Why is this relation transitive? On $A = \{1,2,3,4,5\}$, define the relation $R = \{(1,1),(1,2),(2,1),(2,2),(3,3),(3,4),(4,3),(4,4),(5,5)\}$.
According to my book this relation is transitive, but according this definition there has to be something like $(x,y),(y,z) = (x,z)$. I am just confused here since majority of them are like $(1,1)$ which is $(x,x)$ and not $(x,y)$. I know it is reflexive and symmetric, but I just don't see how it could be transitive. 
Edit:
Is it correct then to say $(1,1),(1,2) = (1,2)$ is transitive?
Edit $2$:
Thanks for answer, I was completely fooled and thought my textbook literally meant $x$ and $y$ had to be distinct and coincidentally Grimaldi only used distinct $x$ and $y$ too and never gave any stuff where $x$ and $y$ were the same.
 A: It's pretty simple. $(x,y)$ doesn't mean that $x$ and $y$ have to both be different. You can have $(1,1)$ so $x=1$ and $y=1$. Then $y=x$, and that's okay. You can have two variables that denote the same value.
For example, take $(1,1)$ and $(1,2)$. You can let $x=1$ and $y=1$ and $z=2$. Then we should be able to find $(x,z)$ as well. That's $(1,2)$, again. So transitivity holds, and we're good.
A: Transitivity means (x,y),(y,z) = (x,z), as you commented. If there are no (x,y) for $x\ne y$, it could still be a transitive relation. 
R = {(1,1),(2,2),(3,3),(4,4),(5,5)} would be reflexive, symmetric and transitive!
A: To see why $R$ is transitive, pick any two ordered pairs in $R$ such that the second number of the first ordered pair matches the first number of the second ordered pair. For example, consider:
$$
(\color{red}{4},\color{green}{3}) \text{ and } (\color{green}{3},\color{blue}{3})
$$
Then $R$ must also contain the ordered pair where the matching number in the middle is removed. Indeed, we find that $R$ contains the following ordered pair:
$$
(\color{red}{4},\color{blue}{3})
$$
Admittedly, it's hard to verify this at a glance. We'd have to check all $9 \cdot 8 = 72$ possible pairs!
