 # Show there is a sequence of rational numbers converging to tapiat0wa4c 2022-06-03 Answered
Show there is a sequence of rational numbers converging to any irrational number.
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The space of real numbers $R$ is the metric completion of $Q$ the space of rational numbers. To construct $R$, start with $Q$, and let $C$ be the space of Cauchy sequences of $Q$. Two elements $\left({x}_{n}\right)$ and $\left({y}_{n}\right)$ are equivalent if and only if $\left({x}_{n}-{y}_{n}\right)$ converges towards $0$, then $R$ is the quotient space for this equivalence relation. You can identify $Q$ with the equivalence classes of constant sequences.

If $\left[{x}_{n}\right]$ is the class of the Cauchy sequence $\left({x}_{n}\right)$, define $\left({y}_{n}^{i}\right)$ by ${y}_{l}^{i}={x}_{i}$ for every $l$. Then $\left(\left[{y}_{n}^{i}\right]\right)$ converges towards $\left[{x}_{n}\right]$ and $\left[{y}_{n}^{i}\right]$ can be identified with the rational ${x}_{i}$.