4023f10ps4a

# 4023f10ps4a - Homework#4 Solutions Due October 4 2010 1 For...

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Homework #4 Solutions Due: October 4, 2010 1. For each of the following relations R on the set X determine whether R is re±exive, symmetric, antisymmetric, or transitive. (a) X = Z aRa if and only if a ± b + 1; (b) X = Z aRb if and only if a + b is even; (c) X = Z aRb if and only if a + b is odd; (d) X = Z aRb if and only if a and b are relatively prime; (e) X = R aRb if and only if a ² b is an integer; (f) X = R aRb if and only if a 2 ± b 2 . You may record your answers (yes/no) in the following table: Relation Re±exive Symmetric Antisymmetric Transitive (a) yes no no no (b) yes yes no yes (c) no yes no no (d) no yes no no (e) yes yes no yes (f) yes yes no yes I Solution. Some comments on the answers recorded in the table: (a) Re±exive : a ± a + 1 is always true, so R is re±exive. Symmetric : Let a = 0, b = 2. Then 0 ± 2 + 1 but 2 0 + 1. That is, aRb but b is not related to a , so R is not symmetric. Antisymmetric : Let a = 1, b = 0. Then a ± b + 1 so aRb and b ± a + 1 so bRa , but a 6 = b , so the relation is not antisymmetric. Transitive : Let a = 1, b = 0, and c = ² 1. Then a ± b + 1 and b ± c + 1, but a ± c + 1, so the relation is not transitive. (b) Re±exive : 2 a is always even, so R is re±exive. Symmetric : a + b = b + a so R is symmetric. Antisymmetric ; a + b = b + a even does not imply that a = b , so R is not antisymmetric. Transitive : If a + b and b + c are both even, then so is ( a + b ) + ( b + c ) ² 2 b = a + c , so R is transitive.

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4023f10ps4a - Homework#4 Solutions Due October 4 2010 1 For...

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