# A little proof for functions

1. Sep 24, 2003

### StephenPrivitera

Let F be the set of all functions whose domain is R. Considering the binary operations + and * on F, show that P10-P12 cannot hold.
P10 is essentially the trichotomy law. In my book, it is written, "For all a in R, a=0, xor a in P, xor -a in P," where P is the set of positive numbers. Um, how does this make ANY sense for functions?! I can't say that the function F is positive! P11 and P12 are closure under addition and multiplication respectively. Again these make no sense in this context. So my answer to this question presently is that P10-P12 make no sense for the operations on F.
Anyone else see some sense in this?

Edit: an afterthought
I can see sense in this if they mean f(x)=0, f(x)>0 xor -f(x)>0 for a given x, but f(x) is not a function - it's a number.

Last edited: Sep 24, 2003
2. Sep 24, 2003

### Hurkyl

Staff Emeritus
Remember that the term "positive number" is defined by the properties listed in P10-12. Basically, the problem seems to be asking you to prove that it is impossible to choose a set P so that P10-12 hold.

3. Sep 24, 2003

### StephenPrivitera

IOW, show that it is impossible to define a set of "positive" functions? I guess I still don't get it. Do they mean it's impossible to divide the set of functions into a partition of three sets? That's essentially what P10 does to the reals.

Last edited: Sep 24, 2003
4. Sep 24, 2003

### Hurkyl

Staff Emeritus
First off, allow me to make sure we're talking about the same axioms; you're refering to the definition of an ordered ring:

An ordered ring (R, P) is a ring R together wtih a subset P (called the set of positive elements) of R such that

(1) 0 is not in P
(2) if a is in R, then either a as in P, a = 0, or -a as in P
(3) if a and b are in P then a + b and a * b are in P

Allow me to do an example and prove that one cannot order the complex numbers.

First off, note that if P exists, then 1 as in P. Apply rule (2) to -1...
case 1: -1 = 0
this cannot be true
case 2: -1 is in P
by rule 3, 1 = (-1) * (-1) is in P
case 3: -(-1) is in P
thus 1 = -(-1) is in P
qed

From this, we can conclude that -1 cannot be in P.

Now, assume we can select a subset P of C that satisfies the above axioms.

By property 2, either i is in P, i = 0, or -i is in P.

case 1: i = 0
this clearly cannot be true
case 2: i is in P
by rule (3), -1 = i * i is in P, which is a contradiction
case 3: -i is in P
by rule (3), -1 = (-i) * (-i) is in P, which is a contradiction

Since all cases lead to contradictions, our assumption that such a P exists must be false, therefore the complex numbers cannot be ordered.

5. Sep 24, 2003

### StephenPrivitera

Ah, see, this is a calculus class. Fortunately, I followed what you said because my professor happened to mention what a ring is . It makes sense what you say because in previous parts of this problem the author asks me to show that F satisfies P1-P9 except P7 (these are the properties of a ring). Then in part c he asks me to show that F does not satisfy P10-P12. The author never tells me what I'm proving though, just that F satisfies some properties and not others. Let me ask you this: in the set F what, which element is 0? F is the set of all functions, but 0 isn't in F... is it? Is the set P any subset? If b is a element of F, what is -b? Is it the additive inverse of b, is it the inverse funtion of b? What is the additive inverse of a set?

6. Sep 24, 2003

### HallsofIvy

Staff Emeritus
You are talking about the "algebra" of functions. The "zero" function, f(x)= 0 for all x, is the additive identity (THAT'S the 0 function you were wondering about). Given a functions f(x), there exist a function -f(x) defined by the property that f(x)+ (-f(x))= 0 for all x (Not the "inverse" function: that name (without the modifier "additive" or "multiplicative" is reserved for the inverse of composition). The functions f(x)= 1 for all x is the multiplicative identity and (if it exists) 1/f(x) is the multiplicative inverse of f(x). Notice that, while all numbers except 0 have a multiplicative inverse, many functions do not have a multiplicative inverse.

The "order" rules that you are talking about assert that THERE EXIST a set P such that
If f and g are both in P, then f+ g is in P
If f and g are both in P, then f*g is in P
For every f one and only one holds:
1) f is in P
2) f is in -P
3) f= 0 (i.e. f(x)= 0 for all x)

Certainly it is possible to partition ANY set into three subsets (well, as long as it has at least 3 members!). The question is whether it is possible to partition the set of all functions into 3 subsets (one of which only contains the 0 function) satisfying those properties.

7. Sep 24, 2003

### Hurkyl

Staff Emeritus
I suspect he means you to use the ordinary definitions of addition and multiplication... that is:

(f + g)(x) = f(x) + g(x)
(fg)(x) = f(g(x))

The zero element in any ring is the additive identity. In this context, it is the function &epsilon; such that

f + &epsilon; = f

(-f) usually (always?) means the additive inverse; here, the function such that:

f + (-f) = &epsilon;

/begin digression

0 is a very polymorphic symbol in mathematics; it is commonly used to represent any additive identity. Some authors, though, will use an alternate symbol, like &epsilon; because it is sometimes difficult to know to which additive identity 0 is referring in some contexts.

For instance, today I was working on a problem and I had the symbol 0 referring to:
the real number 0
a column vector of zeroes (three different sizes)
a row vector of zeroes (three different sizes)
a matrix of zeroes (9 different dimensions)

That is sixteen different meanings for the symbol 0 within a couple square inches on my whiteboard! (to be fair, I put a vector bar over the vector 0's, and a hat over the matrix 0's, so it wasn't quite so bad)

/end digression

Finally, P is not just any subset, P is any subset that satisfies properties P10-12.

(note: for a given ring R, there may exist several subsets P that satisfy P10-P12)

8. Sep 24, 2003

### StephenPrivitera

That rather makes me anrgy. In the chapter, the author makes a big stink about how a function is a set of ordered pairs. He emphasized, "Don't confuse the number f(x) with the function f." And now all of a sudden he defines the set of all functions whose domain is R to be a set of rules which define functions rather than a set of sets of ordered pairs. As far as my mind bends, I see no way to create an additive inverse of a set of ordered pairs. Lousy inconsistent math book.

Side note: I'm thinking maybe I can use the formal definition of a function to work out that other problem I have the the math forum. We'll see how that goes.

"a row vector of zeroes (three different sizes)"
Yesterday, I left my physics workshop early because I was upset that my TA wouldn't believe me that there was such a thing as a zero vector. He refused to believe me because I couldn't tell him what direction it pointed. I told him I thought it pointed in all directions simultaneously. Apparently, to a physicist, the existence of a zero vector is of no importance. But I know from when I studied vector calculus that it does exist. And I take your comment as proof, even though you may not be talking about what I'm talking about.

"today I was working on a problem"
When do you find time??!! You're always here helping me!

Last edited: Sep 24, 2003
9. Sep 24, 2003

### Hurkyl

Staff Emeritus
Let me give an analogy...

does not the definition "x is the unique number such that 3 * x = 6" define x just as well as "x is 2"?

A "pointwise" definition like:

for all x: (f + g)(x) = f(x) + g(x)

is as equally rigorous as

f + g = {(x, y) in R2| there exists a and b in R such that (x, a) is in f, (x, b) is in g, and a + b = y}

(both, technically, still require a proof that f + g is well-defined and is a function, but the proof is roughly identical for each)

However, it is defintely a sign of acuteness that this issue troubled you; it shows you're learning what you're being taught instead of merely accepting it. I would advise taking some time to practice translating definitions to and from ordinary notation and set of ordered pairs notation like I did above with f + g... and play with some things that fail to be functions, like defining g(x) as:

g(x)2 = x

to see just how it fails to translate.

10. Sep 25, 2003

### HallsofIvy

Staff Emeritus
Hey, if you are taking mathematics, your mind is going to have to be a lot more flexible!

If f is the set of ordered pairs {(0,1),(1,3),(2,8),(3,-5)}
then its "additive inverse" is g= {(0,-1),(1,-3),(2,-8),(3,5)}
because addition of functions is defined by "f+g is the function such that (f+g)(x)= f(x)+ g(x)".

In the above case, f+g is {(0,1+(-1),(1,3+(-3)),(2,8+(-8)),(3,-5+(5))}= {(0,0),(1,0),(2,0),(3,0)}, the "0 function" or additive identity.

Defining a function as "a set of ordered pairs such that no two pairs have the same first member" is perfectly correct and is, in fact, the standard definition.

11. Sep 25, 2003

### StephenPrivitera

Sure that makes sense. I was thinking more like there is a set -A such that A+(-A)=0, meaning that by adding two sets we obtain 0.

So in other words, defining a function by its rule and by a set are essentially equivalent. For example, f(x)=0 and f={(x,0) | x in R} are both ok as far as formality goes.

So for proving the original problem, I imagine I should do something more creative than exhausting all the possibilities of the subset P! I have class now, but I'll work on it later and let you know if I get stuck.

Thanks a lot for everything.

12. Sep 25, 2003

### StephenPrivitera

Imagine I have defined a set P so that P10-P12 hold.

Now consider the function F

F(x)=
........ / 0, x [x=] xf
........ \ f, x = xf

F is in P.

Now consider the function G

G(x)=
........ / 0, x [x=] xg
........ \ g, x = xg

G is in P.

xg[x=] xf

By P12, G*F should also be in P. But we see that
(G*F)(x)=G(x)*F(x)=
........ / 0, x [x=] xg, x[x=] xf
........| 0*f, x = xf
........ \ g*0, x = xg
=0, for all x

But 0 is not in P. So P12 cannot hold.

How's that for a start!?

Edits: _____
f and g do not equal 0

Also, suppose you argue that G and F were never really in P to begin with. Then certainly -G and -F are in P by P10 and (-G)*(-F)=G*F=0 and by P12 0 should be in P. But 0 is not in P. A similar argument works if we assert that either G or F are in P but not both. In any case k(G*F)=0 whether k=1 or k=-1. Therefore, P12 cannot hold since if it does, 0 is in P.

Further edit: Since I relied on P10 for my arguement, I should say:
If P10 holds, then P12 cannot hold.

Last edited: Sep 25, 2003
13. Sep 25, 2003

### Hurkyl

Staff Emeritus
Yep, that's right!