(1) Determine whether the relation R on the set of all integers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) ∈ R if and only if (a) x − y = 0 (b) x = |y| (c) x + y is odd (2) Which, if any, of the relations from the previous two problems is an equivalence relation?

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.

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.