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Proof of f(x) = x

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that f(x) = x (for x is rational) if f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y).


    2. Relevant equations



    3. The attempt at a solution
    I substituted y = x to get f(2x) = f(2x), which means that you can pull constants out of the function. Therefore, for all x and y, f(xy)=xy and so f(x) = x and f(y) = y.

    However, I am not sure that this solution works because it seems to simple and illogical in a way. I also didn't prove that x is rational.
     
  2. jcsd
  3. Sep 26, 2009 #2

    gabbagabbahey

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    Is this a typo? Don't you mean f(2x)=2f(x)?
     
  4. Sep 26, 2009 #3
    Sorry, I meant f(2x) = 2f(x). Thanks.
     
  5. Sep 26, 2009 #4

    HallsofIvy

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    I would be inclined to work with the various number systems. For example, you should be able to use induction to show that f(nx)= nf(x)= f(n)f(x) for any positive integer n and thus that f(n)= n.
    Then use the fact that f(n+0)= f(n)+ f(0) to show that f(0)= 0.
    Use the fact that f(-n)= f((-1)(n))= nf(-1) while f(1- 1)= f(1)+ f(-1)= f(0)= 0 to get f(-n)= -n.

    Finally, use the fact that f(1)= f(m(1/m))= mf(1/m)= 1 to get rational numbers.

    By the way, if you add the requirement that f is continuous for all x, then f(x)= x for all real numbers. Without that added requirement, that is not true.
     
  6. Sep 26, 2009 #5

    D H

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    As gabbagabbahey noted, surely you meant f(2x)=2f(x). All that means you can take a factor of 2 out of the argument. It does not necessarily mean you can "pull constants out of the function". BTW, if this were a valid step, why take the extra step of going to f(xy)=xy, etc.? You have already proved the assertion! (Invalidly, of course).
     
  7. Sep 26, 2009 #6
    Thanks for the reply. So in general this would prove that f(x) = x, since I start with the base case (f(0) = 0), then prove it for positive integers using the induction method, then for negative integers, then rational numbers. Aren't the proofs a little redundant? I don't quite see the point in proving negative integers for rational numbers - is it to ascertain that the function holds for negative rational numbers as well? There's also a hint in there to use the fact that rational numbers exist between any number (in order to prove f(x) = x for all x). So wouldn't this have already proven that?
     
  8. Sep 26, 2009 #7
    I had wanted to prove that f(x) = x in general by using that specific example, but I guess I would have to work through all the number systems for that.
     
  9. Sep 26, 2009 #8
    Would it be possible to prove that for irrational numbers? (The exact same thing, except defining n as an irrational number?)
     
  10. Sep 27, 2009 #9
    are you by any chance in regina rotman's class at u of t?
     
  11. Sep 27, 2009 #10

    Office_Shredder

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    You can prove that f(x)=x for all algebraic numbers (essentially, f(a polynomial of x) = the polynomial evaluated at f(x)) but for transcendental numbers it seems you can define f(x) to be something different. There was another thread about the same function a couple days ago that covered the details, I'll see if I can find the link

    EDIT:
    https://www.physicsforums.com/showthread.php?t=339490

    Here it is
     
  12. Sep 27, 2009 #11
    Thanks, I'll try that.
     
  13. Sep 27, 2009 #12

    Hurkyl

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  14. Sep 27, 2009 #13
    This question is part of my assigned homework too, and I can't figure out how to prove that f(x) = x for irrational numbers. The book has a hint that says "use the fact that between any two numbers there is a rational number".

    In previous parts of the question, I've already proved that:
    f(1) = 1,
    f(x) = x if x is rational,
    f(x)>0 if x>0, and
    f(x)>f(y) if x>y.

    So I'm assuming that there is an easy way to prove this for the irrational numbers, using only these four facts, the two properties of the function [ f(xy) = f(x)f(y) and f(x+y) = f(x)+f(y) ] and the hint that between any two numbers, there exists a rational number.

    Any suggestions?
     
  15. Sep 27, 2009 #14

    Hurkyl

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    Can f(pi) be bigger than 4? 3.2?
     
  16. Sep 27, 2009 #15
    Well...

    Assume that f(pi) > 4

    [tex]
    \begin{align*}
    f(\pi) &> 4\\
    f(\pi) &> f(4)\\
    f(\pi-\pi) &> f(4-\pi)\\
    f(0) &> f(4-\pi)\\
    0 &> f(4-\pi)
    \end{align*}
    [/tex]

    but this contradicts the statement that if x > 0, f(x) > 0. So f(pi) cannot be greater than 4.

    But I'm not sure how that helps me. Or if what I just did there is valid.
     
  17. Sep 27, 2009 #16
    I guess we can sort of generalize it that f(pi) will never be larger than f(x) when x is a rational number larger than pi?
     
  18. Sep 27, 2009 #17

    Hurkyl

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    That sounds correct. Can you find a less convoluted way of saying it?
     
  19. Sep 27, 2009 #18

    jgens

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    Your proof can be simplified considerably: since [itex]\pi < 4[/itex] we know that [itex]f(\pi) < f(4) = 4[/itex]. It's really the same argument, it just uses another fact that you've already proven.

    To help you complete the proof, suppose that [itex]f(\alpha) \neq \alpha[/itex] for some [itex]\alpha[/itex] irrational. Then we have that [itex]f(\alpha) = \alpha + \epsilon[/itex] for some [itex]\epsilon \neq 0[/itex]. Can you see how this produces a contradtion?
     
  20. Sep 27, 2009 #19
    is the epsilon considered to be a rational number?
     
  21. Sep 27, 2009 #20
    i'm sorry, i'm still confused. i can't see how it produces a contradiction.
     
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