- #1

- 6

- 0

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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- Thread starter ypatia
- Start date

- #1

- 6

- 0

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

- #2

mathman

Science Advisor

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- 467

How about f(n)=in?

- #3

CRGreathouse

Science Advisor

Homework Helper

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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.

- #4

- 491

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How about

for n>0

f(2n-1)=2n

f(2n)=-2n+1

f(-2n+1)=-2n

f(-2n)=2n-1

f(0)=0

for n>0

f(2n-1)=2n

f(2n)=-2n+1

f(-2n+1)=-2n

f(-2n)=2n-1

f(0)=0

- #5

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Nice, chronon. Nice.

- #6

matt grime

Science Advisor

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(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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