Q. 23.6( 12 Votes )

Relations R1, R2, R3 and R4 are defined on a set A = {a, b, c} as follows :

R1 = {(a, a) (a, b) (a, c) (b, b) (b, c), (c, a) (c, b) (c, c)}

R2 = {(a, a)}

R3 = {(b, a)}

R4 = {(a, b) (b, c) (c, a)}

Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (iii) symmetric (iii) transitive.

Answer :

We have set,

A = {a, b, c}


Here, R1, R2, R3, and R4 are the binary relations on set A.


So, recall that for any binary relation R on set A. We have,


R is reflexive if for all x A, xRx.


R is symmetric if for all x, y A, if xRy, then yRx.


R is transitive if for all x, y, z A, if xRy and yRz, then xRz.


So, using these results let us start determining given relations.


We have


R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}


(i). Reflexive:


For all a, b, c A. [ A = {a, b, c}]


Then, (a, a) R1


(b, b) A


(c, c) A


[ R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]


So, a, b, c A, then (a, a), (b, b), (c, c) R.


R1 is reflexive.


(ii). Symmetric:


If (a, a), (b, b), (c, c), (a, c), (b, c) R1


Then, clearly (a, a), (b, b), (c, c), (c, a), (c, b) R1


a, b, c A


[ R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]


But, we need to try to show a contradiction to be able to determine the symmetry.


So, we know (a, b) R1


But, (b, a) R1


So, if (a, b) R1, then (b, a) R1.


a, b A


R1 is not symmetric.


(iii). Transitive:


If (b, c) R1 and (c, a) R1


But, (b, a) R1 [Check the Relation R1 that does not contain (b, a)]


a, b A


[ R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]


So, if (b, c) R1 and (c, a) R1, then (b, a) R1.


a, b, c A


R1 is not transitive.


Now, we have


R2 = {(a, a)}


(i). Reflexive:


Here, only (a, a) R2


for a A. [ A = {a, b, c}]


[ R2 = {(a, a)}]


So, for a A, then (a, a) R2.


R2 is reflexive.


(ii). Symmetric:


For symmetry,


If (x, y) R, then (y, x) R


x, y A.


Notice, in R2 we have


R2 = {(a, a)}


So, if (a, a) R2, then (a, a) R2.


Where a A.


R2 is symmetric.


(iii). Transitive:


Here,


(a, a) R2 and (a, a) R2


Then, obviously (a, a) R2


Where a A.


[ R2 = {(a, a)}]


So, if (a, a) R2 and (a, a) R2, then (a, a) R2, where a A.


R2 is transitive.


Now, we have


R3 = {(b, a)}


(i). Reflexive:


a, b A [ A = {a, b, c}]


But, (a, a) R3


Also, (b, b) R3


[ R3 = {(b, a)}]


So, a, b A, then (a, a), (b, b) R3


R3 is not reflexive.


(ii). Symmetric:


If (b, a) R3


Then, (a, b) should belong to R3.


a, b A. [ A = {a, b, c}]


But, (a, b) R3


[ R3 = {(b, a)}]


So, if (a, b) R3, then (b, a) R3


a, b A


R3 is not symmetric.


(iii). Transitive:


We have (b, a) R3 but do not contain any other element in R3.


Transitivity can’t be proved in R3.


[ R3 = {(b, a)}]


So, if (b, a) R3 but since there is no other element.


R3 is not transitive.


Now, we have


R4 = {(a, b) (b, c) (c, a)}


(i). Reflexive:


a, b, c A [ A = {a, b, c}]


But, (a, a) R4


Also, (b, b) R4 and (c, c) R4


[ R4 = {(a, b) (b, c) (c, a)}]


So, a, b, c A, then (a, a), (b, b), (c, c) R4


R4 is not reflexive.


(ii). Symmetric:


If (a, b) R4, then (b, a) R4


But (b, a) R4


[ R4 = {(a, b) (b, c) (c, a)}]


So, a, b A, if (a, b) R4, then (b, a) R4.


R4 is not symmetric.


It is sufficient to show only one case of ordered pairs violating the definition.


R4 is not symmetric.


(iii). Transitivity:


We have,


(a, b) R4 and (b, c) R4


(a, c) R4


But, is it so?


No, (a, c) R4


So, it is enough to determine that R4 is not transitive.


a, b, c A, if (a, b) R4 and (b, c) R4, then (a, c) R4.


R4 is not transitive.

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