Relations, functions and ordered pairs.

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April 29th 2010, 02:47 PM
jamieirl

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.
April 29th 2010, 02:54 PM
shenanigans87

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}
April 29th 2010, 03:07 PM
jamieirl

Quote:

Originally Posted by shenanigans87

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.
April 29th 2010, 03:16 PM
Tikoloshe

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.
April 29th 2010, 03:45 PM
shenanigans87

That's true. A function has one unique value of x for every y value.
April 29th 2010, 04:03 PM
jamieirl

Quote:

Originally Posted by Tikoloshe

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.
April 29th 2010, 04:33 PM
Tikoloshe

Quote:

Originally Posted by jamieirl

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 $\{0,2\}$ .
April 29th 2010, 04:40 PM
shenanigans87

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.
April 29th 2010, 04:42 PM
jamieirl

Quote:

Originally Posted by Tikoloshe

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.
April 29th 2010, 05:07 PM
Tikoloshe

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.
April 29th 2010, 05:09 PM
Tikoloshe

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 $A$ .
April 29th 2010, 07:42 PM
jamieirl

Quote:

Originally Posted by Tikoloshe

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 :)