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Defining relations for an n-tuple

  1. Sep 26, 2009 #1
    Given an element (6, 5, 4) of S (that is {(6, 5, 4)} is a subset of S); assuming S is a relation, how exactly do we donate the relations between the elements of the 3-tuple formed in this case; the relation can be of 3 sorts (in a 3-tuple) -
    Out of these 3 how many relations will be valid for (6,5,1) ∈ S.

    I've tired to explain this question but I have a feeling it's still not clear...pls notify if this is so.
  2. jcsd
  3. Sep 26, 2009 #2


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    What does 6S5 mean? S is a 3-ary relation, so you need three inputs. Are you just looking for an alternative to infix notation? You could say S(6, 5, 4) to mean that (6, 5, 4) is in S.

    Or do you want the S in 6S5 to stand for a member of the set on which S is defined?
  4. Sep 27, 2009 #3
    Yes, you got it, THIS is the question...suppose 6, 5 and 4 form a 3-tuple of this 3-ary relation...what will it actually mean.

    For e.g in a binary relation (K) we say that an ordered pair will be a member of K if and only if the second element of the ordered pair is twice of first, then if we're giving an element (2, 4)∈K, then we can say 2 is related to 4 by R...i.e 4 is twice of 2.

    I cannot make such exact relations for n-ary relations...so I'm asking how do we make relations of R(6, 5, 4)?...I mean how exactly are 6, 5 and 4 related through R?
  5. Sep 27, 2009 #4


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    Well, you can say that R is a 3-ary relation on S, and (a, b, c) is in R iff for any a, b, and c in S

    a + b = c
    a < b < c
    a = 2 and b > 3c
    a is the mother of b and c
    a, b, and c are sitting at the same table
    a, b, and c have a common factor
    a, b, and c are equidistant from a point d in S
    there exists some x in T such that a(x) = b(x) = c(x)​
    And so on. You just need to be sure that the object you define this way is indeed a set. Or you can define R recursively, e.g.,

    (1, 2, 3) is in R
    if (a, b, c) is in R, then (a+1, 2b, c2) is in R

    Or if R is finite, you can just list its members.

    R = {(1, 2, 3), (2, 4, 9), (3, 8, 81), (4, 16, frog)}​

    Does this help?
    Last edited: Sep 27, 2009
  6. Sep 27, 2009 #5
    Yes that did...thanks!... that means, in this example of this 3 tuple, we have have the following relations between the elements of this tuple -
    Now it can also be -
    and this tuple will still be a subset of R
    Here S is another relation. I concluded this from what you said here -
    a + b = c

    Have I covered all possibilities with this 3-tuple?
  7. Sep 28, 2009 #6


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    I don't understand. Do you want R to be a binary relation? Let's go back a minute. Or let's start over.

    S is a set. We don't care what its members are.
    R is a 3-ary relation on S, meaning that R is a subset of S3, i.e., the set of all 3-tuples whose members are all in S.
    P is a binary relation on S.
    Q is a binary operation on S, meaning that Q is a subset of S2xS.

    Do you want to define R in terms of P and Q? You can do this, yes. However, saying 6R5S4 only makes sense if R and S are both binary relations, and even then it is not clear whether you mean (6R5)S4 or (6R5)(5S4) or something else. Infix notation isn't as suited to relations or operations involving more than two inputs because there isn't an obvious way to separate things. Prefix or postfix works better: R(a, b, c), (a, b, c)R, (or if you only use atomic symbols) Rabc, abcR.

    And an n-ary relation contains only n-tuples. It doesn't contain any (n-1)-tuples. If I want to define a set of relations sum-of-n that contain tuples whose last member is the sum of the preceding members, I need a different relation for each number of addends that I want. That is, I need a 3-ary sum-of-2 = {(a, b, c) | a + b = c}, a 4-ary relation sum-of-3 = {(a, b, c, d) | a + b + c = d}, etc. But sum-of-2 is not a subset of sum-of-3. They might contain the same information under certain circumstances (so you might define sum-of-2 = {(a, b, c) | (0, a, b, c) is in sum-of-3} where 0 is your additive identity), but an n-ary and m-ary relation do not have the same structure if n != m.

    It is possible for an n-ary relation to include another n-ary realtion, as < includes <.

    Note also that an n-ary operation does not have the same structure as its related (n+1)-ary relation. For example, the binary operation + on N is itself a binary relation. That is what the equal signs signifies. ((a, b), c) is in + iff a + b = c. They are two ways of saying the same thing. Now you can use + to define a 3-ary relation sum-of-2 such that (a, b, c) is in sum-of-2 iff ((a, b), c) is in +. But clearly (a, b, c) and ((a, b), c) do not have the same structure.

    Note also that you can't necessarily go the other way and use an n-ary relation to define an (n-1)-ary operation by just gathering up the first n-1 terms into a tuple because you need to ensure that your operation meets the uniqueness condition. (1, 1, 2) and (1, 1, 3) can both be in the same relation but ((1, 1), 2) and (1, 1), 3) cannot both be in the same operation.

    Does this address what you are trying to do?
  8. Sep 28, 2009 #7
    I think I got the answer from here...so the answer to -

    No it cannot (hopefully) if a + b = c, then it's the property of S to exhibit such relations among it's 3-tuples.
    Considering this example (a + b = c), any other element...like y, k, l such that y+k=l will be applicable as an element of S, but if y+k != l...it will not be applicable. Or in this example, the criteria for a 3-tuple to be a part of the relation S is that the sum of the first 2 elements of this 3 tuple should be equal to the third element of this tupple.

    But Q is a binary operation, how can it be an element?

    Thanks for the info.

    Or in general all operators; relational, equality, urinary etc...


    Ok then...problem's cleared.
  9. Sep 28, 2009 #8


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    I'm not sure what your original string meant, but your explanation here sounds correct.

    An operation can be considered a set. Most objects can be considered sets. But note that I said subset there, not member, and we are apparently still using S to mean different things. Q is a subset of (the Cartesian product) S2xS. Q takes two members of S and returns a member of S, a concept you can capture with the ((a, b), c) structure.
  10. Sep 30, 2009 #9
    Ok...I got that...thanks!
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