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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
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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