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Math Help - Relations, functions and ordered pairs.

  1. #1
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    Relations, functions and ordered pairs.

    Hello all and thanks in advance for all of the help you'll be giving me for the next couple of weeks. I'm just finishing a college algebra class and admittingly I haven't been the best student. The class is mostly online, that coupled with most of the in-class teachings consisting of calculator help (and the teacher stated at the start of the semester that she doesn't know how to use a Ti-89, which I have and I didn't want to buy another calculator after spending 130+ on mine) leaves me basically on my own. She (the professor) was nice enough to give us a review for the final exam (which I'll be asking plenty of questions about in the future) as well as a 3 question take home exam; which is where my first question is coming from. Anyway sorry for the long introduction, my first question from the take home exam:

    "1. One way to define a relation is "A set of ordered pairs".
    a. Give an example
    b. What is the difference between a function and a relation? Is your example in part a) a function? How can you tell?"


    I've scored fairly well on most of the assignments/test/quizzes thus far, but for some reason when I look at this take home exam it looks alien to me. Thanks for any replies.
    Last edited by mr fantastic; April 29th 2010 at 08:49 PM. Reason: Re-titled.
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  2. #2
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    A relation can be a set of ordered pairs, like you stated.

    i.e. {1,1} {3, 5} {4, 14} etc.

    A function is a formula from which you can derive ordered pairs.

    If the function is y=x^2 you can derive infinitely many relations by plugging in numbers. Let's get a relation for 0 < x < 3

    {0, 0} {1, 1} {2, 4} {3, 9}
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  3. #3
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    Quote Originally Posted by shenanigans87 View Post
    A relation can be a set of ordered pairs, like you stated.

    i.e. {1,1} {3, 5} {4, 14} etc.

    A function is a formula from which you can derive ordered pairs.

    If the function is y=x^2 you can derive infinitely many relations by plugging in numbers. Let's get a relation for 0 < x < 3

    {0, 0} {1, 1} {2, 4} {3, 9}
    Thank you for the quick reply. I think I've gotten too used to the online-assignment format, when I have a written question in front of me I go blank. So for part a. I just give any random ordered pairs as an example, and for part b. I just state that no, part a. is not a function and "How can you tell?" I'd answer; "
    A function is a formula from which you can derive ordered pairs." and example a. (the ordered pairs example) is acquired from a function, not a function itself. Am I understanding this correctly? Thanks again.
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  4. #4
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    I believe shenanigans87’s answer is misleading. A relation can be defined as any subset of the cartesian product of two sets (i.e., a set of ordered pairs). A function is a relation (a set of ordered pairs) in which each first coordinate has a unique second coordinate (i.e., if (a,b)\in f and (a,c)\in f, then then b=c).

    These are standard definitions and have nothing to do with having a formula or infinitely many elements.
    Last edited by Tikoloshe; April 29th 2010 at 06:07 PM. Reason: Changed “if (a,b)=(a,c)” to “if (a,b)\in f and (a,c)\in f, then”
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  5. #5
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    That's true. A function has one unique value of x for every y value.
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  6. #6
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    Quote Originally Posted by Tikoloshe View Post
    I believe shenanigans87’s answer is misleading. A relation can be defined as any subset of the cartesian product of two sets (i.e., a set of ordered pairs). A function is a relation (a set of ordered pairs) in which each first coordinate has a unique second coordinate (i.e., if (a,b)=(a,c) then b=c).

    These are standard definitions and have nothing to do with having a formula or infinitely many elements.
    So based on what you're saying I could answer this question like this:

    1. One way to define a relation is "A set of ordered pairs".
    a. Give an example: (0,3)(2,3)
    b. What is the difference between a function and a relation?

    Answer: A relation is any subset of the cartesian product of two sets. While a function is only one type of relation in which the first coordinate has a unique second coordinate.

    Is your example in part a) a function? Answer: No
    How can you tell? Answer: The relation does not meet the criteria required to be a function; a function must be a relation in which the first coordinate has a unique second coordinate (i.e., (a,b)=(a,c) then (b=c)).

    How does that look?
    Also, would this be an example of a function; (0,3),(2,2), since they have unique y's? I think I just confused myself.
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  7. #7
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    Quote Originally Posted by jamieirl View Post
    Also, would this be an example of a function; (0,3),(2,2), since they have unique y's?
    Sure would. Now there is a small point that I think shenanigans87 initially tried to make. That is the definition of a function includes its domain. In some contexts of discrete mathematics there is a distinction between the domain and first set of your cartesian product.

    That is, let f:A\to B be a function from A to B. Then f\subset A\times B such that \forall a\in\mathrm{Dom}(f) there is a unique b\in B such that (a,b)\in f.

    My point is the (possible) distinction between A and \mathrm{Dom}(f). Depending on what definition you choose to find useful, you might demand that the two sets are always equal for a function. In this case, f=\{(0,3),(2,2)\} is not a function on \mathbb{R}, but is a function on the set \{0,2\}.
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  8. #8
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    Not only that but consider the parabola y^2=x
    You could pull infinitely many ordered pairs from that which don't repeat the same x value but that doesn't make it a function with respect to your definition.
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  9. #9
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    Quote Originally Posted by Tikoloshe View Post
    Sure would. Now there is a small point that I think shenanigans87 initially tried to make. That is the definition of a function includes its domain. In some contexts of discrete mathematics there is a distinction between the domain and first set of your cartesian product.

    That is, let f:A\to B be a function from A to B. Then f\subset A\times B such that \forall a\in\mathrm{Dom}(f) there is a unique b\in B such that (a,b)\in f.

    My point is the (possible) distinction between A and \mathrm{Dom}(f). Depending on what definition you choose to find useful, you might demand that the two sets are always equal for a function. In this case, f=\{(0,3),(2,2)\} is not a function on \mathbb{R}, but is a function on the set \{0,2\}.
    I wish I could follow that better because it looks like very useful information, I understand about half of what you said there I think. Hehe.
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  10. #10
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    It might help if I hadn’t made a typo (now fixed). I hope you reread my first post

    Here’s an example of what I mean: let A=\{1,2,3,4,5\} and B=\{6,7,8,9,10\}.

    Suppose I define the following relations:
    f=\{(1,6),(2,9),(3,10),(4,9),(5,7)\}
    g=\{(2,7),(4,6),(5,7)\}
    h=\{(1,10),(2,9),(3,8),(4,7),(5,6),(2,7)\}

    All standard definitions agree that h is not a function since (2,9)\in h and (2,7)\in h, yet 7\neq9. All standard definitions agree that f:A\to B is a function. Some definitions will say that g:A\to B is a function which is simply not defined at 1 or 3. Others will say that g:A\to B is not a function, yet g:\{2,4,5\}\to B is.
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  11. #11
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    Note that no one says that a function must “hit” every point in B. The issue is whether it is defined for every point in A.
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  12. #12
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    Quote Originally Posted by Tikoloshe View Post
    Note that no one says that a function must “hit” every point in B. The issue is whether it is defined for every point in A.
    Thanks for all of your replies, I'm going to let this simmer for a bit and see what happens
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