Is There a Function f: Z -> Z Such That f(f(n))=-n for Every Integer?

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Discussion Overview

The discussion centers around the existence of a function \( f: \mathbb{Z} \to \mathbb{Z} \) such that \( f(f(n)) = -n \) for every integer \( n \). Participants explore various potential forms of such a function and the implications of these forms.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The original poster (OP) expresses skepticism about the existence of such a function and seeks proof of its impossibility.
  • One participant suggests the function \( f(n) = in \), but this is challenged as not being integer-valued.
  • Another participant proposes a specific piecewise definition for \( f \) that includes mappings for positive, negative, and zero integers.
  • There is a positive acknowledgment of the piecewise function's structure, with a visualization of \( f \) as a permutation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of such a function. While some propose specific forms, others remain uncertain about the feasibility and correctness of these proposals.

Contextual Notes

There are unresolved questions regarding the validity of the proposed functions and their adherence to the requirement that \( f \) maps integers to integers. The discussion reflects varying interpretations of the problem and the definitions involved.

ypatia
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Is there any function (if any) f: Z -> Z such that
f(f(n))=-n , for every n belongs to Z(integers) ??


I think that there is not any function like the one described above but how can we prove it. Any ideas??
Thanks in Advance
 
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How about f(n)=in?
 
Not integer-valued. (I assume if the OP meant Gaussian integers that would have been mentioned, since that's the obvious solution.)

I've been thinking about this for a few hours now and I can't see any way to do it, but I can't prove that it's impossible.
 
How about

for n>0
f(2n-1)=2n
f(2n)=-2n+1
f(-2n+1)=-2n
f(-2n)=2n-1

f(0)=0
 
Nice, chronon. Nice.
 
Indeed - it's nice to visualize f as a piecewise permutation

(0)(-2,-1,2,1)(-4,-3,4,3)...(-2n,-2n+1,2n,2n-1)...

and recall that (abcd)^2=(ac)(bd)
 

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