Me and my friends were complaining about one of the exercises in discrete math.
If there are two equivalence relations $R_1$ and $R_2$ on set $X$. Is $R_{1}\setminus R_{2}$ still an equivalence relation?
My example:
If set $X = \{a,b,c\}$
$$\begin{align} R_{1} &= \{(a,a),(b,b),(c,c),(a,b),(b,a)\}\\ R_{2} &= \{(a,a),(b,b),(c,c),(a,c),(c,a)\} \end{align}$$
$R_{1}\setminus R_{2} = \{(a,b),(b,a)\}$ which isn't an equivalence relation.
Some of them say that $R_{1}\setminus R_{2}$ is still an equivalence relation (not necessarily on my example)
Can someone make it clear for me and rest of us?