Let N0 denote the set of nonnegative integers. Find all functions from N0 to itself such that.
\ f(m+f(n)) = f(f(m))+f(n)\qquad\text{for all}\; m, n\in\mathbb{N}_{0}. \
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2 Answers
Shubhodip
·2011-07-16 03:27:58
I just need to prove that 1 is in the range of f.
Some work of mine may help you to help me..[3]
1) f(0) = 0
2) f(f(m)) = f(m) for all m
3) f(m+ f(n)) = f(m) + f(n)= f(f(m)+ f(n))
4) f(k(f(m)) = kf(m)
5)If 1 is in range of f we must have f(1) = 1
So as you can see, we just need to prove f(1) = 1 to get the non-constant solution or f(1) = 0 to get constant solution.
But thats the step where i am stuck...[2]
Shubhodip
·2011-07-16 03:35:31
Helping you some more
Let f(1) = p
We have f(mp) = mp
f(1+mp)= p(m+1) for all m