InterviewSolution
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InterviewSolution
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Determine whether each of the following relations are reflexive, symmetric and transitive : (i) Relation R in the set `A= {1, 2, 3, …, 13, 14}` defined as `" " R = { (x, y ) : 3x - y =0}` (ii) Relation R in the set N of natural numbers defined as `" " R = {(x, y) : y = x + 5 and x lt 4}` (iii) Relation R in the set `A= { 1, 2, 3, 4, 5, 6}` as `" "R = { (x, y) : y ` is divisible by x`}` (iv) Relation R in the set Z of all integers defined as `" " R = {(x,y) : x -y ` is an integer `}` (v) Relation R in the set A of human beings in a town at a particular time given by (a) `R= {(x, y) : and y` work at the same place `}` (b)` R = { (x, y) : x and y ` live in the same locality `}` (c) `R = {(x, y) : x ` is exactly 7 cm taller than y` }` (d) `R = {(x, y) : x ` is wife of `y}` (e) `R= { (x, y) :x ` is father of `y}` |
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Answer» (i) A =` { 1, 2, 3, …, 13, 14}` and `" " R = {(x, y) : 3x -y =0}` For reflexive `(x,x) in R AA x in A` but ` " " 3x -y =0 rArr y = 3x` `therfore " " (x, x) notin R ` if ` x = 2 in A` `rArr R `is not reflexive, For symmetricity `(x, y) in R rArr (y, x) in R AA x, y in R ` Now, `" " (x, y) in R rArr 3 x -y =0` `" " rArr 3y -x ne 0` `" " rArr (y, x) notin R` `therefore R` is not symmetric. e.g., `(1, 3) in R and (3, 1) notin R` For transitivity `(x, y) in R, (y, z) in R rArr (x, z) in R` `therefore (1, 3) in R and (3, 9) in R rArr (1, 9) in R` `rArr R` is not transitive. (ii) `R= {(x,y) : y = x + 5 and x lt 4 }` and N is the set of natural numbers. `rArr R = {(1, 6), (2, 7), (3, 8)}` For reflexive, `(1,1) notin R` `rArr R` is not reflexive. For transitivity, `(x,y) in R (y, z) in R rArr (x, z) in R`. No pair satisfies this condition. `therefore R` is not transitive. (iii) `A= {1, 2, 3, 4, 5, 6}` and `R = {(x, y) : y` is divisible by `x}` `rArr R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5, (6, 6)}` For each `x in A, (x, x) in R` `therefore R ` is reflexive. For each `x, y in R, (x, y) in R cancel(rArr) (y, x) in R` `therefore R` is not symmetric. For each `x, y, z in A` if `(x, y) in R, (y, z) in R` then `(x, z) in R` `therefore R` is transitive. (iv) In the set of all integers Z `" " R = {(x, y) : x -y ` is an integer. } Which is true. `therefore R` is reflexive. For symmetricity, `(x, y) in R rArr (x -y)` is an integer. `" " rArr (y-x) in R` ` therefore R` is symmetric. `therefore ` For transitivity. `(x, y) in R and (y, z) in R` `rArr (x -y) ` is an integer and `(y-z)` is an integer. `rArr (x-y)+ (y-z)` is an integer. `rArr (x, z) in R` `therefore R` is transitive. (v) (a) `R= {(x, y) : x and y` work at the same place } This relation is reflexive, symmetric and transitive. (b) `R= {(x, y) : x and y` live in the same locality}. This relation is reflexive, symmetric and transitive. (c) `R= {(x, y) : x` is exactly 7 cm taller than y } `(x, x) notin R` because x is no exactly 7 cm taller than y `therefore R ` is not reflexive. `(x, y) in R rArr x ` is exactly 7 cm taller than y. `" " cancel ( rArr) y` is exactly 7 cm taller than x. `" " cancel (rArr) (y, x) in R` `therefore R` is not symmetric. `therfore (x, y) in R and (y, z) in R rArr x ` is exactly 7 cm taller than y and y is exactly7 cm taller than z. |
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