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 Properties of $$f: \mathbf{D} \to \mathbf{R}$$ that ensures $$f \; |_{\mathbf{N}} = I_{\mathbf{N}}$$
02-16-2010, 03:40 PM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Properties of $$f: \mathbf{D} \to \mathbf{R}$$ that ensures $$f \; |_{\mathbf{N}} = I_{\mathbf{N}}$$
(1) $$\mathbf{N}\subset \mathbf{D}$$
(2) $$f(\mathbf{N})\subset\mathbf{N}$$
(3) $$f(2)=2$$
(4) $$f(mn) = f(m)f(n)$$
(5) $$(m<n)\Rightarrow (f(m)<f(n))$$
The properties are clearly necessary. We need to prove that they are also sufficient.

By (3)and(4), $$2=f(2)=f(1\cdot 2)=f(1)f(2)=2f(1)$$ and so $$f(1)=1$$

If we consider $$0 \in \mathbf{N}$$, then $$f(0)=0$$ since $$1=f(1)>f(0)=f(0)f(0)=f^{2}(0)\geq 0$$ by $$(2\sim4)$$

If (1)$$\sim$$(5) is not sufficient, let $$m \in \mathbf{N}$$ be the smallest one such that $$f(m) \neq m$$, then $$m$$ is not $$0,1$$ and must not be even. otherwise $$f(m/2) = \frac{1}{2} f(2)f(m/2)=\frac{1}{2}f(m)\neq m/2$$ will contradicts the minimality of $$m$$.

So $$m=2k+1$$ for some $$k \in \mathbf{N}^{+}$$ and $$f(q) = q$$ whenever $$q \in \mathbf{N}$$ and $$q <2K+1$$. Therefore
$m-1 = f(m-1)<f(m)<f(2(k+1))=f(2)f(k+1)=2(k+1)=m+1$This forces $$f(m) = m$$ by (2). and we see there is no such an $$m$$.
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