f2006Solns9 - ECE 190 -- Fall, 2006 Assignment 9 Solutions...

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Unformatted text preview: ECE 190 -- Fall, 2006 Assignment 9 Solutions Section 10.2 1 See text. 4 (a) y 2 z 1 O 0 W 3 (b) R4 is not reflexive: (0, 0) R4. (c) R4 is symmetric. (d) R4 is not transitive: (2, 1) R4 and (1, 3) R4 but (2, 3) R4. 8 .. .. . +..... (a) 0 ........ ............ .. . .. . ...... .. .. + . .. . 1 ..................... ... .. ........ ... ... ... 2 3 (b) R8 is not reflexive: (2, 2) R8. (c) R8 is symmetric. (d) R8 is transitive. 12 See text. 13 C is not reflexive. For x = 0 we have 02 + 02 = 1 so 0 C 0. C is symmetric. If x2 + y2 = 1, then y2 + x2 = 1. C is not transitive. If x = 0, y = 1, z = 0, then x C y and y C z but x C z. 16 F is reflexive. If m = n, m - n = 0 and 5|0 so m F m. F is symmetric. If m F n, then 5|(m - n). Hence 5|(n - m) and thus n F m. F is transitive. If m F n and n F p, then 5|(m - n) and 5|(n - p). Hence k1 , k2 Z such that m - n = 5k1 and n - p = 5k2. Adding both sides of these equations, we have m - p = 5(k1 + k2 ). Since (k1 + k2) Z, 5|(m - p) and thus m F p, as required. 26 See text. Section 10.3 1 See text. 2 (a) See text. (b) {(0, 0), (1, 1), (3, 3), (4.4), (1, 3), (3, 1), (1, 4), (4, 1), (3, 4), (4, 3), (2, 2)} (c) {(0, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4), } 5 See text. 1 19 To prove that it is an equivalence relation, we must show that it is reflexive, symmeteric, and transitive. The proof of this is almost identical to that given in Question 16 of Section 10.2. The equivalence classes are: {x Z|x = 4k, for some integer k} {x Z|x = 4k + 1, for some integer k} {x Z|x = 4k + 2, for some integer k} {x Z|x = 4k + 3, for some integer k} 33 See text. Section 10.5 1 See text. 2 See text. 8 See text. 9 For R to be a partial order relation, it must be reflexive, antisymmetric, and transitive. R is not antisymmetric: (-1, 1) R (because (-1)2 (1)2 ) and (1, -1) R (because (1)2 (-1)2 ). Thus R is not a partial order relation. 16(a) See text. 17 (a) See text. (b) {0, 1, 2} .. . ...... . . . .. ... . .. .. . ... . ..... . ... . ... . .. ... . .. ... . .. . . .. .. . ... . ... .. ... . . .. .... . ... . . .... ... . ... . . . ... .. . ... .. . . . ... ... . . .. ... . .. . ... . ... ... . ... .. . . . ... .. ... ... . . . .. . ... ... ... . . . ... .. . .. ... . .. .. . ... . ... ... . ... . .. ... .... ..... . . ... . . .. .. . ... . .. . .... ... .... . .. . . . ... . ... .. . . .. .. . .... ... ... . . . ... . ... ... . . ... ... . ... . ... ... . . . .. . ... . .... . . .. . . .... ... ... .. . ... . .. . . .. ... ... .. ... ... . . ... . ... . . ... .... ... . . . . . ... . .. ..... ... .. . ..... . . . . ... .. . . .... . . . . . . ... .... ... .. . . ... .... ... .... . . . .. . ... . ... .. .. ... . . ... ... . . . .. . .. .. .. . . .. .. . . .. ... ... ... ... ... . . . . . . . . .. .. . .. .... ... ... . ... . . .... ... . .... .. . . ... . ..... ... . . .. . ... .. . . .. . . ..... ..... .... ... . . .... . .... ... ... . . ... .. . .. . ... .. . .. . . . ... .. . ... .. . . .. ... . ... ... . . .. .... . . . ... . ... .. . ... . ... ... ... . . .. . ... .. ... . . ... ... . ... .. . . . .. ... .. . ... . . ... ... . . .. ... .. . ... . . .... ... .. . . .. ... .. . ... . . ... ... . .... ... . .... . . ... . .. .. . ...... . . ... . {0, 1} {1, 2} {0, 2} {1} {0} {2} 2 ...
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