assignment4_sol - MATH 135 Assignment 4 Solutions Hand-In...

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MATH 135 Fall 2005 Assignment 4 Solutions Hand-In Problems Exercise 2-9: If 3 p 2 = q 2 where p, q Z , show that 3 is a common divisor of p and q . Solution: Since 3 | 3 p 2 , then 3 | q 2 . By Theorem 2.53, 3 | q or 3 | q , so 3 | q . Let q = 3 t , with t Z . Then 3 p 2 = 9 t 2 or p 2 = 3 t 2 . Since 3 | 3 t 2 , 3 | p 2 , and so 3 | p as above. Therefore, 3 is a common divisor of p and q . NOTE: In fact, the only integer values of p and q which satisfy the equation 3 p 2 = q 2 are p = q = 0. Exercise 2-54: Convert (9 A 411) 12 to base 10. ( A is the symbol for ten in base 12.) Solution: (9 A 411) 12 = 9(12) 4 + 10(12) 3 + 4(12) 2 + 1 · 12 + 1 = 186624 + 17280 + 576 + 13 = 204493 = (204493) 10 Exercise 2-58: Convert 433 to base 5. Solution: 433 = 86 · 5 + 3 86 = 17 · 5 + 1 17 = 3 · 3 + 2 3 = 0 · 5 + 3 Hence 433 = (3213) 5 . Check: (3213) 5 = 3(5 3 ) + 2(5 2 ) + 1(5) + 3 = 433. Exercise 2-63: Add and multiply (3130) 4 and (103) 4 together in base 4. Solution: Base 4 Addition Table + (0) 4 (1) 4 (2) 4 (3) 4 (0) 4 (0) 4 (1) 4 (2) 4 (3) 4 (1) 4 (1) 4 (2) 4 (3) 4 (10) 4 (2) 4 (2) 4 (3) 4 (10) 4 (11) 4 (3) 4 (3) 4 (10) 4 (11) 4 (12) 4 Base 4 Multiplication Table · (0) 4 (1) 4 (2) 4 (3) 4 (0) 4 (0) 4 (0) 4 (0) 4 (0) 4 (1) 4 (0) 4 (1) 4 (2) 4 (3) 4 (2) 4 (0) 4 (2) 4 (10) 4 (12) 4 (3) 4 (0) 4 (3) 4 (12) 4 (21) 4 Thus (3130) 4 + (103) 4 (3233) 4 (3130) 4 × (103) 4 (22110) 4 + (313000) 4 (1001110) 4 Check: (3130) 4 = 3(4 3 ) + 1(4 2 ) + 3(4 1 ) = 220 (103) 4 = 1(4 2 ) + 3(4 0 ) = 19 (3233) 4 = 3(4 3 ) + 2(4 2 ) + 3(4 1 ) + 3(4 0 ) = 239 = 220 + 19 (1001110) 4 = 1(4 6 ) + 1(4 3 ) + 1(4 2 ) + 1(4 1 ) = 4180 = 220 · 19 . 1
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Exercise 2-66: If a = (342) 8 and b = (173) 8 , find a - b without converting to base 10. Solution: Mimic the algorithm used when subtracting in Base 10. (342) 8 - (173) 8 (147) 8 (To subtract the “3” from the “2”, we must borrow 1 from the “4”. We then subtract “3” from “12” to get “7”. Having borrowed 1 from the “4”, we now must subtract “7” from “3”. To do this, we must borrow 1 from the leading “3”. We then subtract “7” from “13” to get “4”. Lastly, we subtract “1” from “2” to get “1”.) Check: (342) 8 = 3(8 2 ) + 4(8 1 ) + 2(8 0 ) = 226 (173) 8 = 1(8 2 ) + 7(8 1 ) + 3(8 0 ) = 123 (147) 8 = 1(8 2 ) + 4(8 1 ) + 7(8 0 ) = 103 = 226 - 123 Exercise 2-72: Find lcm(12827 , 20099) . Solution: We compute gcd(12827 , 20099) and then use the fact that lcm( a, b ) = ab gcd( a, b ) . We use the Euclidean Algorithm to find gcd(12827 , 20099): 20099 = 1(12827) + 7272 12827 = 1(7272) + 5555 7272 = 1(5555) + 1717 5555 = 4(1717) + 404 1717 = 4(404) + 101 404 = 4(101) + 0 . By the Euclidean Algorithm, gcd(12827
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This homework help was uploaded on 04/18/2008 for the course LINEAR ALG MATH135 taught by Professor Vanderburgh during the Spring '08 term at Waterloo.

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assignment4_sol - MATH 135 Assignment 4 Solutions Hand-In...

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