Q. 3 C3.5( 4 Votes )

Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive :

R3 on R defined by (a, b) ϵ R3 a2 – 4 ab + 3b2 = 0.

Answer :

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


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


R3 on R defined by (a, b) R3 a2 – 4ab + 3b2 = 0


Check for Reflexivity:


a R,


(a, a) R3 needs to be proved for reflexivity.


If (a, b) R3, then we have


a2 – 4ab + 3b2 = 0


Replace b by a, we get


a2 – 4aa + 3a2 = 0


a2 – 4a2 + 3a2 = 0


–3a2 + 3a2 = 0


0 = 0, which is true.


(a, a) R3


So, a R, (a, a) R3


R3 is reflexive.


Check for Symmetry:


a, b R


If (a, b) R3, then we have


a2 – 4ab + 3b2 = 0


a2 – 3ab – ab + 3b2 = 0


a (a – 3b) – b (a – 3b) = 0


(a – b) (a – 3b) = 0


(a – b) = 0 or (a – 3b) = 0


a = b or a = 3b …(1)


Replace a by b and b by a in equation (1), we get


b = a or b = 3a …(2)


Results (1) and (2) does not match.


(b, a) R3


R3 is not symmetric.


Check for Transitivity:


a, b, c R


If (a, b) R3 and (b, c) R3


a2 – 4ab + 3b2 = 0 and b2 – 4bc + 3c2 = 0


a2 – 3ab – ab + 3b2 = 0 and b2 – 3bc – bc + 3c2


a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0


(a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0


(a – b) = 0 or (a – 3b) = 0


And (b – c) = 0 or (b – 3c) = 0


a = b or a = 3b And b = c or b = 3c


What we need to prove here is that, a = c or a = 3c


Take a = b and b = c


Clearly implies that a = c.


[ if a = b, just substitute a in place of b in b = c. We get, a = c]


Now, take a = 3b and b = 3c


If a = 3b



Substitute in b = 3c. We get



a = 9c, which is not the desired result.


(a, c) R3


R3 is not transitive.


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