If R and R1 are equivalence relations on a set A, then so are the relations
- R-1
- R∪R1
- R∩R1
- All of these
The Correct Option is A, C
Solution and Explanation
The correct answer is/are option(s):
(A): R-1
(C): R∩R1
Given, R1 and R
are equivalence relations on set A.
A relation is called the equivalence relation if it is reflexive, symmetric and transitive.
Since, R1
is a transitive relation.
(1) R1
is reflexive.
i.e. (a,a)∈R1 for all a∈A
(2) R1
is symmetric.
i.e. if (a,b)∈R1, then (b,a)∈R1; a,b∈A
(3) R1
is transitive.
i.e. if (a,b)∈R1, and (b,c)∈R1 then (a,c)∈R1; a,b,c∈A
Similarly, R2
is also an equivalence relation. So,
(1) R
is symmetric.
i.e. if (a,b)∈R then (b,a)∈R; a,b∈A
(2) R
is reflexive.
i.e. (a,a)∈R for all a∈A
(3) R
is transitive.
i.e. if (a,b)∈R, and (b,c)∈R then (a,c)∈R; a,b,c∈A
We have to prove that R1∩R
is an equivalence relation.
Check reflexive:
For all a∈A,(a,a)∈R1 and (a,a)∈R [As both R1 and R2 are reflexive on A]
We know that if an element belongs to set A and also to set B, then the element will also belong to (A∩B).So, as a∈A,(a,a)∈R1 and (a,a)∈R(a,a)∈(R1∩R)
∴(R1∩R) is reflexive on A.
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Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
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