Can we find for any real number $x$ the sequence of rationals $q_n(x)$ with properties:

  • $\lim\limits_{n\to\infty} q_n(x)=x$

  • $q_n(x+y)=q_n(x)+q_n(y)$

  • $q_n(xy)=q_n(x)q_n(y)$


  • 1
    $\begingroup$ Your second and third points ask for $q_n$ to be a ring homomorphism of the reals into the rationals; and the first point ensures that it is a ring homomorphism onto the rationals for all large $n$. $\endgroup$ – Anthony Quas Dec 6 '19 at 19:16
  • $\begingroup$ $q(0)=0$ and it looks like $q_n(1)=1$, mainly by multiplicative property, hence $q_n(-x)=-q_n(x)$ and $q_n(1/x)=1/q_n(x)$ for $x\ne 0$.. $\endgroup$ – Wlod AA Dec 6 '19 at 19:17
  • $\begingroup$ Thus $q_n$ is identity on $\Bbb Q$ for all large $n$. (Earlier, I meant for large $n$ too). $\endgroup$ – Wlod AA Dec 6 '19 at 19:23
  • 4
    $\begingroup$ Every ring homomorphism between fields is injective... $\endgroup$ – YCor Dec 6 '19 at 20:26
  • $\begingroup$ Does a ring homomorphism between fields have to send 1 to 1? $\endgroup$ – Anthony Quas Dec 6 '19 at 21:43

Assume $q(x)\neq 0.$ Then $q(2x)=q(x+x)=q(x)+q(x)=2q(x)$ but also $q(2x)=q(2)q(x)$ so $q_n(2)=2$ for all $n.$ But then $2=q(2)=q(\sqrt{2}^2)=q(\sqrt{2})^2$ leading to $q(\sqrt{2})=\sqrt{2}$ which is not rational.

For any rational $r$ one has $q_n(r)=r$ for all $n.$


$\newcommand{\R}{\mathbb R}$ We can consider $q_n$ as a homomorphism from $\R$ into $\R$. We have $q_n(1)=q_n(1)^2\to1^2=1\ne0$ (as $n\to\infty$). So, for some natural $N$ and all natural $n\ge N$ we have $q_n(1)=1$. So, for such $n$ we have $q_n(r)=r$ for all rational $r$. Also, for such $n$, any real $y\ge0$, and $x=\sqrt y$, we have $q_n(y)=q_n(x)^2\ge0$, which implies that $q_n$ preserves the order on $\R$, which implies that $q_n$ is the identity map of $\R$, which means that the range of $q_n$ cannot be $\mathbb Q$.

Thus, the answer to your question is no.


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