TITLE: What is an example function of a transitive yet non-reflexive and non-symmetric relation?
QUESTION [5 upvotes]: If I were to come up with a function $f: \mathbb{N} \to \mathbb{N}$ that makes relation $R_a$ transitive yet non-reflexive and non-symmetric such that 
$$q \, R_a \, z  \text{ if } f(q) = z,$$ what kind of function should I come up with? I spent the past three hours trying to come up with one but I still don't understand how inputting one number can produce two different results... Do I need to use a piecewise?
Because isn't it transitive when you input $1$ and by some relation, you then get $2$.
Then when you input $2$ into that relation you get $3$. 
Because of transitivity $1$ now should also produce $3$. This makes no sense to me...

REPLY [2 votes]: Suppose $f$ makes $R_a$ transitive. Let $y,z\in \mathbb{N}$ such that  $$f(x)=y, f(y)=z$$ for some $x \in \mathbb{N}.$ Transitivity now implies $f(x)=z.$ So $y=z.$
So $f(y)=y$ for every $y \in f[\mathbb{N}].$
Of course, constants are an example. But they are not the only example. The above argument suggests that you only need to find a function which fixes the elements in the range. For instance

$$f(n)=\begin{cases}1, \text{if }n \text{ is odd}\\2, \text{ if }n \text{ is even}\end{cases}.$$
$f$ is not reflexive because $f(3)\neq3$ and not symmetric because $f(3)=1$ but $f(1)\neq 3.$

Obviously you could have chosen any odd and even number in place of $1$ and $2$ respectively.
You can come up with plenty of other examples in a similar way.