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I was told to find $\def\R{\mathbb R}$$\R$ to $\R$ ring homomorphisms.

I have proved that if $x>0$ then $f(x)>0$ After this it follows $f$ is a strictly increasing function.

Then I attempted to prove continuity for all points in $\R$.

I have in this following way. Please check if it is correct. Let $c$ be an arbitrary point of $\R$. For any $\def\e{\epsilon}\e>0$, by the Archimedean property there exits $n\in\mathbb N$, natural number such that $1/n<\e$ for $|x-c|<1/n$ we have $-1/n <x-c<1/n$ hence $f(-1/n)<f(x-c)<f(1/n)$ and therefore $-1/nf(1)<f(x-c)<1/nf(1)$

$|f(x)-f(c)|< \frac{1}{n}f(1) <\epsilon f(1) $ I will conclude $\displaystyle \lim_{x \to c} f(x)=f(c)$;

is my proof correct? You can provide an independent proof too!

Have a look at this :Ring homomorphisms $\mathbb{R} \to \mathbb{R}$.

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UNM
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You've proven that $f$ is increasing. The next step is to note that $f$ has to fix the rationals. Now suppose that there is some (obviously irrational) $x$ such that $x \neq f(x)$. Then there must be some rational $q$ between $x$ and $f(x)$. Now what happens if $x < q$? Or if $x > q$?

kahen
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