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Homework Help: 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


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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


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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


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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


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


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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


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

    Assume that f(pi) > 4

    f(\pi) &> 4\\
    f(\pi) &> f(4)\\
    f(\pi-\pi) &> f(4-\pi)\\
    f(0) &> f(4-\pi)\\
    0 &> f(4-\pi)

    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


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


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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.
  22. Sep 27, 2009 #21


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    No, it doesn't need to be. Sorry for the confusion, I should have been more clear. Let's try again . . .

    For definiteness, suppose that [itex]f(\alpha) > \alpha[/itex]. Now since there is a rational number between any real numbers, we can find a rational number [itex]a[/itex] such that [itex]f(\alpha) > a > \alpha[/itex]. Now, can you use this to produce a contradiction?
  23. Sep 27, 2009 #22
    Ah yes, I do see now. Much appreciated, thank you!
  24. Sep 27, 2009 #23
    To prove that f(x) > 0 if x > 0, I said that if x > 0, then it can be written as a square of some number a so that x = a^2. As long as that assumption is true, you should be able to figure it out using the properties of the function.

    The problem relies heavily on the information that you gather from the previous parts.
  25. Sep 27, 2009 #24
    Now, I'm able to prove that f(a) cannot be less than a, nor can f(a) be greater than a. Does this imply that f(a) = a? Assuming that f(a) and a exist, then they must be equal, I think. But is it valid to say that?
  26. Sep 27, 2009 #25
    Or how about this?

    Since f(a) is not less than a, and f(a) is not greater than a, then there is no rational number q between a and f(a). Therefore, a = f(a). That sounds a bit more valid...
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