Let A={1,2,3}. Then number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is
1
2
3
4
The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric,
but not transitive is given by: R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈R and (1, 3), (3, 1) ∈R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will
become transitive.
Hence, the total number of desired relations is one.
The correct answer is A (1)
Let \( A = \{0,1,2,\ldots,9\} \). Let \( R \) be a relation on \( A \) defined by \((x,y) \in R\) if and only if \( |x - y| \) is a multiple of \(3\). Given below are two statements:
Statement I: \( n(R) = 36 \).
Statement II: \( R \) is an equivalence relation.
In the light of the above statements, choose the correct answer from the options given below.
If vector \( \mathbf{a} = 3 \hat{i} + 2 \hat{j} - \hat{k} \) \text{ and } \( \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \), then which of the following is correct?