Q. 43.8( 9 Votes )

Let A = {1, 2, 3}

Answer :

We have been given,

A = {1, 2, 3}


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


Let us take R1.


R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}


(i). Reflexive:


1, 2, 3 A [ A = {1, 2, 3}]


(1, 1) R1


(2, 2) R2


(3, 3) R3


So, for a A, (a, a) R1


R1 is reflexive.


(ii). Symmetric:


1, 2, 3 A


If (1, 3) R1, then (3, 1) R1


[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]


But if (2, 1) R1, then (1, 2) R1


[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]


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


a, b A


R1 is not symmetric.


(iii). Transitivity:


1, 2, 3 A


If (1, 3) R1 and (3, 3) R1


Then, (1, 3) R1


[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]


But, if (2, 1) R1 and (1, 3) R1


Then, (2, 3) R1


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


a, b, c A


R1 is not transitive.


Now, take R2.


R2 = {(2, 2), (3, 1), (1, 3)}


(i). Reflexive:


1, 2, 3 A [ A = {1, 2, 3}]


(1, 1) R2


(2, 2) R2


(3, 3) R2


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


R2 is not reflexive.


(ii). Symmetric:


1, 2, 3 A


If (1, 3) R2, then (3, 1) R2


[ R2 = {(2, 2), (3, 1), (1, 3)}]


If (2, 2) R2, then (2, 2) R2


[ R2 = {(2, 2), (3, 1), (1, 3)}]


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


a, b A


R2 is symmetric.


(iii). Transitivity:


1, 2, 3 A


If (1, 3) R2 and (3, 1) R2


Then, (1, 1) R2


[ R2 = {(2, 2), (3, 1), (1, 3)}]


So, if (a, b) R2 and (b, c) R2, then (a, c) R2


a, b, c A


R2 is not transitive.


Now take R3.


R3 = {(1, 3), (3, 3)}


(i). Reflexive:


1, 3 A [ A = {1, 2, 3}]


(1, 1) R3


(3, 3) R3


So, for a A, (a, a) R3


R3 is not reflexive.


(ii). Symmetric:


1, 3 A


If (1, 3) R3, then (3, 1) R3


[ R3 = {(1, 3), (3, 3)}]


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


a, b A


R3 is not symmetric.


(iii). Transitivity:


1, 3 A


If (1, 3) R3 and (3, 3) R3


Then, (1, 3) R3


[ R3 = {(1, 3), (3, 3)}]


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


a, b, c A


R3 is transitive.


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