Exercise 213 8 Prove or disprove that if a bc where a b and c are positive

# Exercise 213 8 prove or disprove that if a bc where a

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Exercise 2.13. #8: Prove or disprove that if a | bc , where a, b , and c are positive integers and a 6 = 0, then a | b or a | c . Proof. This is false: for example, take a = 6, b = 2 and c = 3. Then clearly a divides neither 2 nor 3, though it divides their product, since every number divides itself. (Note, however, that if we assumed a was prime, then this statement would be true!) Exercise 2.14. #10: What is the quotient and remainder when (d) - 1 is divided by 23 (e) - 2002 is divided by 87. Proof. (d) The greatest integer multiple of 23 less than - 1 is - 23. The remainder is 22. (e) The greatest integer multiple of 87 less than - 2002 is - 2088. The remainder is 86. Exercise 2.15. #12: What time does a 24 hour clock read (c) 168 hours after it reads 19 : 00? Proof. (c) It reads 19 : 00 - 168 hours is exactly 24 hours × 7. Exercise 2.16. #14: Suppose that a and b are integers, a 11 mod 19 and b 3 mod 19. Find the integer c with 0 c 18 such that (d) c 7 a + 3 b mod 19 Proof. a 11 mod 19 implies 7 a 77 1 mod 19 , and, likewise, b 3 mod 19 implies 3 b 9 mod 19. It follows that 7 a + 3 b 10 mod 19 . Exercise 2.17. #18: Show that if a is an integer and d is an integer greater than 1, then the quotient and remainder obtained when a is divided by d are b a/d c and a - d b a/d c , respectively. Proof. The quotient is the greatest integer multiple of d less than or equal to a , divided by d , and r is a - qd . Certainly b a d c d is an integer multiple of d less than or equal to a so q ≥ b a d c . Suppose towards contradiction that q > b a d c . Then q > b a d c d + 1 so we have a qd ≥ b a d c d + d,
14 MATH 55 - HOMEWORK 4 SOLUTIONS which gives the inequality a - d ≥ b a d d . Dividing by d and taking floor functions, we get b a - d d c = b a d - 1 c ≥ b a d c , a contradiction. So q = b a d c . Then the remainder is clearly r = a - d b a d c , since r = a - qd . Exercise 2.18. Decide whether each of these integers is congruent to 3 modulo 7: (a) 37 (d) - 67 Proof. (a) No: the difference between 3 and 37 is 34 which is not divisible by 7. (d) Yes: the difference between - 67 and 3 is 70 = 7 × 10. Exercise 2.19. #32: Find each of these values: (c) (7 3 mod 23) 2 mod 31 Proof. First, we calculate the number inside the square: 7 3 7 × 49 7 × 3 21 mod 29 . So the number inside the square is 21. Then we calculate using reduction modulo 31: 21 2 7 × 3 × 21 7 × 63 7 × 1 7 mod 31 . So the answer is 7. Exercise 2.20. #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 . Proof. To show ac bc mod mc we have to show that mc | ( ac - bc ). Since a b mod m , we know m | ( a - b ) so there is some integer d so that md = ( a - b ). It follows then that ( mc ) d = ( a - b ) c = ( ac - bc ) . So we’ve shown for some integer d , ( mc ) d = ( ac - bc ), i.e. mc | ( ac - bc ) so ac bc mod mc . Exercise 2.21. #38: Show that if n is an integer, then n 2 0 or 1 mod 4. Proof. For any integer, we know n 0 , 1 , 2 , or 3 mod 4. It follows that n 2 0 or 1 mod 4 since 2 2 0 mod 4 and 3 2 1 mod 4. Exercise 2.22. #40: Prove that if n is an odd positive integer, then n 2 1 mod 8.

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• Spring '08
• STRAIN
• Math, Natural number, Finite set, Countable set