$a_{n+1} = a_n/2 + 1/a_n$ 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!

Maybe this is overkill, but one approach would be to use the Banach Fixed Point Theorem. On closed intervals, it simplifies to:

Let $f:[a,b]\to[a,b]$ be such that there exists $0\leq K<1$ with $|f(x)-f(y)|\leq K|x-y|$ for all $x,y$. This is called a contraction on $[a,b]$. Then $f$ has a unique fixed point in $[a,b]$ i.e. there is a unique $x^*\in[a,b]$ with $f(x^*)=x^*$. Furthermore, if we let $x_1\in X$ and define $x_{n+1}:=f(x_n)$ for all $n\geq 1$, then $x_n\to x^*$.

Applying this to the interval $[1,2]$, we define $f:[1,2]\to[1,2]$ by $f(x):=\frac{x}{2}+\frac{1}{x}$. Then $|f(x)-f(y)|\leq\frac{1}{2}|x-y|$ gives that $f$ is a contraction. Since $a_{n+1}=f(a_n)$ for all $n$, we have that $a_n$ converges to the unique fixed point of $f$ by BFPT, which one can easily check is $\sqrt{2}$. Since $a_n$ converges, it must be Cauchy and, of course, it converges outside of $\mathbb{Q}$.
