2
$\begingroup$

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?

Thanks

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @mattbiesecker I know, if this function is monotone I can prove this function is the identity. $\endgroup$ – user42912 May 11 '15 at 0:52
  • 1
    $\begingroup$ It is easy to prove on $Q$, $f$ is id. $\endgroup$ – Yimin May 11 '15 at 0:57
4
$\begingroup$

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 $f(x) \geq f(y)$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.