Take log: $\lg f(mn)=\lg f(m)+\lg f(n)$. Let $g(x)=\lg f(x)$, then $g(xy)=g(x)+g(y)$, so $g(x) = \log_a x$ Since: $f(2)=2$, so $g(2)=\lg 2 = \log_a 2$. Hence $a=10$. Therefore: $g(x)=\lg x = \lg f(x)$. Thus $f(x)=x$. Is that correct?

shobber, the function is not real-valued, and you have to use the hypothesis $f(n) < f(n+1)$ otherwise function $f(2^n (2*m+1)) = 2^n$ (for instance) would work. $f(1) = 1$, $f(2^n) = 2^n$, $f(2) < f(3) < f(4)$ so $f(3) = 3$ It's also clear that $f(n) \geq n$ If n is the smallest integer such as $n < f(n)$. If $n=2N$ then $f(n) = 2f(N) = 2N$ contradiction. If $n=2N+1$ then $f(2N)=2f(N)=2N < f(n) < f(2N+2) = 2(N+2)$ and $f(n) = 2N+1=n$ contradiction.

What is real valued? Why my proof was wrong because it is real valued?

We can use Cauchys equations

seamusoboyle wrote: We can use Cauchys equations That is what I did.

Ooops, so you did

shoober, seamusoboyle, do you mean that : $f(mn) = f(n)f(m), \forall m, n \in N_0 \Rightarrow f(n) = n^a$ ? If yes, this is false : Take $f(2n) = f(n)$ and $f(2n+1) = 2n+1$ for instance.

I've never heard of Cauchyequations but this might just do it. Because of (1) and (2) we get $f(0)=0$ and $f(1)=1$. Now from (3) we get $f(4)=f(2\cdot2)=4$, which leaves only one possibility for $f(3)$, that is $f(3)=3$. Continuing this way, we get that $f(n)=n$ is the only possible answer, unless i made a mistake somewhere.

K.W.M.A.N wrote: I've never heard of Cauchyequations but this might just do it. Because of (1) and (2) we get $f(0)=0$ and $f(1)=1$. Now from (3) we get $f(4)=f(2\cdot2)=4$, which leaves only one possibility for $f(3)$, that is $f(3)=3$. Continuing this way, we get that $f(n)=n$ is the only possible answer, unless i made a mistake somewhere. Hi Karsten, welcome to MathLinks! Your solution is correct (or at least it's the same as mine) but I have a small style remark: for making mathematical proofs more clear, it's better not to write "and continue like that" or something alike. (on a competition, the graders cannot know if you really saw the trick and will not give you full marks) When possible, try to word it somehow like this: Quote: First, we will show that $f(2^k)=2^k$ for all $k\in\mathbb{N}$. Proof by induction: $f(1)=1, f(2)=2, f\left(2^k\right)=2f\left(2^{k-1}\right)$. Now, we have that $f(n)=n$ for all $n\in\mathbb{N}$. Between $2^{k-1}$ and $2^k$ we have $2^{k-1}-1$ numbers and $2^{k-1}-1$ different values the function can attain, thus we have a unique way to sort them ascending. As $f(n)=n$ is a valid way, this is our unique solution.

KWMAN are my initials in case you were wondering. It appears that I was not so clear indeed because your proof is slightly different from mine: I'll try to explain what I meant with "continue like this", I use induction of course: Suppose that i have shown already that $f(n)=n$ for $n<2k+1$. Then I will proof that $f(n)=n$ for $n<2(k+1)+1=2k+3$. For $n=2k+2$ is $f(n)=2f(k+1)=2(k+1)$, because $k+1< 2k+1$ ($k \neq0$). But now is $2k=f(2k)<f(2k+1)<f(2k+2)=2k+2$ so $f(2k+1)=2k+1$. I think the difference is that I used this order 2,(0,1),4,3,6,5,8,7,&c while you used this one 2,(0,1),4,3,8,(5,6,7),16,(9,10,11,12,13,14,15),32,&c,

shobber wrote: Let $N_0=\{0, 1, 2 \cdots \}$. Find all functions: $N_0 \to N_0$ such that: (1) $f(n) < f(n+1)$, all $n \in N_0$; (2) $f(2)=2$; (3) $f(mn)=f(m)f(n)$, all $m, n \in N_0$. we can prove that $f(2^k)=2^k,k \in N_0$ and $f(n+1)>f(n)$ $\rightarrow$ we have $f(2^k)<f(2^k+1<...<2^{k+1}-1<2^{k+1} \rightarrow$ we know that $f(n) \in N_0$ $\rightarrow$ the numbers between $f(2^k)$ and $f(2^{k+1}$ should be the numbers between $2^k$ and $2^{k+1}$ $\rightarrow$ $f(n)=n$(?)

jensen wrote: shobber wrote: Let $N_0=\{0, 1, 2 \cdots \}$. Find all functions: $N_0 \to N_0$ such that: (1) $f(n) < f(n+1)$, all $n \in N_0$; (2) $f(2)=2$; (3) $f(mn)=f(m)f(n)$, all $m, n \in N_0$. we can prove that $f(2^k)=2^k,k \in N_0$ and $f(n+1)>f(n)$ $\rightarrow$ we have $f(2^k)<f(2^k+1<...<2^{k+1}-1<2^{k+1} \rightarrow$ we know that $f(n) \in N_0$ $\rightarrow$ the numbers between $f(2^k)$ and $f(2^{k+1}$ should be the numbers between $2^k$ and $2^{k+1}$ $\rightarrow$ $f(n)=n$(?) isn't that exactly what I wrote?

oh yes,i didn't see your solution,sorry.