Question

# Given an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive.

Solution

## (i) Let A = {5, 6, 7}. Define a relation R on A as R = {(5, 6), (6, 5)}. Relation R is not reflexive as {(5, 5), (6, 6), (7, 7)} /∈R. Now, as (5,6)∈R and also (6,5)∈R, R is symmetric. (5,6),(6,5)∈R, but (5, 5) /∈R.Therefore R is not transitive. Hence, relation R is symmetric but not reflexive or transitive.  R = {(a, b): a < b} For any a∈R, we have (a,a)/∈R since a cannot be strictly less than a (itself). Therefore, R is not reflexive. [∵a=a] Now, (1,2)∈R (as 1 < 2) But, (2,1)/∈R  as 2 not less than 1 Now, let (a, b), (b, c) ∈R.  ⇒a<b and b<c⇒a<c⇒(a,c)∈R. Therefore, R is transitive. Hence, relation R is transitive but neither symmetric nor reflexive.  Let A ={4,6,8} Defined a relation R on A as A ={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)} Relation R is reflexive, because (4,4),(6,6), (8,8)∈R. Relation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a, b∈R. Relation R is not transitive. since (4,6),(6,8), ∈R, but (4,8)/∈R. Hence, relation R is reflexive and symmetric but not trasitive.   Define a relation R in R(real numbers) as:  R={(a,b):a3≥b3}  Clearly (a,a) as a3=a3. Therefore, R is reflexive. Now, (2,1)∈R( as 23≥13) But, (1,2)/∈R (as 13<23) Therefore, R is not symmetric . Now, let (a,b),(b,c)∈R ⇒a3≥b3 and b3≥c3⇒a3≥c3⇒(a,c)∈R. Therefore, R is transtive. Hence, relation R is reflexive and transitive but not symmetric. Let A ={-5,-6} Defince a relation R on A as: R ={(-5,-6),(-6,-5),(-5,-5)} Relation R is not reflexive as (−6,−6)/∈R Relation R is symmetric as (−5,−6)∈R and (−6,−5)∈R. And (−5,−6),(−6,−5)∈R⇒(−5,−5)∈R.  Therefore, the relation R is transitive. Hence, relation R is symmetric and transitive but not reflexive. Mathematics

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