Let

begin{align}f(x+y) &= f(x) + f(y), \

f(xy) &= f(x) f(y).

end{align}

(1) Prove that $f(0) = 0$ and that $f(1) = 0$ or $1$.(2) Prove that $f(n) = n f(1)$ for every integer $n$ and then that $f(n/m) = (n/m) f(1)$ for all integers $n,m$ such that $m neq 0$. Conclude that either $f(q) = 0$ for all rational numbers $q$ or $f(q) =q$ for all rational numbers $q$.

(3) Prove that $f$ is nondecreasing, that is to say that $f(x) geq f(y)$ whenever $x geq y$ for any real numbers $x$ and $y$.

**(1)** Let $x=y=0$. Then equation (1) gives us $$f(0)=2f(0)$$ which means $boxed{f(0)=0}$. Now let $x=y=1$. Equation (2) gives us $$f(1)=f(1)^2$$ which means $boxed{f(1)=0 text{ or } 1}$.

**(2)** $f(n) = f(underbrace{1+1+…+1}_textrm{n times}) = underbrace{f(1)+f(1)+…+f(1)}_textrm{n times} = n f(1)$. Also, $f(n/m)=f(n) f(1/m) = n f(1) f(1/m)$, but I don’t know how to show that $f(1/m)=1/m$. **Any hints as to how I should proceed?**

**(3)** **How do I model real numbers in this scenario? Our identity functions for $f(n)$ and $f(m/n)$ only apply for naturals and rationals respectively, and I don’t think I can talk about reals in terms of products of rationals (much less naturals).**