Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴R is reflexive. Now, Let (P, Q) ∈ R
⇒ The distance of point P from the origin is the same as the distance of point Q from the
origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the
origin.
⇒ (Q, P) ∈ R
∴R is symmetric.
Now,
Let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of
points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ R
∴R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes
through point P.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive
but not symmetric.
Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive
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.