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Find a bijection $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $f\left( x\right) \neq x$ for all $x\in\mathbb{N}$, $f^{2}=id$.

My answer is:

Let $f:=\begin{cases} 0,x=1\\ 1,x=0\\ x,x>1\end{cases}$

Thus, $f$ is not identity but $f^2$ is identity.

Can you check my answer?

2 Answers2

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As Catalin pointed out in the comments, your $f$ is not correct.

Let $f(2k+1)=2k$ and $f(2k)=2k+1$.

Fimpellizzeri
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You've got a great idea, but you've failed the requirement that $f(x)$ must always be something different than $x.$ However, you can adapt the approach you've already taken to fix it. How can you fix the fact that $f(2)=2$ and $f(3)=3$ simultaneously? How can you fix the fact that $f(4)=4$ and $f(5)=5$ simultaneously? Does this give you an idea for how to fix everything? Can you come up with a piecewise formula to accomplish this?

Cameron Buie
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  • Let $f(2x+1)=2x$ and $f(2x)=2x+1$. Thus, $f(x)\neq x$ for all $x$ in $\mathbb{N}$. But, $f^2(2x+1)=2x+1$. I.e., $f^2$ is identity. –  Dec 30 '16 at 02:51
  • Is it enough?.. –  Dec 30 '16 at 02:51
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    Well, if you've proved that all natural numbers have either the form $2x+1$ or $2x$ for some natural number $x$ (you should specify that $x$ is a natural number, by the way), then it should be fine. It wouldn't hurt to point out that $f^2(2x)=2x,$ too. Alternately, a piecewise formula of the type $$f(x)=\begin{cases}g(x)&x\in A\h(x)&x\in B\end{cases}$$ makes the rest of the proof almost trivial. All you have to show is that $g(h(x))=x$ for all $x\in B,$ $h(g(x))=x$ for all $x\in A,$ $A\cap B=\emptyset,$ and $A\cup B=\Bbb N.$ In this case, you've probably already proved that last two facts. – Cameron Buie Dec 30 '16 at 03:00
  • Let $g(x)=x$ for all $x$ in {$0,2,4,...,2n$:$n\in\mathbb{N}$}$=A$ and $h(x)=x$ for all $x$ in {$1,3,5,7,...,2n+1:n\in\mathbb{N}$}$=B$. –  Dec 30 '16 at 03:33
  • So, how can I show $g(h(x))=x$ for all $x\in B$, $h(g(x))=x$ for all $x\in A$, $A\cap B=\emptyset$, $A\cup B=\mathbb{N}$? Are these clear? –  Dec 30 '16 at 03:41
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    You seem to have chosen $A$ and $B$ perfectly! Your notation doesn't make sense, though. Instead, you could say "for $x\in{2n+1:n\in\Bbb N},$"for example. Or, you could just say "for $x$ odd." A more dire issue is that you haven't chosen $g$ or $h$ correctly--neither $g(h(x))$ nor $h(g(x))$ is defined, and $f(x)$ is, in fact, the identity function! – Cameron Buie Dec 30 '16 at 03:46
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    As a hint for how to figure out $g,$ start with the formula $f(2n+1)=2n,$ set $x=2n+1,$ and substitute to get $f(x)=g(x)$ for $x=2n+1.$ We can do something similar for the $x=2n$ case. – Cameron Buie Dec 30 '16 at 03:48