Functional analysis?

Question

Find all one-one and onto functions f:A-A such that
f(m+f(n))=f(f(m))+f(n) hols for all m and n in A.

You can not assume n=0 as n E N and 0 need not be in N

U CANNOT CUT AND PASTE SOMEBODY ELSE'S ANSWERS, ENEMY!!!

Answer 1

enemy, if you copy & paste my answers again I will report you.

All right, here's the general answer. In what follows, f²() = f(f()).

Since f is one-on-one and "onto", we can define g as the inverse of f. Therefore,
f(m+n) = f(m+f(g(n)) = f²(m) + f(g(n)) = f²(m) + n for all m, n.

Now let's look at f(n+n+n).
(1) : f(n+n+n) = f²(n+n) + n = f(f²(n) + n) + n. But also,
(2) : f(n+n+n) = f²(n) + (n + n)

The RHS's of (1) and (2) being equal,
f(f²(n) + n) + n = f²(n) + n + n
---> f(f²(n) + n) = f²(n) + n
---> f(n + f²(n)) = f²(n) + n
---> f²(n) + f²(n) = f²(n) + n
---> f²(n) = n for all n.

Now, as we've seen,
f(m+n) = f²(m) +n
---> f(m+n) = m+n.

So. If n is the sum of two elements of A, then f(n)=n.
If f(n)=m (different than n), then three conditions must be met:
(1) : n cannot be the sum of two elements of A.
(2) : f(m)=n to ensure f²(n)=n
(3) : m also cannot be the sum of two elements of A (otherwise we would have f(m)=m, not n).

In short, f must be the identity function except that we may have pairs of elements m, n E A, that are not sums of elements of A, f(m)=n and f(n)=m. For example, if A = {2, 3, 4, 5, 6...} we can have f(2)=3, f(3)=2, f(n)=n for n>3.

If 0 E A, only the identity function works because f(n) = f(n+0) = n+0 = n.

Answer 2

If f(m + f(n)) = f(f(m)) + f(n), then

(1) f(m+n) = f(f(m)) + n

Proof: F being one-to-one and "onto", ther is an y E A such that
n=f(y), so f(m+n) = f(m + f(y)) = f(f(m)) + f(y) = f(f(m)) + n.

Let's look at the case where n=0:
f(m) = f(m + 0)
= f(f(m)) + 0 by (1)
= f(f(m)).

So (2) f(f(m)) = f(m) for all m. Now, for every m there is an x such that f(x)=m, therefore

f(m) = f(f(x)) = f(x) by (2) = m

Therefore the identity function is the only one that will do.