# GATE2015-3-41

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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?

1. Both reflexive and symmetric
2. Reflexive but not symmetric
3. Not reflexive but symmetric
4. Neither reflexive nor symmetric

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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...

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.

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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?
3
we should check ((p,q),(p,q))∈R for reflexivity like (a,a)∈R for two elements, not ((p,p),(p,p))∈R
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

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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!!
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
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.
0
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
0
I don't think I understood the question. :(
((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. 
–1 vote
((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. 

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