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

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

Dick

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

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

Because I was missing the obvious looking too deeply at a trivial case.

--Elucidus

Elucidus

I believe this is how those whacked out discontinuous extensions alluded to earlier are made.

--Elucidus

Dick

Nice to know smart people have been looking at this and haven't found some obvious proof I've been completely blind to.

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?

Dick

Ooops. Guess I had a false recollection. Thanks for getting this thread on the right track again.

Elucidus

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

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

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

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

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

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

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

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).