lean_workbook_plus_14516

Another solutionReplacing $y$ with $xy$ and a bit of manipulation, we get: \n\n

\n\nIn particular, $f(x+y) > f(x)$ for all $x, y\in \mathbb{Q}^+$ , so $f$ is strictly increasing and thus injective. Intertwining between $x$ and $y$ gives us $f(x) + \frac{f(xy)}{f(x)} = f(y) + \frac{f(xy)}{f(y)}$ . Solving this gives us $f(xy) = f(x)f(y)$ whenever $x\neq y$ (as $f$ is injective). To generalize this result, notice that $f(1) f(2) = f(2)$ so $f(1) = 1$ , and for $x\neq 1$ ,: \n\n

\n\nHence, $f(x^2) = f(x)^2$ , so indeed $f(xy) = f(x) f(y)$ for all $x, y\in \mathbb{Q}$ . Equation $(1)$ now becomes: \n\n

\n\nso the function $x\mapsto (f(x) - x^2)$ is additive. Since it's domain is $\mathbb{Q}^+$ , it must be linear; there exists a rational number $c$ such that $f(x) = x^2 + cx = x(x+c)$ for all $x\in \mathbb{Q}^+$ . Since we obtained $f(1) = 1$ , this means $c = 0$ , so $f(x) = x^2 \; \forall x\in \mathbb{Q}^+$ .