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This question already has an answer here:

What can be $f$ so that $f^2(x) = -x$ for all $x\in R$?

I know that if $f^2(x) = -x$ then $f(x)$ is injective and $f$ can not be continuous.

But I can not find an example of discontinuous function so that $f^2(x) = -x$ for all $x\in R$.

Can anyone help me?

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marked as duplicate by Namaste, Mike Earnest, Omnomnomnom, user370967, MrYouMath Oct 4 '17 at 15:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $f^2(x)=f\circ f(x)$ $\endgroup$ – Jorge Fernández Hidalgo Oct 4 '17 at 15:45
  • $\begingroup$ See math.stackexchange.com/questions/312385/… $\endgroup$ – Robert Z Oct 4 '17 at 15:45
  • $\begingroup$ First of all, clarify your notation. Presumably, $f^2(x) = f(f(x))$. Second, how do you know that $f$ cannot be continuous? Finally, what have you tried? $\endgroup$ – Omnomnomnom Oct 4 '17 at 15:46
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    $\begingroup$ @Omnomnomnom Yes, but the answer there provides a discontinuous solution as well $\endgroup$ – Mike Earnest Oct 4 '17 at 15:48
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    $\begingroup$ @MikeEarnest just took a closer look as you commented, thanks $\endgroup$ – Omnomnomnom Oct 4 '17 at 15:48
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Just look at the cycle representation of $f\circ f=-x$, there is one fixed element and a bunch of $2$-cycles, So to build a suitable permutation just pair up the $2$-cycles and for each $(a,b)$ and $(c,d)$ add the $4$-cycle $(a,c,b,d)$.

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  • $\begingroup$ In summary, if $A,B$ is a partition of the positive reals and $g:A\rightarrow B$ is a bijection then the permutation with cycles $(a,g(a),-a,-g(a))$ works. $\endgroup$ – Jorge Fernández Hidalgo Oct 4 '17 at 15:51

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