Explanation
Clearly, the relation is symmetric but it is neither reflexive nor transitive
A relation from P to Q is a subset of P × Q
R = A × B
G.C.D of a and b is 2 and G.C.D of b and a is also 2. So, it is symmetric only
Since R is reflexive relation on A, therefore (a, a) ϵ R for all a ϵ A.
For any a ϵ N, we find that a | a, therefore R is reflexive but R is not symmetric, because aRb does not imply that bRa
The relation is not symmetric, because A ⊂ B does not imply that B ⊂ A. But it is anti – symmetric because A ⊂ B and B ⊂ A = A = B
x is a brother of y, but y is not necessarily of x it may be a sister so it is not symmetric but it is only transitive.
The void relation R on A is not reflexive as (a, a) ϵ for any a ϵ A. The void relation is symmetric and transitive
For any a ϵ R, we have a ≥ a, therefore the relation R is reflexive but it is not symmetric as (2, 1) ϵ R but (1, 2) ∉ R. The relation R is transitive also, because (a, b) ϵ R, (b, c) ϵ R imply that a ≥ b and b ≥ c which is turn imply that a ≥ c.
These are fundamental concepts
We have (a, b) R (a, b) for all (a, b) ϵ N × N
Since a + b = b + a. Hence, R is reflexive.
R is symmetric for we have (a, b) R (c, d)
a + d = b + c
d + a = c + b
⇒ c + d = d + a ⇒ (c, d)R(a, b),
Hence, r is symmetric.
Then by definition of R, we have
a + d + c +f = b + c + d + e or a + f = b + e
Hence (a, b) R (e, f)
Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒(a, b) R (e, f)
Hence, R is transitive
Here (3,3), (6, 6), (9, 9), (12, 12), [Reflexive];
(3, 6), (6, 12), (3, 12), [Transitive].
Hence, reflexive and transitive only.
Given A = {1,2,3,4}
R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}
(2, 3) ϵ R. Hence, R is not symmetric.
R is not reflexive as (1, 1) ∉ R
R is not a function as (2, 4) ϵ R and (2, 3) ϵ R.
R is not transitive as (1, 3) ϵ R and (3, 1) ϵ R but (1, 1) ∉ R.
In option (d), ordered pair (a, d) ∉
A x B. Thus it is not a relation
By definition, a relation is equivalence if it is reflexive, symmetric and transitive
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