For function satisfying given conditions prove that $f(x). f(\frac{1}{x})=1$

Question

If $f(x \times f(y)) =\frac{f(x)}{y}$ for every $x,y \in R$ and $y \neq0$, then prove that $f(x)\times f(\frac{1}{x})=1$

Is it possible to find the function $f(x)$ here using the given data? I tried it by using $x=0$ and $y=k$ ($k$ is any non-zero number), to get $f(0)$ then tried using partial differentiation with respect to x but the $f(y)$ is creating problem. Could someone please help me with it?

Answer 1

Suppose $f(x)\ne 0$ for all $x$. Then exist $x_0$ such that $f(x_0) = a\ne 0$. If we plug this in to the given equation we get:

$$ f(x_0f(y)) = {a\over y}$$ so our function is bijective. Now put $y=1$, $c=f(1)$ and we get $$f(xc) = f(x) \;\;\;\forall x\;\;\;\;\Longrightarrow\;\;\;\; c=1$$ So we have $f(f(y)) =1/y$ for all $y\ne 0$.

So $${1\over xf(y)}=f(f(xf(y))) = f\Big({f(x)\over y}\Big)$$ We put $x=1$ and we get the desired equation.

Answer 2

Write $a:=f(1)$, assumed non-zero for non-triviality; then putting $y=1$ gives $f(ax)=f(x)$, from which follows that $f(a)=a$. Putting $y=a$ gives $f(xa)=f(x)/a$, so $f(a)=1$. Hence $a=1$, that is, $f(1)=1$.

Putting $x=1$ gives $f\circ f(y)=1/y$. Thus $f(1/x)=f\circ f\circ f(x)=1/f(x)$, which is the required relation.