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Thread: Check my proof that f(x+y) = f(x) +f(y) iff f(u) = u

  1. #1 Check my proof that f(x+y) = f(x) +f(y) iff f(u) = u 
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    Additionally the problem states that f(1) = 1. It is fairly easy to show <---

    ----> My basic idea is that f(x) = f(x*1) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = 1 + 1 + ... + 1 = x

    I feel confident in my reasoning, but given that this was a challenge problem for my Algebra class the simplicity of my proof has me thinking that I left something out. Additionally, the problem gives the condition that f: Q ---> Q, an assumption which I do not appear to have used.

    (In the event that my proof is insufficient, please know that I am not at this stage seeking hints. Just need a verification right now.)


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    Would this result fail in the case that f: R ---> R ?


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  4. #3  
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    Ah nevermind, I can't write 1/2 as a sum of 1's.
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  5. #4  
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    I think you're on the right track, but like you've realized, that proof only works for integers.
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  6. #5  
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    So you want to prove that linear function is linear? Kinda redundant IMO but why not. Your proof seems a bit strange.

    edit: sorry didn`t noticed: " I am not at this stage seeking hints. Just need a verification right now.)" Hope I didn`t spoiled it for you

    Last edited by Gere; September 19th, 2013 at 06:37 AM.
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  7. #6  
    KJW
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    The "iff" means that you need to prove both the necessary and the sufficiency condition.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  8. #7  
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    The proof works for all rational numbers: p = f(p) = f(q(p/q)) = f(p/q) + ..... +f(p/q) = qf(p/q). To get to real numbers you need continuity.
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  9. #8  
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    I have pondered about functions like this when I was twelve. I believe I have proved that f(xy) = f(x)f(y) only works if f(x) = 1 for all x, but I'm not sure, maybe that was different.
    This proof only really works for positive integer inputs of x. Here are some interesting facts:
    If f(x + y) = f(x) + f(y), then f(0) = 0 necessarily, for f(y) = f(y + 0) = f(y) + f(0), and cancelling out the f(y) from the far left and right leave 0 = f(0).
    Also necessarily, f(-x) = -f(x), for f(0) = f(x + -x) = f(x) + f(-x) = 0
    Both of these properties work without setting f(1) = 1, and since they are based off the identity element and the inverse element, they might be of some use in abstract algebra.
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  10. #9  
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    Quote Originally Posted by anticorncob28 View Post
    I have pondered about functions like this when I was twelve. I believe I have proved that f(xy) = f(x)f(y) only works if f(x) = 1 for all x, but I'm not sure, maybe that was different.
    This proof only really works for positive integer inputs of x. Here are some interesting facts:
    If f(x + y) = f(x) + f(y), then f(0) = 0 necessarily, for f(y) = f(y + 0) = f(y) + f(0), and cancelling out the f(y) from the far left and right leave 0 = f(0).
    Also necessarily, f(-x) = -f(x), for f(0) = f(x + -x) = f(x) + f(-x) = 0
    Both of these properties work without setting f(1) = 1, and since they are based off the identity element and the inverse element, they might be of some use in abstract algebra.
    When you omit f(1) = 1, then the solution is f(x) = cx for all rational x (c constant = f(1)). As I mentioned before, you need continuity to get it for real numbers.
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  11. #10  
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    Nevermind. I was going to say something, but I realized it doesn't actually work.
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  12. #11  
    KJW
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    If for all and arbitrary constant , then:



    If for all and , then:

    by definition of

    Therefore, since is a constant.

    Therefore, .

    But and therefore , and thus .

    Therefore, ... QED
    anticorncob28 and patangua like this.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  13. #12  
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    KJW, dude, let him figure this out by himself

    Quote Originally Posted by oldfold View Post
    (In the event that my proof is insufficient, please know that I am not at this stage seeking hints. Just need a verification right now.)
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  14. #13  
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    Mmmm...
    Quote Originally Posted by KJW View Post
    If for all and arbitrary constant , then:


    Blah, blah, blah

    Therefore, ... QED
    Why am I not surprised?
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  15. #14  
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    @Guitarist:

    KJW proved both directions, meaning that

    1.) if f(x) = kx, then f(x+y) = f(x) + f(y) for all x,y
    2.) if f(x+y) = f(x) + f(y) holds true for all x,y, then it follows f(x) = k*x for some constant k.

    However, @KJW, you put the assumption that f(x) be differantiable into it. Further on, I think you need to let , since for , it is rather true that

    with
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  16. #15  
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    It is not necessary to assume differentiability, continuity is enough. Show that it goes both ways for all rational numbers. Extend to reals using continuity.

    Without continuity, f(x+y) = f(x) + f(y) will not imply f*x) = kx. The construction is fairly high level, using something called the Hamel basis.
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  17. #16  
    Forum Freshman holysword's Avatar
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    Quote Originally Posted by oldfold View Post
    Additionally the problem states that f(1) = 1. It is fairly easy to show <---

    ----> My basic idea is that f(x) = f(x*1) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = 1 + 1 + ... + 1 = x

    I feel confident in my reasoning, but given that this was a challenge problem for my Algebra class the simplicity of my proof has me thinking that I left something out. Additionally, the problem gives the condition that f: Q ---> Q, an assumption which I do not appear to have used.

    (In the event that my proof is insufficient, please know that I am not at this stage seeking hints. Just need a verification right now.)
    You (almost) showed that if f(x+y) = f(x) +f(y) then f(u) = u.
    Since you need to prove the "iff", now you need to show that f(u) = u implies f(x+y) = f(x) +f(y). It is also a straighforward proof though.
    The Q -> Q is needed to guarantee that f(u)=u is possible at all. If f would be a function from R to N (natural numbers) for instance, f(3.1) could not be 3.1.
    "Nolite arbitrari quia venerim mittere pacem in terram non veni pacem mittere sed gladium"
    Yeshua Ha Mashiach
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  18. #17  
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    One can show that if f is a continuous function such that f(x+ y)= f(x)+ f(y) then f(x)= Cx for some number C. The condition that f(1)= 1 requires that C= 1 so that f(x)= x.

    HOWEVER, there exist non-continuous functions satisfying those conditions. One way to "construct" them is:
    Think of the real numbers as a vector space over the field of rational numbers. Then we can construct a basis for the vector space. Assign values of f(x), for x a "basis" number any way like, then extend it to all real numbers "by linearity". That provides that f(x+ y)= f(x)+ f(y) as well as f(ax)= af(x) for a any rational number.

    Of course, since the set of real numbers is uncountable while the set of rational numbers is irrational, this basis must be uncountable. There is no good way to write it but we can, for instance, take one of the "basis" numbers to be 1 (or any other single rational number), another to be, say, sqrt(2). Defined f(1)= 0, say, (so f(a)= 0 for a any rational number) and f(sqrt(2))= 1. However you define it for "basis" numbers, that will give a function satisfying f(x+ y)= f(x)+ f(y), f(1)= 1, but NOT f(x)= x
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  19. #18  
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    p = f(p) = f(q(p/q)) = f(p/q) + ..... +f(p/q) = qf(p/q)
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