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How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for all re

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    How to show that there exist no function f,g:R[itex]\nearrow[/itex]R satisfying f(x)g(y)=x+y, for all real x,y?


    2. Relevant equations

    Any equation for showing function?

    3. The attempt at a solution

    how about starting with sumthing like let h(x,y)=x+y ??
     
    Last edited: Jan 11, 2013
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  3. Jan 11, 2013 #2

    HallsofIvy

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    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    I would recommend exploring what would have to be true if such a formula were true. For example, if f(x)g(y)= x+ y, then f(0) g(y)= y for all y so if we set a= f(0), and [itex]a\ne 0[/itex], g(y)= y/a. Similarly, f(x)g(0)= x so if b= g(0), and [itex]b\ne 0[/itex], then f(x)= x/b. That means that we must have f(x)g(y)= xy/ab for some numbers a and b, provided f(0) and g(0) are not equal to 0. You will need to determine what happens if f(0)= 0 and g(0)= 0.
     
  4. Jan 11, 2013 #3

    mfb

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    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    You could try to construct f(x) and g(y) via their derivatives and lead this to a contradiction.
    Alternatively, consider f(0)g(a), f(a)g(0) and so on.
     
  5. Jan 11, 2013 #4

    Ray Vickson

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    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    This assumes continuity, differentiability, etc., which was not part of the problem statement.
     
  6. Jan 11, 2013 #5

    mfb

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    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    We know that f(x)g(y) is differentiable. Anyway, we don't need it, as the result of the integral is already there:

    Consider ##x\neq-y## => ##g(y)\neq 0##:
    f(x)=(x+y)/g(y) (you could derive both sides with respect to x here)
    That leads to the same concept as HallsofIvy proposed.
     
  7. Jan 12, 2013 #6
    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    so, i get this...
    a=0 when [itex]b\ne 0[/itex]
    b=0 when [itex]a\ne 0[/itex]
    ab=0 when [itex]a\ne 0[/itex] and [itex]b\ne 0[/itex]
    then, i get nothing here... so, what does it mean? Am i approaching the wrong way?
     
  8. Jan 13, 2013 #7

    mfb

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    Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

    There is your contradiction, if you really know this.
     
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