hence option B is incorrect...

33 votes

Let $R$ be a relation on the set of ordered pairs of positive integers such that $((p,q),(r,s)) \in R$ 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

68 votes

Best answer

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)) \in \mathrel{R} \;\leftrightarrow(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

$((a,b),(c,d))\in \mathrel{R}\rightarrow((c,d),(a,b))\in \mathrel{R}$

$((a,b),(c,d))\in\mathrel{R} \rightarrow(a-d=b-c)$

$((c,d),(a,b))\in\mathrel{R}$

$\because (c-b=d-a)\leftrightarrow (d-a=c-b)\leftrightarrow(-(a-d)=-(b-c))\leftrightarrow(a-d=b-c)$

**So, it is symmetric.**

Hence, **C is the correct option.**

19 votes

- take example for symmetric relation ... for p=6, q=4,r=4, s=6, (6,4)(4,6)

p-s=6-6=0

q-r=4-4=0

so if we will check the condition (means p-s and q-r) for symmetric relation then p-s =q-r will b same always

now we can see the relation (6,4)(4,6) is not reflexive but exist in the relation R. so we can say that R may or may not b reflexive.

so option C is correct Not reflective but symmetric

0

if we take example (1,1) (2,2) then also it will give same value -1 after performing p-s and q-r so now it seems like reflexive. plz correct me

1

no thats not reflexive!

when we say ((p,q)(r,s)) belongs to R means

{(1,1)(1,1) } or{ (2,2)(2,2)} should belong to R;

here first orederd pair is taken as first element and second ordered pair will be taken as second element;

it will not be reflexive as {(1,2)(1,2)} does not belongs to R;

hope it clears ur doubt!!

above relation is not reflexive ,symmetric and not transitive!!

when we say ((p,q)(r,s)) belongs to R means

{(1,1)(1,1) } or{ (2,2)(2,2)} should belong to R;

here first orederd pair is taken as first element and second ordered pair will be taken as second element;

it will not be reflexive as {(1,2)(1,2)} does not belongs to R;

hope it clears ur doubt!!

above relation is not reflexive ,symmetric and not transitive!!

7 votes

Let us first rearrange the equation to make it look simple

p−s=q−r

p+r=q+s // which means sum of first elements is same as sum of last two elements.

Reflexive:-

(p,q)R(p,q) iff p+p=q+q ,which is not true always so not reflexive

Symmetric.

If (p,q)R(r,s) => (r,s)R(p,q)

p+r=q+s => r+p=s+q

which will always hold.Hence symmetric

So, c is correct

p−s=q−r

p+r=q+s // which means sum of first elements is same as sum of last two elements.

Reflexive:-

(p,q)R(p,q) iff p+p=q+q ,which is not true always so not reflexive

Symmetric.

If (p,q)R(r,s) => (r,s)R(p,q)

p+r=q+s => r+p=s+q

which will always hold.Hence symmetric

So, c is correct

1

Why are we assuming that if (a,B) exists then (B,a) must exist? Can't we have a pair of ordered set such as{ (6,5), (3,4) }alone in the relation. This follows the rule above. But it's neither symmeteric nor reflexive.