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Solutions to Exercises 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 2 clear. Suppose, then, that assertion (1) holds for an arbitrary but fixed p N 2 and for all n, m N. Then from the definition of addition and the induction hypothesis, we have2 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 2 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 2 2 (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 2 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 2 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 induc- tion. 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 2 ≥ natural number x N with n + x = m. If m = n, then for x = 0, we have the 2 equality n + x = n + 0 = m. If m > n, then there exists a N, a > 0, such 2 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 ∗···∗ ∗···∗ 2 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 fol- lows 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 j 1 ··· k n N with a a + 1 = 3 n. If now 3 a for some j 1,. .,k , then 2 1 ··· k · j j 2 f g by Lemma 2.4 (ix), we have 3 (a a ), that is, there exists m N with j 1 ··· k 2 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 2 and r 1,2 (j = 1,. .,k), from which follows a = 3 n + (r 1), which j 2 f g 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 ≥ 2 3 a = 43, a = 1807, a = (1807 1) 1806 + 1 = 3263443. With = p P 4 5 6 − · Mn f 2 j p a , we obtain the first six sets, j ng = 2 , = 3 , = 7 , = 43 , M1 f g M2 f g M3 f g M4 f g = 13, 139 , = 3263443 . M5 f g M6 f g We claim that we have the equality a = 5 b + r with b N and r 2,3 n · n n n 2 n 2 f g (n N, n 1). This can be proved by induction on n. If n = 1, the assertion is 2 ≥ clear. We now assume that the assertion holds for arbitrary but fixed n N, 2 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 2 n+1 2 f g as desired. We have thus shown that 5 - a , that is, 5 / , for all n N, n 2 Mn 2 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 2 · that 3 (p + 1). If p = a, we are done. If p < a, then there exists q N, q > 1, j 2 with a = p q. Since 3 (p q + 1), it follows by Exercise 2.5 (b) that 3 (p + 1) · j · j or 3 (q + 1). In the first case, we are done. In the second case, we repeat the j 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 that2 there· are only finitely many prime num- bers in the set 2 + 3 N, we must have p p ,..., p . In particular, we have · 2 f 1 ng p (p p ). However, since we have the divisibility relation p a, we must j 1 ··· n j have p 1 from the divisibility rules, which is a contradiction. j 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 2 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 2 natural numbers a,b N, b odd, with n = a b and 1 a < n, 1 < b n. 2 · ≤ ≤ 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 f 1 ng f 1 mg of a and all divisors of b.
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