1
Rational Numbers
1.1. Evidently, (mp)n = m(np) for all m, n, p ∈ Z. Hence, (mp, np) ∼ (m, n). We need p 6= 0 to make sure that (mp, np) is a fraction. n m = . Then (n, 1) ∼ (m, 1), that is, n · 1 = m · 1. Hence, 1 1 n = m, so ϕ is one-to-one. Furthermore (cf. (1,1) on p. 4 in the book),
1.2. Suppose that
ϕ(m + n) =
m n m+n = + = ϕ(m) + ϕ(n), 1 1 1
and ϕ(m · n) =
m n m·n = · = ϕ(m) · ϕ(n), 1 1 1
for all m, n ∈ Z. 1.3. Straightforward verification of the properties defining a field. 1.4. Suppose that a + b = 0 in F. Then, −a + a + b = −a. By Property A4, b = −a. Hence, −a is a unique additive inverse of a. For a 6= 0, let b be an element of F such that a · b = 1. By Property M3, we have a−1 = 1 · a−1 = a · b · a−1 = b · a · a−1 = b · 1 = b. Hence, a−1 is a unique multiplicative inverse of a.
1.5. By Property D, 0 + 0 = 0 implies a · 0 + a · 0 = a · 0. Hence, a · 0 = −a · 0 + a · 0 + a · 0 = −a · 0 + a · 0 = 0. 1.6. (a) By Property M3, 1 · 1 = 1. By Property M4 and Exercise 1.4, 1−1 = 1. (b) By Property M1, a−1 ·b−1 ·a·b = a−1 ·a·b−1·b = 1. Hence, (a·b)−1 = a−1 ·b−1 . (c) If c · b = a, then c = c · b · b−1 = a · b−1. If c = a · b−1, then c · b = a · b−1 · b = a. (d) By part (b) above, a·c a c = (a · c) · (b · d)−1 = a · c · b−1 · d−1 = a · b−1 · c · d−1 = · . b·d b d (e) By part (d), (f) We have
a·c a c a = · = . b·c b c b
a·d+c·b = (a · d + c · b)(b · d)−1 = (a · d + c · b)(b−1 · d−1 ) b·d = a · d · b−1 · d−1 + c · b · b−1 · d−1 a c = a · b−1 + c · d−1 = + . b d 3
a c = , then (a · b−1 ) · b · d = (c · d−1 ) · b · d, so a · d = b · c. If a · d = b · c, b d a c then (a · d) · b−1 · d−1 = (b · c) · b−1 · d−1 , so = a · b−1 = c · d−1 = . b d (h) Because, clearly, (b−1 )−1 = b, a −1 b = (a · b−1 )−1 = a−1 · b = . b a (g) If
a·d . It is easy to (i) Correction. The right hand side of the equality must be b·c c −1 a c a d a·d d verify that = . Therefore, by part (d) above, ÷ = · = . d c b d b c b·c
1.7. Note that Z[x] is an integral domain (cf. Exercise A.14)) and repeat the steps in the proof of Theorem 1.1. 1.8. Repeat the steps in the proof of Theorem 1.4.
1.9. Suppose that 1 < 0 and let a 6= 0 be a positive element of F. Then a · a−1 = 1 < 0. By part (b) of Definition 1.7, a = a · a−1 · a < a · 0 = 0, a contradiction. 1.10. Correction. The function τ in the proof of Theorem 1.8 is defined only for nonnegative integers, not for all n ∈ Z. For negative integers, we set τ (−n) = −τ (n) for n ∈ N. Thus Exercise 1.10 is, in fact, a part of the definition of the function τ . 1.11. By Definition 1.7(b), ab−1 < cd−1 is equivalent to ad < cb, because b > 0 and d > 0. 1.12. By mathematical induction. The case n = 3 follows immediately from the transitivity property of the linear order < (cf. Definition A.7). By the same property, the case of n = k + 1 follows from the case of n = k. 1.13. First, suppose that x > 0. By the Archimedean Property, there is n ∈ N such that x < n, so the set {k ∈ N : x < k} is nonempty. By the Well-Ordering Principle (cf. Theorem A.6), there is m ∈ N such that m − 1 ≤ x < m. For a non-positive x, note that the claim is trivial if x ∈ Z \ N. If x ∈ / Z \ N, apply the previous argument to −x. 1.14. It is shown in Exercise 1.8 that Z(x) is a field. An order on Z(x) is defined in Example 1.3. To show that Z(x) endowed with this order is an ordered field, repeat the steps in the proofs of Theorems 1.6 and 1.7. 1.15. Suppose to the contrary that D = p2 /q 2 for some p, q ∈ N. We may assume that q is the least natural number satisfying this equation (cf. the proof of Theorem 1.10.). Because D is not a perfect square, there is t ∈ N such that t2 < D < (t + 1)2 . 4
(Why?) It follows that tq < p < (t + 1)q. Let q 0 = p − tq and p0 = Dq − tp. It is easy to verify that 0 < q 0 < q and p0 > 0. We have p02 − Dq 02 = (Dq − tp)2 − D(p − tq)2
= D2 q 2 − 2Dqtp + t2 p2 − Dp2 + 2Dptq − Dt2 q 2
= (D2 q 2 − Dt2 q 2 ) − (Dp2 − t2 p2 ) = Dq 2 (D − t2 ) − p2 (D − t2 ) = (Dq 2 − p2 )(D − t2 ) = 0,
because D = p2 /q 2 . This contradicts the minimality of q. 1.16. Let a = x − y, b = y − z, so x − z = a + b. We need to show that |a + b| ≤ |a| + |b|. By Exercise 1.17(e), −|a| ≤ a ≤ |a| and −|b| ≤ b ≤ |b|. Hence, −(|a| + |b|) ≤ a + b ≤ |a| + |b|. The desired result follows from Exercise 1.17(f). 1.17. We prove only part (h) using the triangular inequality from the proof of Theorem 1.12 and other parts of this exercise. We have |a| = |(a − b) + b| ≤ |a − b| + |b|. Therefore, |a| − |b| ≤ |a − b|. On the other hand, |b| = |a + (b − a)| ≤ |a| + |b − a| = |a| + |a − b|. Hence, |a| − |b| ≥ −|a − b|. The result follows from part (e). 1.18. If a ≤ b ≤ c, then |a − b| = b − a and |b − c| = c − b. Hence, |a − b| + |b − c| = (b − a) + (c − b) = c − a = |a − c|. In the opposite direction, the proof is by contradiction. Suppose that b < a, so also b < c. Then |a − b| + |b − c| = (a − b) + (c − b) = a + c − 2b > a + c − 2a = c − a = |a − c|, a contradiction. Similarly, we obtain a contradiction by assuming that b > c. Therefore, a ≤ b ≤ c. 1.19. Without loss of generality, we may assume that x > y. We have b − a = y − a + x − y + b − x ≥ x − y, because y − a ≥ 0 and b − x ≥ 0. Hence the result. 1.20. Correction. The inequality |ak | > ε in Exercise 1.20 must be replaced with |ak − a| > ε. 5
To form the negation of the statement in Definition 1.12, we must replace the quantifier “for every positive ε” by “there exists ε > 0”, the quantifier “there exists” by “for every”, and the quantifier “for all” by “for some”. 1.21. Standard Calculus exercises. 1.22. Inasmuch as (an ) is Cauchy, for ε > 0 there is N ∈ N such that |am − an | < ε,
for all m, n > N .
By Exercise 1.17(h), |am | − |an | ≤ |am − an | < ε,
for all m, n > N .
Therefore, (|an |) is Cauchy. 1.23. Apply Theorem 1.20 for a constant sequence (bn ) = (k, k, . . .). 1.24. First note that the sequence (an ) is bounded below by a positive rational number. Indeed, because lim a2n = p > 0, there is only a finite number of terms of (a2n ) that are smaller than p/2. Hence, (a2n ) is bounded below by a positive rational number (cf. Exercise 1.25). It follows that there is a positive r ∈ Q such that an > r for all n ∈ N. By Theorem 1.16, (a2n ) is Cauchy, that is, for ε > 0 there is N ∈ N such that |a2m − a2n | < 2r · ε for all m, n > N . We have |am − an | =
2r · ε |a2m − a2n | < = ε, am + an 2r
for all m, n > N .
Therefore, (an ) is a Cauchy sequence. 1.25. The proof is by induction on the cardinality of the set A. The claim is trivial if A consists of one or two elements. Suppose that it holds for every n-element set and let A be a set containing n + 1 elements. For an element x of A, let A0 = A \ {x} and a = min A0 , b = max A0 . If x < a, we set min A = x, max A = b, if x > b, we set max A = x, min A = a. Finally, if a ≤ x ≤ b, we set min A = a, max A = b. It is not difficult to verify that in all these cases, min A ≤ c ≤ max A for every c ∈ A. 1.26. Recall that a tail of the sequence (an ) is a subsequence (am , am+1 , . . .), m ∈ N. For ε > 0 there is N ∈ N such that |an −a| < ε for all n > N . It follows that |an − a| < ε for all n > max{m, N }. Hence, the subsequence converges to the same limit. 1.27. We define the mapping f recursively. By the Well-Ordering Principle (cf. Theorem A.6), the set S has a least element. We define f(1) to be this element. Suppose the values f(1), f(2), . . . , f(n) are defined and satisfy the required inequalities. By the same Principle, the nonempty set S\{f(1), . . . , f(n)} has a least element. We define f(n + 1) to be this element. Clearly, thus defined function f satisfies the required inequalities. 1.28. Straightforward algebra. 6
2
Real Numbers
2.1. Suppose that b 6= b0 are two suprema of a nonempty subset of an ordered field. By Definition 2.1, b < b0 and b0 < b, which contradicts the Trichotomy Property of the relation < (cf. Theorem A.14). A similar argument proves uniqueness of the infimum. 2.2. Let E be a subset of F bounded below and E 0 the set of all lower bounds of E. Because F is Dedekind complete, it has a supremum, c = sup E 0 . By Definition 2.1, c = inf E. 2.3. q − p = (r − p) + (s − r) + (q − s) > 0 + (s − r) + 0 = s − r. 2.4. Suppose that an ≤ an+1 , for all n ∈ N. For a given n, we prove that an ≤ an+k , k ∈ N, by induction on k. The case k = 1 is trivial. Suppose that an ≤ an+k−1 . By transitivity, the inequalities an ≤ an+k−1 and an+k−1 ≤ an+k imply an ≤ an+k . A similar argument proves that bn ≥ bm for n < m. 2.5. By the Archimedean Property (cf. Definition 1.8), for ε > 0 there is N ∈ N such that N ε > 1. Hence, 1/n < ε for all n > N , that is 1/n → 0. In the other direction, suppose that 1/n → 0 in F. For positive x and y in F there is N ∈ N such that 1/n < x/y for all n > N . Hence, (N + 1)x > y, so the Archimedean Property holds. 2.6. Every b ∈ B is an upper bound of A. Hence, sup A ≤ b for every b ∈ B, so sup A is a lower bound of B. It follows that sup A ≤ inf B. 2.7. Let x and y be upper bounds of the sets A and B, respectively. Then max{x, y} is, clearly, an upper bound of A ∪ B, so this union is bounded. Clearly, sup(A ∪ B) ≥ sup A and sup(A ∪ B) ≥ sup B. Therefore, sup(A ∪ B) ≥ max{sup A, sup B}. We may assume that sup A ≤ sup B. For every x ∈ A, x ≤ sup A ≤ sup B and for every x ∈ B, x ≤ sup B. Hence, for every x ∈ A ∪ B, x ≤ sup B = max{sup A, sup B}, so sup(A ∪ B) ≤ max{sup A, sup B}. The result follows. 2.8. Let x = sup A, y = sup B. For a ∈ A and b ∈ B, we have a + b ≤ x + y, so x + y is an upper bound of A + B. By Theorem 2.1, for ε > 0 there are a ∈ A, b ∈ B such that a < x − ε/2 and b < y − ε/2, so a + b < x + y − ε. By the same theorem, x + y is the supremum of A + B. 2.9. We prove that sup(cE) = c inf E for c < 0. The other cases are treated similarly. 7
For a ∈ E, inf E ≤ a. Hence c inf E ≥ ca for all a ∈ E. Thus, c inf E is an upper bound of cE. Let x be an upper bound of cE, that is, x ≥ ca for all a ∈ E. It follows that x/c ≤ a for all a ∈ E, so x/c ≤ inf E. Then x ≥ c inf E, that is, c inf E is the supremum of cE. 2.10. Clearly, inf E 0 ≤ sup E 0 . Because E 0 ⊆ E, sup E is an upper bound of E 0 . It follows that sup E 0 ≤ sup E. A similar argument shows that inf E ≤ inf E 0 . 2.11. (Necessity.) Part (a) follows from Theorem 2.1 with ε = 1/n. Part (b) is trivial. (Sufficiency.) First, we show that b is an upper bound of E. Suppose to the contrary that there is x ∈ E such that x > b and let n be a positive integer such that 1/n < x − b. Then b + 1/n < x which contradicts part (b). Thus b is an upper bound of E. For ε > 0 let n be a positive integer such that 1/n < ε. Since b −1/n > b −ε, by part (a), b − ε is not an upper bound of E. By Theorem 2.1, b = sup E. √ √ 2.12. Suppose that 2 + 3 is a rational number. Then the number √
2−
√
3= √
−1 √ 2+ 3
is rational, as well as the number √ √ √ √ ( 2 + 3) + ( 2 − 3) √ = 2, 2 √ √ contradicting Theorem 1.10. Hence, 2 + 3 is an irrational number. 2.13. Correction. It is assumed that b 6= 0. Straightforward proofs in all four cases are by contradiction. 2.14. Clearly, lim(an − an ) = 0, so the relation ∼ is reflexive. Also, lim(an − bn ) = 0 implies lim(bn − an ) = 0, so the relation ∼ is symmetric. To prove transitivity of ∼, suppose that lim(an −bn ) = 0 and lim(bn −cn ) = 0. Then lim(an − cn ) = lim[(an − bn ) + (bn − cn )] = lim(an − bn ) + lim(bn − cn ) = 0. Hence the result. 2.15. By definition (cf. (2.1) on page 39 in the book), (an ) ∈ [(a)] if and only if lim(an − a) = 0. Clearly, the latter condition is equivalent to lim an = a. 2.16. Let a = (x + y)/2 and ε = (y − x)/2. By Definition 2.6, there is r ∈ E such that x+y y −x r− < . 2 2 8
Elementary algebra shows that the displayed inequality is equivalent to the chain inequality x < r < y. 2.17. By Lemma 2.12, (e an ) is a Cauchy sequence in F̃, and, by Theorem 2.8, en → [(am )]. Hence there is N ∈ N such that [(am )] − e a an < ε for all n > N . We obtain the desired result by setting a = aN+1 .
2.18. (a) Because a < b if and only if ϕ(a) < ϕ(b) for all a, b ∈ F, the mapping ϕ is one-to-one. (b) We have ϕ(1) = ϕ(1 + 0) = ϕ(1) + ϕ(0). Hence, ϕ(0) = 0. By part (a), ϕ(1) 6= 0. We have ϕ(1) = ϕ(1 · 1) = ϕ(1) · ϕ(1). Therefore, ϕ(1) = 1. (c) Straightforward verification of the properties defining an ordered field.
2.19 (Necessity.) Suppose that ϕ(an ) → 0 in G. For ε > 0 in F, ϕ(ε) > 0 in G. There is N ∈ N such that −ϕ(ε) < ϕ(an ) < ϕ(ε),
for all n > N .
Because ϕ is an embedding, −ε < an < ε,
for all n > N .
Hence, an → 0 in F. (Sufficiency.) Let an → 0 in F and ε > 0 in G. By the density property, there is δ > 0 in F such that 0 < ϕ(δ) < ε. Because an → 0, there is N ∈ N such that −δ < an < δ, for all n > N . Then −ε < ϕ(−δ) < ϕ(an ) < ϕ(δ) < ε,
for all n > N .
It follows that ϕ(an ) → 0 in G. 2.20. If F is not an Archimedean field, then there are positive elements x and y in F such that nx ≤ y for all n ∈ N, that is, x + · · · + x ≤ y for all n ∈ N. {z } | n
Then β(x) + · · · + β(x) ≤ β(y). Note that β(x) > 0 and β(y) > 0. It follows | {z } n
e is not Archimedean. that F
2.21. Let ϕ : F → G be an isomorphism and x, y positive elements in G. Because F is Archimedean, there is n ∈ N such that nϕ−1 (x) > ϕ−1 (y). Since ϕ is an isomorphism, nx > y (cf. solution 2.20). Hence, G is Archimedean. 2.22. Let (an ) be a Cauchy sequence in Q. For eny ε > 0 in R there is ε0 ∈ Q such that 0 < ε0 < ε (cf. Exercise 2.16). Because (an ) is Cauchy in Q, there is N ∈ N such that |am − an | < ε0 < ε,
for all m, n > N . 9
Hence, (an ) is Cauchy in R. Clearly, every Cauchy sequence of rational numbers in R is also Cauchy in the field Q. 2.23. Hint. Use the approach from Exercise 2.22. 2.24. (Necessity.) Theorem 2.14. (Sufficiency.) Suppose that Q is dense is F and let x and y be positive elements of F. We my assume that x < y. Let p and q be positive rational numbers such that p < x and q > y (cf. Exercise 2.16). Because Q is Archimedean, there is n ∈ N such that np > q. We have nx > np > q > y. Hence, F is Archimedean. 2.25. Clearly, both sets F and G are closed under operations of addition and multiplication. Verifying ordered field properties is a tedious but straightforward exercise. Suppose that ϕ : F → G is an isomorphism. It is not difficult to see that ϕ is the identity map on Q (which is an ordered subfield of both F and √ G).√ We have ( 2)2 = 2 in √ F. Because√ϕ is an isomorphism, we must have (ϕ( 2))2 = ϕ(2) = 2. Let ϕ( 2) = a + b 3. If ab 6= 0, then √ √ (a + b 3)2 = a2 + 2 3 ab + 3b2 = 2, which implies
√
2 − a2 − 3b2 , 2ab a contradiction, because the number on the right hand side p is rational. If a = 0, then b2 = 2/3, which is not possible, because 2/3 is not a rational number (prove it!). If b = 0, then a2 = 2, a contradiction. It follows that F and G are not isomorphic. 3=
2.26. It suffices to show that every open interval (a, b) in R contains an irrational number (cf. Definition 2.6). Because Q is dense in R (cf. Theorem 2.14), there is a rational number p such that √ a < p < b. By the Archimedean Prop2 1√ erty, there is n ∈ N such that n > , which is equivalent to p + 2 < b. b−p n √ Clearly, p + 2/n is an irrational number that belongs to (a, b). 2.27. Let (A, B) be a cut of F. Suppose that c < c0 are two cut points for (A, B). The point d = (c + c0 )/2 satisfies c < d < c0 . Because c and c0 are cut points, d must belong to both sets A and B which is impossible. 2.28. Let c be a cut point for a cut (A, B). Clearly, c is an upper bound of the set A (cf. Definition 2.11). Any other upper bound of A must be in the set B. Hence, c is the least upper bound of A. Similarly, c is the greatest lower bound of the set B.
10