1if f(n)=1-n ,
then f(f(n))=f(1-n)=1-(1-n)=n. so (i) holds.
f(f(n+2)+2)=f((-n-1)+2=f(1-n)=n so (ii) is satisfied
its easy to see that (iii) is also satisfed.
so the given function satisfies the given conditions.
Prove that f(n) = 1−n is the only integer-valued function
defined on the integers that satisfies the following
conditions.
(i) f(f(n)) = n, for all integers n;
(ii) f(f(n + 2) + 2) = n for all integers n;
(iii) f(0) = 1.
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5 Answers
62(i) f(f(n)) = n, for all integers n;
(ii) f(f(n + 2) + 2) = n for all integers n;
(iii) f(0) = 1.
f(f(0)) = 0
f(1) = 0
f(f(n+2)+2) = n
f(f(2)+2) = 0
f(f(f(n + 2) + 2)) = f(n + 2) + 2
thus, f(n) = f(n + 2) + 2
f(n+1) = f(n+3) + 2
f(n+2) = f(n+4) + 2
now put n=0 you will get the solution :)
1
9no bhargav u r not giving proof that its the only sol possible
u r only confirming it
continuing from #2 ( sir dint finish it and ans isnt obvious enough ...... donno wats in his mind )
f(n) = f(n+2) +2
s0
f(n) = f(n+2r) +2r
put n=0
f(2r)=1-2r
and
f(n+1)=f(n+1+2r)+2r
put n=0
f(2r+1)=1 - (2r+1)
so always
f(n)=1-n
62arrey exactly this was in my mind.. and that is why i said put n=0 :P
9but sir
thus, f(n) = f(n + 2) + 2
f(n+1) = f(n+3) + 2
f(n+2) = f(n+4) + 2
now put n=0 you will get the solution :)
how does it prove f(n) =1-n directly ??
