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Consider the following relations:

  • $R_1\:(a,b)$ iff $(a+b)$ is even over the set of integers
  • $R_2 \:(a,b)$ iff $(a+b)$ is odd over the set of integers
  • $R_3 \:(a,b)$ iff $a.b > 0$ over the set of non-zero rational numbers
  • $R_4\:(a,b)$ iff $|a-b| \leq 2$ over the set of natural numbers

Which of the following statements is correct?

  1. $R_1$ and $R_2$ are equivalence relations, $R_3$ and $R_4$ are not
  2. $R_1$ and $R_3$ are equivalence relations, $R_2$ and $R_4$ are not
  3. $R_1$ and $R_4$ are equivalence relations, $R_2$ and $R_3$ are not
  4. $R_1, R_2, R_3$ and $R_4$ all are equivalence relations
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$R1)$ 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 definitely $(b,a)$

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: $|a-a|\leq2$

Symmetric: if $|a-b|\leq 2$ definitely $|b-a|\leq2$ when $a,b$ are natural numbers

Transitive: $|a-b|\leq2$ and $|b-c|\leq2$, does not imply $|a-c|\leq2$

Ex: $|4-2|\leq2$ and $|2-0|\leq2$ , but $|4-0|>2$ ,

Hence, $R2$ and $R4$ are not equivalence.

Answer is $B.$
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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.
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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

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  1. R1(a,b)
    • Reflexive : Yes, because (a+a) is even.
    • Symmetric : Yes, (a+b) is even ⟹ (b+a) is even.
    • Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.
      So R1 is equivalence relation.

       
  2. R2(a,b)
    • Reflexive : No, because (a+a) is even.
      So R2 is not equivalence relation.

       
  3. R3(a,b)
    • Reflexive : Yes, because a.a > 0.
    • Symmetric : Yes, a.b > 0 ⟹ b.a > 0.
    • Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.
      So R3 is equivalence relation.

       
  4. R4(a,b)
    • Reflexive : Yes, because |a-a| ≤ 2.
    • Symmetric : Yes, |a-b| ≤ 2 ⟹ |b-a| ≤ 2.
    • Transitive : No, because |a-b| ≤ 2 and |b-c| ≤ 2 ⇏ (a-c) is even.
      So R4 is not equivalence relation.

So option (b) is correct.. 

Answer:

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