GATE CSE 2015 Set 3 | Question: 41

Question

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 $R$?

• A. Both reflexive and symmetric
• B. Reflexive but not symmetric
• C. Not reflexive but symmetric
• D. Neither reflexive nor symmetric

Comments

To differentiate between option B and C, we can take empty relation which satisfies the given condition but empty relation is not a reflexive relation

hence option B is incorrect...

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Reflexive → NO.

Irreflexive → YES.

Symmetric → YES.

Antisymmetric → NO.

Asymmetric → NO.

Transitive → NO.

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@raja11sep It's not transitive . 

 

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Yes, NOT transitive but you have some mistakes in calculation.

Answer 1

p – s = q – r

=> p + r = q + s

=> sum of first terms = sum of second terms of ordered pair (p,q) and (r,s).

we shall now take examples to reach the answer.

Is (1,3)R(1,3) ? No because 1+1 != 3+3. Therefore the Relation is not reflexive.

consider (2,3) and (3,2). Now (2,3) R (3,2) because 2+3=3+2. Also (3,2) R (3,2).This relation is symmetric.

From the same example we can see that the  transitive pair would be (2,3) and (3,2) which again satisfies the condition, therefore is transitive.

Hence option C is a matching with our conclusion.

Answer 2

Option C

Given, $((p,q),(r,s)) \in R$ iff $p-s=q-r$

Simple check:

Let $(p,q) = a \ \& \ (r,s)  =b \implies aRb$

For Reflexive:

$a R a? \implies (p,q) R (p,q)?$

$p-q \neq q-p, hence\ not \ reflexive.$

For Symmetric: 

If $aRb$ then $bRa \ ?$ 

$(p,q)R(r,s) → (r,s)R(p,q) ? $

LHS : Is given $((p,q),(r,s)) \in R$ iff $p-s=q-r$

RHS: $r-q = s-p \implies p-s = q-r$

$LHS = RHS , hence \ symmetric$

Comments

@Abhrajyoti00

The relation you considered here for proving symmetricity is wrong ,

here the (p,q) & (r,s) is one relation.

consider (p,q) = a and (r,s)=b then it is like aRb , 

so to prove or disprove reflexivity we can take as     a R a [ i.e (p,q) R (p,q)]  

but in case of symmetricity we have to take  aRb → bRa [i.e (p,q)R(r,s) → (r,s)R(p,q)].

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@Pranavpurkar Thanks :) Edited the answer.

Answer 3

((p, q), (r, s)) ∈ R if and only if p–s = q–r

(p, q) is not related to (p, q)

as p-q is not same as q-p.

The relation is symmetric because if p–s = q–r, then s-q = s-p.