prove it, use induction:!, and assuming n! n n, we have (n+)! n!(n+) n n (n+) (n+) n (n+) (n+) n+. This establishes n! n n by induction. For the second part, write n! 2 n n (n ), so (n!) 2 n (n ) 2 n or (n!) 2 equals the product of k(n + k) over all k n. Because k(n + k) n k 2 + (n + )k n (k n)(k ),

8 274 A Solutions the minimum of k(n + k) over all k n equals n. Thus (n!) 2 n n n n n. More explicitly, what we are doing is using ( n ) ( n ) n a(k) b(k) a(k)b(k), n, k k which is established by induction for a : N R and b : N R, then inserting a(k) k and b(k) n + k, k n The inequality ( + a) n + (2 n )a is clearly true for n. So assume it is true for n. Then (+a) n+ (+a) n (+a) (+(2 n )a)(+ a) + 2 n a + (2 n )a 2 + (2 n+ )a because a 2 a. Hence it is true for all n. The inequality ( + b) n + nb is true for n. So suppose that it is true for n. Because + b, then ( + b) n+ ( + b) n ( + b) ( + nb)( + b) + (n + )b + nb 2 + (n + )b. Hence the inequality is true for all n. k Solutions to Exercises.4.4. Because x max(x, x), x x. Also a < x < a is equivalent to x < a and x < a, hence to x max(x, x) < a. By definition of intersection, x > a and x < a is equivalent to {x : x < a} {x : x > a}. Similarly, x > a is equivalent to x > a or x > a: x lies in the union of {x : x > a} and {x : x < a}..4.2 Clearly. If x, then x > or x >, hence x max(x, x) >. If x > and y >, then xy xy x y. If x > and y <, then xy is negative, so xy (xy) x( y) x y and similarly, for the other two cases..4.3 If n, the inequality is true. Assume it is true for n. Then a + + a n + a n+ a + + a n + a n+ a + + a n + a n+ by the triangle inequality and the inductive hypothesis. Hence it is true for n +. By induction, it is true for all naturals n..4.4 Assume first that a >. Let S {x : x and x 2 < a}. Because S, S is nonempty. Also x S implies x x x 2 < a, so S is bounded above. Let s sups. We claim that s 2 a. Indeed, if s 2 < a, note that ( s + ) 2 s 2 + 2s n n + n 2

9 A. Solutions to Chapter 275 s 2 + 2s n + n s2 + 2s + < a n if (2s + )/n < a s 2 or, equivalently, if n > (2s + )/(a s 2 ). Because s 2 < a, b (2s+)/(a s 2 ) is a perfectly well-defined, positive real. Because supn, such a natural n > b can always be found. This rules out s 2 < a. If s 2 > a, then b (s 2 a)/2s is positive. Hence there is a natural n satisfying /n < b which implies s 2 2s/n > a. Hence ( s ) 2 s 2 2s n n + n 2 > a, so (by Exercise.2.6), s /n is an upper bound for S. This shows that s is not the least upper bound, contradicting the definition of s. Thus we are forced to conclude that s 2 a. Now, if a <, then /a > and / /a is a positive square root of a. The square root is unique by Exercise By completing the square, x solves ax 2 + bx + c iff x solves (x + b/2a) 2 (b 2 4ac)/4a 2. If b 2 4ac <, this shows that there are no solutions. If b 2 4ac, this shows that x b/2a is the only solution. If b 2 4ac >, take the square root of both sides to obtain x + b/2a ±( b 2 4ac)/2a..4.6 If a < b, then a n < b n is true for n. If it is true for n, then a n+ a n a < b n a < b n b b n+. Hence by induction, it is true for all n. Hence a b implies a n b n. If a n b n and a > b, then by applying the previous, we obtain a n > b n, a contradiction. Hence for a, b, a b iff a n b n. For the second part, assume that a >, and let S {x : x and x n a}. Because x, x n. Hence x x xx n x n < a. Thus s sups exists. We claim that s n a. If s n < a, then b s n (2 n )/(a s n ) is a well-defined, positive real. Choose a natural k > b. Then s n (2 n )/k < a s n. Hence by Exercise.3.22, ( s + ) n ( s n + ) n k sk ( ) s n + 2n s n + sn (2 n ) < a. sk k Hence s + /k S. Hence s is not an upper bound for S, a contradiction. If s n > a, b ns n /(s n a) is a well-defined, positive real, so choose k > b. By Exercise.3.22, ( s ) n ( s n ) n k ( sk s n n ) sk s n sn n k > a.

10 276 A Solutions Hence by the first part of this exercise, s /k is an upper bound for S. This shows that s is not the least upper bound, a contradiction. We conclude that s n a. Uniqueness follows from the first part of this exercise..4.7 If t k/(n 2) is a rational p/q, then 2 (kq)/(np) is rational, a contradiction..4.8 Let (b) denote the fractional part of b. If a, the result is clear. If a, the fractional parts {(na) : n } of {na : n } are in [, ]. Now, divide [, ] into finitely many subintervals, each of length, at most, ǫ. Then the fractional parts of at least two terms pa and qa, p q, p, q N, must lie in the same subinterval. Hence (pa) (qa) < ǫ. Because pa (pa) Z, qa (qa) Z, we obtain (p q)a m < ǫ for some integer m. Choosing n p q, we obtain na m < ǫ..4.9 The key issue here is that 2n 2 m 2 n 2 f(m/n) is a nonzero integer, because all the roots of f(x) are irrational. Hence 2n 2 m 2. But 2n 2 m 2 (n 2 m)(n 2 + m), so n 2 m /(n 2 + m). Dividing by n, we obtain 2 m n ( 2 + m/n)n 2. (A..) Now, if 2 m/n, the result we are trying to show is clear. So let us assume that 2 m/n. In this case, 2+m/n 2 2+(m/n 2) Inserting this in the denominator of the right side of (A..), the result follows..4. By Exercise.4.5, the (real) roots of f(x) are ±a. The key issue here is that m 4 2m 2 n 2 n 4 n 4 f(m/n) is a nonzero integer, because all the roots of f(x) are irrational. Hence n 4 f(m/n). But, by factoring, f(x) (x a)g(x) with g(x) (x + a)(x ), so a m n n 4 g(m/n). (A..2) Now, there are two cases: If a m/n, we obtain a m/n /n 4 because /n 4. If a m/n, then < m/n < 3. So from the formula for g, we obtain < g(m/n) 5. Inserting this in the denominator of the right side of (A..2), the result follows with c /5..4. First, we verify the absolute value properties A,B, and C for x, y Z. A is clear. For x, y Z, let x 2 k p and y 2 j q with p, q odd. Then xy 2 j+k pq with pq odd, establishing B for x, y Z. For C, let i min(j, k) and note x + y or x + y 2 i r with r odd. In the first case, x + y 2, whereas, in the second, x + y 2 2 i. Hence x + y 2 2 i max(2 j, 2 k ) max( x 2, y 2 ) x 2 + y 2. Now, using B for x, y, z Z, zx 2 z 2 x 2 and zy 2 z 2 y 2. Hence zx/zy 2 zx 2 / zy 2

11 A. Solutions to Chapter 277 z 2 x 2 / z 2 y 2 x 2 / y 2 x/y 2. Hence 2 is well-defined on Q. Now, using A,B, and C for x, y Z, one checks A,B, and C for x, y Q. Solutions to Exercises.5.5. If (a n ) is increasing, {a n : n } and {a n : n N} have the same sups. Similarly, for decreasing. If a n L, then a n ց L. Hence a n+n ց L and a n ր L. Hence a (n+n) ր L. We conclude that a n+n L..5.2 If a n ր L, then L sup{a n : n }, so L inf{ a n : n }, so a n ց L. Similarly, if a n ց L, then a n ր L. If a n L, then a n ց L, so ( a n ) a n ր L, and a n ր L, so ( a n ) a n ց L. Hence a n L..5.3 First, if A R +, let /A {/x : x A}. Then inf(/a) implies that, for all c >, there exists x A with /x < /c or x > c. Hence supa. Conversely, supa implies inf(/a). If inf(/a) >, then c > is a lower bound for /A iff /c is an upper bound for A. Hence supa < and inf(/a) / supa. If / is interpreted as, we obtain inf(/a) / supa in all cases. Applying this to A {a k : k n} yields b n /a n, n. Moreover, A is bounded above iff inf(/a) >. Applying this to A {b n : n } yields sup{b n : n } because a n ց. Hence b n. For the converse, b n implies sup{b n : n }. Hence inf{a n : n }. Hence a n..5.4 Because k n n, a n a kn a n. Because a n L means a n L and a n L, the ordering property implies a kn L. Now assume (a n ) is increasing. Then (a kn ) is increasing. Suppose that a n L and a kn M. Because (a n ) is increasing and k n n, a kn a n, n. Hence by ordering, M L. On the other hand, {a kn : n } {a n : n }. Because M and L are the sups of these sets, M L. Hence M L..5.5 From the text, we know that all but finitely many a n lie in (L ǫ, L+ǫ), for any ǫ >. Choosing ǫ L shows that all but finitely many terms are positive..5.6 The sequence a n n + n /( n + + n) is decreasing. Because a n 2 /n has limit zero, so does (a n ). Hence a n a n and a n a a..5.7 We do only the case a finite. Because a n a, a n is finite for each n N beyond some N. Now, a n sup{a k : k n}, so for each n, we can choose k n n, such that a n a kn > a n /n. Then the sequence (a kn ) lies between (a n) and (a n /n), hence converges to a. But (a kn ) may not be a subsequence of (a n ) because the sequence (k n ) may not be strictly increasing. To take care of this, note that, because k n n, we can choose a

12 278 A Solutions subsequence (k jn ) of (k n ) which is strictly increasing. Then (a p : p k jn ) is a subsequence of (a n ) converging to a and similarly (or multiply by minuses), for a..5.8 If a n L, then by definition, either a L or a L. For definiteness, suppose that a L. Then from Exercise.5.7, there is a subsequence (a kn ) converging to a. From.5, if 2ǫ L a >, all but finitely many of the terms a kn lie in the interval (a ǫ, a +ǫ). Because ǫ is chosen to be half the distance between a and L, this implies that these same terms lie outside the interval (L ǫ, L + ǫ). Hence these terms form a subsequence as requested..5.9 From Exercise.5.7, we know that x and x are limit points. If (x kn ) is a subsequence converging to a limit point L, then because k n n, x n x kn x n for all n. By the ordering property for sequences, taking the limit yields x L x..5. If x n L, then x x L. Because x and x are the smallest and the largest limit points, L must be the only one. Conversely, if there is only one limit point, then x x because x and x are always limit points..5. If M <, then for each n, the number M /n is not an upper bound for the displayed set. Hence there is an x n (a, b) with f(x n ) > M /n. Because f(x n ) M, we see that f(x n ) M, as n ր. If M, for each n, the number n is not an upper bound for the displayed set. Hence there is an x n (a, b) with f(x n ) > n. Then f(x n ) M..5.2 Note that 2 ( a + 2 a ) 2 ( a 2 2), a >. (A..3) 2a Because e 2 2, e. By (A..3), e n+ as soon as e n. Hence e n for all n by induction. Similarly, (A..3) with a d n plugged in and d n 2, n, yield e n+ e 2 n/2 2, n..5.3 If f(a) /(q + a), then f(a) f(b) f(a)f(b) a b. This implies A. Now, A implies x x n x n x x n x nx n x x n, where x(k) n denotes x n with k layers peeled off. Hence x x n x n x n x n x(n ) n, n. (A..4) Because x (n ) n /q n, (A..4) implies B. For C, note that, because q n, x /[ + /(q 2 + )] (q 2 + )/(q 2 + 2). Let a (c + )/(c + 2). Now, if one of the q k s is bounded by c, (A..4) and C imply x x n a, as soon as n is large enough, because all the factors in (A..4) are bounded by. Similarly, if two of the q k s are bounded by c, x x n a 2, as soon as n is large enough. Continuing in this manner, we obtain D. If q n, we are done, by B. If not, then there is a c, such that q k c for infinitely many n.

13 A. Solutions to Chapter 279 By D, we conclude that the upper and lower limits of ( x x n ) lie between and a N, for all N. Because a N, the upper and lower limits are. Hence x x n. Solutions to Exercises.6.6. First, suppose that the decimal expansion of x is as advertised. Then the fractional part of n+m x is identical to the fractional part of m x. Hence x( n+m m ) Z, or x Q. Conversely, if x m/n Q, perform long division to obtain the digits: From Exercise.3.9, m nd +r, obtaining the quotient d and remainder r. Similarly, r nd 2 + r 2, obtaining the quotient d 2 and remainder r 2. Similarly, r 2 nd 3 + r 3, obtaining d 3 and r 3, and so on. Here d, d 2, are digits (because < x < ), and r, r 2, are zero or naturals less than n. At some point the remainders must start repeating, and, therefore, the digits also..6.2 We assume d N > e N, and let x.d d 2.e e 2. If then x y. If y.d d 2 d N (e N + ), z.e e 2 e N 99, then z x. Because.99, z y. Hence x y and x z. Because x y, d N e N + and d j for j > N. Because x z, e j 9 for j > N. Clearly, this happens iff N x Z..6.3 From Exercise.3.22, ( + b) n + nb for b. In this inequality, replace n by N and b by /N(n + ) to obtain [ /N(n + )] N /(n + ) n/(n + ). By Exercise.4.6, we may take Nth roots of both sides, yielding A. B follows by multiplying A by (n+) /N and rearranging. If a n /n /N, then by B, a n a n+ (n + )/N n /N n /N (n + ) /N N(n + ) (N )/N n /N (n + ) /N N(n + ). +/N Summing over n yields C..6.4 Because e 2 2 and e n+ e 2 n /2 2, e 2 e 2 /2 2 (3 2 2)/ 2. Similarly,

14 28 A Solutions e 3 e (3 2 2) ( ). Now, assume the inductive hypothesis e n+2 2n. Then e (n+)+2 e n+3 e2 n e2 n+2 ( 2n) 2 2 n+. Thus the inequality is true by induction..6.5 Because [, 2] 2[, ], given z [, ], we have to find x C and y C satisfying x + y 2z. Let z.d d 2 d 3. Then for all n, 2d n is an even integer satisfying 2d n 8. Thus there are digits, zero or odd (i.e.,,, 3, 5, 7, 9), d n, d n, n, satisfying d n + d n 2d n. Now, set x.d d 2d 3 and y.d d 2d 3. Solutions to Exercises.7.7. Because B is countable, there is a bijection f : B N. Then f restricted to A is a bijection between A and f(a) N. Thus it is enough to show that C f(a) is countable or finite. If C is finite, we are done, so assume C is infinite. Because C N, let c minc, c 2 minc \ {c }, c 3 minc \ {c, c 2 }, and so on. Then c < c 2 < c 3 <. Because C is infinite, C n C \{c,, c n } is not empty, allowing us to set c n+ min C n, for all n. Because (c n ) is strictly increasing, we must have c n n for n. If m C \ {c n : n }, then by construction, m c n for all n, which is impossible. Thus C {c n : n } and g : N C given by g(n) c n, n, is a bijection..7.2 With each rational r, associate the pair f(r) (m, n), where r m/n in lowest terms. Then f : Q N 2 is an injection. because N 2 is countable and an infinite subset of a countable set is countable, so is Q..7.3 If x n A n, then x A n for some n. Let n be the least such natural. Now, because A n is countable, there is a natural m such that x is the mth element of A n. Define f : n A n N N by f(x) (m, n). Then f is an injection. Because N N is countable, we conclude that n A n is countable. To show that Q Q is countable, let Q (r, r 2, r 3, ), and, for

15 A. Solutions to Chapter 28 each n, set A n {r n } Q. Then A n is countable for each n. Hence so is Q Q, because it equals n A n..7.4 Suppose that [, ] is countable, and list the elements as a.d d 2 a 2.d 2 d 22 a 3.d 3 d 32. Let a.d d 2, where d n is any digit chosen, such that d n d nn, n (the diagonal in the above listing). Then a is not in the list, so the list is not complete, a contradiction. Hence [, ] is not countable..7.5 Note that i + j n + implies i 3 + j 3 (n + ) 3 /8 because at least one of i or j is (n + )/2. Sum (.7.6) in the order of N 2 given in.7: m 3 + n 3 i 3 + j 3 (m,n) N 2 n i+jn+ 8 (n + ) 3 n n i+jn+ 8n (n + ) 3 8 n 2 <. n.7.6 Because n s <, the geometric series implies n s n s n s m (n s ) m. Summing over n 2, ( n s ) n sm n2 m ( ) n ms n2 m n2 Z(ms). m

16 282 A Solutions.7.7 Because a n and b n converge, from the text, we know that their Cauchy product converges. Thus with c n i+jn+ a ib j and by the triangle inequality, c n a i b j <. n n n i+jn+ i+jn+ Hence n c n converges absolutely. Now, in the difference N D N ( N ) N a i b j a i b j there is cancellation, with the absolute value of each of the remaining terms occuring as summands in a i b j. nn+ i+jn+ Thus D N, as N ր. On the other hand, D N n c n ( a i )( b j ), hence the result..7.8 Let ã n ( ) n+ a n and b n ( ) n+ b n. If c n is the Cauchy product of ã n and bn, then c n ã i bj ( ) i+ ( ) j+ a i b j ( ) n+ c n, i+jn+ i+jn+ i where c n is the Cauchy product of a n and b n..7.9 As in Exercise.5.3, x n x m x n x nx n x n (n ), for m n. Because x n (n ) /q n, this yields x n x m /q n, for m n. Hence if q n, (/q n ) is an error sequence for (x n ). Now, suppose that q n. Then there a c with q n c for infinitely many n. Hence if N n is the number of q k s, k n + 2, bounded by c, lim nր N n. Because (q k+2 + )/(q k+2 + 2), the first inequality above implies that x (k) n x n x m j ( ) Nn c +, m n. c + 2 Set a (c + )/(c + 2). Then in this case, (a Nn ) is an error sequence for (x n ). For the golden mean x, note that x + /x; solving the quadratic, we obtain x ( ± 5)/2. Because x >, we must take the +.

17 A.2 Solutions to Chapter Because a x and a, a n xa n+ is true for n. Now assume it is true for n; then a n+ a n+2 a n+ a n a n+ (a n /a n+ ) x which equals x by the previous exercise. This derives the first identity. The second identity is clearly true when n, so assume it is true for n. Then a n+ a n /x (a /x n )/x a /x n+..7. From the previous problem, a n a /x n /x n, hence the series is a geometric series a 2 n x 2(n ) (x 2 ) n n n n x2 x 2 x 2 x. A.2 Solutions to Chapter 2 Solutions to Exercises By the theorem, select a subsequence (n k ) such that (a nk ) converges to some a. Now apply the theorem to (b nk ), selecting a subsubsequence (n km ) such that (b nkm ) converges to some b. Then clearly (a nkm ) and (b nkm ) converge to a and b, respectively, hence (a n, b n ) subconverges to (a, b) For simplicity, we assume a, b. Let the limiting point be L.d d 2. By the construction of L, it is a limit point. Because x is the smallest limit point (Exercise.5.9), x L. Now, note that if t [, ] satisfies t x n for all n, then t any limit point of (x n ). Hence t x. Because changing finitely many terms of a sequence does not change x, we conclude that x t for all t satisfying t x n for all but finitely n. Now, by construction, there are at most finitely many terms x n.d. Hence.d x. Similarly, there are finitely many terms x n.d d 2. Hence.d d 2 x. Continuing in this manner, we conclude that.d d 2 d N x. Letting N ր, we obtain L x. Hence x L. Solutions to Exercises If lim x c f(x), there is at least one sequence x n c with x n c, n, and f(x n ). From Exercise.5.8, this means there is a subsequence (x kn ) and an N, such that f(x kn ) /N, n. But this means

18 284 A Solutions that, for all n, the reals x kn are rationals with denominators bounded in absolute value by N. hence N!x kn are integers converging to N!c. But this cannot happen unless N!x kn N!c from some point on, which implies x kn c from some point on, contradicting x n c for all n. Hence the result Let L inf{f(x) : a < x < b}. We have to show that f(x n ) L whenever x n a+. So suppose that x n a+, and first assume (x n ) is decreasing. Then (f(x n )) is decreasing. Hence f(x n ) decreases to some limit M. Because f(x n ) L for all n, M L. If d >, then there is an x (a, b) with f(x) < L + d. Because x n ց a, there is an n with x n < x. Hence M f(x n ) f(x) < L + d. Because d > is arbitrary, we conclude that M L or f(x n ) ց L. In general, if x n a+, x n ց a. Hence f(x n) ց L. But x n x n, hence L f(x n ) f(x n). So by the ordering property, f(x n ) L. This establishes the result for the inf. For the sup, repeat the reasoning, or apply the first case to g(x) f( x) Assume f is increasing. If a < c < b, then apply the previous exercise to f on (c, b), concluding that f(c+) exists. Because f(x) f(c) for x > c, we obtain f(c+) f(c). Apply the previous exercise to f on (a, c) to conclude that f(c ) exists and f(c ) f(c). Hence f(c) is between f(c ) and f(c+). Now, if a < A < B < b and there are N points in [A, B] where f jumps by at least δ >, then we must have f(b) f(a) Nδ. Hence given δ and A, B, there are, at most, finitely many such points. Choosing A a + /n and B b /n and taking the union of all these points over all the cases n, we see that there are, at most, countably many points in (a, b) at which the jump is at least δ. Now, choosing δ equal, /2, /3,..., and taking the union of all the cases, we conclude that there are, at most, countably many points c (a, b) at which f(c+) > f(c ). The decreasing case is obtained by multiplying by minus Choose any c (a, b), and consider the partition with just two points a x < x < x 2 < x 3 b, where x c and x 2 x. Then the variation corresponding to this partition is f(x) f(c). Hence f(x) f(c) I for x > c. Similarly, f(x) f(c) I for x < c. Hence f(x) is bounded by I + f(c) If f is increasing, then there are no absolute values in the variation (2.2.) which, therefore, collapses to f(x n ) f(x ). Because f is also bounded, this last quantity is bounded by some number I. Hence f is of bounded variation. Now, note f and g of bounded variation implies f and f + g are also of bounded variation, because the minus does not alter the absolute values, and the variation of the sum f + g, by the triangle inequality, is less or equal to the sum of the variation of f and the variation of g. This implies the second statement.

19 A.2 Solutions to Chapter If a < x < y < b, every partition a x < x < x 2 < < x n x < x n+ b with x n x yields, by adding the point {y}, a partition a x < x < x 2 < < x n x < x n+ y < x n+2 b with x n+ y. Because v(x) and v(y) are the sup of the variations over all such partitions, respectively, this and (2.2.) yield v(x) + f(y) f(x) v(y). Hence v is increasing and, throwing away the absolute value, v(x) f(x) v(y) f(y); hence v f is increasing. Thus f v (v f) is the difference of two increasing functions. Finally, because v I, both v and v f are bounded Look at the partition x, x, x 2 /2, x 2,, x n /n, x n, where x i is an irrational between x i and x i+, i n (for this to make sense, take x n+ ). Then the variation corresponding to this partition is 2s n, where s n is the nth partial sum of the harmonic series. But s n ր. Solutions to Exercises Let f be a polynomial with odd degree n and highest-order coefficient a. Because x k /x n, as x ±, for n > k, it follows that f(x)/x n a. Because x n ±, as x ±, it follows that f(± ) ±, at least when a >. When a <, the same reasoning leads to f(± ). Thus there are reals a, b with f(a) > and f(b) <. Hence by the intermediate value property, there is a c with f(c) By definition of µ c, f(x) f(c) µ c (δ) when x c < δ. Because x c < 2 x c for x c, choosing δ 2 x a yields f(x) f(c) µ c (2 x c ), for x c. If µ c (+), setting ǫ µ c (+)/2, µ c (/n) 2ǫ. By the definition of the sup in the definition of µ c, this implies that, for each n, there is an x n (a, b) with x n c /n and f(x n ) f(c) ǫ. Because the sequence (x n ) converges to c and f(x n ) f(c), it follows that f is not continuous at c. This shows A implies B Let A f((a, b)). To show that A is an interval, it is enough to show that (inf A, supa) A. By definition of inf and sup, there are sequences m n inf A and M n supa with m n A, M n A. Hence there are reals c n, d n with f(c n ) m n, f(d n ) M n. Because f([c n, d n ]) is a compact interval, it follows that [m n, M n ] A for all n. Hence (inf A, supa) A. For the second part, if f((a, b)) is not an open interval, then there is a c (a, b) with f(c) a max or a min. But this cannot happen: Because f is strictly monotone we can always find an x and y to the right and to the left of c such that f(x) and f(y) are larger and smaller than f(c). Thus f((a, b)) is an open interval Let a supa. Because a x for x A and f is increasing, f(a) f(x) for x A, or f(a) is an upper bound for f(a). Because a supa, there is a sequence (x n ) A with x n a (Theorem.5.4). By continuity, f(x n )

20 286 A Solutions f(a). Now, let M be any upper bound for f(a). Then M f(x n ), n. Hence M f(a). Thus f(a) is the least upper bound for f(a). Similarly, for inf Let y n f(x n ). From Exercise 2.3.4, f(x n) sup{f(x k ) : k n} y n, n. Because x n x and f is continuous, y n f(x n) f(x ). Thus y f(x ). Similarly, for lower stars Remember x r is defined as (x m ) /n when r m/n. Because [ (x /n ) m] n [( (x /n ) mn x /n) n] m x m, x r (x /n ) m also. With r m/n Q and p in Z, (x r ) p [ (x /n ) m] p (x /n ) mp x mp/n x rp. Now, let r m/n and s p/q with m, n, p, q integers and nq. Then [(x r ) s ] nq (x r ) snq x rsnq x mp. By uniqueness of roots, (x r ) s (x mp ) /nq x rs. Similarly, (x r x s ) nq x rnq x snq x rnq+snq x (r+s)nq (x r+s ) nq. By uniqueness of roots, x r x s x r+s We are given a b sup{a r : < r < b, r Q}, and we need to show that a b c, where c inf{a s : s > b, s Q}. If r, s are rationals with r < b < s, then a r < a s. Taking the sup over all r < b yields a b a s. Taking the inf over all s > b implies a b c. On the other hand, choose r < b < s rational with s r < /n. Then c a s < a r a /n a b a /n. Taking the limit as n ր, we obtain c a b In this solution, r, s, and t denote rationals. Given b, let (r n ) be a sequence of rationals with r n b. If t < bc, then t < r n c for n large. Pick one such r n and call it r. Then s t/r < c and t rs. Thus t < bc iff t is of the form rs with r < b and s < c. By Exercise 2.3.6, (a b ) c sup{(a b ) s : < s < c} sup{(sup{a r : < r < b}) s : < s < c} sup{a rs : < r < b, < s < c} sup{a t : < t < bc} a bc Because f(x+x ) f(x)f(x ), by induction, we obtain f(nx) f(x) n. Hence f(n) f() n a n for n natural. Also a f() f( + ) f()f() af(). Hence f(). Because f(n n) f(n)f( n) a n f( n), we obtain f( n) a n for n natural. Hence f(n) a n for n Z. Now, (f(m/n)) n f(n(m/n)) f(m) a m, so by uniqueness of roots f(m/n) a m/n. Hence f(r) a r for r Q. If x is real, choose rationals r n x. Then a rn f(r n ) f(x). Because we know that a x is continuous, a rn a x. Hence f(x) a x Given ǫ >, we seek δ >, such that x < δ implies (/x) < ǫ. By the triangle inequality, x (x ) + x. So δ < and x < δ implies x > δ and x x x < δ δ.

21 A.2 Solutions to Chapter Solving δ/( δ) ǫ, we have found a δ, δ ǫ/( + ǫ), satisfying < δ < and the ǫ-δ criterion Let A n be the set of all real roots of all polynomials with degree d and rational coefficients a, a,, a d, with denominators bounded by n and satisfying a + a + + a d + d n. Because each polynomial has finitely many roots and there are finitely many polynomials involved here, for each n, the set A n is finite. But the set of algebraic numbers is n A n, hence is countable Let b be a root of f, b a, and let f(x) (x b)g(x). If b is rational, then the coefficients of g are necessarily rational (this follows from the construction of g in the text) and f(a) (a b)g(a). Hence g(a). But the degree of g is less than the degree of f. This contradiction shows that b is irrational Write f(x) (x a)g(x). By the previous exercise, f(m/n) is never zero. Because n d f(m/n) is an integer, or n d f(m/n) n d m/n a g(m/n), a m n n d g(m/n). (A.2.) Because g is continuous at a, choose δ > such that µ a (δ) < ; that is, such that x a < δ implies g(x) g(a) µ a (δ) <. Then x a < δ implies g(x) < g(a) +. Now, we have two cases: either a m/n δ or a m/n < δ. In the first case, we obtain the required inequality with c δ. In the second case, we obtain g(m/n) < g(a) +. Inserting this in the denominator of the right side of (A.2.) yields a m n n d ( g(a) + ), which is the required inequality with c /[ g(a) + ]. Now, let ( ) c min δ,. g(a) + Then in either case, the required inequality holds with this choice of c Let a be the displayed real. Given a natural d, we show that a is not algebraic of order d. This means, given any ǫ >, we have to find M/N such that a M N < ǫ N d. (A.2.2)

22 288 A Solutions Thus the goal is: given any d and ǫ >, find M/N satisfying (A.2.2). The simplest choice is to take M/N equal to the nth partial sum s n in the series for a. In this case, N n!. The question is, which n? To figure this out, note that k! n!k when k n +, and n! 2, so a s n kn+ ( n! ) n+ k! k kn+ n!k ( n! ) k ( n! ) n+ k 2 k ( n! ) n 2 n!. Thus if we select n satisfying n d and 2/ n! < ǫ, we obtain a M N < ǫ N n ǫ N d hence s n M/N satisfies (A.2.2) From.6, we know that n r converges when r + /N. Given s > real, we can always choose N with + /N < s. The result follows by comparison To show that b log a c c log a b, apply log a obtaining log a b log a c from either side. For the second part, /5 log 3 n /n log 3 5 which converges because log 3 5 > Such an example cannot be continuous by the results of 2.3. Let f(x) x + /2, x < /2, and f(x) x /2, /2 x <, f(). Then f is a bijection, hence invertible First, assume f is increasing. Then by Exercise 2.2.3, f(c ) f(c) f(c+) for all except, at most, countably many points c (a, b), where there are, at worst, jumps. Hence f is continuous for all but at most countably many points at which there are, at worst, jumps. If f is of bounded variation, then f g h with g and h bounded increasing. But, then f is continuous wherever both g and h are continuous. Thus the set of discontinuities of f is, at most, countable with the discontinuities at worst jumps Let M sup f((a, b)). Then M >. By Theorem.5.4, there is a sequence (x n ) in (a, b) satisfying f(x n ) M. Now, from Theorem 2..2, (x n ) subconverges to some x in (a, b) or to a or to b. If (x n ) subconverges to a or b, then (f(x n )) subconverges to f(a+) or f(b ), respectively. But this cannot happen because M > f(a+) and M > f(b ). Hence (x n ) subconverges to some a < x < b. Because f is continuous, (f(x n )) must subconverge to f(x). Hence f(x) M. This shows that M is finite and M is a max Fix y and set h(x) xy f(x). Then by superlinearity

23 A.2 Solutions to Chapter ( h( ) lim (xy f(x)) lim x y f(x) ) ; x x x similarly h( ). Thus Exercise applies and the sup is attained in the definition of g(y). Now, for x > fixed and y n, g(y n ) xy n f(x). Hence g(y n ) y n x f(x) y n. It follows that the lower limit of (g(y n )/y n ) is x. Because x is arbitrary, it follows that the lower limit is. Hence g(y n )/y n. Because (y n ) was any sequence converging to, we conclude that lim y g(y)/ y. Similarly, lim y g(y)/ y. Thus g is superlinear Suppose that y n y. We want to show that g(y n ) g(y). Let L L be the upper and lower limits of (g(y n )). For all z, g(y n ) zy n f(z). Hence L zy f(z). Because z is arbitrary, taking the sup over all z, we obtain L g(y). For the reverse inequality, let (y n) be a subsequence of (y n ) satisfying g(y n ) L. Pick, for each n, x n with g(y n ) x n y n f(x n). From 2., (x n) subconverges to some x, possibly infinite. If x ±, then superlinearity (see previous solution) implies the subconvergence of (g(y n )) to. But L L g(y) >, so this cannot happen. Thus (x n) subconverges to a finite x. Hence by continuity, g(y n) x ny n f(x n) subconverges to xy f(x) which is g(y). Because by construction, g(y n ) L, this shows that L g(y). Hence g(y) L L g(y) or g(y n ) g(y) Note that f(x) and f(x) iff x Z. We are supposed to take the limit in m first, then n. If x Q, then there is an N N, such that n!x Z for n N. Hence f(n!x) for n N. For such an x, lim m [f(n!x)] m, for every n N. Hence the double limit is for x Q. If x Q, then n!x Q, so f(n!x) <, so [f(n!x)] m, as m ր, for every n. Hence the double limit is for x Q In the definition of µ c (δ), we are to maximize (/x) (/c) over all x (, ) satisfying x c < δ or c δ < x < c+δ. In the first case, if δ c, then c δ. Hence all points x near and to the right of satisfy x c < δ. Because lim x + (/x), in this case, µ c (δ). In the second case, if < δ < c, then x varies between c δ > and c + δ. Hence x c x c xc is largest when the numerator is largest ( x c δ) and the denominator is smallest (x c δ). Thus { δ/(c 2 cδ), < δ < c, µ c (δ), δ c.

24 29 A Solutions Now, µ I (δ) equals the sup of µ c (δ) for all c (, ). But, for δ fixed and c +, δ c eventually. Hence µ I (δ) for all δ >. Hence µ I (+), or f is not uniformly continuous on (, ) Follow the proof of the uniform continuity theorem. If µ(+) >, set ǫ µ(+)/2. Then because µ is increasing, µ(/n) 2ǫ for all n. Hence for each n, by the definition of the sup in the definition of µ(/n), there is a c n R with µ cn (/n) > ǫ. Now, by the definition of the sup in µ cn (/n), for each n, there is an x n R with x n c n < /n and f(x n ) f(c n ) > ǫ. Then by compactness ( 2.), (x n ) subconverges to some real x or to x ±. It follows that (c n ) subconverges to the same x. Hence ǫ < f(x n ) f(c n ) subconverges to f(x) f(x), a contradiction If 2 2 is rational, we are done. Otherwise a 2 2 is irrational. In this case, let b 2. Then a b ( 2 2) is rational. A.3 Solutions to Chapter 3 Solutions to Exercises Because a >, f(). If a, we already know that f is not differentiable at. If a <, then g(x) (f(x) f())/(x ) x a /x satisfies g(x) as x, so f is not differentiable at. If a >, then g(x) as x. Hence f () Because 2 is irrational, f( 2). Hence (f(x) f( 2))/(x 2) f(x)/(x 2). Now, if x is irrational, this expression vanishes whereas, if x is rational with denominator d, q(x) f(x) f( 2) x 2 d 3 (x 2). By Exercise.4.9 it seems that the limit is zero, as x 2. To prove this, suppose that the limit of q(x) is not zero, as x 2. Then there exists at least one sequence x n 2 with q(x n ). It follows that there is a δ > and a sequence x n 2 with q(x n ) δ. But this implies all the reals x n are rational. If d n is the denominator of x n, n, we obtain, from Exercise.4.9, q(x n ) d2 n d 3 n c d n c. Because d n (Exercise 2.2.), we conclude that q(x n ), contradicting our assumption. Hence our assumption must be false, hence lim x 2 q(x) or f ( 2).

25 A.3 Solutions to Chapter Because f is superlinear (Exercise 2.3.2), g(y) is finite, and the max is attained at some critical point x. Differentiating xy ax 2 /2 with respect to x yields y ax, or x y/a for the critical point, which, as previously said, must be the global max. Hence g(y) (y/a)y a(y/a) 2 /2 y 2 /2a. Because f (x) ax and g (y) y/a, it is clear they are inverses Suppose that g (R) is bounded above. Then g (x) c for all x. Hence g(x) g() g (z)(x ) cx for x >, which implies g(x)/x c for x >, which contradicts superlinearity. Hence g (R) is not bounded above. Similarly, g (R) is not bounded below To show that f (c) exists, let x n c with x n c for all n. Then for each n there is a y n strictly between c and x n, such that f(x n ) f(c) f (y n )(x n c). Because x n c, y n c, and y n c for all n. Because lim x c f (x) L, it follows that f (y n ) L. Hence (f(x n ) f(c))/(x n c) L. Because (x n ) was arbitrary, we conclude that f (c) L To show that f (c) f (c+), let x n c+. Then for each n, there is a y n between c and x n, such that f(x n ) f(c) f (y n )(x n c). Because x n c+, y n c+. It follows that f (y n ) f (c+). Hence (f(x n ) f(c))/(x n c) f (c+) which implies f (c) f (c+) and similarly, for f (c ) If a x < x < x 2 < < x n+ b is a partition, the mean value theorem says f(x k ) f(x k ) f (z k )(x k x k ) for some z k between x k and x k, k n +. Because f (x) I, we obtain f(x k ) f(x k ) I(x k x k ). Summing over k n +, we see that the variation corresponding to this partition is I(b a). Because the partition was arbitrary, the result follows Let c Q. We have to show that, for some n, f(c) f(x) for all x in (c /n, c + /n). If this were not the case, for each n, we can find a real x n satisfying x n c < /n and f(x n ) > f(c). But, by Exercise 2.2., we know that f(x n ) because x n c and x n c, contradicting f(c) >. Hence c must be a local maximum If f is even, then f( x) f(x). Differentiating yields f ( x) f (x), or f is odd and similarly, if f is odd. 3.. Let g(x) (f(x) f(r))/(x r), x r, and g(r) f (r). Then g is continuous and f(r) iff f(x) (x r)g(x). 3.. As in the previous exercise, set