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

# Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a-b| ≤ 1. Then, R is

A
reflexive and symmetric but not transitive
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
reflexive and transitive but not symmetric
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
symmetric and transitive but not reflexive
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
an equivalence relation.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A reflexive and symmetric but not transitive(i) |a−a|=0≤1 is always true (ii) a R b ⇒ |a−b| ≤1⇒|−(a−b)|≤1⇒|b−a|≤1⇒b R a. So, R is symmetric (iii) 2R 1 and 1 R 12 But, 2 is not related to 12. So, R is not transitive.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Surface Area of Solids
MATHEMATICS
Watch in App
Join BYJU'S Learning Program