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Homework Help: Beginner's mathematical proof / composition of relations

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose r and s are two positive real numbers. Let Dr and Ds be defined as in part 3 of Example 4.3.1. What is [tex]D_r \circ D_s[/tex]? Justify your answer with a proof. (Hint: In your proof, you may find it helpful to use the triangle inequality.)

    2. Relevant equations
    Example 4.3.1:
    For each positive real number r, let Dr = [tex]\left\{(x,y)\in \mathbb{R} \times \mathbb{R} | ( |x - y| < r \right\}[/tex]. Then Dr is a relation on [tex]\mathbb{R}[/tex].

    [tex]\circ[/tex] = composition

    3. The attempt at a solution

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

    Last edited: Mar 11, 2010
  2. jcsd
  3. Mar 11, 2010 #2
    Your textbook probably talks about composition of relations, but in case it doesn't, you can read about the idea on Wikipedia.
  4. Mar 11, 2010 #3


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

    Can you state the definition of [itex] D_s\circ D_r[/itex]? Once you do that, you may see what the triangle inequality has to do with this problem.
  5. Mar 11, 2010 #4
    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 [tex]D_s \circ D_r[/tex] would be...
  6. Mar 11, 2010 #5


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    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.
  7. Mar 11, 2010 #6
    I have been using the definition, and I still have no idea what [tex]D_s \circ D_r[/tex] 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.
  8. Mar 11, 2010 #7
    If one relation is on [tex]X\times Y[/tex] and the other is on [tex]Y\times Z[/tex], is there any reason why you can't have [tex]X=Y=Z=\mathbb{R}[/tex]?
  9. Mar 11, 2010 #8

    I'm still not really following though. I just cannot figure out how to combine the two relations to create a composition. =/
  10. Mar 11, 2010 #9
    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.
  11. Mar 11, 2010 #10
    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 [tex]D_s \circ D_r[/tex] would be? Maybe then I would understand how it works.
  12. Mar 11, 2010 #11
    [tex](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[/tex].

    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 [tex]F\circ S[/tex] 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 [tex]S\circ F[/tex] 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.
  13. Mar 11, 2010 #12
    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.
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