No union of two equivalence relations may not be an equivalence relation because of transitive dependency.
Equivalence relation: satisfy Reflexive, symmetric, and transitive property
- Reflexive $\cup$ Reflexive = Reflexive
- Symmetric $\cup$ Symmetric = Symmetric
- Transitive $\cup$ Transitive $\neq$ Transitive why
Example $: R = \left \{ \left ( 1,2 \right ), \left ( 3,4 \right ), \left ( 1,4 \right ) \right \},\;S$ $=$ ${ (2, 3)}$
Union $= \left \{ \left ( 1,2 \right ), \left ( 3,4 \right ), \left ( 1,4 \right ), (2, 3) \right \}$ which is not transitive i.e. $(1, 3)$ and $(2, 4)$ is missing.
So, False is the answer.