Functions

Algebra Level 3

Let Q > 0 \mathbb{Q}_{>0} denote the set of all positive rational numbers. If f : Q > 0 Q > 0 f:{ \mathbb{Q} }_{ >0 } \rightarrow {\mathbb{Q} }_{ >0 } satisfies

f ( x 2 f ( y ) 2 ) = f ( x ) 2 f ( y ) , \large f\left( x^{ 2 }{ f\left( y \right) }^{ 2 } \right) = {f(x)}^2f\left( y \right) ,

evaluate f ( 2019 ) . f(2019).


The answer is 1.

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1 solution

Tristan Goodman
Aug 4, 2019

Solving the given equation algebraically results in f(x)=sqrt(x) but this does not satisfy the condition of all input giving a rational output, since for example sqrt(2) is irrational, so the function has to be either 1 or 0, but the conditions state that the function must have domain and range greater than zero; it must therefore be a constant function equal to 1.

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