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Administrator
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Let A be a set, and let R be a relation on A. Define the relation R'
on A by R' = (A x A) - R.
(1) If R reflexive, is R' necessarily reflexive, necessarily not
reflexive or not necessarily either?
Ans. R' is necessarily not reflexive.
Proof:
Let A be a non empty set. Let R be a reflexive relation on A. Consider
any x ∈ A. Since R is reflexive, (x, x) ∈ R. And R' = (A x A) - R
implies (x, x) ∉ R'. Hence R' is necessarily not reflexive.
(2) If R symmetric, is R' necessarily symmetric, necessarily not
symmetric or not necessarily either?
Ans. R' is necessarily symmetric.
Proof:
Let A be a non empty set. Let R be a symmetric relation on A. Let (x,
y) ∈ R'. We want to show that (y, x) ∈ R'.
(x, y) ∈ R' implies that (x, y) ∉ R. Clearly, (y, x) ∉ R because
if (y, x) ∈ R then, by symmetry of R, we get (x, y) ∈ R which is a
contradiction. Hence, (y, x) ∉ R. Thus, (y, x) ∈ R'.
(3) If R transitive, is R' necessarily transitive, necessarily not
transitive or not necessarily either?
Ans. R' is not necessarily either.
Proof:
Proof of R' is not necessarily transitive: A = {x, y, z}, R = {(x,
y)}
Clearly in this example, R is transitive but R' is not. Since, (x, z)
∈ R' and (z, y) ∈ R' but (x, y) ∉ R'.
Proof of R' is not necessarily not transitive:: A = {x, y}, R ={(x,
x), (x, y)}, R' ={(y, y), (y, x)}
Clearly in this example both R and R' are transitive.
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Locked
Apr 05, 2013; 8:30am
)...but as I saw your explanation...it's get very clear that we Can't...even on can check it with setting up a arbitrary set-A & work accordingly with given conditions...