If $\forall x,y \in B$ if $(x,y) \in S$ and if $x+y$ is even then $x=y$ then each class of $S$ has at most $2$ elements 
Set $B = \{1,2,3,4,5\}$, $S$ - equivalence relation. It is given that for all $x,y \in B$ if $(x,y)\in S$ and if $x+y$ is an even number then $x = y$. In such case is it true that:
  
  
*
  
*the number of elements in each equivalence class of $S$ is at most $2$
  
*any relation $S$ would have an equivalence class made up of just one even number.
  

As far as I understand $S$ could be only of such form:
$$
S = \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5 \end{pmatrix}
$$
because all pairs of $(x,y), x \neq y$ which are both odd numbers can't be in $S$ as well as all pairs $(x,y), x \neq y$ which are both even numbers for example:
$$
\begin{pmatrix}1\\3 \end{pmatrix}, \begin{pmatrix}2\\4 \end{pmatrix}
$$
because their sum will be even but $x \neq y$.
In addition $(x,y), x\neq y$ where one of them is odd and one is even also can't be in $S$ because then the relation will not be transitive and hence will not be an equivalence relation. 
In such case I think the statement 1 is false because all equivalence classes are exactly of size $1$ and statement 2 is true because we have for example the equivalence class $\{2\}$ which is one even number.
I'm not sure about my logic because the question is quite tricky.

Assume that $S$ has an equivalence class with $2$ even numbers, namely $2$ and $4$. Then, $(x,y)\in S$ and $x+y=6$ which is even, but this contradicts $x=y$. Hence, two even numbers cannot be in the same equivalence class.
Now, assume that there exists an equivalence class with two odd numbers $x\neq y$. Then $x+y=\text{even},$ but again $x=y$ is not satisfied. Hence, if $x\neq y$ and $x,y$ odd, then they cannot be in the same equivalence class. 
So, you have that any equivalence class in $S$ contains at most one even and at most one odd number. Since, there are five elements in $B$, this leaves you with a class that necessarily has one element. But I do not see why this has to be an even number. So, I agree with the first statement but disagree with the second statement (the even numnber part).
