Question #171214

RELATION.

Given the following set:

1. X = {1, 2, 3, 4, 5} defined by the rule (x, y) ∈ R if x + y ≤ 6 


a. List the elements of R

b. Find the domain of R

c. Find the range of R

d. Draw the digraph

e. Properties of the Relation 


1
Expert's answer
2021-03-16T08:19:25-0400

Given the X={1,2,3,4,5}X = \{1, 2, 3, 4, 5\} define the relation RR by the rule (x,y)R(x, y) \in R if x+y6x + y \le 6  


a. Let us find the list the elements of R:R:


R={(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(1,4),(4,1),(2,3),(3,2),(1,5),(5,1),(2,4),(4,2),(3,3)}R=\{(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(1,4),(4,1),(2,3),(3,2),(1,5), \newline (5,1),(2,4),(4,2),(3,3)\}


b. Let us find the domain of R:R:


dom(R)={xX  (x,y)R}={1,2,3,4,5}dom(R)=\{x\in X\ |\ (x,y)\in R\}=\{1,2,3,4,5\}



c. Let us find the range of R:R:


range(R)={yX  (x,y)R}={1,2,3,4,5}range(R)=\{y\in X\ |\ (x,y)\in R\}=\{1,2,3,4,5\}


d. Let us draw the digraph of R:R:




e. Let us study some properties of the relation R:R:


Since (4,4)R(4,4)\notin R, the relation RR is not reflexive.


If (x,y)R(x, y) \in R, then x+y6x + y \le 6, and hence y+x6y + x \le 6. It follows that (y,x)R(y, x) \in R, and the relation is symmetric.


Taking into account that (5,1)R(5,1)\in R and (1,5)R(1,5)\in R, but (5,5)R(5,5)\notin R, we conclude that RR is not a transitive relation.


It follows that RR is not an equivalence relation.  



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