Construct a cauchy sequence $(r_n)$ such that $r_n\in\mathbb{Q}$ $ \forall n\in\mathbb{N}$ , $(r_n)$ is a Cauchy sequence but $\lim_{n \to \infty}r_n $ does not belong to $\mathbb{Q}$
Can anyone let me know where to start? I'm completely at sea with this so just an idea of the bare essentials to prove this would be helpful.
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What about sequence defined by $u_0 = 1$ and $u_{n+1} = \frac{u_n + \frac{2}{u_n}}{2}$ for $n\in\mathbf{N}$ which is convergent in $\mathbf{R}$ to a square root of $2$ so that it is cauchy in $\mathbf{R}$ and also in $\mathbf{Q}$, and whose limit can't be rational. It is an elementary example as, whatevers tastes may be, it is quicker to prove its convergence and the irrationality of its limit (in $\mathbf{R}$) than to do it for $\left(1+\frac{1}{n}\right)^n$...
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Pick your favorite real number. Call it $r$. Then consider the sequence of intervals $I_n = (r-\frac{1}{n}, r+\frac{1}{n})$ for $n=1, 2, ...$
Use the axiom of choice to choose a rational number from each interval. The resulting sequence converges to $r$.
Alternatively you can look at the decimal expansion of $r$ and take a sequence ${r_n}$ where $r_n$ is just the first $n$ terms after the decimal point.
Hint Consider $r_n = \left(1+\frac{1}{n}\right)^n \in \Bbb Q$.