Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive
but not symmetric.
R = {(a, b); a ≤ b}
Clearly (a, a) ∈ R as a = a.
∴R is reflexive.
Now,
(2, 4) ∈ R (as 2 < 4)
But, (4, 2) ∉ R as 4 is greater than 2.
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
Then,
a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R
∴R is transitive.
Hence, R is reflexive and transitive but not symmetric.
LIST I | LIST II | ||
A. | Range of y=cosec-1x | I. | R-(-1, 1) |
B. | Domain of sec-1x | II. | (0, π) |
C. | Domain of sin-1x | III. | [-1, 1] |
D. | Range of y=cot-1x | IV. | \([\frac{-π}{2},\frac{π}{2}]\)-{0} |
What is the Planning Process?
Relation is said to be empty relation if no element of set X is related or mapped to any element of X i.e, R = Φ.
A relation R in a set, say A is a universal relation if each element of A is related to every element of A.
R = A × A.
Every element of set A is related to itself only then the relation is identity relation.
Let R be a relation from set A to set B i.e., R ∈ A × B. The relation R-1 is said to be an Inverse relation if R-1 from set B to A is denoted by R-1
If every element of set A maps to itself, the relation is Reflexive Relation. For every a ∈ A, (a, a) ∈ R.
A relation R is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R, for all a & b ∈ A.
A relation is said to be transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A
A relation is said to be equivalence if and only if it is Reflexive, Symmetric, and Transitive.