Does following funtion exists?

Question

Does there exists two real valued functions $f$ and $g$ on $R$ such that $f \circ g = x^{2018}$ and $ g\circ f = x^{2019}$ ?

My attempt : since $g \circ f $ is bijective thus $f$ is one one and $g$ is onto. Now $f \circ g(-x) = f \circ g(x) $ imply $ g(-x) = g(x) $ (because $ f $ is one one) thus g is even function. Now i dont know how to proceed from here any hint will be helpfull for me.... (I know no such map exists as answer)

Bijayan Ray · Answer 1 · 2019-01-08T18:07:56.890

0

$$g(f(x))=\ x^{2019}$$ $$=> f(g(f(x)))=f(\ x^{2019})$$ $$=> (f(x))^{2018}=f(x^{2019})$$ (by associativity of function composition)

there exist no such f and g because$ f(1)=1, f(-1)=1 => g(1)=1$ and $ g({(-1)}^{2018})=-1$

contradiction hence proved

well one may argue f(1)=0

in that case consider $f(0),f(1),f(-1) $will have either 1 or 0 value and by pigeon hole principle two of them will have atleast same value hence we are done

edited Jan 08 '19 at 18:07

answered Jan 08 '19 at 17:47

Bijayan Ray

608

What if $f(1) = 0$? – fleablood Jan 08 '19 at 18:00
hmm you are correct – Bijayan Ray Jan 08 '19 at 18:02
consider then f(0),f(-1),f(1) they can be 0 or 1 – Bijayan Ray Jan 08 '19 at 18:04
by pigeon hole principle atleast two of them will have same value so we are done – Bijayan Ray Jan 08 '19 at 18:05

Does following funtion exists?

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