Find all the functions satisfying $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for all real x,y [duplicate]

I will first show the work that I have done and I will then explain what I have left to prove. I have shown that for $\sigma\in Aut_{\mathbb{Q}}(\mathbb{R})$, then if $x,y\in\mathbb{R}$, with $x<y$ then $\sigma(x)<\sigma(y)$.

I need to prove that $-\frac{1}{m} < x-y < \frac{1}{m}$ implies that $-\frac{1}{m} < \sigma(x) - \sigma(y) < \frac{1}{m}$ for every positive integer $m$. Conclude that $\sigma$ is continuous on $\mathbb{R}$. I then need to prove that any continious map on $\mathbb{R}$ that fixed $\mathbb{Q}$ is the identity map, whence $Aut_{\mathbb{Q}}(\mathbb{R}) = \{e\}$.

Suppose that $x>0$ for some $x\in\mathbb{R}$, then $x = y^2$ for some $y\in\mathbb{R}$. Since $\sigma$ is a ring homomorphism we have that $\sigma$ is multiplivative in the sense that $\sigma(x) = \sigma(y^2) = \sigma(y)\sigma(y)$. Also, $\sigma(y)\sigma(y) = \sigma(y)^2 > 0$. This proves that if $x>0$ then $\sigma(x)>0$. In other words, $\sigma$ takes the positive reals to the positive reals.

Now choose $x<y$ with $x,y\in\mathbb{R}$, $x>0$, $y>0$. So, $\sqrt{x}<\sqrt{y}$. We know $\sigma(x)=\sigma(\sqrt{x})^2$ and $\sigma(y)=\sigma(\sqrt{y})^2$. Since $y-x>0$, then $\sigma(y-x)>0$. Since $\sigma$ is a ring homomorphism, then $\sigma(y-x) = \sigma(y) - \sigma(x)$, and since $y-x>0$, then $\sigma(y-x)>0$. Therefore, $\sigma(y)-\sigma(x) > 0$ or $\sigma(x)<\sigma(y)$, as we wanted to prove.

I am having trouble with $-\frac{1}{m} < x-y < \frac{1}{m}$ implies that $-\frac{1}{m} < \sigma(x) - \sigma(y) < \frac{1}{m}$ for every positive integer $m$. I can certainly reduce $-\frac{1}{m} < x-y < \frac{1}{m}$ to $-\frac{1}{\sigma(m)} <\sigma(y)-\sigma(x)<\frac{1}{\sigma(m)}$. But then to answer this question it seems like I have to prove that $\sigma(m) = m$, which I don't know how to do. Can you please help?

Solution 1:

Using $x<y \Rightarrow \sigma(x)<\sigma(y)$ alone you can show that $\sigma = id$. Let $y\in \mathbb R$. If $\sigma(y)>y$, let $p\in \mathbb Q$ and $y< p< \sigma(y)$. Then $\sigma(y)< \sigma(p)=p$. If $\sigma(y)<y$, let $q\in \mathbb Q$ such that $\sigma(y) < q< y$. Then $q = \sigma(q)< \sigma(y)$ which is also impossible. Thus $\sigma(y) = y$.