$a_{n+1} = a_n/2 + 1/a_n$ is Cauchy but has no limit in $\mathbb{Q}$ I want to show that the sequence recursively defined by

$a_{n+1} = \frac{a_n}{2} + \frac{1}{a_n}, \:\: a_1=1$

is a Cauchy sequence that does not converge in $\mathbb{Q}$.
My idea was to show that this is a bounded sequence with bounds $1\leq a_n \leq 2$ and with a monotone tail, so that the monotonicity principle implies convergences in $\mathbb{R}$. But then how do I show that the limit is not in $\mathbb{Q}$?
Is there a way to use the definition of a Cauchy sequence directly?
Any help would be much appreciated!

I could obtain $a_n$ in a closed form.
$$a_{n+1} = \frac{a_n}2 + \frac1{a_n}$$
Substituting $\,b_n=a_n-\sqrt2$ (so that $\,\lim b_n\rightarrow0$),
$$b_{n+1}+\sqrt2=\frac{b_n+\sqrt2}2 + \frac1{b_n+\sqrt2}$$
$$b_{n+1}=\frac{b_n-\sqrt2}2 + \frac1{b_n+\sqrt2}=\frac{{b_n}^2}{2(b_n+\sqrt2)}$$
Taking reciprocal on both sides:
$$\frac1{b_{n+1}}=\frac2{b_n}+\frac{2\sqrt2}{{b_n}^2}$$
Adding a constant $C$:
$$\frac1{b_{n+1}}+C=\frac{2\sqrt2}{{b_n}^2}+\frac2{b_n}+C=2\sqrt2\left(\frac1{b_n}+C\right)^2\rightarrow C=\frac{\sqrt2}{4}$$
$$2\sqrt2\left(\frac1{b_{n+1}}+\frac{\sqrt2}{4}\right)=(2\sqrt2)^2\left(\frac1{b_n}+\frac{\sqrt2}{4}\right)^2$$
Substituting $\,c_n=2\sqrt2\left(\frac1{b_n}+\frac{\sqrt2}{4}\right)$,
$$c_{n+1}={c_n}^2={c_{n-1}}^4=\cdots={c_1}^{2^n}$$
$$c_n={c_1}^{2^{n-1}}=\left(\frac{2\sqrt2}{1-\sqrt2}+1\right)^{2^{n-1}}=(-3-2\sqrt2)^{2^{n-1}}$$
$$\therefore\,\,a_n=b_n+\sqrt2=\frac{1}{\frac{c_n}{2\sqrt2}-\frac{\sqrt2}{4}}+\sqrt2=\sqrt2\left(\frac{2}{c_n-1}+1\right)\\
=\sqrt2\left(\frac{2}{(-3-2\sqrt2)^{2^{n-1}}-1}+1\right)$$
