December 1999 QUESTION 2 Total Marks: 15 Marks |
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SUGGESTED SOLUTIONS |
(a) |
Consider the following relations on the set {1 ,2 ,3 ,4 }
R 2 ={(1 ,1 ),(1 ,2 ),(2 ,1 )} R 3 ={(1 ,1 ),(1 ,2 ),(1 ,4 ),(2 ,1 ),(2 ,2 ),(3 ,3 ),(4 ,1 ),(4 ,4 )} R 4 ={(2 ,1 ),(3 ,1 ),(3 ,2 ),(4 ,1 ),(4 ,2 ),(4 ,3 )} R 5 ={(1 ,1 ),(1 ,2 ),(1 ,3 ),(1 ,4 ),(2 ,2 ),(2 ,3 ),(2 ,4 ),(3 ,3 ),(3 ,4 ),(4 ,4 )} R 6 ={(3 ,4 )}
The relations R 3 and R 5 are reflexive since they both contain all pairs of the form (a ,a ),for all elements a . (ii)Which of these relations are symmetric?Explain.[2 marks ] The relations R 2 and R 3 are symmetric because in each case (b ,a )belongs to the relation whenever (a ,b )does.
The relations R 4,R 5 and R 6 are all antisymmentric because for each of these relations there is no pair of elements a and b with a .b such that both (a ,b ) and (b ,a )belong to the relation.
The relations
R 4, R 5 and R 6 are transitive. For each of these relations, if (a
,b ) and (b ,c ) belong to the relation,then so does
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[8] |
(b) |
Let A ,B and C be sets.Show,using algebraic laws,that
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[4] |
(c) |
Why is f not a function from R to R in the following equations?
f (0 )is not defined.
f (x ) is not defined for x <0.g
f (x ) is not well –defined since there are two distinct values assigned to each x . |
[3] |