# Finding a key values of a modification of Cauchy’s functional equation

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).

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