Four Relations on $\{1,2,3\}$ 
Define four relations on $\{1, 2, 3\}$.

*

*Both transitive and symmetric.

*Transitive and not symmetric.

*Symmetric but not transitive.

*Neither transitive nor symmetric.


I have:

*

*$R= \{(1,2),~(2,1),~(2,2),~(1,1)\}$ Transitive and Symmetric

*$R= \{(1,1),~(2,2),~(1,2)\}$ Transitive and Not Symmetric

*$R= \{(1,2),~(2,1),~(2,3),~(3,2)\}$ Not Transitive and Symmetric

*$R= \{(1,1),~(1,2),~(2,3)\}$ Not Transitive and Not Symmetric

Am I understanding what the question was asking correctly and are my relations correct?
 A: Overall, as already stated, your work is perfectly fine and correct, and you understand what is necessary. However, I would like to present a bit of a visual intuition for understanding these relations and properties, and how your relations meet the desired criteria.

It is easiest to see these properties by means of a directed graph, such as the below one for your first relation:

In a graph of this form, node $x$ points to node $y$ if and only if $(x,y) \in R$ (or, alternatively stated, $xRy$).
In graphs like these, then, your properties can be seen as so:

*

*Reflexivity: Every node has an arrow pointing to itself.

*Symmetry: If an arrow goes from $x$ to $y$, there is an arrow going in the opposite direction: $y$ to $x$. Basically, between two distinct nodes, there is either no "loop" or a closed loop.

*Transitivity: If node $x$ points to $y$, and $y$ points to $z$, then $x$ points to $z$. Essentially, for any three distinct nodes, if two pairs are connected, the third pair must also be connected, to form a triangle.

So let's look at your examples:

First, your initial $R$:

You can see this is symmetric because there is the closed loop between $1$ and $2$. Neither point to $3$, so that's okay. (We only require that a given path allow you to backtrack between distinct nodes. Stray nodes are okay.)
Transitivity follows from the "self-pointing" arrows. Walking from $1$ to $2$ to $1$, is the same as staying at $1$ (following the loop pointing to and from $1$). Same for the arrow for $2$.

For your second example $R$:

This is not symmetric, since the loop $1 \to 2$ is not closed in the other way.
This is transitive since the only case to consider is $1 \to 1 \to 2$, and since $1 \to 2$, this follows trivially. Perhaps it's easier to see this if we have multiple $2$ and $1$ nodes to avoid the self-pointing arrows and rely more on the triangle analogy:

Perhaps it's clearly with this one why any two sides of a "triangle" in the directed graph always have a third side completed as well.

For your third example $R$:

In this case, we have symmetry since we only have closed loops and no one-way paths.
We don't have transitivity, however, since the top-left side of either triangle ($1 \to 3$ resulting from $3 \to 2 \to 1$, nor $3 \to 1$ resulting from $1 \to 2 \to 3$) is not present. (Some arguments could also be made through the lack of self-pointing arrows -- for instance, $1 \to 2 \to 1$ should imply $1 \to 1$ exists if $R$ is transitive -- but the triangle bit is easiest to visualize.)

For your final relation:

Here, we have no symmetry since there are some one-way paths instead of closed loops. Similarly, transitivity is not present since the completion of this triangle ($3 \to 1$) is not present.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
