Assessment: The following question was given to ChatGPT. Determine what's correct, fix all the mistakes, and assign a letter grade: A,B,C,D,F. Let us consider some new definitions. (a) Irreflexive
AAxnot(xRx)
(b) Antitransitivity:
AAx,y,z(xRy^(^())yRz->not(xRz))
Give a relation R on a set A such that it is neither reflexive, symmetric, transitive, antisymmetric, irreflexive nor antitransitive. To create a relation
R
on a set
A
such that it is neither reflexive, symmetric, transitive, antisymmetric, irreflexive, nor antitransitive, we can follow these steps: Reflexive:
R
is not reflexive if there exists at least one element
ainA
such that
(a,a)!inR
. Symmetric:
R
is not symmetric if there exist elements
a,binA
such that
(a,b)inR
but
(b,a)!inR
. Transitive:
R
is not transitive if there exist elements
a,b,cinA
such that
(a,b)inR
and
(b,c)inR
but
(a,c)!inR
. Antisymmetric:
R
is not antisymmetric if there exist elements
a,binA
such that
(a,b)inR
and
(b,a)inR
but
a!=b
. Irreflexive:
R
is not irreflexive if there exists at least one element
ainA
such that
(a,a)inR
.
◻
Antitransitive:
R
is not antitransitive if there exist elements
a,b,cinA
such that
(a,b)inR
and
(b,c)inR
and
(a,c)inR
. Consider the set
A={1,2,3}
. Define the relation
R
on
A
as follows:
R={(1,2),(2,3),(3,1),(1,1)}
Let's check each property: Reflexive:
R
is not reflexive because
(2,2)!inR
and
(3,3)!inR
. Symmetric:
R
is not symmetric because
(1,2)inR
but
(2,1)!inR
. Transitive:
R
is not transitive because
(1,2)inR
and
(2,3)inR
, but
(1,3)!inR
. Antisymmetric:
R
is not antisymmetric because
(1,1)inR
and
(1,1)inR
but
1=1
, so we need another pair to violate this, but for the given pairs, it doesn't. Irreflexive:
R
is not irreflexive because
(1,1)inR
. Antitransitive:
R
is not antitransitive because
(1,2)inR,(2,3)inR
, and
(1,3)!inR
, but we have
(3,1)inR
. Thus, the relation
R={(1,2),(2,3),(3,1),(1,1)}
on the set
A={1,2,3}
satisfies the condition of being neither reflexive, symmetric, transitive, antisymmetric, irreflexive, nor antitransitive.