Question

let A={1, 2, 3, 4} and R be a relation on A given by R = {(x, y):|x-y| isa multiple of 2} check weather R is reflexive symmetric and transitive
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

REFLEXIVE ::

Since For every x ∈ A

|x-x| = 0 is a multiple of 2

So For every x ∈ A , ( x, x ) ∈ R

R is Reflexive

SYMMETRIC ::

Let x, y ∈ A

Also suppose that ( x, y ) ∈ R

Now ( x, y ) ∈ R

➙ |x-y| is a multiple of 2

➙ |y - x | isa multiple of 2

➙ ( y , x ) ∈ R

So ( x, y ) ∈ R implies ( y , x ) ∈ R

So R is Symmetric

TRANSITIVE ::

Let x, y, z ∈ A

Also suppose that ( x, y ) & ( y , z ) ∈ R

Now ( x, y ) & ( y , z ) ∈ R

➙ |x-y| is a multiple of 2 & |y - z | is a multiple of 2

➙ | x - z | is a multiple of 2

➙ ( x , z ) ∈ R

So ( x, y ) & ( y , z ) ∈ R implies ( x, z ) ∈ R

So R is Transitive

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