A relation on a set $D$ is a set of pairs $R$ of pairs
$(x,y)$ where $x$ and $y$ are in $D$.
Example 1 is the relation $a \equiv b~(mod~n)$ considered
as the set of pairs $\{ (a,b) : a-b$
is divisible by $n \}$ is a relation on the set of integers.
Example 2 was an example from the set of U.S. States.
$\{ (x, y) : $ where $x$ and $y$ are U.S. states that share a land border $\}$.
For example this relation looks like the set of pairs $\{$ (California, Nevada), (California, Oregon),
(California, Arizona), (Oregon, California), (Oregon, Washington),
(Oregon, Idaho), (Washington, Idaho), (Washington, Oregon), ...$\}$
(but the list is long and I can't easily list all of them)
and Hawaii doesn't appear in the relation even though it is in the set $D$.
A relation $R$ is called reflexive if $(a,a) \in R$ for all $a \in D$.
A relation $R$ is called symmetric if $(a,b) \in R$, then $(b,a) \in R$.
A relation $R$ is called transitive if $(a,b) \in R$ and $(b,c) \in R$,
then $(a,c) \in R$.
A relation $R$ is called an equivalence relation if
$R$ is reflexive, symmetric and transitive.
Then consider, which of the following relations on the set of integers
(that is $D={\mathbb Z}$) are reflexive, symmetric
and transitive:
We discussed $R_1$ through $R_5$ in class, and I left $R_6$ as an exercise.
$R_1$ is not reflexive because, for instance, $(1,1) \notin R_1$.
$R_1$ is not symmetric because, $(1,-1) \in R_1$, but $(-1, 1) \notin R_1$.
$R_1$ is transitive (and this one is harder to explain because we call it
"vacuously true") because there are no pairs of the form $(a,b)$ and $(b,c)$ in $R_1$.
$R_2$ is not reflexive because, for instance, $(1,1) \notin R_2$.
$R_2$ is symmetric because, $(1,-1) \in R_2$ and $(-1, 1) \in R_2$
$R_2$ is not transitive because $(1,-1)$ and $(-1,1) \in R_2$,
but $(1,1) \notin R_2$.
$R_3$ would be reflexive if it was a relation on $D = \{ 1, -1\}$, but these
relations are supposed to be relations on the set of integers and so, for instance,
$(2,2) \notin R_3$ and is hence not reflexive. $R_3$ however is symmetric
since $(1,-1)$ and $(-1,1)$ are both in $R_3$. $R_3$ is also transitive.
Checking that $R_3$ is transitive is hard to do because you need to check that for
every pair of pairs of the form $(a,b)$ and $(b,c)$, that $(a,c)$ is also in $R_3$.
Here is an exhaustive list:
$(1,-1), (-1, 1) \in R_3$ and $(1,1) \in R_3$
$(1,-1), (-1, -1) \in R_3$ and $(1,-1) \in R_3$
$(-1,1), (1, -1) \in R_3$ and $(-1,-1) \in R_3$
$(-1,1), (1, 1) \in R_3$ and $(-1,1) \in R_3$
$(1,1), (1, -1) \in R_3$ and $(1,-1) \in R_3$
$(1,1), (1, 1) \in R_3$ and $(1,1) \in R_3$
$(-1,-1), (-1, 1) \in R_3$ and $(-1,1) \in R_3$
$(-1,-1), (-1, -1) \in R_3$ and $(-1,-1) \in R_3$
$R_4$ is reflexive because $(x,x) \in R_4$ for all $x \in {\mathbb Z}$.
$R_4$ is symmetric because if $(x,y) \in R_4$, then $(y,x) \in R_4$.
$R_4$ is transitive because if $(x,y)$ and $(y,z) \in R_4$, then
$(x, z) \in R_4$.
$R_5$ is not reflexive because, for instance, $(1,1) \notin R_5$.
$R_5$ is symmetric because if $(x,-x) \in R_5$, then $(-x,x) = (-x, -(-x)) \in R_5$.
$R_5$ is not transitive however since $(1,-1)$ and $(-1,1)$ are both
in $R_5$, but $(1,1) \notin R_5$.
Do $R_6$ as an exercise. That is, tell me if $R_6$ is reflexive, symmetric and
transitive and why.