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Function Proof . . .

  1. Aug 9, 2009 #1

    jgens

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    1. The problem statement, all variables and given/known data

    Prove that if [itex]f[/itex] is the function which is not always zero, that satisfies [itex]f(x+y)=f(x)+f(y)[/itex] and [itex]f(xy)=f(x)f(y)[/itex], we have that [itex]f(x)>0[/itex] if [itex]x>0[/itex]

    2. Relevant equations

    So far I've managed to prove that [itex]f(x)=x[/itex] if [itex]x \in \mathbb{Q}[/itex] and that [itex]f[/itex] must be odd.

    3. The attempt at a solution

    Suppose not, then if [itex]x > 0[/itex] and irrational we have that [itex]f(-x) > 0 > f(x)[/itex]. Since any rational number [itex]b > 0[/itex] can be expressed as the sum of two irrational numbers - [itex]x + (b-x)[/itex] for instance - we have that [itex]b = x + y > 0 [/itex] where [itex]x,y[/itex] are irrational. This implies that,

    [itex]b = f(b) = f(x+y) = f(x) + f(y) > 0[/itex]

    Clearly, both [itex]x,y[/itex] cannot be negative since this would imply that [itex]x+y < 0[/itex] a contradiction. We also have [itex]x,y[/itex] cannot both be positive since this would imply that [itex]f(x)+f(y)<0[/itex] another contradiction.

    I'm not positive that any of this is correct (probably isn't) and I would appreciate any corrections along with suggestions on how to complete the proof. Thanks!
     
  2. jcsd
  3. Aug 9, 2009 #2

    tiny-tim

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    Hi jgens! :smile:

    Hints: what is f(√x)?

    If f(y) = 0, what is f(x/y)?
     
  4. Aug 9, 2009 #3
    I can't really follow your logic in 3). You first assume we have one positive irrational number x which is a counterexample to the problem statement. You then choose an arbitrary positive rational number b, and consider the irrational number y=b-x (about which you know very little). You say that x, y can't both be negative which is correct (since b is positive and also since by definition x > 0 so that can't be negative), but I don't see why we couldn't have y < 0 < x. x+y could still be positive if the absolute value of x is greater than that of y, and while f(x) < 0 you don't necessarily know that f(y) < 0 (remember you only assumed that there was a contradiction at one point x, not at all irrational points).

    I think you're over complicating the problem slightly. If x > 0 we can write it as [itex]x = a^2[/itex] for some a > 0. We then have:
    [tex]f(x) = f(aa) = f(a)^2[/tex]
    Also note that if [itex]a \not = 0[/itex] then a has a multiplicative inverse so:
    [tex]1 = f(1) = f(a)f(1/a)[/tex]
    Try to see if you can show f(x) > 0 from this. You don't need a complete formula for f(x).
     
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