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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
in Set Theory & Algebra by Veteran (105k points)
edited by | 3.7k views
0
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...

5 Answers

+63 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.

by Active (2.3k points)
edited by
0
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
+18 votes
  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

 

by Active (5k points)
edited by
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
0
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!!
+6 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
by Boss (25.3k points)
+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. :(
+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. 
by Boss (10k points)
–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. 
by Boss (10k points)

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