Understanding Reflexive Relations I'm reviewing some problems to try and get a better understanding of relations. I get how a reflexive relation works on a defined relation with numbers, but not so much when its done with a set builder of sorts.
Let $R$ = {$(a,b)|a + b$ is odd} be a relation on all integers.
My professor's answer to $R$ being reflexive is that "Let $a$ =2, 2 + 2 = 4 and 4 is not odd. Therefore, ℛ is not reflexive."
For a function to be reflexive if for all $x\in A$, $(a,a)\in R$, this is where my flaw of understand this comes in but doesn't that mean that unless the relation is defined then all relations are reflexive?
EDIT: for a relation being defined I meant this $R$ on {1,2,3} given by $R$ = {(1,1), (2,2), (2,3), (3,3)}. 
 A: No.  The condition for reflexivity for a relation $R$ is that for all $a$ in the domain of $R$, $(a, a) \in R$.  If there is a single element $a$ in the domain such that $(a, a) \not\in R$, $R$ is not reflexive.
A: Consider R = {(a,b)| a < b} be the relation "is less than".  This is a relation.  Any two numbers are either related this way or they are not.  This relation is not reflexive because 1 is not less than 1.
Actually, your professor just gave you a good counter-example (2,2) $\notin$ R whether R ={(a,b)| a + b = odd} or R = {(a,b)| a < b}.
Now consider R = {(a,b)| a + b = even} then R is reflexive because for all a, a+a = 2a is even.
Or consider R = {(a,b)| a = m*b where m is integer}.  Then 4 R 2 but 2 notR 4.  But for any a; a = 1*a so a R a, so R is reflexive (but not symmetric).
And here's an example where 3 R 3 but 2 not R 2.  R = {(a,b)|a and b odd}. (So not reflexive.)
And finally yours were R is reflexive and "nothing else" R = {(a,b)| a = b} = {(a,a)}
A: Here is some serious hand-waving aiming at giving some intuition.
See the cartesian product $A \times A$ as some sort of "generalized grid", i.e. There is one row for all $a \in A$ as well as and one column for all $a \in A$.
If $(a,b) \in R$ then blacken the entry at row $a$ and column $b$.
$R$ is reflexive if and only if all entries on the main diagonal are blackened. Bingo !
