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Simple doubt with relations

  1. Apr 15, 2012 #1
    Hi. I've starting working with vectors in linear algebra but I need to have previous knowledge of equivalence relations so I started studying that but I have a simple doubt with the following relation:

    $$R = \{ (a,b)/a,b \in A,{\text{ a - b is an integer multiple of 2}}\} $$

    In this case could be for example that 1R1 or 2R2? because zero is multiple of every integer. The thing is that the relation is defined by a and b so I don't know that if a relation is defined by (a,b) then a has to be different to b because in this following one they have to be equal:
    $$E = \{ (a,a)/a \in A\} $$

    Thanks!
     
  2. jcsd
  3. Apr 15, 2012 #2

    Dick

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    Yes, 0 is a multiple of 2. And 1R1 and 2R2. Nothing there says a has to be different from b.
     
  4. Apr 15, 2012 #3
    Alright, and in the case of relation E do they have to be necessarily equal because it says (a,a)?... so if I say for example 5Ex I know that x=5, is that right?

    Thank you Dick!
     
  5. Apr 15, 2012 #4

    HallsofIvy

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    Yes, your "relation E" is just the "Equality" relation. That is one relation (and the prototype equivalence relation).
     
  6. Apr 15, 2012 #5
    OKay, and to define the integers could I write the relation as follows?:

    $$\mathbb{Z} = \{ ((a,b),(c,d)) \in N \times N,a + d = b + c\} $$

    I don't understand the definition of integers using relations.... how could an ordered pair represent a natural number?
     
    Last edited: Apr 15, 2012
  7. Apr 15, 2012 #6

    HallsofIvy

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    That's poorly phrased. For one thing ((a, b), (c, d)) is NOT in N X N. It is a pair of pairs of integers while NXN is pairs of integers.

    Rather, the set of integers is the set of equivalence classes of pairs of positive integers where the equivalence is defined by "(a, b)~ (c, d) if and only if a+ d= b+ c.

    What other pairs of integers are in the same equivalence class as, say, (1, 3)? What is the simplest way to characterize those pairs? What about the equivalence class containing (3, 1)?

    Now, to actually define the integers in that way, you would also have to define "order", addition, and multiplication. Apparently you haven't done that yet.
     
  8. Apr 15, 2012 #7
    So you're defining a relation with another relation inside? I still don't understand it comparing it to the other simple relations I mentioned above. Could you explain me comparing it and showing me an example of a integer using that relation?

    It says that it is defined over NxN the relation (a,b)~(c,d) iff..... but what does that mean? That the pair (a,b) and (c,d) each belongs to natural numbers?

    Thank you very much!
     
    Last edited: Apr 15, 2012
  9. Apr 15, 2012 #8
    Alright I understood that the equivalence is done in the cartensian product of N (NxN) given that condition. What I don't get is how do you know that each pair represents the substraction of the first and second? Example:

    $$[(4,7)]=[(2,5)]=[(5,8)]=[(1,4)]$$

    They're all equivalents, but what does tell me that they represent the integer -3 i.e. that I have to substract the second from the first? A relation only defines a set of ordered pairs.. not numbers
     
    Last edited: Apr 15, 2012
  10. Apr 15, 2012 #9

    Dick

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    Halls is right. To show your set of equivalence classes represent the integers you need to define some sort of addition operation (at least) in the ordered pairs. Have you defined that (a,b)+(c,d)=(a+c,b+d)?
     
  11. Apr 15, 2012 #10
    Hmmm I see.. I haven't defined anything. I was asking because the books only says after defining the relation: "Now you can see that each pair will represent an integer which results from substracting the second term from the first". But I clrearly don't see it until they give me a definition. I know it's true.. but how can I define that substraction?


    Thanks!

    PD: Let the next message be your 20.000th Dick! Congratulations! :D
     
  12. Apr 15, 2012 #11

    Dick

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    Thanks! It is a milestone. The book seems to be saying "Identify (a,b) with the integer a-b". And I think you do want to assume that they are taking the addition operation to be (a,b)+(c,d)=(a+c,b+d). Does that help?
     
  13. Apr 15, 2012 #12
    I don't understand what you're telling me with that addition operation... does that operation you defined enables us to finally identify the integer with the pairs? Is your addition operation sth like this?:

    [tex]\begin{cases} ~[(n,0)]=n \\ ~[(0,n)]=-n \end{cases}[/tex]
     
    Last edited: Apr 15, 2012
  14. Apr 15, 2012 #13

    Dick

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    Yes, but (n,0) isn't in your set of ordered pairs, because 0 isn't in N. I'm defining addition as (a,b)+(c,d)=(a+b,c+d), since you book seems have neglected to do so. If n is in N, a representation of n in the integers among the ordered pairs is (n+1,1).
     
  15. Apr 15, 2012 #14
    But isn't 0 a natural?
    But what is your definition for? The reusult is still a pair of ordered numbers, not a number. So I could simply say (a,b)=a-b but I don't think is that simple, is it?

    thanks!
     
  16. Apr 15, 2012 #15

    Dick

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    You said the books says, "Now you can see that each pair will represent an integer which results from subtracting the second term from the first". I would read that as saying (a,b) in the ordered pairs corresponds to a-b in the integers. And my definition of N is {1,2,3,4,...}.
     
  17. Apr 15, 2012 #16
    Yes the book does say that but they didn't define that, that's why I coudln't identify the integer just because they tell me so with words. So would this be the definition?:

    $$\begin{cases} ~[(n+1,1)]=n \\ ~[(1,n+1)]=-n \end{cases}$$ for all n in N.
    Edit: this would be better:
    $$\begin{cases} ~[(n+m,m)]=n \\ ~[(m,n+m)]=-n \end{cases}$$

    If it is, then why did you define that (a,b)+(c,d)=(a+c,b+d) just reading what the book says? the book says (a,b)= the integer a-b, not what you wrote, but is it the same ? Because I'm not realising... tranks! If you are tired don't bother answering, I almost got it and I don't to cause you trouble. I just want to understand everything at 100% even if I'm at 99%.. hehe i'm just that kind of person
     
    Last edited: Apr 15, 2012
  18. Apr 15, 2012 #17

    Dick

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    There are two steps here. One is defining what 'addition' means for ordered pairs. And I think that's implicitly (a,b)+(c,d)=(a+c,b+d). You need that. The second step is how to identify an equivalence class of ordered pairs with an integer. What they are asking you to show is that addition in the equivalence classes is consistent with addition in the identified integers.
     
  19. Apr 15, 2012 #18
    Why do you say that I need the addition for ordered pairs? Just having the equivalence classes couldn't I just identify the corresponding integer by defining?:

    $$\begin{cases} ~[(n+m,m)]=n \\ ~[(m,n+m)]=-n \end{cases}$$

    (4,7)=(-3+7,7)=-3
    (4,7)=(4,3+4)=3

    In fact I just could say that
    (n+m,m)]=n
    and I still will have all the integers


    PD: Perhaps you're ahead trying to define the operations to define the integers. I don't want to do that yet, I just want to identify the integer with the equivalence classes. I think we are misunderstaning each other
     
    Last edited: Apr 15, 2012
  20. Apr 16, 2012 #19

    Dick

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    Ok. That's fair. Then you've shown your mapping is onto the integers. Then all you need to check is that it is well defined on your equivalence classes, i.e. if (a,b) is equivalent to (c,d) then a-b=c-b, if you haven't already done so.
     
    Last edited: Apr 16, 2012
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