in Set Theory & Algebra edited by
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17 votes

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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3 Answers

23 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 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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8 Comments

nice
0
In R1 both a and b should be either odd or even to get sum as even.

If one is odd and the other is even then we get odd number as sum. So why can't we say that it is not symmetric.
0
arjun sir

can you give me the example for R3 and R4 by using set (123) .
0
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 ???
0

@akshayaK Pair (-4/3,-1/2) can also belong to R3. 

Why do you think that pair should not be present?

 

0

condition is mentioned there to be in relation as a.b > 0.

$\left ( -4/3, -1/2 \right )\Rightarrow \left ( a *b \right ) \ngtr 0$

0
2/3=0.667 >0
0
:) yaa , i was interpreting 0.66 as 0 (int value in c prog). tx
0
11 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.
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

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