I think it's important you understand how reflexivity, symmetry, and transitivity apply to relations.
Relations
Before talking about the reflexivity, symmetry, transitivity of a relation, first let's talk about binary relations. A binary relation on a set $A$ is simply a subset of the Cartesian product $A \times A$. We choose what goes into that subset when we define the relation. Usually, the relation is defined by some mathematical rule or model, such as the one in your question, but it need not be so. It could just be a random selection of elements in $A \times A$ with no apparent coherent pattern or relationship. Fortunately this discussion will only consider relations defined by rules.
So a relation on $A$ is like a filtering of elements in $A \times A$ according to the rules. If $(a, b) \in A \times A$ "passes" the filter (which is the relation $R$), then we say that $a$ is related to $b$. Note that $(a, b)$ is an ordered pair, which means that $a$ coming before $b$ has significance and $(a, b)$ is not equivalent to $(b, a)$, as we'll see below.
The following statements are equivalent with respect to a relation $R$:
- $a$ is related to $b$
- $a R b$
- $(a, b) \in R$
Reflexivity
This property is pretty straight forward. Is $a$ related to itself under the "rules" of the relation. Or, in other words, is $(a, a)$ in the set $R$? For example, if the relation is defined as
$$ R = \{(a, b) \mid a \ge b \quad a, b \in \mathbb{Z} \} \tag{1}$$
then simply apply $(a, a)$ to the definition of the relation: $a \ge a$ is true, so $a$ is related to itself. It is reflexive.
There's no reason that a relation be reflexive. Take for example the relation
$$a R b \iff a > b \tag{2}$$.
Here, $a \not > a$, so $a$ is not related to $a$, hence this relation is not reflexive.
Symmetry
The symmetry of a relation indicates that if $a$ is related to $b$, then $b$ is related to $a$. It is easy to see that neither $(1)$ nor $(2)$ is symmetric. According to $(1)$, $2 R 1$ because $2 \ge 1$. However, $1 \not \ge 2$, so $(1, 2) \not \in R$. That $(2, 1) \in R$ but $(1, 2) \not \in R$ implies that $(1)$ is not symmetric. The same can be said about $(2)$. All it takes is finding counter examples.
A real life example could be:
$$ (a, b) \in R \iff b \mbox{ is } a\mbox{'s spouse} \tag{3}$$
Since marriage is a two way relationship, then $(3)$ is a symmetric relation.
Anti-symmetry
The definition of an anti-symmetric relation is: for all $a, b$, if $a$ is related $b$ and $b$ is related to $a$, then it is necessary that $a = b$. Formally,
$$\forall a, b \in X, \; (a, b) \in R \wedge (b, a) \in R \Rightarrow a = b \tag{4}$$
Let's look at $(1)$. If $a \ge b$ and $b \ge a$ then obviously it is necessary that $a = b$. Hence $(1)$ is anti-symmetric.
What about $(2)$? Well, the contraposition of the implication in $(4)$ states that if $R$ is not reflexive, it can't be anti-symmetric. Since $(2)$ is not reflexive, it's not anti-symmetric.
"A relation can be both symmetric and antisymmetric (e.g., the equality relation)" (Wikipedia).
Transitivity
"If $a$ is related to $b$ and $b$ is related to $c$, then $a$ is related to $c$."
Both $(1)$ and $(2)$ are evidently transitive. When purely monogamous marriages are considered, $(3)$ is transitive too, but you'll be hard pressed to find distinct $a$, $b$, and $c$.
Consider the relation $a$ is related to $b$ if $a$ and $b$ have a parent in common. This is not a transitive relation because it could be that $a$ is the child of parents $A$ and $B$, $b$ is the child of $B$ and $C$, and $c$ is the child of $C$ and $D$, where $A$, $B$, $C$, and $D$ are distinct. $a$ is related to $b$ since they share a common parent $B$. Similarly, $b$ is related to $c$ because $C$ is their common parent. However, it is not the case that $a$ is related to $c$ since $a$'s parents are $A$ and $B$ and $c$'s parents are $C$ and $D$, and we've established that all four parents are unique. Since $a$ and $c$ are not related, then the transitivity of this relation does not hold. Note, however, that this relation is both reflexive and symmetric. It is not anti-symmetric.
Best Answer
How do you come up with a counterexample to transitivity?
If you are having trouble, one thing to try is to see why you cannot prove transitivity. The roadblock to the proof may point you in the direction of a counterexample.
Suppose $a,b,c$ are real numbers, $a\leq b^2$, and $b\leq c^2$. When can you prove $a\leq c^2$, or what prevents you from proving it?
If $a\leq 0$, there’s not going to be a problem. So let’s assume $a\gt 0$.
If $b\leq 0$, then $b\leq c^2$ certainly holds. Here is where we run into problems, because $c^2$ can be pretty small and still be larger than $b$, even if $b^2$ is larger than $a$. So... what if we pick a $b\lt 0$ but with large enough square, and then pick a $c$ with very small square, smaller than $a$? Will that work?