# Rational representation of reals [closed]

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

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

• $$q_n(x+y)=q_n(x)+q_n(y)$$

• $$q_n(xy)=q_n(x)q_n(y)$$

?

• 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 $n$. – Anthony Quas Dec 6 '19 at 19:16
• $q(0)=0$ 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 $x\ne 0$.. – Wlod AA Dec 6 '19 at 19:17
• Thus $q_n$ is identity on $\Bbb Q$ for all large $n$. (Earlier, I meant for large $n$ too). – Wlod AA Dec 6 '19 at 19:23
• Every ring homomorphism between fields is injective... – YCor Dec 6 '19 at 20:26
• Does a ring homomorphism between fields have to send 1 to 1? – 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 $$\mathbb Q$$.