f : Q → Q f(x + f(y)) = y + f(x) ∀ x;y ∈ Q


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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

	$I demote by f^{n} : f \circ . . . \circ f n times$ $I wont precise it, but when a value (x, y, z . . .) is not defined it means it can be any element of Q$ $let I be the identity function$ $I am assuming that Q here stands for Q the set of rationnals$ $but note that this proof works on any group were - 2, - 1, 0, 1, 2 are all distinct$ $I am also assuming that the question is :$ $‘ ‘ Find all f such that {[y o u r p r o p r e t y]}^{″}$ $evaluating for x = 0 we get :$ $f^{2} (y) = y + f (0)$ $applying f over that equality we get :$ $f^{3} (y) = f (y + f (0)) = 0 + f (y)$ $hence : f^{3} = f (1)$ $f (x + f (y)) = y + f (x)$ $f (- f (y) + f (y)) = y + f (f (y))$ $i . e . f (0) - y = f^{2} (y)$ $so f^{3} (y) = f (- y + f (0)) = f (- y)$ $i . e . f^{3} = f \circ (- I)$ $hence f = f \circ (- I) (2)$ $y + f (x) = y + f (- x) = f (- x + f (y)) = f (x + f (y))$ $f (0) = f (2 f (y))$ $f (2 f (y)) = y + f (y) = f (0)$ $so f^{2} = f (3)$ $f (x + f (y)) = y + f (x)$ $f^{2} (y) = f (y) + f (0) s o f (0) = 0$ $so f (y) = - y$ $but then y! = 0 \Rightarrow f (- y)! = f (y)$ $this contradicts (2)$ $therefore there is no such {function}_{◼}$
	Commented by mathdanisur last updated on 20/Sep/21

	$Perfect solution S er, thanks$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum