Find all functions $f : N → N$ such that
(a) $f(2) = 2$
(b) $f(mn) = f(m)f(n)$ for all $m, n ∈ N$
(c) $f(m) < f(n)$ for $m < n$
First, I substituted $m=1,n=2$ to get $f(1)=1$. Next, we can easily notice that all powers of $2$ will be equal to themselves. That is $f(4)=4,f(8)=8$, and so on. Now, the next step I'm not so sure is correct. As $f(4)>f(3)>f(2)$, and $f : N → N$, I think $f(3)$ can only be $3$ but again I'm not so sure. If it is so, then I believe the only possible function is $f(x)=x$.
Now for the next part of the problem-
What happens if the third condition is not given to us?
Unfortunately I don't even have the answer to the problem let alone a solution. Any hints would also be helpful, thanks.
More easily :
if $f(1)=1$ and $f(2^n)=2^n$, and because you have $$1 =f(1) < f(2) < f(3) < f(4) < ... < f(2^n)=2^n$$
the only possibility is that $f(2)=2$, $f(3)=3$, $f(4)=4$ and so on.