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Question Number 154647 by mathdanisur last updated on 20/Sep/21

f : Q \to Q

f (x + f (y)) = y + f (x)

\forall x; y \in Q

Answered by TheHoneyCat last updated on 20/Sep/21

I demote by f^{n} : f \circ \dots \circ f n times

I wont precise it, but when a value (x, y, z \dots) is not defined it means it can be any element of Q

let I be the identity function

I am assuming that Q here stands for Q the set of rationnals

but note that this proof works on any group were - 2, - 1, 0, 1, 2 are all distinct

I am also assuming that the question is :

“ Find all f such that {[y o u r p r o p r e t y]}^{″}

evaluating for x = 0 we get :

f^{2} (y) = y + f (0)

applying f over that equality we get :

f^{3} (y) = f (y + f (0)) = 0 + f (y)

hence : f^{3} = f (1)

f (x + f (y)) = y + f (x)

f (- f (y) + f (y)) = y + f (f (y))

i . e . f (0) - y = f^{2} (y)

so f^{3} (y) = f (- y + f (0)) = f (- y)

i . e . f^{3} = f \circ (- I)

hence f = f \circ (- I) (2)

y + f (x) = y + f (- x) = f (- x + f (y)) = f (x + f (y))

f (0) = f (2 f (y))

f (2 f (y)) = y + f (y) = f (0)

so f^{2} = f (3)

f (x + f (y)) = y + f (x)

f^{2} (y) = f (y) + f (0) s o f (0) = 0

so f (y) = - y

but then y! = 0 \Rightarrow f (- y)! = f (y)

this contradicts (2)

therefore there is no such {function}_{}

Commented by mathdanisur last updated on 20/Sep/21

Perfect solution S er, thanks