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+8 votes

Consider the following relations:

  • R1 $(a,b)$ iff $(a+b)$ is even over the set of integers
  • R2 $(a,b)$ iff $(a+b)$ is odd over the set of integers
  • R3 $(a,b)$ iff $a.b > 0$ over the set of non-zero rational numbers
  • R4 $(a,b)$ iff $|a-b| \leq 2$ over the set of natural numbers

Which of the following statements is correct?

  1. R1 and R2 are equivalence relations, R3 and R4 are not
  2. R1 and R3 are equivalence relations, R2 and R4 are not
  3. R1 and R4 are equivalence relations, R2 and R3 are not
  4. R1, R2, R3 and R4 all are equivalence relations
asked in Set Theory & Algebra by Veteran (68.8k points)
retagged by | 677 views

4 Answers

+11 votes
Best answer
R1) Reflexive : a+a=2a always even

 Symmetric: either (a,b) both must be odd or both must be even to have sum as even

therefore, if(a,b) then definately (b,a)

Transitive: if(a,b) and (b,c) , then both of them must be even pairs or odd pairs and therefore (a,c) is even

R2) Reflexive : a+a=2a cant be odd ever

R3) Reflexive: a.a>0

Symmetric: if a,b>0 then both must be +ve or -ve, which means b.a >0 also exists

Transitive : if a.b>0 and b.c>0 then to have b as same number, both pairs must be +ve or -ve which implies a.c>0

R4) Reflexive: |a-a|<=2

Symmetric: if |a-b|<=2 definately |b-a|<=2 when a,b are natural numbers

Transitive: |a-b|<=2 and |b-c|<=2, doesnt imply |a-c|<=2

ex: |4-2|<=2 and |2-0|<=2 , but |4-0|>2 ,


hence, R2 and R4 are not equivalence

answered by Junior (977 points)
selected by
+8 votes
ans is B.

for R1 and R3 are Reflexive symmetric and transitive.

R2 is not transitive. ex: (1,2) (2,3) (1,3), which is not satisfying condition.

R4 is not transitive . ex: (1,3) (3,5) (1,5), which is not satisfying condition.
answered by Boss (8.4k points)
0 votes

EVEN= {even + even }  OR { odd + odd}     // it will be reflexive,symmetric,transitive 

ODD = {even +odd }                                      // it will be symmetric only

Ex :-  A{1,2,3}

For even(R1)  :-  { (1,1) (2,2) (3,3) (1,3) (3,1) }            here we can't add (1,2) & (2,3) bcoz it won't be even

For odd (R2) :-  { (1,2) (2,3) (2,1) (3,2) }                      Here we can't add (11),(2,2) etc bcoz it will be either even + even or odd+ odd

answered ago by Boss (9.5k points)
–2 votes

The correct answer is,(B) R1 and R3 are equivalence relations, R2 and R4 are not

answered by Veteran (15.4k points)

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