State whether the following statement is TRUE or FALSE:
The union of two equivalence relations is also an equivalence relation.
A. TRUE
B. FALSE
GATE 1987 · Discrete Mathematics · Relations · medium
Answer: The statement is FALSE: the union of two equivalence relations need not be transitive, as the counterexample shows.
Check reflexivity and symmetry: If R1 and R2 are both reflexive, then (a,a) is in R1 for all a, hence also in R1 union R2: reflexivity is preserved. If R1 is symmetric: (a,b) in R1 implies (b,a) in R1, so (b,a) also in the union whenever (a,b) is; same for R2. Thus the union is always symmetric.
Counterexample for transitivity: Let S = {1,2,3}. R1 = {(1,1),(2,2),(3,3),(1,2),(2,1)} is an equivalence (partition: {1,2} and {3}). R2 = {(1,1),(2,2),(3,3),(2,3),(3,2)} is an equivalence (partition: {1} and {2,3}). R1 union R2 = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Now (1,2) and (2,3) are in the union but (1,3) is NOT. Transitivity fails.