Question #115546
a) Provide an example of a relation T on set A = {1, c, 3} that is irreflexive and satisfies trichotomy. (What is Trichotomy)

b) Let S= {(1, 2), (3, 4), (2, 3)} be a relation on set B = {1, 2, 3, 4}.
Is S functional? Motivate your answer

c) Let C = {1, 2, a, b} and let R = {(1, 1), (a, b), (b, 2)} and S = {(2, 1), (a, 1), (b, b), (b, 2), (2, a)} be two relations on C
i) Determine R o S (S;R)
ii) Which ordered pairs must be added to R to make it a reflexive relation?
1
Expert's answer
2020-05-14T18:05:16-0400

a) On the set X={1,c,3}X = \{1,c,3\}, the relation R={(1,c),(1,3),(c,3)}R = \{ (1,c), (1,3), (c,3)\} is irreflexive because xX ⁣:(x,x)R\forall x\in X\colon (x,x)\notin R, and trichotomous (see first example in [1], it is the same). RR is trichotomous (see first property in [1]) because it is asymmetric, (x,y)R(y,x)∉R(x,y)\in R \rightarrow (y,x)\not\in R, and semiconnex, xyX ⁣:(x,y)R(y,x)R\forall x \ne y\in X\colon (x,y)\in R \, \lor \, (y,x)\in R.


b) Yes, because it is one-to-one


c) i) S;RS;R means that firstly SS then RR. Hence

RS={(2,1),(a,1),(b,2),(2,b)}R\circ S=\{(2,1),(a,1),(b,2),(2,b)\}

ii) (1,1),(2,2),(a,a),(b,b)(1,1), (2,2), (a,a), (b,b)


[1] https://en.wikipedia.org/wiki/Trichotomy_(mathematics)


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