1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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}

A
T is an equivalent realtion on R but S is not
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Neither S nor T is an equivalence relation on R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
S is an equivalence relation on R but T is not
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Both S and T are an equivalence relation on R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Types of Relations
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program