Symmetric and Reflexive relation Let R be a relation on the set of ordered pairs of positive integers such that ((p,q),(r,s))∈R if and only if p−s=q−r. Which one of the following is true about R?
Both reflexive and symmetric
Reflexive but not symmetric
Not reflexive but symmetric
Neither reflexive nor symmetric
Solution:
The key trick here is to realize that the relation is of the form :
{ordered pair, ordered pair} and not simply ordered pair.
Ok, so for reflexive
$\forall_{a,b}\, if((a,b),(a,b)) \in \mathrel{R} \rightarrow \text{reflexive}$
$((a,b),(a,b))∈R↔(a−b=b−a)$  (not possible for any postive integers b and a)
But that is a contradiction hence it is not reflexive.
Now, for symmetric
$((a,b),(c,d))∈R→((c,d),(a,b))∈R$
$((a,b),(c,d))∈R→(a−d=b−c)$
$((c,d),(a,b))∈R$
$∵(c−b=d−a)↔(d−a=c−b)↔(−(a−d)=−(b−c))↔(a−d=b−c)$
This is the solution given in from trustworthy source of technical examiniation.
But my doubt is If we are saying that
$((a,b),(c,d))∈R→((c,d),(a,b))∈R$ to be symmetric right?
Why are we checking symmetric relation on ordered pair and reflexive relation on only two pairs $((a,b),(a,b))∈R$
I know Reflexive relation is said to be reflexive if $xRx  \forall_{x}\ $ in $R$
 A: It looks like you're being confused by the fact that there are two different levels of ordered pairs around: First we have pairs of numbers -- each such pair is one of the items we're relating. Then, the technical modeling of the relation involves pairs of those items.
Perhaps it would makes it clearer for you to use different notation for the two kinds of pairs?
Let's write the pairs of numbers (the "items") as column vectors:
$$[^a_b] \quad\text{ instead of }\quad (a,b)$$
And then let's write the relation itself with an infix symbol rather than a pair and an $\in$:
$$x \sim y \quad\text{ instead of }\quad (x,y) \in R$$
The definition of your relation is now
$$ [^a_b] \sim [^c_d] \iff a-d=b-c $$
And the properties you're asked about are:

*

*$\sim$ is reflexive iff for all $a$ and $b$ it is true that $[^a_b] \sim [^a_b]$.

*$\sim$ is symmetric iff for all $a$, $b$, $c$, $d$ it is true that $[^a_b] \sim [^c_d]$ implies $[^c_d]\sim[^a_b]$.

Does this make it easier to relate the model solution to the standard definitions of "reflexive" and "symmetric" (where there a single variables that stand for an entire item)?
A: Intuitively, a relation is reflexive if everything is related to itself, so $aRa$.  It only mentions one item in the set.  In your case, where the items are ordered pairs (as you noted) we require $(a,b)R(a,b)$.  The ordered pair $(a,b)$ is one item in our universe of discourse.  We are checking reflexive on only one pair but using it twice as the definition requires.
Symmetry says it doesn't matter which order you take two items in and ask if they are related.  You have applied the definition correctly to your case, saying it requires $(a,b)R(c,d) \implies (c,d)R(a,b)$.  You are checking symmetry on a pair of ordered pairs  because symmetry requires a pair of items and your items are ordered pairs.
A: Back to the basics.
A relation $R$ on a set $A$ is reflexive iff $(x,x)\in R, \forall x\in A.$
It is symmetric iff $(x,y)\in R\Rightarrow (y,x)\in R, \forall x,y\in A.$
The point is that reflexivity is not a conditional statement, while symmetry is. Besides, reflexivity looks for ordered pairs with identical entries, but in symmetric ones, we look for the reverse of an ordered pair, especially those with different first and second entries.
