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This is my first post on Math SE soo... I was reading over a thread a while ago that claims the only solution to $f(f(f(x))) = x$ for $f : \mathbb{R} \to \mathbb{R}$ is $f(x) = x$, but.. I seemed to have found a counterexample: $$f(x) = \frac{1}{1-x}$$ Here, we have $$f(f(f(x))) = \frac{1}{1-\frac{1}{1-\frac{1}{1-x}}} = \frac{1}{1+\frac{1-x}{x}} = \frac{1}{\frac{1}{x}} = x$$ Well... looking over the original problem statement, would we say that my function satisfies $f : \mathbb{R} \to \mathbb{R}$? Or would it be something more like $f : \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{0\}$?

Arkyter
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  • yes i think the domain $\text{dom} (f)$ is the issue – 311411 Dec 14 '21 at 20:41
  • Your proposed function does not work since, as noted, its domain is not $\Bbb R$ but $\Bbb R \setminus {1}$. – PrincessEev Dec 14 '21 at 20:41
  • got it! :3 thanks. – Arkyter Dec 14 '21 at 20:44
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    It also doesn’t work because it’s not strictly increasing - $f(0) > f(2)$, for example. The problem stated that if $f$ is strictly increasing, then $f$ is the identity function. – Mark Saving Dec 14 '21 at 20:45
  • What is $f(1)?$ – Thomas Andrews Dec 14 '21 at 20:47
  • Yeah, I see where my point breaks down now. $f(1)$ is undefined which doesn't satisfy the original problem proposed. (it's also not strictly increasing as Mark Saving pointed out) Thanks guys! – Arkyter Dec 14 '21 at 20:48
  • And what is $f(f(0))?$ – Thomas Andrews Dec 14 '21 at 20:48
  • yeah, that's also undefined. – Arkyter Dec 14 '21 at 20:49
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    If $f$ is assumed to be continuous, then one can prove that $f$ is identity. Otherwise the question makes no sense as one can choose any partition of $\Bbb R$ into subsets of three elements and define an arbitrary function. – WhatsUp Dec 14 '21 at 20:50
  • if I have a follow-up question that I'd like to ask would I make another post for it? or would I ask in the comments section? – Arkyter Dec 14 '21 at 20:50
  • But there are a lot of functions of this form (Möbius transformations) $X\to X,$ where $X$ is $\mathbb R\cup {\infty}$ or $\mathbb C\cup {\infty}.$ If $a+d=-1, ad-bc=1,$ then $$f(x)=\frac{ax+b}{cx+d}$$ satisfies this condition. These are continuous on these “projective lines.” $a=0,b=-1,c=1,d=-1$ gives tour case. – Thomas Andrews Dec 14 '21 at 20:53
  • An example of a function $f: \mathbb R \to \mathbb R$ satisfying $f(f(f(x))) = x$ is $$ f(x) = \cases{x - 2 & if $\lfloor x \rfloor$ is divisible by $3$\cr x+1 & otherwise}$$ – Robert Israel Dec 14 '21 at 20:53
  • that's an interesting example... thanks for the notes guys! – Arkyter Dec 14 '21 at 20:59

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