Proving equivalence relations in special symbols

Question

For a function $f: A\to B$, I have a relation $@$ on $A$ described by

$(\forall x,y \in A)\quad x @ y \Leftrightarrow f(x) + f(y)$

Is there any way to show that $@$ is an equivalence relation?

Answer 1

To check if a relation @ is an equivalence relation, you need to verify three things:

• @ is reflexive
• @ is symmetric
• @ is transitive

For your particular relation, reflexivity is apparent as $f(x) = f(x)$, (so $x@x$). Symmetry is also not very difficult. If $x@y$, do we have $y@x$? Of course, $x@y$ means $f(x)=f(y)\leftrightarrow f(y)=f(x) \leftrightarrow y@x$. Transitivity is also very easy to see, because $=$ itself is a transitive relation. Explicitly if $x@y \land y@z$, this means that $f(x)=f(y) \land f(y)=f(z)$. So, in particular, by transitivity of $=$, we have that $f(x) = f(z) \leftrightarrow x@z$.