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...

---

Reflexive → NO.

Irreflexive → YES.

Symmetric → YES.

Antisymmetric → NO.

Asymmetric → NO.

Transitive → NO.

---

@raja11sep It's not transitive . 

 

---

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

Answer 1

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.

Comments

If we take all symmetric pairs such as {((1,1)(1,1)),((2,2),(2,2)) and so on... according to que p-s=q-r will always be 0. Then how it is not reflexive?

---

we should check ((p,q),(p,q))∈R for reflexivity like (a,a)∈R for two elements, not ((p,p),(p,p))∈R

Answer 2

1. 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

 

Comments

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

---

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!!

Answer 3

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

Comments

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.

---

Why {5,6} and {3,4} is not there? If you define on 3,4,5,6 then you should take all the pairs which follow this property

---

I don't think I understood the question. :(

Answer 4

((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.