Adv Alegbra HW Solutions 44

Adv Alegbra HW Solutions 44 - R = { ( x , x ) : x 1 2 } ....

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44 (ii) If B Y , prove that f 1 ( B 0 ) = f 1 ( B ) 0 , where B 0 denotes the complement of B . Solution. Absent. 2.18 Let f : X Y be a function. De f ne a relation on X by x x 0 if f ( x ) = f ( x 0 ) . Prove that is an equivalence relation. If x X and f ( x ) = y , the equivalence class [ x ] is denoted by f 1 ( y ) ; it is called the fber over y . Solution. Absent. 2.19 Let X ={ rock , paper , scissors } . Recall the game whose rules are: paper dominates rock, rock dominates scissors, and scissors dominates paper. Draw a subset of X × X showing that domination is a relation on X . Solution. Absent. 2.20 (i) Find the error in the following argument which claims to prove that a symmetric and transitive relation R on a set X must be re- F exive; that is, R is an equivalence relation on X .I f x X and xRy , then symmetry gives yRx and transitivity gives xRx . Solution. There may not exist y X with x y . (ii) Give an example of a symmetric and transitive relation on the closed unit interval X =[ 0 , 1 ] which is not re F exive. Solution. De f
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Unformatted text preview: R = { ( x , x ) : x 1 2 } . Now R is the identity on Y = [ , 1 2 ] , so that it is symmetric and transitive. However, R does not contain the diagonal of the big square X X , and so R is not a re F exive relation on X . For example, 1 1. 2.21 True or false with reasons. (i) The symmetric group on n letters is a set of n elements. Solution. False. (ii) If S 6 , then n = 1 for some n 1. Solution. True. (iii) If , S n , then is an abbreviation for . Solution. True. (iv) If , are cycles in S n , then = . Solution. False. (v) If , are r-cycles in S n , then is an r-cycle. Solution. False. (vi) If S n is an r-cycle, then 1 is an r-cycle for every S n . Solution. True....
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