Solutions to Exercises

In this chapter we present solutions to some exercises along with hints for helping to solve others.

Solutions to Exercises in Chapter I

Exercise 1.8. We prove ﬁrst the validity of the associative law of addition, that is, the equality

n + (m + p) = (n + m) + p (1) for all n, m, p N. We do this by induction on p. For p = 0, the assertion is ∈ clear. Suppose, then, that assertion (1) holds for an arbitrary but ﬁxed p N ∈ and for all n, m N. Then from the deﬁnition of addition and the induction hypothesis, we have∈ the equality

(n + m) + p∗ = ((n + m) + p)∗ = (n + (m + p))∗ = n + (m + p)∗ = n + (m + p∗), as desired. This proves the associativity of addition. The ﬁrst distributive law, that is, the equality

(n + m) p = n p + m p (2) · · · for all n, m, p N, can be proved using the associativity and commutativity ∈ of addition along with induction on p. For p = 0, the assertion is clear. Sup- pose now that (2) holds for an arbitrary but ﬁxed p N and for all n, m N. Then we have ∈ ∈

(n + m) p∗ = (n + m) p + (n + m) = (n p + m p) + (n + m) · · · · = (n p + n) + (m p + m) = n p∗ + m p∗, · · · · as desired. This proves the asserted distributivity. Using the ﬁrst distributive law, one can then prove the commutativity of multiplication by induction. This then also yields the validity of the second distributive law. Finally, one can prove the associativity of multiplication by induction.

Exercise 1.10. Let m, n N. If m = 0 or n = 0, then one immediately has ∈ the equality m n = 0 by Deﬁnition 1.5 (2) and the commutativity of mul- · © Springer International Publishing AG 2017 J. Kramer and A.-M. von Pippich, From Natural Numbers to Quaternions, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-69429-0 248 Solutions to Exercises tiplication. To prove the converse, we assume m = 0 and n = 0 and prove 6 6 the inequality m n = 0. Since m = 0 and n = 0, there exist a,b N with · 6 6 6 ∈ m = a∗ = a + 1 and n = b∗ = b + 1. Therefore, m n = m b∗ = (m b) + m = (m b) + (a + 1) = (m b + a) + 1 = (m b + a)∗, · · · · · · that is, the natural number m n is the successor of the natural number · m b + a. By the third Peano axiom, we must then have m n = 0. · · 6 Exercise 1.14. The power law from Lemma 1.13 can by proved by induction.

Exercise 1.17. The proof of properties (i), (ii), and (iii) of Remark 1.16 are left to the reader.

Exercise 1.20. Properties (i) and (ii) of Remark 1.19 can be proved by induction.

Exercise 1.23. We leave it to the reader to come up with suitable examples.

Exercise 1.25. Let m, n N with m n. We ﬁrst prove the existence of a ∈ ≥ natural number x N with n + x = m. If m = n, then for x = 0, we have the ∈ equality n + x = n + 0 = m. If m > n, then there exists a N, a > 0, such ∈ that m = n∗···∗ (a times). Then for x = 0∗···∗ (a times), we have the equality n + x = n + 0 = n = m. To prove uniqueness, let y N be another ∗···∗ ∗···∗ ∈ natural number with n + y = m. If x < y, then by Remark 1.19 (i) and the commutativity of addition, we have the inequality

m = n + x = x + n < y + n = n + y = m, that is, we conclude that m < m, a contradiction. If x > y, then there follows analogously the contradiction m > m. Therefore, we must have x = y, proving the asserted uniqueness.

Exercise 2.5. (a) By assumption we have 3 (a a + 1), that is, there exists | 1 ··· k n N with a a + 1 = 3 n. If now 3 a for some j 1,. . .,k , then ∈ 1 ··· k · | j ∈ { } by Lemma 2.4 (ix), we have 3 (a a ), that is, there exists m N with | 1 ··· k ∈ a a = 3 m. We therefore have the equality 1 ··· k · 1 = 3 n a a = 3 n 3 m = 3 (n m), · − 1 ··· k · − · · − a contradiction. Thus none of the numbers a1,..., ak is divisible by 3. (b) We assume that none of the numbers a1 + 1,. . ., ak + 1 is divisible by 3. One ﬁrst shows that then one must have a + 1 = 3 n + r for certain n N j · j j j ∈ and r 1,2 (j = 1,. . .,k), from which follows a = 3 n + (r 1), which j ∈ { } j · j j − for j = 1,. . .,k implies the equality rj = 2, since by part (a), no aj is divisible by 3. By multiplying out, one shows that the number Solutions to Exercises in Chapter I 249

k a1 ak 1 = ∏(3 nj + 1) 1 ··· − j=1 · − is divisible by 3. But this contradicts the assumption that a a + 1 is di- 1 ··· k visible by 3. Therefore, at least one of the numbers a1 + 1,. . ., ak + 1 must be divisible by 3.

Exercise 2.12. To make the proceedings clear, we calculate first, with a1 = 2 and a = (a 1) a + 1 for n N, n 1, the numbers a = 3, a = 7, n+1 n − · n ∈ ≥ 2 3 a = 43, a = 1807, a = (1807 1) 1806 + 1 = 3263443. With = p P 4 5 6 − · Mn { ∈ | p a , we obtain the first six sets, | n} = 2 , = 3 , = 7 , = 43 , M1 { } M2 { } M3 { } M4 { } = 13, 139 , = 3263443 . M5 { } M6 { } We claim that we have the equality a = 5 b + r with b N and r 2,3 n · n n n ∈ n ∈ { } (n N, n 1). This can be proved by induction on n. If n = 1, the assertion is ∈ ≥ clear. We now assume that the assertion holds for arbitrary but fixed n N, ∈ n 1. Then we have ≥ a = (a 1) a + 1 = (5 b + r 1) (5 b + r ) + 1 n+1 n − · n · n n − · · n n = 5 (5b2 + 2b r b ) + r2 r + 1. · n n n − n n − n If r = 2, then r2 r + 1 = 3; if r = 3, then r2 r + 1 = 5 + 2. Therefore, in n n − n n n − n both cases, a is of the form 5 b + r with b N and r 2,3 , n+1 · n+1 n+1 n+1 ∈ n+1 ∈ { } as desired. We have thus shown that 5 - a , that is, 5 / , for all n N, n ∈ Mn ∈ n 1. ≥ Exercise 2.13. Let us assume, contrary to the assertion, that there are only N finitely many prime numbers p1,..., pn in 2 + 3 . We then consider the natural number · a := 3 p p 1. · 1 ··· n − We have that a > 1, and by Lemma 2.9, a has a prime divisor p. Since 3 - a, it follows that p = 3. We now show that p 2 + 3 N; that is, we must show 6 ∈ · that 3 (p + 1). If p = a, we are done. If p < a, then there exists q N, q > 1, | ∈ with a = p q. Since 3 (p q + 1), it follows by Exercise 2.5 (b) that 3 (p + 1) · | · | or 3 (q + 1). In the first case, we are done. In the second case, we repeat the | process for a prime divisor of q. Finally, after finitely many steps, we obtain a prime divisor p of a with p 2 + 3 N. We now proceed as in Euclid’s proof, for on the assumption that∈ there· are only finitely many prime numbers in the set 2 + 3 N, we must have p p ,..., p . In particular, we have · ∈ { 1 n} p (p p ). However, since we have the divisibility relation p a, we must | 1 ··· n | have p 1 from the divisibility rules, which is a contradiction. | Exercise 2.15. We prove the contrapositive of the asserted implication. 250 Solutions to Exercises

(i) Suppose that n is not a prime. Then there exist natural numbers a,b N ∈ with n = a b and 1 < a,b < n. We thus obtain · n a b a a (b 1) a (b 2) a 2 1 = 2 · 1 = (2 1) (2 · − + 2 · − + + 2 + 1). − − − · ··· Since 1 < 2a 1 < 2n 1, it follows that 2a 1 is a nontrivial divisor of 2n 1, which− proves that− 2n 1 is not prime. − − − (ii) Let n N, n > 0, with n not a power of 2. Then n > 2, and there exist ∈ natural numbers a,b N, b odd, with n = a b and 1 a < n, 1 < b n. ∈ · ≤ ≤ Since b is odd, we obtain

n a b a a (b 1) a 2 + 1 = 2 · + 1 = (2 + 1) (2 · − 2 + 1). · ∓ · · · − Since 1 < 2a + 1 < 2n + 1, it follows that 2a + 1 is a nontrivial divisor of 2n + 1, which proves that 2n + 1 is not prime.

Exercise 2.18. Let a ,..., a and b ,..., b denote the sets of all divisors { 1 n} { 1 m} of a and all divisors of b. Since a and b are relatively prime, the set of all divisors of a b is equal to a b j = 1,. . ., n; k = 1,. . ., m . It follows that · { j · k | } n m S(a) S(b) = (a1 + + an) (b1 + + bm) = ∑ ∑ aj bk = S(a b). · ··· · ··· j=1 k=1 · ·

This completes the proof of the assertion.

Exercise 2.19. The assertion can be proved by induction on m.

Exercise 2.20. (a) We have the equalities

S(220) 220 = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284, − S(284) 284 = 1 + 2 + 3 + 71 + 142 = 220. − Therefore, we have S(220) = 220 + 284 = S(284), which proves that the numbers 220 and 284 are amicable. (b) We must show that S(a) = a + b = S(b). Since x,y,z are distinct odd primes, it follows that

S(a) = S(2n x y) = S(2n) S(x) S(y) = (2n+1 1)(x + 1)(y + 1), · · · · − S(b) = S(2n) S(z) = (2n+1 1)(z + 1), · − where we have used the well-known equality S(2n) = 2n+1 1. A direct − calculation shows that x y = 9 22n 1 9 2n 1 + 1, and therefore, x y + · · − − · − · x + y = z. This implies S(a) = (2n+1 1)(z + 1) = S(b). Finally, we calculate − Solutions to Exercises in Chapter I 251

n n 2n n 1 2n 1 n+1 a + b = 2 (x y + z) = 2 (9 2 9 2 − ) = 2 − 9 (2 1) · · · · − · · · − = (z + 1) (2n+1 1) = S(a) = S(b), · − which proves that the numbers a and b are amicable.

Exercise 3.2. We obtain the prime factorizations 720 = 24 32 5, 9797 = 97 101 and 360360 = (23 32 5)360 = 21080 3720 5360. Finally,· on using· the third· binomial formula four· times,· we obtain· ·

232 1 = (22 1) 22 + 1 24 + 1 28 + 1 216 + 1 − − · · · · = 3 5 17 257 65537. · · · · Exercise 3.7. Let a = 232 1 and b = 255 with prime factorizations (see Ex- ercise 3.2) −

a = ∏ pap = 3 5 17 257 65537, b = ∏ pbp = 3 5 17; p P · · · · p P · · ∈ ∈ here a = 1, a = 1, a = 1, a = 1, a = 1, a = 0 for all p P 3 5 17 257 65537 p ∈ \ 3,5,17,257,65537 and b = 1, b = 1, b = 1, b = 0 for all p P 3,5,17 . { } 3 5 17 p ∈ \{ } Therefore, b a for all p P, which by the criterion of Lemma 3.5, proves p ≤ p ∈ that b a. | Exercise 4.6. With the help of Theorem 4.3, we obtain (3600, 3240) = 360, (360360, 540180) = ((23 32 5)360,(22 33 5)180) = 2360 3540 5180, (232 1, 38 28) = 5, where for· the· last equality,· · we used the prime· · factorization− from− Exercise 3.2 and the prime factorization 38 28 = (32 22) (32 + 22) (34 + 24) = 5 13 97. − − · · · · Exercise 4.13. We have (2880, 3000, 3240) = (120, 3240) = 120 and [36, 42, 49] = [252, 49] = 1764.

Exercise 4.15. For example, the numbers a1 = 6, a2 = 10, a3 = 15 are relatively prime, since (a1, a2, a3) = (6,10,15) = (2,15) = 1. The numbers a1, a2, a3, however, are not pairwise relatively prime, since we have (a1, a2) = 2. N Exercise 4.17. Let a1,..., an . We leave to the reader the proof of the following equivalence: ∈

(a ,..., a ) [a ,..., a ] = a a a ,..., a pairwise relatively prime. 1 n · 1 n 1 ··· n ⇐⇒ 1 n This proves the desired criterion.

Exercise 5.2. We obtain 773 = 2 337 + 99. Further, we calculate 25 34 52 = (22 32) (23 32 52) = (5 7 + 1)· (23 32 52) = 7 (23 32 53) + (23· 32· 52). Since· 216· + 1·= 4·8 + 1, it follows· · that· 232 · 1 = (2·16 ·1)(4·8 + 1) + 0.· · − − 252 Solutions to Exercises

Exercise 5.4. This process can be carried out for arbitrary natural numbers g > 1. One obtains the unique representation

` ` 1 1 0 n = q` g + q` 1 g − + + q1 g + q0 g · − · ··· · · with natural numbers 0 q g 1 (j = 0,. . ., `) and q = 0, called the g-adic ≤ j ≤ − ` 6 representation of the natural number n.

Solutions to Exercises in Chapter II

Exercise 1.3. For a, b, c , one has the equalities ∈ Rn (a b) c = (a + b) c = ( (a + b) + c), ⊕ ⊕ Rn ⊕ Rn Rn a (b c) = a (b + c) = (a + (b + c)). ⊕ ⊕ ⊕ Rn Rn Rn Division with remainder of a + b and b + c by n yields the uniquely determined numbers q , q N such that 1 2 ∈ a + b = q n + (a + b) and b + c = q n + (b + c), 1 · Rn 2 · Rn whence follows

( (a + b) + c) = (a + b + c q n) = (a + b + c) Rn Rn Rn − 1 · Rn = (a + b + c q n) = (a + (b + c)). Rn − 2 · Rn Rn We have thereby shown that the operation is associative. Analogously, one can prove that the operation is associative.⊕

Exercise 1.4. (a) The set 2 N = 2 n n N of even natural numbers is a · { · | ∈ } nonempty subset of N. If 2 m, 2 n 2 N, then · · ∈ · 2 m + 2 n = 2 (m + n) 2 N, (2 m) (2 n) = 2 (m 2 n) 2 N. · · · ∈ · · · · · · · ∈ · Thus both + and are operations on 2 N. Since the operations + on N and on N are associative,· it follows that in· particular, the operations + on 2 N and· on 2 N are associative. Therefore, both (2 N, +) and (2 N, ) are· semigroups.· · · · · The set 2 N + 1 = 2 n + 1 n N of odd natural numbers is a · { · | ∈ } nonempty subset of N. If 2 m + 1 and 2 n + 1 are in 2 N + 1, then · · · (2 m + 1) + (2 n + 1) = 2 (m + n + 1) 2 N, · · · ∈ · (2 m + 1) (2 n + 1) = 2 (m 2 n + m + n) + 1 2 N + 1. · · · · · · ∈ · Solutions to Exercises in Chapter II 253

Therefore, while is an operation on 2 N + 1, we see that + is not an operation on 2 N + 1. Therefore,· (2 N + 1, +)· is not a semigroup. The operation on N is associative,· and so in particular,· the operation on 2 N + 1 is asso-· ciative. Therefore, (2 N + 1, ) is a semigroup. · · · · (b) Let k N, k > 1. The set k N = k n n N is a nonempty subset of ∈ · { · | ∈ } N. One shows, as in (a), that both (k N, +) and (k N, ) are semigroups. · · · Exercise 1.5. Because of the inequality

3 3 2 3 2 6 9 2 (2 3) 2 = (2 ) 2 = (2 ) = 2 · = 2 = 2 = 2 (3 ) = 2 (3 2), ◦ ◦ ◦ 6 ◦ ◦ ◦ the operation on N is not associative, and therefore (N, ) is not a semigroup. ◦ ◦

Exercise 1.8. If A = a is a one-element set, then 1 { 1} map(A ) = id , 1 { } where the mapping id : A A is given by the assignment a a . The 1 −→ 1 1 7→ 1 semigroup (map(A ), ) is abelian. If A = a , a ,... is an arbitrary set 1 ◦ 2 { 1 2 } that contains at least two elements a = a , then 1 6 2 map(A ) = id, f , g,... , 2 { } where the mapping id : A A is given by a a (a A ), the mapping 2 −→ 2 j 7→ j j ∈ 2 f : A A by a a , a a (a A a ), and the mapping g : A 2 −→ 2 1 7→ 2 j 7→ j j ∈ 2 \{ 1} 2 −→ A by a a , a a (a A a ). But then we have 2 2 7→ 1 j 7→ j j ∈ 2 \{ 2} ( f g)(a ) = f (g(a )) = f (a ) = a = a = g(a ) = g( f (a )) = (g f )(a ), ◦ 1 1 1 2 6 1 2 1 ◦ 1 whence (map(A ), ) is a nonabelian semigroup. 2 ◦

Exercise 1.12. Let e` be a left identity element and er a right identity element of H. Then e = e e = e , ` ` ◦ r r where the ﬁrst equality follows from the fact that er is a right identity element of H, and the second from the fact that e` is a left identity element of H.

Exercise 1.14. (a) By Exercise 1.4, (2 N,+) and (2 N, ) are semigroups. It remains to show that there exists an· additive identity· · element in 2 N. By the deﬁnition of addition, 0 is this element. Since 1 2 N, there is· no multiplicative identity element, so that (2 N, ) is only a6∈ semigroup.· · · (b) We leave it to the reader to ﬁnd other examples of semigroups that are not monoids. 254 Solutions to Exercises

Exercise 2.3. (a) Suppose that g0 and g00 are two inverse elements to an element g G. Then ∈ g0 = g0 e = g0 (g g00) = (g0 g) g00 = e g00 = g00 , ◦ ◦ ◦ ◦ ◦ ◦ where the second equality follows from the fact that g00 is in particular a right inverse to g, and the fourth equality follows from the fact that g0 is in particular a left inverse to g.

(b) Let g`0 be a left inverse and gr0 a right inverse to an element g G. Then it follows that ∈

g0 = g0 e = g0 (g g0 ) = (g0 g) g0 = e g0 = g0 , ` ` ◦ ` ◦ ◦ r ` ◦ ◦ r ◦ r r analogously to part (a).

Exercise 2.6. (a) Let g 1 G be the inverse element to g G. Then − ∈ ∈ 1 1 g g− = e = g− g. ◦ ◦ 1 1 1 Thus g is the inverse element to g− , that is, (g− )− = g. (b) Let g 1 G be the inverse element to g G and h 1 G the inverse − ∈ ∈ − ∈ element to h G. Then ∈ 1 1 1 1 1 1 (h− g− ) (g h) = h− (g− g) h = h− e h = h− h = e, ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 1 1 1 1 1 1 (g h) (h− g− ) = g (h h− ) g− = g e g− = g g− = e. ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ Thus h 1 g 1 is the inverse element to g h, that is, (g h) 1 = h 1 g 1. − ◦ − ◦ ◦ − − ◦ − The calculational rules (c) and (d) follow directly from the deﬁnition.

Exercise 2.9. We compare only the groups that have the same numbers of elements. The Cayley tables of the groups ( 4, ), ( 5 0 , ), and (D4, ) have, reading from left to right, the followingR ⊕ form:R \{ } ◦

0 1 2 3 1 2 3 4 d d s s ⊕ ◦ 0 1 0 1 0 0 1 2 3 1 1 2 3 4 d0 d0 d1 s0 s1 1 1 2 3 0 2 2 4 1 3 d1 d1 d0 s1 s0 2 2 3 0 1 3 3 1 4 2 s0 s0 s1 d0 d1 3 3 0 1 2 4 4 3 2 1 s1 s1 s0 d1 d0 One may conclude from the Cayley tables that all three groups under consideration are abelian. We now determine the smallest nonzero natural number n such that gn = e for g = e. In the group ( , ), we have e = 0 and 6 R4 ⊕ 12 = 2, 13 = 3, 14 = 0; 22 = 0; 32 = 2, 33 = 1, 34 = 0.

In the group ( 0 , ), we have e = 1 and R5 \{ } Solutions to Exercises in Chapter II 255

22 = 4, 23 = 3, 24 = 1; 32 = 4, 33 = 2, 34 = 1; 42 = 1.

Thus in each group there are two elements with n = 4 and one element with n = 2. In the group (D , ), however, d2 = s2 = s2 = e with e = d , that is, 4 ◦ 1 0 1 0 there is no element with n = 4. The Cayley tables for ( 6, ) and (D6, ), reading from left to right, have the following form: R ⊕ ◦

0 1 2 3 4 5 d d d s s s ⊕ ◦ 0 1 2 0 1 2 0 0 1 2 3 4 5 d0 d0 d1 d2 s0 s1 s2 1 1 2 3 4 5 0 d1 d1 d2 d0 s2 s0 s1 2 2 3 4 5 0 1 d2 d2 d0 d1 s1 s2 s0 3 3 4 5 0 1 2 s0 s0 s1 s2 d0 d1 d2 4 4 5 0 1 2 3 s1 s1 s2 s0 d2 d0 d1 5 5 0 1 2 3 4 s2 s2 s0 s1 d1 d2 d0 From these tables, one can see that the group ( , ) is abelian. The group R6 ⊕ (D , ) is nonabelian, since s s = d = d = s s . We again determine 6 ◦ 0 ◦ 1 2 6 1 1 ◦ 0 the smallest nonzero natural number n such that gn = e for g = e. In ( , ), 6 R6 ⊕ we have e = 0 and

12 = 2, 13 = 3, 14 = 4, 15 = 5, 16 = 0; 22 = 4, 23 = 0; 32 = 0; 42 = 2, 43 = 0; 52 = 4, 53 = 3, 54 = 2, 55 = 1, 56 = 0.

There exist, therefore, in ( , ) two elements with n = 6, two elements R6 ⊕ with n = 3, and one element with n = 2. In (D , ), we have e = d and 6 ◦ 0 2 3 2 3 2 2 2 d1 = d2, d1 = d0; d2 = d1, d2 = d0; s0 = d0, s1 = d0, s2 = d0.

There exist, therefore, in (D , ) no element with n = 6, two elements with 6 ◦ n = 3, and three elements with n = 2.

Exercise 2.10. (a) One shows using the Cayley table

1 2

1 1 2 2 2 1 for ( 3 0 , ) and the Cayley table from Exercise 2.9 for ( 5 0 , ) that (R \{ 0} , ) and ( 0 , ) are groups. R \{ } R3 \{ } R5 \{ } (b) We leave to the reader the task of verifying the assertions of Exam- ple 2.8 (iii) regarding the dihedral group (D , ). 2n ◦ (c) Let n 3. We consider the elements ≥ 1 2 3 n 1 2 3 n π = ··· , π = ··· 1 2 1 3 n 2 3 1 2 n ··· ··· 256 Solutions to Exercises of Sn, where for n > 3, each of the elements 4,. . .,n is mapped to itself. Then 1 2 3 n 1 2 3 n π π = ··· = ··· = π π , 1 ◦ 2 3 2 1 n 6 1 3 2 n 2 ◦ 1 ··· ··· where for n > 3, each of the elements 4,. . .,n is mapped to itself. This proves that (S , ) for n 3 is nonabelian. n ◦ ≥ Exercise 2.13. This is proved by induction on n.

Exercise 2.19. We have S = π , π , π , π , π , π with 3 { 1 2 3 4 5 6} 1 2 3 1 2 3 1 2 3 π = , π = , π = , 1 1 2 3 2 2 3 1 3 3 1 2 1 2 3 1 2 3 1 2 3 π = , π = , π = . 4 1 3 2 5 3 2 1 6 2 1 3

We calculate ord(π1) = 1, ord(π2) = ord(π3) = 3, and ord(π4) = ord(π5) = ord(π6) = 2.

Exercise 2.23. Since dk = d (k = 0,. . .,n 1) and dn = d , we have d = 1 k − 1 0 h 1i d0,..., dn 1 . Using the subgroup criterion, one can show that the nonempty { − } subset g = ..., (g 1)2, g 1, g0 = e, g1 = g, g2,... G is a subgroup of G h i { − − } ⊆ for every group G. In particular, d1 = d0,..., dn 1 is a subgroup of D2n. h i { − }

Exercise 2.26. We have S3 = π1, π2, π3, π4, π5, π6 with πj (j = 1,. . .,n) as in Exercise 2.19. We have the{ cyclic subgroups }

π = π , π = π ,π ,π = π , h 1i { 1} h 2i { 1 2 3} h 3i π = π ,π , π = π ,π , π = π ,π , h 4i { 1 4} h 5i { 1 5} h 6i { 1 6} and the subgroup S3 itself, which is not cyclic. One can see that S3 has no other subgroups.

Exercise 3.3. We have S = π , π , π , π , π , π with π (j = 1,. . .,n) as 3 { 1 2 3 4 5 6} j in Exercise 2.19 and D6 = d0, d1, d2, d0 s0, d1 s0, d2 s0 , where s0 is re- ﬂection in the median of the{ side joining◦ vertices◦ 1 and◦ 2.} By the deﬁnition of the group homomorphism f : D S , we have 6 −→ 3

f (d0) = π1, f (d1) = π3, f (d2) = π2, f (d s ) = π , f (d s ) = π , f (d s ) = π , 0 ◦ 0 6 1 ◦ 0 5 2 ◦ 0 4 which proves that f is bijective and therefore in fact a group isomorphism.

k1 k2 Exercise 3.5. Let dj1 s0 and dj2 s0 with j1, j2 0,. . ., n 1 and k1, k2 ◦ ◦ ∈ { 1 − } ∈ 0, 1 be two elements of D . Since d s = s d− , we have { } 2n j ◦ 0 0 ◦ j Solutions to Exercises in Chapter II 257  d d , if k = 0, k = 0;  j1 j2 1 2  ◦ d d s0, if k1 = 0, k2 = 1; k1 k2 j1 ◦ j2 ◦ (dj1 s0 ) (dj2 s0 ) = 1 ◦ ◦ ◦ dj d− , if k1 = 1, k2 = 1;  1 ◦ j2  1 dj d− s0, if k1 = 1,k2 = 0. 1 ◦ j2 ◦ It follows that

k k k k sgn d s 1 d s 2 = k k = sgn d s 1 sgn d s 2 . j1 ◦ 0 ◦ j2 ◦ 0 1 ⊕ 2 j1 ◦ 0 ⊕ j2 ◦ 0 Therefore, sgn is a group homomorphism, and we have im(sgn) = and R2 ker(sgn) = d j = 0,. . ., n 1 . { j | − } Exercise 3.7. By Lemma 3.6, we have f injective ker( f ) = e . It there- ⇐⇒ { G} fore sufﬁces to prove, under the assumption G < ∞, the equivalence f in- | | jective f surjective. We have ⇐⇒ f injective g = h for g,h G implies f (g) = f (h) ⇐⇒ 6 ∈ 6 f (G) = G ⇐⇒G <∞ | | | | | | for every g G there exists h G with f (h) = g ⇐⇒ ∈ ∈ f surjective. ⇐⇒ This proves the assertion.

Exercise 3.8. If g G and ord(g) = n, then e = f (e) = f (gn) = f (g)n, which ∈ implies ord f (g) n = ord(g). ≤ Exercise 3.9. Suppose that f : D S is a group isomorphism. Then for 24 −→ 4 each g D , we must have the equality ∈ 24 ord(g) = ord( f (g)).

Since ord(d ) = 12, we must have that f (d ) S is an element of order 2 2 ∈ 4 12. We ﬁrst determine the orders of the elements of S4. We obtain the nine elements of order 2,

1 2 3 4 1 2 3 4 1 2 3 4 , , , 2 1 3 4 3 2 1 4 4 2 3 1 1 2 3 4 1 2 3 4 1 2 3 4 , , , 1 3 2 4 1 4 3 2 1 2 4 3 1 2 3 4 1 2 3 4 1 2 3 4 , , , 2 1 4 3 3 4 1 2 4 3 2 1 the eight elements of order 3, 258 Solutions to Exercises

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 , , , , 2 3 1 4 3 1 2 4 2 4 3 1 4 1 3 2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 , , , , 3 2 4 1 4 2 1 3 1 3 4 2 1 4 2 3 and the six elements of order 4,

1 2 3 4 1 2 3 4 1 2 3 4 , , , 2 3 4 1 2 4 1 3 3 4 2 1 1 2 3 4 1 2 3 4 1 2 3 4 , , . 3 1 4 2 4 3 1 2 4 1 2 3

The group S4 has, therefore, only elements of orders 1, 2, 3, and 4. There can therefore be no group isomorphism between D24 and S4.

Exercise 3.11. (a) If f : ( , ) ( , ) is a group homomorphism, then R4 ⊕ −→ R4 ⊕ f (0) = 0. Since = 1 , it follows that R4 h i f (2) = f (1 1) = f (1) f (1), f (3) = f (1 1 1) = f (1) f (1) f (1), ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ that is, f is uniquely determined by specifying the image of f (1). Therefore, there are precisely four distinct group homomorphisms, f1, f2, f3, f4, given by the assignments

f1(0) = 0, f1(1) = 0, f1(2) = 0, f1(3) = 0, f2(0) = 0, f2(1) = 1, f2(2) = 2, f2(3) = 3, f3(0) = 0, f3(1) = 2, f3(2) = 0, f3(3) = 2, f4(0) = 0, f4(1) = 3, f4(2) = 2, f4(3) = 1, whence follows ker( f ) = , ker( f ) = 0 , ker( f ) = 0,2 , ker( f ) = 0 , 1 R4 2 { } 3 { } 4 { } im( f ) = 0 , im( f ) = , im( f ) = 0,2 , im( f ) = . This shows in par- 1 { } 2 R4 3 { } 4 R4 ticular that f2 and f4 are bijective. (b) Since = 1 , every group homomorphism g : ( , ) ( , ) Rp h i Rp ⊕ −→ Rn ⊕ is uniquely determined by specifying the image g(1). One now shows that only g(1) = 0 is possible, since n and p are relatively prime. There is, therefore, only one group homomorphism g, which is given by the assignment g(m) = 0 (m ). We have ker(g) = and im(g) = 0 . ∈ Rp Rp { } Exercise 4.3. (a) The veriﬁcation of the statement of Example 4.2 is left to the reader. (b) The order relation is not an equivalence relation on N, since is not symmetric. ≤ ≤ (c) The relation is not an equivalence relation on N, since is not transi- tive. ∼ ∼ Solutions to Exercises in Chapter II 259

Exercise 4.6. Let M be a set with an element m M. The equivalence class ∈ of m with respect to equality “=” is M = m M m = m , that is, the m { 0 ∈ | 0 } set of all elements of M that are equal to m. We leave it to the reader to ﬁnd additional equivalence relations and determine the associated equivalence classes.

Exercise 4.11. We leave the solution of this exercise to the reader.

Exercise 4.12. Let U = π = π , π . The left coset of an element π h 4i { 1 4} ∈ S with respect to U is given by π U = π π , π π . Therefore, the 3 ◦ { ◦ 1 ◦ 4} following are all left cosets of S3 with respect to U:

π U = U = π U, 1 ◦ 4 ◦ π U = π , π = π U, 2 ◦ { 2 6} 6 ◦ π U = π , π = π U. 3 ◦ { 3 5} 5 ◦ Exercise 4.15. (a) If g G, then ord(g) = U for U = g G, which implies ∈ | | h i ≤ by Lagrange’s theorem that ord(g) G . | | | (b) Suppose G = p for a prime number p. If g G, g = e, then ord(g) > 1, | | ∈ 6 and therefore, by part (a), we must have the equality ord(g) = p, which implies G = g . h i (c) Let G = 4, and write G = e, a, b, c . If G has an element g G with | | { } ∈ ord(g) = 4, then G = g , that is, G is cyclic, and therefore isomorphic to the h i group ( , ). If G has no element of order 4, then every element g G, R4 ⊕ ∈ g = e, has order 2, that is, a2 = b2 = c2 = e. Since G is a group, it follows 6 that a b = c = b a, a c = b = c a, and b c = a = c b. Therefore, G is ◦ ◦ ◦ ◦ ◦ ◦ isomorphic to the group (D4, ). Therefore, up to group isomorphism, there are precisely two groups of order◦ 4, given by the following Cayley tables:

e a b c e a b c ◦ ◦ e e a b c e e a b c a a b c e a a e c b b b c e a b b c e a c c e a b c c b a e Both groups are abelian. One shows further that up to group isomorphism, the only groups of order 6 are ( , ) and (D , ). Thus every abelian group of order 6 is isomorphic R6 ⊕ 6 ◦ to ( , ), and every nonabelian group of order 6 is isomorphic to (D , ). R6 ⊕ 6 ◦ Exercise 4.19. Exercises 4.11 and 4.12 can be solved analogously for right cosets.

Exercise 4.21. Let U = π = π , π . Then U π = π , π = π , π = h 4i { 1 4} ◦ 2 { 2 5} 6 { 2 6} π U, which implies that U is not a normal subgroup of S . Analogously, 2 ◦ 3 one can show that π and π are not normal subgroups of S . h 5i h 6i 3 260 Solutions to Exercises

Exercise 4.24. (a) For every element h H, one has h H = H = H h. Now ∈ ◦ ◦ let g G H. Then g H = H and H g = H. Because of [G : H] = 2, we ∈ \ ◦ 6 ◦ 6 obtain the disjoint decomposition H ˙ (g H) = G = H ˙ (H g) of G. It ∪ ◦ ∪ ◦ then follows that g H = H g must hold. Altogether, one has the equality ◦ ◦ g H = H g for all g G. This proves that H is a normal subgroup of G. ◦ ◦ ∈ (b) The mapping f : G , given by −→ R2 ( 0, if g H; f (g) = ∈ 1, if g / H, ∈ is a surjective group homomorphism.

Exercise 4.26. Since ker( f ) is a subgroup, indeed a normal subgroup, of S3, it must be the case that ker( f ) is one of the groups π , A , S . If ker( f ) = { 1} 3 3 π , then f is injective, which is impossible because of 6 = S > = 3. { 1} | 3| |R3| If ker( f ) = A , then ord(π ) < ord( f (π )), which is impossible because { 3} 4 4 of Exercise 3.8. The only possibility is then ker( f ) = S3, which implies that f (π) = 0 for all π S . ∈ 3 Exercise 5.10. Let f : G be the surjective group homomorphism of −→ R2 Exercise 4.24. Then ker( f ) = H, and by Corollary 5.8, we have an isomorphism G/H ∼= 2. This isomorphism can also be read off from the Cayley table R H g H • ◦ H H g H ◦ g H g HH ◦ ◦ for the group G/H = H, g H , where g G H is arbitrary, and H = { ◦ } ∈ \ e H. We note here that (g H) (g H) = H must hold, since otherwise, G ◦ ◦ • ◦ we would have from (g H) (g H) = (g g) H = g H the equalities ◦ • ◦ ◦ 1◦ ◦ g g h = g h g h = h g = h h− for certain h , h H, and ◦ ◦ 1 ◦ 2 ⇐⇒ ◦ 1 2 ⇐⇒ 2 ◦ 1 1 2 ∈ thus the contradiction g H. ∈ Exercise 5.11. It is clear that from H E G, K E G, and K H, one has K E H. ⊆ That H/K is a normal subgroup of G/K comes immediately from the proof of isomorphism. To this end, we deﬁne the mapping f : G/K G/H by the assignment −→ g K g H. ◦ 7→ ◦ Since K H, it follows that f is well deﬁned. On account of K E G and ⊆ H E G, it is clear that f is a homomorphism. Furthermore, we have

ker( f ) = g K f (g K) = H = g K g H = H { ◦ | ◦ } { ◦ | ◦ } = g K g H = H/K. { ◦ | ∈ } Solutions to Exercises in Chapter III 261

This proves in particular that H/K E G/K. Since f is surjective, it follows from the homomorphism theorem that

(G/K)/(H/K) = (G/K)/ker( f ) ∼= G/K, as claimed.

Exercise 6.4. (a) Let A = a , a ,... be a set with a = a . Then map(A) = { 1 2 } 1 6 2 id, f , g,... , whereby the mapping id : A A is given by the assignment { } −→ a a (a A), the mapping f : A A by a a , a a (a A a ), j 7→ j j ∈ −→ 1 7→ 2 j 7→ j j ∈ \{ 1} and the mapping g : A A by a2 a1, aj aj (aj A a2 ). It then follows that −→ 7→ 7→ ∈ \{ }

( f g)(a ) = f (a ) = a = ( f id)(a ), ( f g)(a ) = f (a ) = a = ( f id)(a ), ◦ 1 1 2 ◦ 1 ◦ 2 1 2 ◦ 2 which proves f g = f id; however, we have g = id, which proves that in ◦ ◦ 6 the semigroup (map(A), ), the ﬁrst cancellation law is invalid. One may show analogously◦ that the second cancellation law is also invalid. (b) We leave it to the reader to ﬁnd further examples of semigroups that are not regular.

Exercise 6.6. (a) The solution to this exercise is left to the reader. (b) On (2 N + 1) (2 N + 1) = (a, b) a, b 2 N + 1 we deﬁne the relation · × · { | ∈ · }

(a, b) (c, d) a d = b c (a, b, c, d 2 N + 1). ∼ ⇐⇒ · · ∈ · If we write a for the equivalence class [a, b] of (a,b) (2 N + 1) (2 N + 1), b ∈ · × · then the group G := ((2 N + 1) (2 N + 1))/ can be identiﬁed with the · a × · ∼N set of all fractions of the form b , where a,b 2 + 1 and a,b are relatively prime. We leave the detailed construction from∈ · Theorem 6.5 to the reader.

Exercise 7.6. The generalization to Z of the addition and multiplication rules in Remark 1.19 of Chapter I is left to the reader.

Exercise 7.9. The veriﬁcation of the assertions of this example are left to the reader.

Solutions to Exercises in Chapter III

Exercise 1.2. The proof of the calculational laws from Lemma 1.1 are left to the reader. 262 Solutions to Exercises

Exercise 1.5. The proof of Theorem 1.4 is left to the reader.

Exercise 1.9. We give an idea of the proof. We assume that

a = e p p = e q q · 1 ··· r · 1 ··· s for e 1 and for prime numbers p ,..., p , q ,...,q (r N, s N), not ∈ { } 1 r 1 s ∈ ∈ necessarily distinct. Since now we have p a and therefore p e q q , it 1 | 1 | · 1 ··· s follows with the help of Euclid’s lemma, Lemma 1.7, that p q for some j = 1 | j 1,. . .,s. Since p1 is prime, we must have p1 = qj. Without loss of generality, we may assume (by renumbering if necessary) that p1 = q1. Application of the cancellation law implies the equality

p p = q q . (3) 2 ··· r 2 ··· s

Since p2 divides the left-hand side of (3), p2 must also divide the right-hand side. As in the ﬁrst step, we conclude that p2 = q2. Proceeding in this way, we obtain the equalities r = s and pj = qj for j = 1,. . .,r, which proves the asserted uniqueness.

Exercise 2.5. We leave it to the reader to prove that the polynomial ring (R[X],+, ) is a commutative ring if and only if (R,+, ) is commutative. · · Exercise 2.6. Let A be a nonempty set and (R,+ , ) a ring. Then, 0 R, R ·R R ∈ and therefore, the mapping 0 : A R, a 0 , is an element of map(A, R). −→ 7→ R Hence, the set map(A, R) is not empty. We now show that + on map(A, R) is associative. To this end, let f , g, h map(A, R). Then for all a A, we have ∈ ∈

(( f + g) + h)(a) = ( f + g)(a) +R h(a) = ( f (a) +R g(a)) +R h(a) def of + def of +

= f (a) +R (g(a) +R h(a)) +R associative, since R is a ring

= f (a) +R (g + h)(a) = ( f + (g + h))(a), def of + def of + which proves that + is associative. The mapping 0 : A R is the identity −→ element with respect to +, since for all f map(A, R), we have the equality ∈

(0 + f )(a) = 0(a) +R f (a) = 0R +R f (a) = f (a) def. of + def. of 0 0R id. el. w.r.t. +R and analogously the equality ( f + 0)(a) = f (a) for all a A. The other ring ∈ properties of map(A, R) are proved similarly using the ring properties of R.

Exercise 2.7. The solution of this exercise is left to the reader. Solutions to Exercises in Chapter III 263

Exercise 2.11. If n > 1 is not prime, then there exist natural numbers a, b ∈ , a > 1, b > 1, with a b = n. But then we have a b = 0. Therefore, a and Rn · b are zero divisors of . Rn Exercise 2.12. We show here only that the lack of zero divisors in (R,+, ) · implies the lack of zero divisors in (R[X],+, ). We assume that the ring R[X] · has zero divisors and show that then the ring R must have zero divisors. Let, then, f (X) = a Xn + + a X + a (a = 0) and g(X) = b Xm + + n · ··· 1 · 0 n 6 m · ··· b X + b (b = 0) be zero divisors in R[X], so that 1 · 0 m 6 f (X) g(X) = (a b ) Xn+m + + (a b + a b ) X + a b = 0, · n · m · ··· 1 · 0 0 · 1 · 0 · 0 where 0 denotes the zero element of R[X], i.e., the zero polynomial. In particular, we must have a b = 0, which proves that R has zero divisors. n · m Exercise 2.13. The ring (map(A, R),+, ) from Exercise 2.6 has as its zero · element 0 : A R, a 0 . If now, for example, we have R = ( , , ), −→ 7→ R R6 ⊕ then for f : A R, a 2 (a A) and g : A R, a 3 (a A), we have the equality −→ 7→ ∈ −→ 7→ ∈

( f g)(a) = f (a) g(a) = 2 3 = 0 = 0(a), · def of def of f , g · that is, f and g are zero divisors of (map(A, R),+, ). Therefore, the ring · (map(A, R),+, ) also has zero divisors if R has zero divisors. We note that · the ring (map(A, R),+, ) can possess zero divisors even if R has no zero divisors. ·

Exercise 2.19. In the polynomial ring (Z[X],+, ), the unit element is given · by the unit polynomial 1. Let f (X) = a Xn + + a X + a (a = 0) be a n · ··· 1 · 0 n 6 unit in Z[X]. Then there exists a polynomial g(X) = b Xm + + b X + m · ··· 1 · b (b = 0) in Z[X] such that 0 m 6 f (X) g(X) = (a b ) Xn+m + + (a b + a b ) X + a b = 1. · n · m · ··· 1 · 0 0 · 1 · 0 · 0 If n > 0, then we must have in particular that a b = 0, in contradiction n · m to the fact that Z has no zero divisors. Therefore, f and hence g must be of the form f (X) = a and g(X) = b . The equality a b = 1 shows that f 0 0 0 · 0 and g are units if and only if a 1, 1 and a = b . The polynomial ring 0 ∈ { − } 0 0 (Z[X],+, ) has therefore only the units 1, 1 . · { − } Exercise 2.20. That the units of a ring (R,+, ) with unit element 1 form a group under multiplication with multiplicative· identity element 1 follows directly from how units are deﬁned.

Exercise 2.21. The group of units of 5 is ( 5 0 , ) and is therefore isomorphic to the group ( , ). The groupR ofR units\{ of} is ( 1, 3, 5, 7 , ). R4 ⊕ R8 { } 264 Solutions to Exercises

We have 3 3 = 1, 5 5 = 1, 7 7 = 1, which shows that this group is isomorphic to (D4, ). The group of units of 10 is ( 1, 3, 7, 9 , ). We have 3 7 = 1, 9 9 = 1,◦ which shows that this groupR is{ isomorphic} to ( , ) R4 ⊕ and therefore to ( 5 0 , ). The group of units of 12 is ( 1, 5, 7, 11 , ). We have 5 5 =R1, 7\{ 7}= 1, 11 11 = 1, which showsR that{ this group} is isomorphic to (D , ) and therefore to ( 1, 3, 5, 7 , ). 4 ◦ { } Exercise 2.25. For n N, we have that (nZ,+, ) is a subring of (Z,+, ). ∈ · · Exercise 2.26. We leave it to the reader to show that (R,+, ) is a subring of · the polynomial ring (R[X],+, ). ·

Exercise 3.2. (a) The mapping f1 is a ring homomorphism. (b) The mapping f2 is not a ring homomorphism, since for g1(X) = X and g (X) = 1, we have f (g (X) g (X)) = f (1 X) = 1 = 0 = 1 0 = f (X) 2 2 1 · 2 2 · 6 · 2 · f (1) = f (g (X)) f (g (X)). 2 2 1 · 2 2 (c) The mapping f3 is a ring homomorphism if and only if r = 0. (d) The mapping f4 is a ring homomorphism. (e) The mapping f5 is a ring homomorphism.

Exercise 3.5. The proof of Lemma 3.4 is left to the reader.

Exercise 3.6. We obtain

n j o ker( f1) = ∑ aj X aj R, a0 = 0 , im( f1) = R. j N · ∈ ∈

The mapping f2 is not a ring homomorphism. Furthermore, we have ker( f ) = map(A, R), im( f ) = 0 (if r = 0); 3 3 { } ker( f ) = g map(A, R) g(a) = 0 , im( f ) = g(a) g map(A, R) ; 4 { ∈ | } 4 { | ∈ } ker( f ) = f (X) R[X] f (r) = 0 , im( f ) = f (r) f (X) R[X] = R, 5 { ∈ | } 5 { | ∈ } for the ring homomorphisms under consideration.

Exercise 3.13. To prove the equality a = R, we must show that a R. To this ⊇ end, let r R. Since 1 a and a is an ideal, it follows that r 1 = r a. This ∈ ∈ · ∈ proves that R a. ⊆ Exercise 3.14. No, since for every subring U of (Z,+, ), we have in particu- · lar that (U,+) is a subgroup of (Z,+). Therefore, we must have U = nZ for some n N, which proves that (U,+) is an ideal of Z. ∈ Exercise 3.15. For example, Z is a subring of the polynomial ring (Z[X],+, ) · that is not an ideal of Z[X]. Solutions to Exercises in Chapter III 265

Exercise 3.16. The principal ideals of Z[X] are of the form h f f Z[X] { · | ∈ } for some h Z[X]. We leave it to the reader to show that the ideal ∈ a := 2 f + X g f , g Z[X] { · · | ∈ } is not a principal ideal of (Z[X],+, ). · Exercise 3.18. We have ker( f ) = (X) and ker( f ) = (X r). If the ring R 1 5 − possesses a unit element, then ker( f3) = (1).

Exercise 3.27. Let a Z. Then the mapping f : Z[X] Z given by the as- ∈ −→ signment f (X) f (a) is a surjective ring homomorphism with ker( f ) = 7→ (X a) by Exercises 3.2, 3.6, and 3.18. Invoking Corollary 3.25, we see that − Z[X]/(X a) = Z is a ring isomorphism, as desired. − ∼ Exercise 3.28. An analogy to the group isomorphism from Exercise 5.11 of Chapter II can also be formulated and proved for rings by replacing “subgroup” and “normal subgroup” throughout with “ideal” and observing that the group homomorphisms that arise are also ring homomorphisms. We leave this task to the reader.

Exercise 4.4. This is impossible, since one can show that every skew field with finitely many elements is in fact a field.

Exercise 5.3. The proofs of associativity of , commutativity of , and the second distributive law are left to the reader.⊕

Exercise 5.5. We consider the ring homomorphism f : K Quot(K) given −→ by a [a, 1]. The ring homomorphism f is well defined, since K is a field, 7→ and therefore, we have 1 K. Moreover, f is injective, which can be shown ∈ as follows: Assume that [a, 1] = [b, 1] for a, b K. Then (a, 1) (b, 1), which ∈ ∼ implies a 1 = 1 b and hence a = b. We now show that f is also surjective. · · To this end, let [a, b] Quot(K) with a, b K, b = 0, be an arbitrary element. ∈ ∈ 1 6 Since K is a field, there exists an inverse b− K to b. Therefore, we have 1 ∈ a b− K and · ∈ 1 1 f (a b− ) = [a b− , 1] = [a, b], · · where the last equality follows from (a b 1, 1) (a, b) a b 1 b = · − ∼ ⇐⇒ · − · 1 a a = a. Therefore, f is a ring isomorphism, and we have the ring · ⇐⇒ isomorphism K ∼= Quot(K). Exercise 6.3. Let r = s/t for s Z and t N 0 . Then r corresponds to the ∈ ∈ \{ } element [s, t] Z (Z 0 ). If now d = (s,t) > 0 is the greatest common ∈ × \{ } divisor of s and t, then we may write s = d a and t = d b with a Z, · · ∈ b N 0 , and a, b relatively prime. From the equality s b = (d a) b = ∈ \{ } · · · (d b) a = t a we infer the equality [s, t] = [a, b]. This proves the existence · · · 266 Solutions to Exercises of the claimed representative. To prove the uniqueness of the representative, we assume that [c, d] with c Z, d N 0 , and c, d relatively prime is also ∈ ∈ \{ } an element such that [s, t] = [c, d]. Then in particular, we have [a, b] = [c, d], and therefore a d = b c. · · If p is a prime number such that p a, then we must have p b c and conse- | | · quently, since p is prime, p b or p c. But since a and b are relatively prime, | | we must have p c. Conversely, one can show that for an arbitrary prime p | with p c, we must also have p a. From this it follows that a c and also c a, | | | | which, since a and c have the same sign, proves the equality a = c. We conclude by an analogous argument the equality b = d, completing the proof of uniqueness.

Exercise 6.4. Since one usually learns this proof in a ﬁrst course in real anal- ysis, we leave this exercise to the reader.

Exercise 6.7. The generalizations of the addition and multiplication rules from Remark 1.19 of Chapter I are left to the reader.

Exercise 7.4. We obtain in Z[X] the decompositions 20X = 22 5 X and · · 10X2 + 4X 6 = 2 (X + 1) (5X 3). Therefore, 2 is the greatest common − · · − divisor, and 22 5 X (X + 1) (5X 3) = 100X3 + 40X2 60X the least · · · · − − common multiple, of the polynomials 20X and 10X2 + 4X 6 in Z[X]. − Exercise 7.9. We leave it to the reader to come up with relevant examples.

Exercise 7.15. Part (ii) of the proof of Lemma 7.14 is left to the reader.

Exercise 7.25. For example, the polynomial ring K[X,Y] in two variables over a ﬁeld K is not a principal ideal domain, since the ideal

a := X f + Y g f , g K[X,Y] { · · | ∈ } is not principal.

Exercise 7.38. (a) We obtain (123456789,555555555) = 9. (b) We calculate

X4 + 2X3 + 2X2 + 2X + 1 = (X + 1) (X3 + X2 X 1) + (2X2 + 4X + 2), · − − 1 1 X3 + X2 X 1 = X (2X2 + 4X + 2) + 0. − − 2 − 2 · We thereby obtain (X4 + 2X3 + 2X2 + 2X + 1, X3 + X2 X 1) = 2X2 + − − 4X + 2. Solutions to Exercises in Chapter IV 267

Solutions to Exercises in Chapter IV

1 1 1 Exercise 1.6. (a) We obtain the decimal expansions 5 = 0.2, 3 = 0.3, 16 = 1 1 0.0625, 11 = 0.09, and 7 = 0.142857. (b) One shows that a reduced fraction a (a, b Z; b = 0) has a terminating b ∈ 6 decimal expansion if and only if b = 2k 5l with k, l N. · ∈ (c) We consider the fraction 1 for m N, m = 0, with 2 - m and 5 - m. One m ∈ 6 shows that m 1 is the maximal period length of the decimal expansion −1 of the fraction m , considering that there can be at most m 1 remainders 1 − on division by m. The fraction 7 , for example, has a period length that is maximal. (d) Without loss of generality, we may assume that the periodic decimal fraction has the form

0.q 1 ... q v q (v+1) ... q (v+p) − − − − with natural numbers v 0, p > 0. Then ≥ ∑v v j ∑p p j a j=1 q j10 − 1 j=1 q (v+j)10 − = − + − . b 10v 10v · 10p 1 − For example, for 0.123, one obtains the fraction

123 123 41 = = . 103 1 999 333 − Exercise 2.2. (a) Without loss of generality, we may assume that e Q, 0 < ∈ e < 1. For n N, we then have ∈

1 1 e < e − < n. n + 1 ⇐⇒ e

If we set N(e) := [(1 e)/e], where [x] is the greatest integer less than or − equal to x, then for all n N with n > N(e), we have the inequality ∈

1 < e. n + 1 This proves that the sequence 1 is a rational null sequence. Using n+1 n 0 the inequality ≥ n 1 < 2n n + 1 N n for n , n 5, one shows analogously that 2n n 0 is also a rational null sequence.∈ ≥ ≥ 268 Solutions to Exercises

1 (b) Other examples of rational null sequences are the sequences k (n+1) n 0 with k N, k 2. ≥ ∈ ≥

Exercise 2.18. Let α = (an) + n, β = (bn) + n be two real numbers with α < β, that is, there exist q Q, q > 0, N(q) N with b a > q for all n ∈ ∈ n − n ∈ N with n > N(q). To prove the asserted independence of the choice of the representing rational Cauchy sequences, we assume that (an) + n = (an0 ) + n and (bn) + n = (bn0 ) + n for rational Cauchy sequences (an0 ) and (bn0 ). Then in particular, we must have (an0 ) = (an) + (cn) = (an + cn) and (bn0 ) = (bn) + (dn) = (bn + dn) for rational null sequences (cn) and (dn). This yields

b0 a0 = (b a ) + (d c ) (b a ) d c . (4) n − n n − n n − n ≥ n − n − | n − n| Since, moreover, (d c ) is a rational null sequence, there exists for e := n − n q/2 Q an Ne(e) N such that for all n N with n > Ne(e), we have the ∈ ∈ ∈ Q inequality dn cn < e. So if we set q0 := q e = q/2 , we have q0 > 0, and with (4)| we− obtain| the inequality − ∈

b0 a0 > q e = q0 n − n − N for all n with n > N(q0) := max(N(q), Ne(e)). This proves the claimed independence∈ of representative.

Exercise 2.22. We leave the proof of Lemma 2.21 to the reader.

Exercise 2.25. Real null sequences whose elements are irrational numbers √2 N include, for example, the sequences k with k , k 1. (n+1) n 0 ∈ ≥ ≥ Exercise 2.30. We begin by considering that for a rational number a Q, 0 ∈ a0 > 0, we have the following equivalences with an error of δ0:

a + δ = √2 (a + δ )2 = 2 2a δ + δ2 = 2 a2 0 0 ⇐⇒ 0 0 ⇐⇒ 0 0 0 − 0 2 2 2 a0 δ0 δ0 = − . ⇐⇒ 2a0 − 2a0 So if we set 2 2 2 a0 2 + a0 a1 := a0 + − = , 2a0 2a0 2 then because a0 > 0, we have the inequality a > 2, that is, a1 > √2. This 1 implies 2 a2 /(2a ) < 0, and we therefore have for − 1 1 2 2 2 a1 2 + a1 a2 := a1 + − = 2a1 2a1 Solutions to Exercises in Chapter IV 269

2 √ both a1 > a2 and a2 > 2, that is, a2 > 2. We consider now the rational se- Q quence (an)n 0 with a0 , a0 > 0, arbitrary and ≥ ∈ 2 2 + an N an+1 := (n ,n 1). (5) 2an ∈ ≥ One ﬁrst shows by induction that both a > a for all n N, n 1, and n n+1 ∈ ≥ a2 > 2, that is, a > √2, for all n N, n 1. Using these inequalities, one n n ∈ ≥ then shows in a second step that (an)n 0 is a rational Cauchy sequence. This rational Cauchy sequence has limit α ≥ R, α > 0. Because of the recurrence formula (5), we see that α satisﬁes the∈ equation

2 + α2 α = α = √2, 2α ⇐⇒ as desired.

Exercise 3.12. (a) The decimal representation 0.101001000100001 . . . is neither terminating nor periodic. Therefore, this number cannot be rational. One can ﬁnd analogous examples, such as 0.121331222133331. . ..

(b) Using the rational Cauchy series (an)n 0 constructed in Exercise 2.30 ≥ for calculating √2, one obtains √2 1.4142135623, accurate to ten decimal ≈ places, by choosing, for example, a0 = 1 and iterating four times.

Exercise 4.2. We consider, for example, the sequence

n2 + 2 (an)n 0 := n . ≥ 2 n 0 ≥ We have a = 2 and a = a = 3/2. For n N, n 3, one shows the inequality 0 1 2 ∈ ≥ 2(n2 + 2) > (n + 1)2 + 2. Since

2 2 2 2 n + 2 (n + 1) + 2 2(n + 2) > (n + 1) + 2 > an > a + ⇐⇒ 2n 2n+1 ⇐⇒ n 1 N for n , n 3, it follows that the sequence (an)n 0 is monotonically decreasing,∈ but≥ not strictly. One shows analogously that≥ the sequences

3 1 n + 3 n+1 12 , n n 0 3 n 0 ≥ ≥ are strictly monotonically decreasing and that the sequence

n3 2 2 − n 2 n 0 − ≥ 270 Solutions to Exercises is monotonically increasing. The sequence

1 n n+1 n 0 ≥ is neither monotonically increasing nor monotonically decreasing. How- ever, the sequence

1 n n+1 n 4 ≥ is strictly monotonically decreasing.

Exercise 4.6. The subset M Q consisting of all sequence terms a (n > 0) ⊆ n in the rational Cauchy sequence (an)n 0 constructed in Exercise 2.30, which ≥ is bounded below, has greatest lower bound √2 / Q. ∈ Exercise 4.7. The greatest lower bound of the set √x x x Q, x 0 is at- { | ∈ ≥ } tained when x = 0 and is equal to zero. The least upper bound is √e e.

Exercise 5.3. The solution of this exercise is left to the reader.

Exercise 6.7. We leave the completion of the details in the sketch of the proof of Theorem 6.5 to the reader.

Solutions to Exercises in Chapter V

Exercise 1.1. The solution of this exercise is left to the reader.

Exercise 1.8. The completion of the proof of Theorem 1.7 is left to the reader.

C Exercise 1.10. Let α = α1 + α2i , α = 0. If α2 = 0 and α1 > 0, then √α1 are ∈ 6 p the solutions to the equation x2 = α. If α = 0 and α < 0, then α i are 2 1 | 1| the solutions to the equation x2 = α. It remains to consider the case α = 0. 2 6 Let β = β + β i C with β = 0. We then have the equivalence 1 2 ∈ 1 6 β2 = α (β2 β2) + (2β β )i = α + α i β2 β2 = α , 2β β = α . ⇐⇒ 1 − 2 1 2 1 2 ⇐⇒ 1 − 2 1 1 2 2

If we substitute the second equation, β2 = α2/(2β1), in the ﬁrst equation, we obtain the equation 4β4 4α β2 α2 = 0. Setting y := β2, we obtain the 1 − 1 1 − 2 1 quadratic equation 4y2 4α y α2 = 0, which has the solutions − 1 − 2 q 2 2 α1 α1 + α2 α α y = = 1 | | . 1,2 2 2 Solutions to Exercises in Chapter V 271

Since now β R, we need consider only the nonnegative solution y . With 1 ∈ 1 β = √y and β = α /(2√y ), we obtain the solution formula 1 1 2 2 1 p α1 + α α2i β = | | q . √2 2α + α 1 | | Altogether, we obtain for the solutions to the equation x2 = α the following solution formula:   √α1 , if α1 > 0, α2 = 0;   p  α1 i , if α1 < 0, α2 = 0; q| | q x1,2 = α +α α α | | 1 + | |− 1 , if α > 0;  2 2 i 2  q q  α +α α α  | | 1 | |− 1 i , if α < 0. 2 − 2 2 From this it follows that the solutions to the equation x2 = i are

1 + i x1,2 = , √2 those to the equation x2 = 2 + i are p p √5 + 2 √5 2 x1,2 = + − i , √2 √2 and those to the equation x2 = 3 2i are − p p √13 + 3 √13 3 x1,2 = − i . √2 − √2

Exercise 1.11. Since by Exercise 1.10 we have the equality ((1 + i)/2)2 = i/2, we obtain, on completing the square,

1 + i 2 i x2 + (1 + i) x + i = 0 x + + = 0. · ⇐⇒ 2 2 If we now substitute y := x + (1 + i)/2, we obtain the quadratic equation y2 = i/2. With the solution formula from Exercise 1.10, we obtain the solution− 1 + i 1 i 1 + i x = y = , 1,2 1,2 − 2 2 ∓ 2 − 2 2 that is, x1 = i and x2 = 1 are the solutions of x + (1 + i) x + i = 0, which can be easily− checked by− substitution. One can prove analogously· that the equation x2 + (2 i) x 2i = 0 has the solutions x = i and x = 2. − · − 1 2 − 272 Solutions to Exercises

Exercise 1.14. (a) First, one veriﬁes the equality α β = α β for all α, β C. One then has · · ∈

α β 2 = (α β) α β = α α β β = α 2 β 2, | · | · · · · · · | | · | | which proves the assertion. (b) Let α , α , β , β N. The product rule from part (a) with α = α + α i 1 2 1 2 ∈ 1 2 and β = β1 + β2i yields

(α2 + α2) (β2 + β2) = (α β α β )2 + (α β + α β )2. 1 2 · 1 2 1 1 − 2 2 1 2 2 1 This implies the assertion.

Exercise 2.3. The statement of Remark 2.2 results immediately from use of the product rule from Exercise 1.14.

Exercise 2.5. Let A M (R) be an invertible matrix. One may convince one- ∈ 2 self that the mapping f : (C, +, ) (M (R), +, ) given by A · −→ 2 · α1 α2 1 α = α1 + α2i A A− 7→ · α2 α1 · − is an injective ring homomorphism. In particular, f induces an isomorphism C C R ∼= im( f ), that is, is isomorphic to the subring im( f ) of M2( ). Exercise 2.8. The solution of this exercise is left to the reader.

Exercise 2.12. We ﬁrst observe that for two complex numbers α, β ∈ C 0 with polar coordinate representations α = α (cos(ϕ) + i sin(ϕ)) \{ } | | · and β = β (cos(ψ) + i sin(ψ)), one has the following multiplication formula: | | ·

α β = α β cos(ϕ)cos(ψ) sin(ϕ)sin(ψ) · | || | · − + i sin(ϕ)cos(ψ) + cos(ϕ)sin(ψ) = α β cos(ϕ + ψ) + i sin(ϕ + ψ) , | || | · where for the second equality we have invoked the addition theorems for sine and cosine. From this follows for m N the equality ∈ αm = α m cos(mϕ) + i sin(mϕ) . | | · Likewise, for n N with n = 0, we obtain the equality ∈ 6 1 1 ϕ ϕ α n = α n cos + i sin . | | · n n This completes the proof of the general formula. Solutions to Exercises in Chapter VI 273

Exercise 4.4. Let p be a prime number. Then f (X) = X2 p is a quadratic − polynomial with integer coefﬁcients, and we have f (√p) = 0. We now assume that there exists a linear polynomial g(X) = aX + b (a, b Z, a = 0) ∈ 6 with g(√p) = 0. But then we must have √p = b/a Q, a contradiction. − ∈ We have therefore shown that √p is algebraic of degree 2.

Exercise 4.12. We leave it to the reader to ﬁnd additional transcendental numbers following the pattern of the Liouville number.

Exercise 5.8. The calculation of better approximations to e is left to the reader.

Solutions to Exercises in Chapter VI

Exercise 1.6. For example, the quadratic polynomial X2 + 1 H[X] has zeros ∈ H i, j, k and every purely imaginary quaternion α2i + α3j + α4k Im( ), 2 2 2 ∈ that satisﬁes the condition α2 + α3 + α4 = 1.

Exercise 1.7. It is clear that R Z(H). We therefore have to show that ⊆ Z(H) R. To this end, let α = α + α i + α j + α k Z(H). For each β = ⊆ 1 2 3 4 ∈ β + β i + β j + β k H, we then have α β = β α. Since 1 2 3 4 ∈ · · α β = (α β α β α β α β ) + (α β + α β + α β α β )i · 1 1 − 2 2 − 3 3 − 4 4 1 2 2 1 3 4 − 4 3 + (α β α β + α β + α β )j + (α β + α β α β + α β )k 1 3 − 2 4 3 1 4 2 1 4 2 3 − 3 2 4 1 and

β α = (β α β α β α β α ) + (β α + β α + β α β α )i · 1 1 − 2 2 − 3 3 − 4 4 1 2 2 1 3 4 − 4 3 + (β α β α + β α + β α )j + (β α + β α β α + β α )k, 1 3 − 2 4 3 1 4 2 1 4 2 3 − 3 2 4 1 however, we have

α β = β α 2(α β α β )i + 2( α β + α β )j + 2(α β α β )k = 0 · · ⇐⇒ 3 4 − 4 3 − 2 4 4 2 2 3 − 3 2 α β = α β α β = α β α β = α β . ⇐⇒ 3 4 4 3 ∧ 4 2 2 4 ∧ 2 3 3 2 1 If α2 = 0, then from the third equality, it follows that β3 = (α3α2− ) β2 for every6 β H. This contradiction implies that we must have α = 0. Similarly,· ∈ 2 one shows that we must also have α3 = 0 and α4 = 0. Altogether, therefore, we have that α = α R. This proves the inclusion Z(H) R. 1 ∈ ⊆

Exercise 1.14. (a) Let α = α1 + α2i + α3j + α4k. We then calculate 274 Solutions to Exercises

α2 = (α2 α2 α2 α2) + 2α α i + 2α α j + 2α α k 1 − 2 − 3 − 4 1 2 1 3 1 4 = (α2 + α2 + α2 + α2) + 2α α, − 1 2 3 4 1 which yields the desired result. (b) This can be established by a direct calculation.

Exercise 1.15. Let α = Im(α) i with Im(α) = (α ,α ,α ) and β = Im(β) i · 2 3 4 · with Im(β) = (β2, β3, β4). We calculate

α β = (α i + α j + α k) (β i + β j + β k) · 2 3 4 · 2 3 4 = ( α β α β α β ) + (α β α β )i − 2 2 − 3 3 − 4 4 3 4 − 4 3 + ( α β + α β )j + (α β α β )k − 2 4 4 2 2 3 − 3 2 = Im(α)t,Im(β)t + Im(α)t Im(β)t i, − × · which proves the assertion.

Exercise 1.18. Let α, β H. We calculate ∈ 2 α, β = 2 Re(α β) = α β + α β = α β + β α. · h i · · · · · · Multiplication of this equality on the right by β yields

2 α, β β = α β β + β α β, · h i · · · · · which on account of β β = β, β R proves the result. · h i ∈ Exercise 1.19. (a) The equality α β = β α can be veriﬁed by a direct calculation. · · (b) Using part (a), one obtains

α β 2 = (α β) α β = α (β β) α = β 2 (α α) = β 2 α 2 = α 2 β 2, | · | · · · · · · | | · · | | · | | | | · | | which proves the assertion. N (c) Let α1, α2, α3, α4, β1, β2, β3, β4 . Using the product rule from part (a), we obtain ∈

(α2 + α2 + α2 + α2) (β2 + β2 + β2 + β2) 1 2 3 4 · 1 2 3 4 = (α β α β α β α β )2 + (α β + α β + α β α β )2 1 1 − 2 2 − 3 3 − 4 4 1 2 2 1 3 4 − 4 3 + (α β + α β + α β α β )2 + (α β + α β + α β α β )2. 1 3 3 1 4 2 − 2 4 1 4 4 1 2 3 − 3 2 This implies the assertion.

Exercise 1.21. The completion of the proof of Theorem 1.20 is left to the reader. Solutions to Exercises in Chapter VI 275

Exercise 1.25. Veriﬁcation of the assertions of this problem is left to the reader.

Exercise 1.27. The solution to this problem is left to the reader.

Exercise 2.3. The statement of Remark 2.2 is an immediate result of the product rule from Exercise 1.19.

Exercise 2.5. Let A M (C) be an arbitrary invertible matrix. One may con- ∈ 2 vince oneself that the mapping f : (H, +, ) (M (C), +, ) given by · −→ 2 · α1 + α2i α3 + α4i 1 α = α1 + α2i + α3j + α4k A A− 7→ · α3 + α4i α1 α2i · − − is an injective ring homomorphism. In particular, f induces an isomorphism H H C ∼= im( f ), that is, is isomorphic to the subring im( f ) of M2( ). R Exercise 2.6. We show that f is an -linear mapping. Let α = α1 + α2i + H H R α3j + α4k and β = β1 + β2i + β3j + β4k . For arbitrary µ,ν , one has ∈ ∈ ∈ f (µα + νβ) = f ((µα1 + νβ1) + (µα2 + νβ2)i + (µα3 + νβ3)j + (µα4 + νβ4)k) (µα + νβ ) + (µα + νβ )i (µα + νβ ) + (µα + νβ )i = 1 1 2 2 3 3 4 4 (µα3 + νβ3) + (µα4 + νβ4)i (µα1 + νβ1) (µα2 + νβ2)i − − α + α i α + α i β + β i β + β i = µ 1 2 3 4 + ν 1 2 3 4 α + α i α α i β + β i β β i − 3 4 1 − 2 − 3 4 1 − 2 = µ f (α) + ν f (β), that is, f is R-linear. The veriﬁcation of the remaining assertions are left to the reader.

Exercise 2.9. The solution to this problem is left to the reader.

Exercise 3.4. The solution to this problem is left to the reader.

Exercise 3.6. We ﬁrst prove that every R-linear mapping v A v (v R3) 7→ · ∈ with A SO (R) is an orientation-preserving rotation of R3 about an axis ∈ 3 passing through the origin. To this end, we begin by noting that for A R ∈ SO3( ), on account of

det(A E) = 1 det(A E) = det(At) det(A E) − · − · − = det(At (A E)) = det(E At) = det(E A)t · − − − = det(E A) = ( 1) det(A E), − − · − 276 Solutions to Exercises we have the equality det(A E) = 0, which proves that 1 is an eigenvalue − of A. Now let a1 denote a normalized (to have length 1) eigenvector of A associated with the eigenvalue 1. We shall see that the mapping v A v 7→ · (v R3) describes a rotation about an axis passing through the origin that ∈ R3 is determined by a1. For this, we extend a1 to an orthonormal basis of by choosing an additional vector a R3, normed to have length 1, that is 2 ∈ perpendicular to a , and then setting a := a a . If now S M (R) denotes 1 3 1 × 2 ∈ 2 a matrix with S (1,0,0)t = a , S (0,1,0)t = a , and S (0,0,1)t = a , then we · 1 · 2 · 3 must have S SO (R). We further obtain ∈ 3 1 t 1 1 t S− A S (1,0,0) = S− A a = S− a = (1,0,0) , · · · · · 1 · 1 which implies the equality

1 0 0 1 S− A S = 0 α β · · 0 γ δ for certain α, β, γ, δ R. But since because of A,S SO (R), we have also ∈ ∈ 3 the inclusion S 1 A S SO (R) and thereby − · · ∈ 3 α β SO (R), γ δ ∈ 2 it follows that there exists a uniquely determined ϕ [0,2π) with ∈ 1 0 0  1 S− A S = 0 cos(ϕ) sin(ϕ) =: Dϕ. · · 0 sin(ϕ) −cos(ϕ)

The mapping v D v (v R3) is an orientation-preserving rotation in the 7→ ϕ · ∈ x2, x3-plane through the angle ϕ about the x1-axis (counterclockwise if a1 points toward the observer). Altogether, we have shown that the mapping v A v (v R3) is an orientation-preserving rotation in the a , a -plane 7→ · ∈ 2 3 through the angle ϕ about the a1-axis (counterclockwise if a1 points toward the observer). We now describe, conversely, an arbitrary orientation-preserving rotation of R3 through the angle ϕ [0,2π) about an axis passing through the origin ∈ t that is determined by the vector (normed to have length 1) a1 := (ν1,ν2,ν3) R3. To this end, we ﬁrst consider the matrices ∈

 ν ν  q  1 0 3 2 2 √ν2+ν2 √ν2+ν2 ν1 + ν3 ν2 0  1 3 1 3   q  R D :=  0 1 0 , D2 :=  2 2  SO3( ). 1 ν2 ν + ν 0  ν3 ν1   − 1 3  ∈ 2 2 0 2 2 − √ν1 +ν3 √ν1 +ν3 0 0 1 Solutions to Exercises in Chapter VI 277

If in the ﬁrst step we apply D1 to a1, we rotate a1 about the x2-axis in such a way that D a lies in the x , x -plane. If in the second step we apply D to 1 · 1 1 2 2 D a , then we rotate D a about the x -axis, so that ﬁnally, D D a is 1 · 1 1 · 1 3 2 · 1 · 1 parallel to the x1-axis. Altogether, the orientation-preserving rotation under 3 1 1 discussion is given by the mapping v A v (v R ) with A := D− D− 7→ · ∈ 1 · 2 · D D D . Multiplying out yields ϕ · 2 · 1  ν2µ + cos(ϕ) ν ν µ ν sin(ϕ) ν ν µ + ν sin(ϕ) 1 1 2 − 3 1 3 2 A = ν ν µ + ν sin(ϕ) ν2µ + cos(ϕ) ν ν µ ν sin(ϕ), 1 2 3 2 2 3 − 1 ν ν µ ν sin(ϕ) ν ν µ + ν sin(ϕ) ν2µ + cos(ϕ) 1 3 − 2 2 3 1 3 where we have set µ := 1 cos(ϕ); this can now be easily decomposed as − A = E + sin(ϕ) N + (1 cos(ϕ)) N2 · − · with  0 ν ν  − 3 2 N :=  ν3 0 ν1, ν ν −0 − 2 1 as asserted. Selected Literature

The following list of books on (elementary) number theory and algebra can serve to ﬁll in some of the gaps in this book’s presentation. Some of these books will take the reader much deeper into various topics. The literature on the concept of number and the representation of numbers is of cultural- historical signiﬁcance, while the works of a historical nature provide insight into the historical development of algebra and number theory. Finally, we offer the interested reader two books on approaches to the teaching of algebra and number theory. Selected literature for the appendices is listed at the end of the respective appendix.

Literature on Number Theory

[1] D. Burton: Elementary number theory. McGraw-Hill Education, 7th edition, 2010. [2] W. A. Coppel: Number theory. An introduction to mathematics. Springer, Berlin Heidelberg New York, 2nd edition, 2009. [3] G. H. Hardy, E. M. Wright: An introduction to the theory of numbers. Ox- ford University Press, 6th edition, 2008. [4] H. Hasse: Number theory. Translated from the 3rd German edition. Sprin- ger, Berlin Heidelberg New York, 1980. [5] L. -K. Hua: Intoduction to number theory. Translated from the Chinese original by P. Shiu. Springer, Berlin Heidelberg New York, 1982. [6] K. Ireland, M. Rosen: A classical introduction to modern number theory. Springer, Berlin Heidelberg New York, 2nd edition, 1990. [7] F. Jarvis: Algebraic number theory. Springer, Cham Heidelberg New York Dordrecht London, 2014. [8] G. A. Jones, J. M. Jones: Elementary number theory. Springer, London, 1998. [9] M. B. Nathanson: Elementary methods in number theory. Springer, Berlin Heidelberg New York, 2000. [10] I. Niven, H. S. Zuckerman, H. L. Montgomery: An introduction to the theory of numbers. John Wiley & Sons, Hoboken, NJ, 5th edition, 2008. [11] D. Redmond: Number theory. Marcel Dekker, New York, 1996. [12] K. H. Rosen: Elementary number theory and its applications. Pearson, Bos- ton, 6th edition, 2010.

© Springer International Publishing AG 2017 J. Kramer and A.-M. von Pippich, From Natural Numbers to Quaternions, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-69429-0 280 Selected Literature

[13] W. Sierpinski: Elementary theory of numbers. Elsevier, Amsterdam, PWN, Warsaw, 2nd edition, 1988. [14] A. Weil: Basic number theory. Springer, Berlin Heidelberg New York, 3rd edition, 1995.

Literature on Abstract Algebra

[15] M. Artin: Algebra. Pearson, Boston, 2nd edition, 2017. [16] R. Cooke: Classical algebra: its nature, origins, and uses. John Wiley & Sons, Hoboken, NJ, 2008. [17] D. S. Dummit, R. M. Foote: Abstract algebra. John Wiley & Sons, Hobo- ken, NJ, 3rd edition, 2003. [18] B. Fine, A. M. Gaglione, G. Rosenberger: Introduction to abstract algebra. Johns Hopkins University Press, Baltimore, MD, 2014. [19] J. Gallian: Contemporary abstract algebra. Brooks Cole, 9th edition, 2016. [20] R. S. Irving: Integers, polynomials, and rings. Springer, Berlin Heidelberg New York, 2004. [21] S. Lang: Algebra. Springer, Berlin Heidelberg New York, 3rd edition, 2002. [22] F. Lorenz: Algebra. Volume I. Fields and Galois theory. Translated from the 1987 German edition by S. Levy. Springer, Berlin Heidelberg New York, 2006. [23] W. K. Nicholson: Introduction to abstract algebra. John Wiley & Sons, Hoboken, NJ, 4th edition, 2012. [24] J. J. Rotman: A ﬁrst course in abstract algebra. Pearson, Boston, 3rd edition, 2005. [25] L. H. Rowen: Algebra: groups, rings and ﬁelds. A K Peters, Wellesley, MA, 1994. [26] J. Stillwell: Elements of algebra: geometry, numbers, equations. Springer, Berlin Heidelberg New York, 1994. [27] B. L. van der Waerden: Algebra. Volume I. Springer, Berlin Heidelberg New York, 9th edition, 1993.

Literature on the Concept of Number

[28] J. H. Conway, R. K. Guy: The book of numbers. Springer Copernicus, New York, 1996. [29] L. Corry: A brief history of numbers. Oxford University Press, Oxford, 2015. [30] H. Ebbinghaus et al.: Numbers. Translated from the 2nd German 1988 edition by H. L. S. Orde. Springer, Berlin Heidelberg New York, 1991. [31] G. Ifrah: From one to zero: a universal history of numbers. Translated from the French original by L. Bair. Penguin Books, New York, 1987. 281

[32] K. Menninger: Number words and number symbols: a cultural history of numbers. Translated from the German revised edition by P. Broneer. Dover, New York, 1992. [33] R. Taschner: Numbers at work: a cultural perspective. Translated from the 2005 German original by O. Binder and D. Sinclair-Jones. A K Peters, Wellesley, MA, 2007.

Literature on the History of Algebra and Number Theory

[34] I. G. Bashmakova, G. S. Smirnova: The beginnings and evolution of algebra. Translated from the Russian original by A. Shenitzer. Mathematical As- sociation of America, Washington, DC, 2000. [35] V.J. Katz, K. H. Parshall: Taming the unknown: a history of algebra from an- tiquity to early twentieth century. Princeton University Press, Princeton, NJ, 2014. [36] A. Weil: Number theory. Birkhäuser Boston, Boston, MA, 1984.

Literature on the Teaching of Algebra and Number Theory

[37] A. Arcavi, P. Drijvers, K. Stacey: The learning and teaching of algebra. IM- PACT Series, Routledge, 2016. [38] J. D. Sally, P.J. Sally: Integers, fractions, and arithmetic: a guide for teachers. American Mathematical Society, Providence, RI, 2012. Index

associative operation, 45 element identity, 47 bounded set, 160 inverse, 48, 101 irreducible, 120 left identity, 47 Cauchy sequence left inverse, 48, 101 of an ordered field, 168 prime, 121 rational, 145 right identity, 47 real, 153 right inverse, 48, 101 Cayley table, 51 unit, 98 Cayley’s octonions, 231 zero, 98 characteristic, 101 equivalence class, 57 completeness equivalence relation, 57 axiom of geometric, 166 Euclid’s lemma, 23, 96 of an ordered field, 168 Euclidean algorithm, 126 of real numbers, 155 extended, 127 completeness principle, 162 Euclidean domain, 125 complex numbers, 184 exponential function, 197 absolute value, 187 complex conjugate, 186 field, 110 imaginary part, 184 absolute value, 168 modulus, 187 algebraically closed, 193 purely imaginary, 184 archimedean, 169 real part, 184 order, 168 complex plane, 184 field of fractions, 117 coset, 61 fraction, 76, 118 left, 59 fundamental theorem right, 61 of algebra, 191 of arithmetic, 22, 96 decimal, 155 genuine, 159 group, 48 infinite, 155 abelian/commutative, 49 terminating, 155 alternating, 68 decimal expansion cyclic, 52 period, 143 dihedral, 50 periodic, 143 symmetric, 51 purely periodic, 143 group homomorphism, 54 Dedekind cuts, 167 group isomorphism, 54 division with remainder, 30, 95, 125 divisor, 15, 94, 120, 121 Hamilton’s quaternions, 219 common, 15, 95, 120 homomorphism theorem greatest common, 25, 97, 120, 122 for groups, 66 proper, 17 for rings, 108 trivial, 17, 95 domain, 100 ideal, 104 © Springer International Publishing AG 2017 J. Kramer and A.-M. von Pippich, From Natural Numbers to Quaternions, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-69429-0 284 Index

divisibility, 121 octonions, 231 greatest common divisor, 122 order intersection, 122 of a group, 52 least common multiple, 122 of an element, 53 principal, 105 orthogonal group R sum, 122 O2( ), 189 R unit, 105 O3( ), 228 zero, 105 image Peano axioms, 9 of a group homomorphism, 55 predecessor, 9 of a ring homomorphism, 104 prime number, 17, 95 index of a subgroup, 61 Fermat, 19 induction, 9 Mersenne, 19 inﬁmum, 161 principal ideal domain, 124 inﬁmum principle, 162 integers, 74 quaternions, 219 absolute value, 76 conjugate quaternion, 221 decimal representation, 141 imaginary part, 220 difference, 75 imaginary space, 221 order, 75 modulus, 221 product, 93 purely imaginary, 220 quotient, 118 real part, 220 integral domain, 100 quotient group, 65 quotient ring, 108 kernel R of a group homomorphism, 55 -algebra, 222 of a ring homomorphism, 104 associative, 222 commutative, 222 dimension, 222 least common multiple, 26, 97, 120, 122 division algebra, 222 lower bound, 160 R-subalgebra, 223 R-algebra homomorphism, 223 monoid, 47 rational numbers, 118 multiplicative group of a ring, 111 absolute value, 119 decimal representation, 143 natural numbers, 9 order, 119 decimal representation, 31 real number line, 164 difference, 14, 75 real numbers, 151 order, 13 absolute value, 153 product, 11 decimal expansion, 159 sum, 10 decimal representation, 159 nested intervals, 162 order, 152 nested intervals principle, 162 real sequence normal subgroup, 61 (strictly) monotonically decreas- null sequence ing, 160 rational, 145 (strictly) monotonically increasing, number 159 algebraic, 193 convergence, 153 amicable, 21 limit, 153 Euler, 197 relatively prime, 28 irrational, 159 pairwise, 28 Liouville, 196 ring, 97 perfect, 20 commutative, 98 transcendental, 194 factorial, 123 Index 285

polynomial, 99 supremum principle, 162 zero, 98 ring homomorphism, 103 theorem ring isomorphism, 103 Gauss’s, 123 Lagrange’s, 60 semigroup, 45 Liouville’s, 194 abelian/commutative, 46 of Euclid, 18 regular, 69 skew ﬁeld, 110 unique factorization domain, 123 special orthogonal group unit, 101 R SO2( ), 189 imaginary, 184 R SO3( ), 228 unitary group C special unitary group U2( ), 225 C SU2( ), 225 upper bound, 160 subgroup, 53 subgroup criterion, 53 well-ordering principle, 14 subring, 102 subring criterion, 102 zero divisor, 100 successor, 9 left, 100 supremum, 161 right, 100