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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?

- $R_1$ and $R_2$ are equivalence relations, $R_3$ and $R_4$ are not
- $R_1$ and $R_3$ are equivalence relations, $R_2$ and $R_4$ are not
- $R_1$ and $R_4$ are equivalence relations, $R_2$ and $R_3$ are not
- $R_1, R_2, R_3$ and $R_4$ all are equivalence relations

## 3 Answers

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 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 $B.$

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 $B.$

### 8 Comments

for transitivity in R3,

if we take pairs (-4/3, -8/3), (-8/3, -1/2) & (-4/3, -1/2)

for these pairs, it is not satisfying transitivity rule.

[acc to transitivity, if (-4/3, -8/3) & (-8/3, -1/2) are included in relation, pair (-4/3, -1/2) should also be there for transitivity to hold, which is not there.]

am i wrong somwhere ???

if we take pairs (-4/3, -8/3), (-8/3, -1/2) & (-4/3, -1/2)

for these pairs, it is not satisfying transitivity rule.

[acc to transitivity, if (-4/3, -8/3) & (-8/3, -1/2) are included in relation, pair (-4/3, -1/2) should also be there for transitivity to hold, which is not there.]

am i wrong somwhere ???

**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