find F: R->R satisfying F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y)


by iN10SE
Tags: analysis, function, real analysis
Dick
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#19
Sep23-09, 11:24 PM
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Quote Quote by Elucidus View Post
I've managed to prove that if F(1) = 0 then F is periodic with period 1 and is therefore F identically 0.

I also have that if F(1) is not 0 then F(x) = x for all rationals. But I haven't made the final jump to all real numbers yet (it may not be possible).

--Elucidus
If F(1)=0 then F(x)=F(1*x)=F(1)*F(x)=0*F(x)=0. Why do you need periodic?
Office_Shredder
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#20
Sep23-09, 11:26 PM
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Using the properties given, F(1) non-zero means F(1)=1 and from that you can prove all algebraic numbers are mapped to themselves. I'm almost 100% sure you can't glean anything from the transcendentals this way, because all you can do is form rational polynomials with addition and multiplication, and the algebraic numbers are exactly those that you can find as roots to rational polynomials.

New idea: Suppose F(pi) = c. What other values of x have we determined F(x) for? Obviously any rational multiple of pi, and any linear combination of integral powers of pi. I think that's it. So you can probably choose a basis of R over Q (here's the axiom of choice), make an equivalence relation based on which guys are multiples of each other, and then on the set of equivalence classes define F to be whatever the hell you want by defining F(basis vector)
Elucidus
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#21
Sep23-09, 11:27 PM
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Quote Quote by Dick View Post
If F(1)=0 then F(x)=F(1*x)=F(1)*F(x)=0*F(x)=0. Why do you need periodic?
Because I was missing the obvious looking too deeply at a trivial case.

--Elucidus
Elucidus
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#22
Sep23-09, 11:32 PM
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Quote Quote by Office_Shredder View Post
New idea: Suppose F(pi) = c. What other values of x have we determined F(x) for? Obviously any rational multiple of pi, and any linear combination of integral powers of pi. I think that's it. So you can probably choose a basis of R over Q (here's the axiom of choice), make an equivalence relation based on which guys are multiples of each other, and then on the set of equivalence classes define F to be whatever the hell you want by defining F(basis vector)
I believe this is how those whacked out discontinuous extensions alluded to earlier are made.

--Elucidus
Dick
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#23
Sep23-09, 11:34 PM
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Quote Quote by Elucidus View Post
Because I was missing the obvious looking too deeply at a trivial case.

--Elucidus
Nice to know smart people have been looking at this and haven't found some obvious proof I've been completely blind to.
Hurkyl
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#24
Sep27-09, 03:01 PM
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Oh bah, I should have been following this thread.

Quote Quote by Elucidus View Post
Additioinally for [tex]x \geq 0, F(x) \geq 0.[/tex]

I'm not sure how to leverage this to a solution but I feel it's nearby but eluding me. I'm not even sure if any of this is relevant.
Okay, so you've proven F maps positive numbers to positive numbers.

Since you can say something about positive numbers, doesn't that suggest you can you say something about the ordering of the real numbers?
Dick
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Sep27-09, 04:33 PM
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Quote Quote by Hurkyl View Post
Oh bah, I should have been following this thread.


Okay, so you've proven F maps positive numbers to positive numbers.

Since you can say something about positive numbers, doesn't that suggest you can you say something about the ordering of the real numbers?
Ooops. Guess I had a false recollection. Thanks for getting this thread on the right track again.
Elucidus
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#26
Sep27-09, 09:01 PM
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Sheesh. I've been so wrapped up in looking for a simple proof by contradiction that I completely missed monotonicity. This was that last piece I needed to complete the proof.

--Elucidus
Caesar_Rahil
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#27
Sep28-09, 03:08 AM
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It has been proven that F(0)=0 and F(1)=1 or 0
put x=-y;
F(x)+F(-x)=0
so F(x) is odd

BELOW PROOF WORKS ONLY IF X IS RATIONAL:

now, if x is Natural number
F(x)
=F(1+1+1+... x times)=F(1)+F(1)+....x times
=x F(1)
=x or 0
since F(x) is odd; so is true for all negatives natural numbers
let x=p/q
p and q are integers
F(p)
=F(p/q*q)
=F(p/q)+F(p/q)+....q times
=q * F(p/q)
F(p/q) = F(p)/q
F(p/q)=p/q or 0
so F(x) = x or 0

I am trying for irrationals
Hurkyl
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#28
Sep28-09, 03:19 AM
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Quote Quote by Dick View Post
Ooops. Guess I had a false recollection. Thanks for getting this thread on the right track again.
That's because it's a sneaky problem! If you know the pattern of generating counterexamples in similar situations, then it's just so utterly obvious that it works here. It just happens to be utterly false.
JG89
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#29
Sep28-09, 08:02 PM
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Spivak gives the following approach for the problem:

1) Prove f(1) = 1
2) Prove f(x) = x for rational x
3) Prove f(x) > 0 if x > 0
4) Prove if x > y then f(x) > f(y)
5) Prove f(x) = x for irrational x (Use the fact that the rationals are dense on R)

The question isn't insanely hard if you follow the above procedure.
Elucidus
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#30
Sep28-09, 11:05 PM
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Quote Quote by JG89 View Post
Spivak gives the following approach for the problem:

1) Prove f(1) = 1
2) Prove f(x) = x for rational x
3) Prove f(x) > 0 if x > 0
4) Prove if x > y then f(x) > f(y)
5) Prove f(x) = x for irrational x (Use the fact that the rationals are dense on R)

The question isn't insanely hard if you follow the above procedure.
Yeah. I knew I needed (5) but blew it on (4). Monotonicity was staring at me, taunting me. I went and looked back at my scratch work and right there on page 2 is

[tex]u \leq v \Rightarrow f(u) \leq f(v)[/tex]

I will mention though that (1) and (2) should be

1) Prove that f(1) = 0 or 1.
2) Prove that if f(1) = 0 then f(x) = 0 for all x, and
that if f(1) = 1 then f(x) = x for all rational x.

--Elucidus
Dick
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#31
Sep28-09, 11:19 PM
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I screwed up most of all by throwing a monkey wrench into peoples thinking by suggesting it might not be true. Mea culpa again.
ashok vardhan
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#32
Aug13-11, 04:46 AM
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intuitively, the solution is f(kx)=kf(x). i am able to prove it for all integers k.but for a rational number k i am facing problem.so i assumed k to be p/q and later that i cant proced.would you give me a small hint
HallsofIvy
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#33
Aug13-11, 06:05 AM
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Given that f(x+ y)= f(x)+ f(y), you can prove, by induction on the positive integer n, that f(nx)= nf(x), x any real number. From that, for n a non-zero integer, f(n/n)= nf(1/n)= f(1)= 1. Thus, f(1/n)= 1/f(n). You can get the case for f(m/n) from that.

This is an entirely different problem from before, isn't it? Requiring that f be continuous at x= 0 rather than that f(xy)= f(x)f(y).


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