Question

Show that the relation R in the set of natural number N given by R ={(x,y):x is odd and y=2x}is an equivalence relation ​

Answer 1

Answer:

Here x is odd number so

R = {(1,2), (3,6), (5,10),....}

so this relation is not reflexive because (x,x) does not belongs to R.

this relation is also not symmetric because (x,y) belongs to R but (y,x) does not belongs to R.

this relation is also not transitive because (x,y) belongs to R but (y,z) and (x,z) does not belongs to R.

so this relation is not equivalence.

MAY YOUR QUESTION IS WRONG

Answer 2

R is not an equivalence relation in the set of natural numbers N.

Given :

R = {(x,y) : x is odd and y = 2x}

Set = Natural number set = N

To prove :

R is an equivalence relation

Solution :

For R to be equivalence relation on the set N, it has to reflexive, symmetric, and transitive on the set N

We have R = {(x,y) : x is odd and y = 2x}

When x = 1, y = 2

x = 3, y = 6

x = 5, y = 10

x = 7, y = 14,..........

Therefore, R = {(3,6), (5,10), (7,14),.......}

N = {1, 2, 3, 4, 5.........}

1. For R to be reflexive, for every a ∈ N, (a,a) ∈ R

  But here 3 ∈ N, but (3,3) ∉ R

  Thus, R is not reflexive.

2. For R to be symmetric, for every (a,b) ∈ R ⇒ (b,a) ∈ R

    But here (3,6) ∈ R, but (6,3) ∉ R

    Thus, R is not symmetric.

3. For R to transitive, for every (a,b) and (b,c) ∈ R, (a,c) ∈ R

   But here R cannot be transitive since a is always odd, and b is always even.

   Thus, R is not transitive.

Hence proved that R cannot be an equivalence relation in the set of natural numbers N.

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