Equivalence Relation and Associated Partition

[chocolatelover]

Homework Statement

(proof) Determine whether or not (x,y)~(w,z) if and only if y=w is an equivalence relation. If it is, then describe the associated partition.

Homework Equations

The Attempt at a Solution

Let x be an element of the reals. It is known that a relation on a set X is said to be reflexive if x~x for all x is an element of X. Since y and w are equal and x is an element of the reals and X is an element of the reals, this relation is reflexive.

It is known that a relaion on a set X is symmetric if for all x, y is an element of X, whenever x~y, then y~x. Since y=w, which is the same thing as saying, x~y and y~x, this relation is symmetric.

The relation is transitive if for all x, y, z is an element of X, if x~y and y~z, then x~z. It is also transitive because y=w.

Could someone please show me where to go from here?
Thank you very much

 

[quasar987]

It is reflexive if for any (x,y) in R², (x,y)~(x,y).

Is this really so?

 

[chocolatelover]

Is it not reflexive because (2,5) is not the same thing as (5,2)? I'm not really sure how to prove whether it is or it isn't. Could you please show me how?

This one is reflexive, right? (x,y)~(w,z) if and only if x^2=w^2 because they are both squared. 2^2=2^2
-2^2=2^2

It is also symmetric and transitive, right? Could you show me how to prove it?(

(x,y)~(w,z) if and only if xw=yz is not reflexive, right? because 2(5)=6(8) is not true

Thank you very much

 

[HallsofIvy]

Homework Statement

(proof) Determine whether or not (x,y)~(w,z) if and only if y=w is an equivalence relation. If it is, then describe the associated partition.

Is it not reflexive because (2,5) is not the same thing as (5,2)? I'm not really sure how to prove whether it is or it isn't. Could you please show me how?

No, this is not reflexive because (2,5)~(2,5) is wrong: (x,y)= (2,5) so y= 5. (w,z)= (2,5) so w= 2. 5 is not equal to 2 so y is not equal to w.

This one is reflexive, right? (x,y)~(w,z) if and only if x^2=w^2 because they are both squared. 2^2=2^2
-2^2=2^2

Which one? If you are going to define a new relation, by "(x,y)~(w,z) if and only if x^2=w^2", please define it first, then ask the question! Yes, this is reflexive. No, it is not because of the squares but because x and w are both the first members of the pair. For any (x,y), (x,y)~(x,y) because x²= x². (Notice that in that last equation I am using the fact that "=" itself is reflexive: a= a.)

It is also symmetric and transitive, right? Could you show me how to prove it?

You prove something like that by showing that it satisfies the definition:
A relation is symmetric if and only if a~b implies b~a. Suppose (x,y)~(w,z). Then x²= w². But then, because w²= x², we have (w,z)= (x,y). Do you see how I used the fact that "=" itself is symmetric?

To show that this relation is transitive, we must show that "if (x,y)~(w,z) and (w,z)~(u,v) then (x,y)~(u,v)", the definition of "transitive". Okay, if (x,y)~(w,z) then x²= w² and if (w,z)~(u,v) then w²= u². From that (and the fact that "=" itself is transitive) we have x²= u² which tells us that (x,y)~(u,v) and we are done.

The "equivalence classes" are sets of point (x,y) where x² is always the same. Since x²= w² if x= w or x= -w, those are pairs of vertical lines, x= a and x= -a, on a graph.

(x,y)~(w,z) if and only if xw=yz is not reflexive, right? because 2(5)=6(8) is not true

Where did you get "2(5)" and "6(8)"? The definition of reflexive is "a~a" for any a in the set. Here, your set is R² or pairs of numbers. To show any relation on R² is reflexive you must show (x,y)~(x,y) for any x and y. There are only two numbers involved, not 4. If (x,y)~(w,z) means xw= yz (first member times first member equals second member times second member), then (x,y)~(x,y) means x²= y². That is not always true. (1, 2)~(1,2) is false because 1² is not equal to 2².

Thank you very much

You are welcome.

 

[chocolatelover]

Thank you very much

(No, it is not because of the squares but because x and w are both the first members of the pair. For any (x,y), (x,y)~(x,y) because x2= x2. (Notice that in that last equation I am using the fact that "=" itself is reflexive: a= a.))

I don't understand this part. In the previos example x and w were both the first members of the pair, but it wasn't reflexive. Could you please explain this to me?

To show that this relation is transitive, we must show that "if (x,y)~(w,z) and (w,z)~(u,v) then (x,y)~(u,v)", the definition of "transitive". Okay, if (x,y)~(w,z) then x2= w2 and if (w,z)~(u,v) then w2= u2. From that (and the fact that "=" itself is transitive) we have x2= u2 which tells us that (x,y)~(u,v) and we are done.

The "equivalence classes" are sets of point (x,y) where x2 is always the same. Since x2= w2 if x= w or x= -w, those are pairs of vertical lines, x= a and x= -a, on a graph.

Now that I know that it is an equivalence class, the partition could be {{1}, {2}, {3}, {4}, {5},} right?

Also, when you see (x,y)~(w,z) if and only if y=w and you are trying to prove whether or not it is reflexive, it means that if (x,y) for all x they are an element of X, right? If not, could you explain it to me, please?

Thank you

 

[HallsofIvy]

Thank you very much

(No, it is not because of the squares but because x and w are both the first members of the pair. For any (x,y), (x,y)~(x,y) because x2= x2. (Notice that in that last equation I am using the fact that "=" itself is reflexive: a= a.))

I don't understand this part. In the previos example x and w were both the first members of the pair, but it wasn't reflexive. Could you please explain this to me?

In the previous example, the relation was (x,y)~(w,v) if y= w. y and w are not both the first members of the pair (nor both the second members). In particular, (1, 2) is not "related" to itself in this relation because if (x,y)= (w,v)= (1, 2) y= 2 and w= 1 so y is NOT equal to w.

To show that this relation is transitive, we must show that "if (x,y)~(w,z) and (w,z)~(u,v) then (x,y)~(u,v)", the definition of "transitive". Okay, if (x,y)~(w,z) then x2= w2 and if (w,z)~(u,v) then w2= u2. From that (and the fact that "=" itself is transitive) we have x2= u2 which tells us that (x,y)~(u,v) and we are done.

The "equivalence classes" are sets of point (x,y) where x2 is always the same. Since x2= w2 if x= w or x= -w, those are pairs of vertical lines, x= a and x= -a, on a graph.

Now that I know that it is an equivalence class, the partition could be {{1}, {2}, {3}, {4}, {5},} right?

No, no, no! This is an equivalence relation on R² Equivalence classes are sets of pairs of numbers. Since x2[/sup] is always the same, Equivalence classes would be, as I said, sets of numbers where y can be anything but x is fixed: vertical lines.

Also, when you see (x,y)~(w,z) if and only if y=w and you are trying to prove whether or not it is reflexive, it means that if (x,y) for all x they are an element of X, right? If not, could you explain it to me, please?

Thank you

I have no idea what you mean by "if (x,y) for all x". Saying (x,y)~(w,v) means that the relation is defined on pairs of things- presumably pairs of numbers.

 

[chocolatelover]

Thank you

Now that I know that it is an equivalence class, the partition could be {{1}, {2}, {3}, {4}, {5},} right?

No, no, no! This is an equivalence relation on R2 Equivalence classes are sets of pairs of numbers. Since x_{2[/sup] is always the same, Equivalence classes would be, as I said, sets of numbers where y can be anything but x is fixed: vertical lines.}

<sub>I'm looking for the partition. Could you please tell me how you would find that?

Thank you</sub>

 

[HallsofIvy]

Each set in a partition using an equivalence relation (called "equivalence classes") consists of all members of the original set that are equivalent. Since in this particular problem (x,y)~(u,v) if and only if x= u, equivalence classes consist of pairs of numbers in which the pairs have the same first member: {(a, y)|y is any real number}. Geometrically, as I said before, the eqivalence classes are vertical lines.

 

[chocolatelover]

Thank you very much

Regards