# Which of the following sets qualify as functions.

1. Jul 22, 2013

### WK95

R1 = { (1,2) , (1,3) , (1,4) (1,5) , (1,6) } No
R2 = { (x,y) in R x R | x = sin(y) }
R3 = { (x,y) in Z x Z | y2 = x }
R4 = { (Φ, {Φ}) , ({Φ},Φ) , (Φ,Φ) , ({Φ},{Φ}) } No
R5 = { (x,y) in N x Z | 0<x<1, 3<y<4 }

A x B means Cartesian product. That much I know. What I don't know is how to solve them with regards to the above problems.

I know that
If A = {a1, a2} and B = {b1, b2}
Then A x B = {(a1, b1),(a1, b2),(a2, b1),(a2, b2)}

2. Jul 23, 2013

### CompuChip

What is the (your :-)) definition of a function?

3. Jul 23, 2013

### HallsofIvy

Staff Emeritus
Compuchip's question is the crucial one. You cannot prove "f is a function" if you don't know the definition of "function". And definitions in mathematics are "working definitions". You use the precise words of the definition in proofs.

So the first thing you should do is look up the definition of "function" in your textbook.

4. Jul 23, 2013

### WK95

"A function f is a rule that assigns to each element x in a set D exactly one element, called f(x), in a set E." - Stewart's Calculus 7th Edition

With that, I ruled out R1 and R4 since for those ones the x value maps out to more than one element f(x).

I don't think R5 is a function either since there are no natural numbers ( N ) for 0 < x < 1 and there are no integers ( Z ) for 3 < x < 4.

What I'm stuck at is R2 and R3. I don't quite understand how to apply Cartesian products to them. For R2, x = sin (y) on it's own is a function. For R3, y^2=x on it's own is not a function.

Last edited: Jul 23, 2013
5. Jul 23, 2013

### CompuChip

That, and the fact that in the case of "function" definitions often differ slightly in different courses or textbooks.

6. Jul 23, 2013

### CompuChip

Yes. Since R2 assigns to every element x in R another element, called sin(x), and no other, it is a function. Personally I think this is a slightly weird definition, since it basically defines the sine function as a relation in terms of the sine function defined... in another way.

Anyway, you are right that R3 is not a function because it is basically y² = x. Can you find a specific counter-example of two elements of R3 that show this? I.e. two pairs (x, y) and (x', y') of which you can show that they are elements of R3 but they violate the "A function f is a rule that..." definition?

7. Jul 23, 2013

### WK95

(4,2) and (4,-2)? Basically (a²,±a)

8. Jul 23, 2013

### verty

R3 = { (x,y) in Z x Z | y2 = x }

This means, all pairs (x,y) such that y^2 = x. We know that (-1)^2 = 1, so (1,-1) is accepted. And we know 1^2 = 1, so (1,1) is accepted. This is a problem, 1 maps to both -1 and 1.

Apply this same thinking to the other examples.

9. Jul 23, 2013

### HallsofIvy

Staff Emeritus
That was not the problem. You were given that x= sin(y). And sin(0)= sin(pi)= 0 so that (0, 0) and (0, pi) are both in the set. That is NOT a function.

10. Jul 23, 2013

### Mandelbroth

Meaning $y$ is not a function of $x$, of course. $x$ is a function of $y$. :tongue:

Joking aside, I think that R5 is a function by your definition, since every $x$ in the domain $D=\emptyset$ is mapped to exactly one element of the codomain $y\in E$. This is an example of something called an empty function.

11. Jul 23, 2013

### WK95

For R5 = { (x,y) in N x Z | 0<x<1, 3<y<4 }, I know N refers to Natural numbers and Z refers to integers. Asides from solving the problem, in this context, how do I go about understanding what { (x,y) in N x Z | 0<x<1, 3<y<4 } actually goes to state? My initial confusion with solving the problem, which in retrospect isn't a hard one, is due to not fully understanding what the equations are actually trying to say.

To start, how do I read this {(x,y) in N x Z | 0<x<1, 3<y<4}?

12. Jul 24, 2013

### CompuChip

Good catch Halls, thanks!

One way to do this is take a square piece of paper and try to visualize it. For example, you can draw
$\{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid 0 < x \le 1, 3 < y \le 4 \}$
by drawing two dotted lines at x = 0 and y = 3, and two solid lines at x = 1 and y = 4. Then you can shade the area that they enclose. Any point in this area, including any point on one of the solid lines will be in the set; but not any point on the dotted lines.

For discrete sets, you will not get a shaded area but a discrete lattice. For example: the set $\mathbb{N} \times \mathbb{N}$ consists of all pairs of points whose coordinates are non-negative integers. So if you draw this on graph paper (with the little blue squares) you will get a dot at every intersection and all those numbers will be in the set. A set like $\{ (x, y) \in \mathbb{N} \times \mathbb{N} \mid 1 \le x \le 4, 1 \le y \le 3 \}$ is a subset of $\mathbb{N} \times \mathbb{N}$ - so you can start by drawing that. But if you draw four solid lines (two horizontal, two vertical) at x = 1, x = 4, y = 1 and y = 3 only those dots that are on the border of the rectangle or inside it will be in your set.

Now can you try to plot R5?

13. Jul 24, 2013

### WK95

What is a discrete lattice? Sorry if my math language and terminology is not up to par. Looking back as I just graduate high school, AP Calculus BC is seriously watered down from serious math (At least this seems very tough and serious to me).

When plotting R5, I have two vertical dotted lines at x=0 and x=1, and two horizontal dotted lines at y=3 and y=4. But for 0<x<1, there are no natural numbers so I'm not sure how to deal with that.

14. Jul 24, 2013

### LCKurtz

That's right. There aren't any of those ordered pairs in the defined region. That's what Mandelbroth was talking about when he mentioned the empty function. There's nothing there.

15. Jul 24, 2013

### CompuChip

A lattice is a grid, a set of points which are spaced equally apart like the intersections of the lines on standard math paper often are (http://oxforddictionaries.com/definition/english/lattice?q=lattice [Broken]). In mathematics it is mostly used for (subsets of) sets like Z x Z.

You are right: there are no points in the region. And we have agreed that in mathematics, any statement like "for all x in X ... " is true by definition if X is the empty set. We call that "vacuously true" (where vacuous is related to vacuum: the statement is true but it doesn't say anything useful). So indeed the statement "f is a rule that assigns to each element x in a set D ..." is true if D has no elements is vacuously true and hence f is a function.

Last edited by a moderator: May 6, 2017
16. Jul 24, 2013

### WK95

Well, I'm embarrassed. Lattice turned out to be such an easy thing.

Could you further simplify (if that is even possible) what you mean by "And we have agreed that in mathematics, any statement like 'for all x in X ... ' is true by definition if X is the empty set". I don't understand how this works. If X is an empty set, there wouldn't be anything in there so that should mean that there is no x's in X. Could you give an example of this statement, "And we have agreed that in mathematics, any statement like 'for all x in X ... ' is true by definition if X is the empty set", in action?

Last edited by a moderator: May 6, 2017
17. Jul 24, 2013

### WK95

Oh I think I get it now. So as per the definition of a function, R5 is a function because nothing maps to nothing?

18. Jul 24, 2013

### LCKurtz

It is a function because every $x$ in its domain maps to exactly one element $y$ in its codomain.

If you don't believe me, demonstrate an $x$ in the domain that isn't mapped thusly.

Similarly, it's also true that every $x$ in the domain is greater than 50. That can only be falsified by demonstrating an $x$ that isn't.

That's the idea about vacuously true statements. They are vacuous.

19. Jul 24, 2013

### WK95

But since it has nothing, how is it true that every x in the domain maps to exactly one element in y if there is nothing to map at all?

I have a bit of an obsession with trying to gain perfect comprehension or at the least, understand something intuitively. I really have to be able to envision exactly how and why somethign works or I'm not satisfied.

But yikes, this set theory stuff is tricky. AP Calculus (BC no less!) did nothing to prepare me for this and I got a 5 on that test. I am in awe at the intelligence of the members here.

Can you recommend me some resources to study up on this sort of stuff? I'm sure it'll come up again once I start college this fall.

Last edited: Jul 24, 2013
20. Jul 24, 2013

### Mandelbroth

If something has to be true for everything in a set, and the set has nothing in it, then it is automatically true because nothing in that set can act as a counterexample.

Believe me, after seeing what some people on here can do, I'm still impressed.

You can get there eventually, though. Math takes practice to get used to, and there's a lot of "deep" logical reasoning to go with it.

Look in your browser for a site called PhysicsForums. You should be able to find it here. :tongue: