# Thread: Relations, functions and ordered pairs.

1. ## 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.

2. 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 $\displaystyle y=x^2$ 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}

3. 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 $\displaystyle y=x^2$ 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.

4. 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 $\displaystyle (a,b)\in f$ and $\displaystyle (a,c)\in f$, then then b=c).

These are standard definitions and have nothing to do with having a formula or infinitely many elements.

5. That's true. A function has one unique value of x for every y value.

6. 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.

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.

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

My point is the (possible) distinction between $\displaystyle A$ and $\displaystyle \mathrm{Dom}(f)$. 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, $\displaystyle f=\{(0,3),(2,2)\}$ is not a function on $\displaystyle \mathbb{R}$, but is a function on the set $\displaystyle \{0,2\}$.

8. Not only that but consider the parabola $\displaystyle y^2=x$
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.

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

My point is the (possible) distinction between $\displaystyle A$ and $\displaystyle \mathrm{Dom}(f)$. 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, $\displaystyle f=\{(0,3),(2,2)\}$ is not a function on $\displaystyle \mathbb{R}$, but is a function on the set $\displaystyle \{0,2\}$.
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.

10. 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 $\displaystyle A=\{1,2,3,4,5\}$ and $\displaystyle B=\{6,7,8,9,10\}$.

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

All standard definitions agree that h is not a function since $\displaystyle (2,9)\in h$ and $\displaystyle (2,7)\in h$, yet $\displaystyle 7\neq9$. All standard definitions agree that $\displaystyle f:A\to B$ is a function. Some definitions will say that $\displaystyle g:A\to B$ is a function which is simply not defined at 1 or 3. Others will say that $\displaystyle g:A\to B$ is not a function, yet $\displaystyle g:\{2,4,5\}\to B$ is.

11. Note that no one says that a function must “hit” every point in $\displaystyle B$. The issue is whether it is defined for every point in $\displaystyle A$.

12. Originally Posted by Tikoloshe
Note that no one says that a function must “hit” every point in $\displaystyle B$. The issue is whether it is defined for every point in $\displaystyle A$.
Thanks for all of your replies, I'm going to let this simmer for a bit and see what happens