Let $f:\mathbb R\to \mathbb R$ be a function such that $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for every $x,y\in \mathbb R$ and $f(1)=1$. In order to prove this function is 1-1, I just need to prove this function is monotonic.

Anyone has some ideas how to proceed?


  • $\begingroup$ Monotonic only implies one-to-one if the function is continuous. $\endgroup$ – Jack M May 11 '15 at 0:45
  • $\begingroup$ @JackM If this function is monotonic I can prove this one is $1-1$. $\endgroup$ – user42912 May 11 '15 at 0:47
  • $\begingroup$ You have enough info to prove this is the identity function $\endgroup$ – matt biesecker May 11 '15 at 0:50
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    $\begingroup$ @mattbiesecker I know, if this function is monotone I can prove this function is the identity. $\endgroup$ – user42912 May 11 '15 at 0:52
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    $\begingroup$ It is easy to prove on $Q$, $f$ is id. $\endgroup$ – Yimin May 11 '15 at 0:57

We'll show that $f$ is monotone increasing.

Notice that if $x\geq 0$ then $f(x)=f(\sqrt{x})^2\geq 0$.

Thus if $x\geq y$, then $x-y \geq 0$, so $f(x)-f(y) = f(x-y) \geq 0$, so that $f(x) \geq f(y)$.

  • $\begingroup$ I just realized, my proof only shows that $f$ is monotone increasing on $(0,\infty)$, but extending the proof to all of $\mathbb{R}$ is simply a consequence of the fact that $f(0)=0$. $\endgroup$ – Shalop May 11 '15 at 1:06
  • $\begingroup$ $x\gt y\implies x-y\gt 0\implies f(x-y)\gt 0\implies f(x)-f(y)\gt 0\implies f(x)\gt f(y)$ $\endgroup$ – user42912 May 11 '15 at 1:08
  • $\begingroup$ Actually, your proof is better than the one I posted, so I updated it to resemble yours :) $\endgroup$ – Shalop May 11 '15 at 1:33
  • $\begingroup$ No problem :) thanks again! $\endgroup$ – user42912 May 11 '15 at 1:37

It is not necessary, as data of this question, the condition f(1) = 1 which is easily deduced as well as f(0) = 0. This is important to get f(-x) = -f(x) and f(1/x) = 1/f(x) and justify the steps f(x-y) = f(x) - f(y) and f (x/y) = f(x)/f(y)

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    $\begingroup$ No, it is also possible that $f(1)=0$ if the condition $f(1)=1$ was not given. For example, the zero function also satisfies the given constraints, except $f(1)=1$. $\endgroup$ – Shalop May 11 '15 at 1:48
  • $\begingroup$ Blessed zero function!!!! $\endgroup$ – Piquito May 11 '15 at 13:27

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