# 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
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P: 24,454
 Quote by Elucidus 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?
 Mentor P: 4,182 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)
P: 286
 Quote by Dick 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
P: 286
 Quote by Office_Shredder 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
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 Quote by Elucidus 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.
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Oh bah, I should have been following this thread.

 Quote by Elucidus Additioinally for $$x \geq 0, F(x) \geq 0.$$ 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?
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 Quote by Hurkyl 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.
 P: 286 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
 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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 Quote by Dick 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.
 P: 720 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.
P: 286
 Quote by JG89 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

$$u \leq v \Rightarrow f(u) \leq f(v)$$

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
 HW Helper Sci Advisor Thanks P: 24,454 I screwed up most of all by throwing a monkey wrench into peoples thinking by suggesting it might not be true. Mea culpa again.
 P: 19 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
 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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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