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A relation $R$ is defined on the set of integers as $xRy$ iff $(x + y)$ is even. Which of the following statements is true?

  1. $R$ is not an equivalence relation
  2. $R$ is an equivalence relation having $1$ equivalence class
  3. $R$ is an equivalence relation having $2$ equivalence classes
  4. $R$ is an equivalence relation having $3$ equivalence classes
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5 Answers

Best answer
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47 votes
$R$ is reflexive as $(x+x)$ is even for any integer.

$R$ is symmetric as if $(x + y)$ is even $(y+x)$ is also even.

$R$ is transitive as if $(x + (y +z))$ is even, then $((x+y) + z)$ is also even.

So, $R$ is an equivalence relation.

For set of natural numbers, sum of even numbers always give even, sum of odd numbers always give even and sum of any even and any odd number always give odd. So, $R$ must have two equivalence classes -one for even and one for odd.

$\left\{\dots,-4,-2,0,2,4, \dots \right\}, \left\{\dots, -3,-1,1,3, \dots, \right\}$

Correct Answer: $C$.
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6 votes

For simplicity Let the set S = {1, 2, 3, 4}

Now the equivalence relation R = { (1,1), (1,3), (2,2), (2,4), (3,3), (3,1), (4,4), (4,2)}

so, the relation R is equivalence relation.

Now we've to find equivalence classes.

equivalence class [x] = { y | y belongs to S & (x,y) belongs to R} for all x belongs to S

i.e we find equivalence classes w.r.t every element of the set.

[1] = [3] = {1,3}

[2] = [4] = {2,4}

so there are two equivalence classes or partitions {1,3} & {2,4}

Option C

1 votes
1 votes
Answer will be (c) x+y = even only when x and y are both even or when x and y are both odd.

reflexive----> odd+odd = even ex : 3 +3 so, (3,3) pair is present similarly (2,2) can also be present. so reflexive

symmetric---> (2,4) and (4,2) or(3,7) and (7,3) are present so symmetric

transitive-----> If (2,4) and (4,8) present then (2,8) should also be present similarly we can do this for odd number pairs...

so it is an equivalence relation..

There are 2 equivalence class i.e., (even,even) and (odd,odd)
Answer:

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