Math Help Forum - View Single Post

math sucks · #5 November 20th, 2007, 07:21 AM

Let $n$ be a positive integer. Then
$f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^n$
So
$f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}$
Now let $\frac{m}{n}$ be a positive rational number. Then
$f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f(1)^{\frac{m}{n}}$
Now let $x$ be a positive real number. By continuity,
$f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x$
where the last result is established by our definition of $f$ over the rational numbers.
Assume $f(1)\neq0$ , then it is trivial that $f(0)=1$ . Thus, for any positive real number $y$ ,
$f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}$
Therefore, $f(x)=f(1)^x$ for all $x$ or assume $f(1)=0$ , then it is trivial that $f(x)=0$ for all $x$