Answer:
Step-by-step explanation:
1)
a) x-y = 0
Reflexive :
x-x=0, where x and 0 are whole numbers is valid for all x
In this way, x-x is a number
Which infers that (x,x) ∈ R
In this way, R is reflexive
Symmetric :-
(x-y) is a whole number
Which infers - (x-y) is likewise a whole number
Which infers (y-x) is a number
In this way, (x-y) is a number infers that (y-x) is likewise a whole number
i.e If (x,y)∈R at that point (y,x)∈R
In this way, R is Symmetric.
Against Symmetric:-
(x,y)∈R
x-y=0, where x and y are whole numbers
Which infers x=y
Assume, (y,x)∈R
y-x=0, where x and y are whole numbers
Which infers y=x
Consequently, (x-y) is a whole number suggests (y-x) is additionally a whole number
for example In the event that (x,y)∈R at that point (y,x)∈R, where x=y
R is Anti-Symmetric.
Transitive:-
Given,
(x-y)∈R and (y-z)∈R - (I)
To demonstrate that,
(x,z)∈R
from (I)
x-y= 0 and y-z=0
Including both, we get
(x-y)+(y-z)=0
x-z=0
Consequently demonstrated that (x,y)∈R and (y,z)∈R infers that (x,z)∈R
Since the above connection is symmetric, reflexive advertisement transitive. Consequently, it is an identicalness connection.
b) x=|y|
Reflexive:-
x isn't reflexive.
Since for - 2
- 2 ≠|-2|, where - 2 is a whole number.
- 2≠|2|
Consequently, (x,x) isn't Reflexive.
Symmetric:-
(x,y)∈R, infers
x=|y|, where x and y are whole numbers ...(i)
Let x =-2 and y= - 2
- 2≠|-2|
- 2≠|2|
Which infers (x,y) itself doesn't have a place with R.
Consequently R isn't symmetric.
Against Symmetric:-
(x,y)∈R, suggests
x=|y|, where x and y are numbers
let x= - 2 and y =-2
- 2 ≠ |-2|
- 2≠|2|
R isn't hostile to symmetric.
Transitive:-
x=|y| and y=|z| ∈ R
let x=2, y=-2 and z= 2
infers that
2=|-2| and - 2≠|2|
for example 2=|2| and - 2∉|2|
In this way, R isn't transitive.
c) x+y is odd
Reflexive:-
x+x is even, where x is a whole number.
In this way, (x,x)∉R.
R isn't Reflexive.
Symmetric:-
(x,y)∈R
x+y is odd, where x and y are whole numbers
Additionally, (y,x) is odd
infers that y+x is additionally odd
In this way, (y,x)∈R
R is Symmetric.
Against Symmetric:-
(x,y)∈R
infers x+y is odd, where x and y are whole numbers
infers that y+x is additionally odd
In this way, (y,x)∈R
R is Anti-Symmetric.
Transitive:-
Given,
(x,y)∈R and (y,z)∈R, where x,y and z are whole numbers ...(i)
To demonstrate that
(x,z)∈ R
from (I)
x+y is odd and y+z is odd''
Including both we get
x+2y+z is odd
which infers (x,z)∉R
In this way, R isn't transitive.
2) x-y=0 is reflexive, symmetric and transitive in this way it is an equality connection.
while x=|y| and x+y is odd are not equality relations.
