A function $f: \mathbb{Q}^+ \cup \{0\} \to \mathbb{Q}^+ \cup \{0\}$ is defined such that $$ f(x) + f(y) + 2xyf(xy) = \frac{f(xy)}{f(x+y)}$$ Then what is the value of $\left[f(1)\right]$ (where $[.]$ denotes the greatest integer function)?

I proceeded this way:
Putting $x=y=0$ I got $f(0) = \frac{1}{2}$ (assuming $f(0) \neq 0$)
Again putting $y=0$ I got $f(x) + f(0) = \frac{f(0}{f(x)}$ which gave 2 values of $f(x)$ as $-1$ and $\frac{1}{2}$.
As $f(0)$ was equal to $\frac{1}{2}$ so I assumed $f(x)$ as a constant function having value $\frac{1}{2}$ for all $x$.
When $f(0) = 0$ then I got $f(x) = 0$ for all $x$.
But the answer was given to be equal to $1$ which means $f(1) \in [1,2)$. Where am I wrong?

  • $\begingroup$ Well if you put $x=y=0$, then $f(0)+f(0)=\frac{f(0)}{f(0)}$ so if $f(0)\neq 0$, then we can say that $\frac{f(0)}{f(0)}=1$ and $f(0)=\frac{1}{2}$. If $f(0)=0$, then you cannot use the equation because $\frac{0}{0}$ is not defined. $\endgroup$ – T. Haddad May 4 '17 at 14:55
  • $\begingroup$ Yes I thought so but then I couldn't find any other way to do it. $\endgroup$ – Sourav Suman May 4 '17 at 14:58
  • $\begingroup$ Also if I take the case for $f(0) = 0$ then we will have $f(x) = 0$ for all $x$. $\endgroup$ – Sourav Suman May 4 '17 at 15:05

Your logic is fine. You can rule out $f(0)=0$ because of the divide by zero it produces on the right. You can rule out $f(x)=-1$ because range of the function is given as $\Bbb Q \cup \{0\}$. Then you can say for $x=y=1$ the defining equation asserts $\frac 12+\frac 12+2\cdot 1 \cdot 1 \cdot \frac12=\dfrac{(\frac 12)}{(\frac 12)}$, which is false. There is no such function.

  1. Assume $f(0) \neq 0$.

    Then $f(0) + f(0) = \frac{f(0)}{f(0)} \rightarrow f(0) = \frac{1}{2}$,

    and $f(x) + f(0) = \frac{f(0)}{f(x)} \rightarrow f(x)^2 + \frac{1}{2}f(x) -\frac{1}{2} = 0.$

    1. Assume $f(1) \neq 0$.

      We also know that $f(1) + f(1) + 2f(1) = \frac{f(1)}{f(2)}$. Or $f(2) = \frac{1}{4}$.

      However, $(\frac{1}{4})^2 + \frac{1}{2}\cdot\frac{1}{4} - \frac{1}{2} \neq 0$. Contradiction.

    2. Assume $f(1) = 0$.

      $0^2 + \frac{1}{2}\cdot 0 - \frac{1}{2} \neq 0$. Contradiction.

Therefore we have $f(0) = 0$. However then we have for any $x$:

$$f(x) = \frac{0}{f(x)}$$

If $f(x) \neq 0$ then the above formula implies $f(x)^2 = 0$ which is a contradiction.

So the function must be $f(x) = 0$ everywhere however if this is the case then the 'definition' is ill-defined (division by zero) everywhere, therefore I argue that $f$ itself is ill-defined.


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