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Find F: R>R satisfying F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y) 
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#19
Sep2309, 11:24 PM

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#20
Sep2309, 11:26 PM

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Using the properties given, F(1) nonzero 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) 


#21
Sep2309, 11:27 PM

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Elucidus 


#22
Sep2309, 11:32 PM

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Elucidus 


#23
Sep2309, 11:34 PM

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#24
Sep2709, 03:01 PM

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Oh bah, I should have been following this thread.
Since you can say something about positive numbers, doesn't that suggest you can you say something about the ordering of the real numbers? 


#25
Sep2709, 04:33 PM

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#26
Sep2709, 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 


#27
Sep2809, 03:08 AM

P: 69

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 


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Sep2809, 03:19 AM

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#29
Sep2809, 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. 


#30
Sep2809, 11:05 PM

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[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 


#31
Sep2809, 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.



#32
Aug1311, 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



#33
Aug1311, 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 nonzero 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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