# Reflexive, Symmetric, Transitive - Prove related problem

## Homework Statement

Let A=RxR=the set of all ordered pairs (x,y), where x and y are real numbers. Define relation P on A as follows: For all (x,y) and (z,w) in A, (x,y)P(z,w) iff x-y=z-w

## Homework Equations

R is reflexive if, and only if, for all x ∈ A,x R x.
R is symmetric if, and only if, for all x,y∈A,if xRy then yRx.
R is transitive if, and only if, for all x,y,z∈A, if xRy and yRz then xRz.

## The Attempt at a Solution

I am supposed to prove that P is reflexive, symmetric and transitive.
To show if P is reflexive do I just state that since y-x=w-z then L is reflexive
To show it is symmetric do I just state y-x=w-z and x-y=z-w, since they are equivalent then they are symmetric
To show if Transitive if (the book gives this answer for a similar problem) since x-y and z-w are integers then x-z=(x-y) + (x-z) is the sum of two integers. Therefore x-z is an integer. Assuming this is correct, are my two answers for reflexive and symmetric incorrect?

Furthermore, how would I list five elements in [(2,6)] or [(5,5)]?

Thanks for the help! :)

Mark44
Mentor

## Homework Statement

Let A=RxR=the set of all ordered pairs (x,y), where x and y are real numbers. Define relation P on A as follows: For all (x,y) and (z,w) in A, (x,y)P(z,w) iff x-y=z-w

## Homework Equations

R is reflexive if, and only if, for all x ∈ A,x R x.
R is symmetric if, and only if, for all x,y∈A,if xRy then yRx.
R is transitive if, and only if, for all x,y,z∈A, if xRy and yRz then xRz.

## The Attempt at a Solution

I am supposed to prove that P is reflexive, symmetric and transitive.
To show if P is reflexive do I just state that since y-x=w-z then L is reflexive
What do you get for (x, y)P(x, y)?
eseefreak said:
To show it is symmetric do I just state y-x=w-z and x-y=z-w, since they are equivalent then they are symmetric
You need to show that (x, y) P (z, w) is the same as (z, w) P (x, y)
eseefreak said:
To show if Transitive if (the book gives this answer for a similar problem) since x-y and z-w are integers then x-z=(x-y) + (x-z) is the sum of two integers.
You can't assume the numbers are integers. The relations are defined on sets of real ordered pairs.
eseefreak said:
Therefore x-z is an integer. Assuming this is correct, are my two answers for reflexive and symmetric incorrect?

Furthermore, how would I list five elements in [(2,6)] or [(5,5)]?
Use the definition of your relation.
eseefreak said:
Thanks for the help! :)

Last edited:
What do you get for (x, y)P(x, y)?
You need to show that (x, y) P (z, w) is the same as (z, w) P (x, y)

Use the definition of your relation.

How can I show that they are the same? It seems easy to show they are the same if there are variables in place but how would I show they are the same for all real numbers?

btw thanks Mark44 for the help

What is (z,w)P(x,y)? my guess would be having to prove that x-y=y-x

Fredrik
Staff Emeritus
Gold Member
Just say "Let w,x,y,z be arbitrary elements of A such that (x,y)P(z,w)". Then prove that (z,w)P(x,y), without making any assumptions about w,x,y,z.

Mark44
Mentor
What is (z,w)P(x,y)? my guess would be having to prove that x-y=y-x
The original definition is this: (x, y) P (z, w) iff x - y = z - w.
(x, y) P (z, w) and x - y = z - w are statements with identical truth values. If either one is true, the other is true. If either one is false, the other is false.For the relation P, the ordered pairs (x, y) and (z, w) satisfy the relation if and only if x - y = z - w.

For the various parts of this problem, substitute the given variables in the equation on the right, making sure to put them in the right places.