Question 1 Section 4.1 11. What time does a 12-hour clock read a) 80 hours after it reads 11:00? b) 40 hours before it reads 12:00? c) 100 hours after it reads 6:00? I don't really understand this question and what the question ask me for. a. (11 + 80) mod 12 = 7 b. (12 + 40) mod 12 = 4 c. (100 + 6) mod 12 = 10 Section 4.2 27. Use Algorithm 5 to find 32003 mod 99. Binary expansion of 644 is: (0111 1101 0011) So using Algorithm 5 requires to define: a = 0111 1101 0011 i = 0, 1, 2,, 10 x = 1 power = 3 A table below shows the algorithm: i a i x Power 0 a 0 = 1 x = 1 x 3 mod 99 x = 3 1 a 1 = 1 x = 3 x 9 mod 99 Power = 3² mod 99 Power = 9 Power = 9² mod 99 2 a 2 = 0 ² mod 99 Power = 27 3 a 3 = 0 Power = 27² mod 99 Power = 36 4 a 4 = 1 x 36 mod 99 x = 81 Power = 36² mod 99 Power = 9 5 a 5 = 0 x = 81 Power = 9² mod 99 6 a 6 = 1 x = 81 x 81 mod 99 7 a 7 = 1 x 27 mod 99 x = 36 ² mod 99 Power = 27 Power = 27² mod 99 Power = 36 8 a 8 = 1 x = 36 x 36 mod 99 Power = 36² mod 99
9 a 9 = 1 x = 9 x 9 mod 99 x = 81 x = 9 Power = 9 10 a 10 = 1 x = 81 x 81 mod 99 Power = 9² mod 99 Using result, we know that 3² ⁰⁰⁸ mod 99 = 27. Question 2 Exercise 34 asks to show that if a is equivalent to b (mod m) and c is equivalent to d (mod m), where a, b, c, d and m are integers with m greater or equal to 2, then a - c is equivalent to b - d (mod m). How can this be shown? Using theorem 3, we get: a b (mod m) a (mod m) = b (mod m). The same can be said for c d (mod m), that is, c (mod m) = d (mod m). So subtracting both equations, we get: a (mod m) c (mod m) = b (mod m) d (mod m). Using Corollary 2 of Theorem 5, we get: (a c) (mod m) = (b d) (mod m) (a c) (b d) (mod m) Question 3 Could you please solve problem 19 of 4.1? Recall the division algorithm, a = mq + r, 0 < r < m. Taking the modulus of this equation: a (mod m) = (mq + r) (mod m) a (mod m) = mq (mod m) + r (mod m) a (mod m) = r Since by definition 0 < r < m, then: 0 < a (mod m) < m If a (mod m) < m/2, then the smallest absolute value congruent should be a (mod m). In the other case, a (mod m) > m/2, then the smallest absolute value congruent should be (a (mod m)) m.
Question 4 Buenas, mi pregunta es sobre el ejercicio 38 de la seccion 4.1. El mismo dice: Show that if n is an integer then n2 0 or 1(mod 4). Lo que he hecho: Caso 1: si n es un entero par entonces se puede expresar como n=2k n^2 = 4m^2 n^2 (mod 4) = 4m^2 (mod 4) n^2(mod 4) = 0 Caso 2: si n es un entero impar entonces se puede expresar como n=(m+1 n^2=(m^2 + 4m +1) n^2(mod 4) = (m^2 + 4m +1) (mod 4) The missing step is: n² (mod 4) = (m²) mod 4 + (4m) mod 4 + (1) mod 4 Therefore, n² (mod 4) = 1. Question 5 Ejercicio 43 - Show that Zm with multiplication modulo m, where m(greater than or equal to)2 is an integer, satisfies the closure, associative, and commutatively properties, and 1 is multiplicative identity. Dividing this proof by cases, we have: Case 1: Closure We know that a x m b = (a x b) mod m, and since (a x b) mod m belongs to Z m. If a and b belong to Z m, then a x m b belongs to Z m. Therefore this proves the closure property. Case 2: Associative (a x m b) x m c = a x m b x m c (a x m b) x m c = a x m (b x m c) If a, b and c are in Z m, then this proves the associative property. Similar procedures can be taken for proving commutative multiplication and multiplication identity. Question 6 En el ejercicio 13 el cual dice: Suppose that a and b are integers, a 4 (mod 13), and b 9 (mod 13). Find the integer c with 0 c 12 such
that: a) c 9a (mod 13). b) c 11b (mod 13). c) c a + b (mod 13). d) c 2a + 3b (mod 13). e) c a2 + b2 (mod 13). f ) c a3 b3 (mod 13). a. Using theorem 3, c (mod 13) = 9a (mod 13) c (mod 13) = 9 (4 mod 13) (mod 13) c (mod 13) = (36 mod 13) (mod 13) c (mod 13) = 10 mod 13 We can conclude c = 10. For the rest of the problems, the procedure is similar. b. c = 8 c. c = 0 d. c = 9 e. c = 6 f. c = 11 Question 7 Question 19 from section 14.3 says: "Show that if (2^n) -1 is prime, then n is prime. [Hint: Use the identity 2^ab -1=(2^a -1)*(2^(a(b-1)) +2^(a(b-2)) +...+2^a +1)]." This is also Mersenne prime [(2^n)-1 is a prime]. Equaling n=ab would give me 2^ab -1. Therefore I may use the identity. From then on I'm not entirely sure how to proceed since this identity is mostly the sum of products. How can I prove this? Proof by contraposition: Assuming n is not prime, then it is composite. That means n is of the form n = ab. Then, 2 n 1 = 2 ab 1. We know that this must be divisible by 2 a 1, because: 2 ab 1 = (2 a 1)(2 a(b-1) + 2 a(b-2) + + 2 a + 1) So that means that if n is composite, then 2 n - 1 is also composite. That completes the proof by contraposition.
Question 8 sec. 4.1 36. Show that if a, b, c, and m are integers such that m 2, c > 0, and a b (mod m), then ac bc (mod mc). Using theorem 3, we know that a (mod m) = b (mod m). Now assume c and d integers, where d > 1. The remainder of x/y is: x (mod y) = x y (x/y) Such that: a m(a/m) = b m(b/m) Now, multiplying both sides by c, we obtain: c ( a m(a/m) ) = c ( b m(b/m) ) ac mc(a/m) = bc mc(b/m) ac mc (ac/mc) = bc mc(bc/mc) Therefore, ac mod mc = bc mod mc So, ac bc (mod mc)