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Question

Consider R is real number and S and R are subsets of R x R defines as :

Which one of the following is true ?

S={(x,y):y=x+1 and 0 <x<2

T={(x,y):x−y is an integer}

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Solution

The correct option is **A** T is an equivalent realtion on R but S is not

1. S = {(x, y) :y = x + l and 0 < x < 2}

• Check for Reflexive Relation:

(x,x):x=x+1 but x≠x+1

Hence cannot be reflexive S is not equivalence relation on R.

2. T ={(x, y): x-y is an integer}

• Check for Reflexive Relation:

(x,x):x−x is integer x−x=0 and 0 ϵ integer

So, T is reflexive.

• Check for Symmetric Relation:

(x , y) : x - y is integer and (y, x) : y - x also an integer.

So. T is symmetric relation.

• Check to Transitive Relation:

(x - y ) : x - y is integer and (y, z) : y - z is integer then (x : z ) : x - z is also integer.

So, T is transitive.

Hence T is equivalence relation but S is not.

1. S = {(x, y) :y = x + l and 0 < x < 2}

• Check for Reflexive Relation:

(x,x):x=x+1 but x≠x+1

Hence cannot be reflexive S is not equivalence relation on R.

2. T ={(x, y): x-y is an integer}

• Check for Reflexive Relation:

(x,x):x−x is integer x−x=0 and 0 ϵ integer

So, T is reflexive.

• Check for Symmetric Relation:

(x , y) : x - y is integer and (y, x) : y - x also an integer.

So. T is symmetric relation.

• Check to Transitive Relation:

(x - y ) : x - y is integer and (y, z) : y - z is integer then (x : z ) : x - z is also integer.

So, T is transitive.

Hence T is equivalence relation but S is not.

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