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A little proof for functions

  • #1
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.
 
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Answers and Replies

  • #2
Hurkyl
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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
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.
 
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  • #4
Hurkyl
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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
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
HallsofIvy
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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
Hurkyl
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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 ε such that

f + ε = f

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

f + (-f) = ε

/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 ε 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
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!
 
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  • #9
Hurkyl
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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. :smile: 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
HallsofIvy
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originally posted by StephenPrivitera
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.
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
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
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.
 
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  • #13
Hurkyl
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Yep, that's right!
 

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