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This question already has an answer here:

Prove that there's no continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(f(x))=-x $.

Any hints?

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marked as duplicate by dxiv, Did, mrf, vrugtehagel, Parcly Taxel May 15 '17 at 8:10

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    $\begingroup$ Any thoughts on your own before we destroy this nice problem? $\endgroup$ – mickep May 15 '17 at 7:35
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    $\begingroup$ After 20+ months on the site and 70+ questions asked by the OP, I found the lack of context of the present question appalling, but when I looked at the previous ones, I understood the play goes on for a long time... $\endgroup$ – Did May 15 '17 at 7:52
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Hint: $f \circ f \circ f=f^{-1}$ so the function is bijective, since the inverse exists. Every continuous bijection $\mathbb R \to \mathbb R$ is monotone.

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  • $\begingroup$ Then show that $f(0)=0$ and$ f(-1)=-f(1)$. But then the strict monotonicity of $f$ implies that $f(1)>f(0)=0$ is contradictory, but $f(-1)>0=f(0)$ is also contradictory...............+1 $\endgroup$ – DanielWainfleet May 15 '17 at 7:51
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How about $f(x)=ix$. Then $f(ix)=i(ix)=i^2x=-x$.

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    $\begingroup$ The function is defined on real numbers. $\endgroup$ – VanDerWarden May 15 '17 at 7:41
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    $\begingroup$ $ix\notin\Bbb R$ for $x\in\Bbb R$ $\endgroup$ – Zelos Malum May 15 '17 at 7:50
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    $\begingroup$ Sorry for misreading.. $\endgroup$ – Wuestenfux May 15 '17 at 7:57

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