Prove that R is an equivalence relation on F

Question

A relation $R$ is defined on the set $F = \{f: \Bbb R \to \Bbb R\}$

$$fRg \iff f(0) = g(0).$$

My approach:

This is reflexive because: $f(0) = f(0)$ is same as $f(0) = g(0)$

This is symmetric because: $f(0) = g(0)$ is same as $g(0) = f(0)$

This is transitive because: $f(0) = g(0)$ implies $g(0) = f(0)$ which implies $f(0) = f(0)$

Am anywhere close to being correct?

Thanks

Answer 1

Transitivity is not correct. It should be: f(0) = g(0),and g(0) = h(0) then f(0) = h(0).

Answer 2

It's transitive because: $f(0)=g(0)$ and $g(0)=h(0) \Rightarrow f(0) = h(0)$.

Check your definition of transitivity!