Approximating $\sqrt{2}$ in rational numbers Let a sequence of rational numbers be defined recursively as $x_{n+1} = (\frac{x_n}{2} + \frac{1}{x_n})$ with $x_1$ some arbitrary positive rational number.
We know that, in the universe of real numbers, this sequence converges to $\sqrt{2}$. But suppose we don't know anything about real numbers. How do we show that that ${x_n}^2$ gets arbitrarily close to $2$? 
I've already shown that $x_n > 2$ and that the sequence is decreasing. But I'm having difficulty showing that ${x_n}^2$ gets as close to $2$ as we want using nothing but inequalities. Since we're assuming no knowledge of real numbers, I don't want to use things like the monotone convergence theorem, the least upper bound property etc.
This exercise is of interest to me because it can it can help explain the development of irrational numbers to a student who knows nothing about them.

Square both sides to obtain
$$x_{n+1}^2 = \frac{x_n^2}{4} + 1 + \frac{1}{x_n^2}$$
Therefore,
$$x_{n+1}^2 - 2 = \frac{x_n^2}{4} - 1 + \frac{1}{x_n^2}$$
We can manipulate this a bit to obtain
$$x_{n+1}^2 - 2 = \frac{x_n^2 - 2}{4} - \frac{x_n^2 - 2}{2x_n^2}$$
$$x_{n+1}^2 - 2 = \frac{(x_n^2 - 2)^2}{4x_n^2}$$
Since the RHS is nonnegative, so is the left, so from this we get $x_{n+1}^2 \geq 2$. Applying this to the denominator of the RHS gives us (for $n > 1$):
$$x_{n+1}^2 - 2 \leq \frac{(x_n^2 - 2)^2}{8}$$
From this we can conclude that as long as $(x_n^2 - 2) < 1$ for some $n$, we will have $x_{n+1}^2 - 2 < 1/8$, and so by induction, $x_{n+k}^2 - 2 < 1/8^k$.
