Functional relations : Trouble seeing transitivity Given the following domain: $\;\{1,2,3,4\}$
And the following relation:
$$\{(1,1),(1,3),(1,5),(2,2),(2,4),(3,1), (3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(5,5)\}$$
It states that this is an equivalence relation, but to my understanding for that to be true the relation needs to be reflexive, symmetric and transitive. 
I can see that the relation is both reflexive and symmetric, but I can't see that it's transitive, which is making me question if it's an equivalence relation.
I'm pretty new to this topic so I wouldn't say I have a concrete understanding of it as it stands.
Thanks in advance
 A: Transitivity means that whenever $(a, b) \in R,$ and also $(b, c) \in R,$ then $(a, c) \in R$.
Can you see that for every ordered pair in your relation, whenever $(a, b), (b, c) \in R,$ then so is $(a, c)$?
For example: 


*

*we see that $(1, 3) \in R$ and $(3, 5)\in R$. So if the relation is transitive, we have to have that $(1, 5) \in R$. And we see that $(1, 5) \in R$ $\large \checkmark$! 

*Similarly, we see    that $(2, 4), (4, 2) \in R$, and so is $(2, 2)
   \in R$ $\large    \checkmark$.


If there are NOT any pairs where $(a, b)\in R, (b, c) \in R,\;$ but $(a, c) \notin R,\;$ then the relation is transitive.
A: The easiest way is to look whether the sets
$$[a]=\{x\in X:a\mathbin{R}x\}$$
form a partition of $X=\{1,2,3,4,5\}$, where $R$ is your relation:
\begin{align}
[1]&=\{1,3,5\}=[3]=[5],\\
[2]&=\{2,4\}=[4].\\
\end{align}
So yes, it's an equivalence relation.
Why does this proves transitivity? Let's suppose $a$, $b$ and $c$ satisfy
$$a\mathbin{R}b\text{ and }b\mathbin{R}c.$$
Then either $a\in\{1,3,5\}$ or $a\in\{2,4\}$. Suppose the second holds: then $b=2$ or $b=4$. In both cases we must have that $c=2$ or $c=4$; therefore $a\mathbin{R}c$.
Similarly, but more tedious, in the case when $a\in\{1,3,5\}$.
