Beginner's mathematical proof / composition of relations

[Testify]

Homework Statement

Suppose r and s are two positive real numbers. Let D_r and D_s be defined as in part 3 of Example 4.3.1. What is D_r \circ D_s? Justify your answer with a proof. (Hint: In your proof, you may find it helpful to use the triangle inequality.)

Homework Equations

Example 4.3.1:
For each positive real number r, let D_r = \left\{(x,y)\in \mathbb{R} \times \mathbb{R} | ( |x - y| < r \right\}. Then D_r is a relation on \mathbb{R}.

\circ = composition

The Attempt at a Solution

I don't know what the composition of D_r and D_s would be, so I am not even close to being able to prove anything.

thanks

 

[Tinyboss]

Your textbook probably talks about composition of relations, but in case it doesn't, you can read about the idea on Wikipedia.

 

[LCKurtz]

You have (x,y) in D_r if |x - y| < r and (x,y) in D_s if |x - y| < s.

Can you state the definition of D_s\circ D_r? Once you do that, you may see what the triangle inequality has to do with this problem.

 

[Testify]

I've read everything my book has to say about the composition of relations, and I cannot figure out what to do with this problem. With the other problems, it would just say, for example, that A was a relation from X to Y and B was a relation from Y to Z, then the composition of B and A would be (x,z), etc.

I just cannot figure out what the definition of D_s \circ D_r would be...

 

[LCKurtz]

I've read everything my book has to say about the composition of relations, and I cannot figure out what to do with this problem. With the other problems, it would just say, for example, that A was a relation from X to Y and B was a relation from Y to Z, then the composition of B and A would be (x,z), etc.

I just cannot figure out what the definition of D_s \circ D_r would be...

You don't have to "figure out the definition". Click on the link Tinyboss gave you. It gives you the definition. Write down what that definition says for your particular problem.

 

[Testify]

I have been using the definition, and I still have no idea what D_s \circ D_r would be.

I just don't understand how you can use composition with two relations over the same elements. If one of them were (x,y) and the other one were (y,z) then I might be able to figure this out on my own.

 

[Tinyboss]

If one relation is on X\times Y and the other is on Y\times Z, is there any reason why you can't have X=Y=Z=\mathbb{R}?

 

[Testify]

no.

I'm still not really following though. I just cannot figure out how to combine the two relations to create a composition. =/

 

[Tinyboss]

You need to talk to your TA or professor, then. You have access to the definition, which is straightforward, but can't make sense of it--you might not have adequate preparation for the class you're in.

 

[Testify]

I'm using Velleman's "How to Prove it" and I'm doing this self study. I suppose I am having some trouble understanding some of these definitions. I guess I will just have to re-read a few chapters...

Could you just tell me what D_s \circ D_r would be? Maybe then I would understand how it works.

 

[Tinyboss]

(x,y)\in D_s\circ D_r\Leftrightarrow\exists z\in\mathbb{R}:(x,z)\in D_s\text{ and }(z,y)\in D_r.

An example for intuition: take the relations S (for "sibling") and F (for "friend"). These are both relations on the set "people x people". Then I am related to you by F\circ S if there's some person P such that P is my sibling and your friend, i.e. you're a "friend of a sibling" to me. Likewise, I'm related to you by S\circ F if there's a person Q such that Q is my friend and your sibling, i.e. you're a "sibling of a friend" to me. You can see how the definition is mirrored in the way you read the composition.

 

[Testify]

How would I be able to prove that using the triangle inequality? I thought that the answer would be something along the lines of |x-z|< r and s or something like that... I know that is the "answer" but I thought the answer would be more... numerical.

Maybe I just totally misunderstood what the question was asking. =/ I'm sorry that I'm being confusing.