Thread: Busted functions View Single Post
2006-02-22, 12:50   #4
Dougy

Aug 2004
Melbourne, Australia

23·19 Posts

Quote:
 Originally Posted by Dougy ...for example, $f(x)=\frac{3}{2}x+\frac{1}{2}$, if $m=3$ then $f^n(m)$ will still always be a natural number.
After I posted it I noticed that this was incorrect. So I'll give another example of a function $f$ which would not be included if $f:\mathbb{N} \rightarrow \mathbb{N}$. Eg. $f(x)=2^{x-2}+1$ has the property that $f(1)$ is not a natural number, while $f(x) \in \mathbb{N}$ when $x \in \mathbb{N}$ and $x>1$.

It is possible to create a related function $g:\mathbb{N} \rightarrow \mathbb{N}$ such that $g(x)=f(x)$ whenever $f(x) \in \mathbb{N}$, and $g(x)=1$ whenever $f(x)$ is not a natural number. But then this would be asking a different question.

Last fiddled with by Dougy on 2006-02-22 at 12:52