relations

[Mauli]

can someone please help me with this?
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?
(2) If R symmetric, is R' necessarily symmetric, necessarily not
symmetric or not necessarily either?
(3) If R transitive, is R' necessarily transitive, necessarily not
transitive or not necessarily either?

[Sumit]

Acc. to me: R' will always be reflexive, symmetric & transitive...It doesn't matter whether R is reflexive, symmetric & transitive or not.....R' will remain reflexive, symmetric & transitive...Bcoz R' is the union of all the Relations on Set A except R..So all the element exists in R will also be in R'...due to the union of all the relations except R...What  do you think????

[Mauli]

no, but the answer says
a) it is  necessarily not reflexive .
b) necessarily symmetric.
c) not necessarily either.
how does one get this..no clue:(

[Sumit]

What yours answers come out to be???...same as me(by same logic)?????

[Amit Goyal]

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.

[Sumit]

thanks sir.....I get them all wrong bcoz I was in doubt with R'...I was including all the element that is including in R to R' as well.(very major mistake )...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...

Caution: check 3rd one transitive wala First, taking 3 elements in Set-A then 4 element in set-A..

[Mauli]

thankyou sir :)
thankyou sumit for your inputs.:)