Approximating $\sqrt{2}$ in rational numbers

Question

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.

Answer 1

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, $x_{n+k}^2 - 2 < 1/8^k$.

Answer 2

The question raised by Vishal is interesting, so I have decided to give my contribution by answering in a detailed way a slight generalization of it. Precisely, I want to show how to approximate by rational numbers the square root of every positive rational number $a$ by proving that the sequence $\langle x_n \rangle_{n\in\mathbb{N}}$, defined as $$ x_{n+1}= \begin{cases} x_1 & n=0\\ \\ \dfrac{1}{2}\left(x_n+\dfrac{a}{x_n}\right) & n\geq 1 \end{cases} $$ is such that $x_n^2\to a$ as $n \to +\infty$ for any choice of the positive rational number $x_1$, i.e. for any $x_1>0$, $x_1\in\mathbb{Q}$.

First of all, by squaring both sides of defining equation of $x_{n+1}$ for all $n\geq 1$ we obtain $$ x_{n+1}^2 = \frac{x_n^2}{4} + \frac{a}{2} + \frac{a^2}{4x_n^2}$$ Subtracting $a$ from both of its two sides, we get $$ \begin{split} x_{n+1}^2 - a & = \frac{x_n^2}{4} - \frac{a}{2} + \frac{a^2}{4x_n^2}\\ & = \frac{1}{4}\left(x_n^2 - 2a + \frac{a^2}{x_n^2}\right)\\ & = \frac{1}{4}\left(x_n - \frac{a}{x_n} \right)^2 = \frac{(x_n^2 - a)^2}{4x_n^2}\geq 0 \end{split} $$ Specializing the equation for $n=1, 2$ we obtain $$ x_2^2 - a = \frac{(x_1^2 - a)^2}{4x_1^2} $$ and $$ \begin{split} x_3^2 - a & = \frac{(x_2^2 - a)^2}{4x_2^2} = \frac{(x_2^2 - a)}{4x_2^2}(x_2^2-a)\\ & = \frac{1}{4}\frac{(x_1^2 - a)^2}{4x_2^2x_1^2}(x_2^2-a) = \frac{1}{4}\frac{(x_1^2 - a)^2}{(x_1^2+a)^2}(x_2^2-a) \\ & \leq \frac{1}{4}(x_2^2-a) = \frac{1}{16}\frac{(x_1^2 - a)^2}{x_1^2} \end{split} $$

From the calculations above, it seems plausible to suppose that the following estimate holds: $$ 0 \leq x_n^2 - a \leq \frac{1}{2^{2(n-1)}} \frac{(x_1^2 - a)^2}{x_1^2}\quad \forall n\geq 2 $$ This is true, as it can be easily shown by generalized induction. Proceeding to do so, we first note that for $n=2$ the estimate holds as we have already shown above: then, assuming it being true for $n$ we obtain $$ \begin{split} x_{n+1}^2 - a & = \frac{(x_n^2 - a)^2}{4x_n^2} = \frac{(x_n^2 - a)}{4x_n^2}(x_n^2-a)\\ & = \frac{1}{4}\frac{(x_{n-1}^2 - a)^2}{4x_n^2x_{n-1}^2}(x_n^2-a) = \frac{1}{4}\frac{(x_{n-1}^2 - a)^2}{(x_{n-1}^2+a)^2}(x_n^2-a) \\ & \leq \frac{1}{4}(x_n^2-a) \leq \frac{1}{4\cdot 2^{2(n-1)}}\frac{(x_1^2 - a)^2}{x_1^2} =\frac{1}{2^{2(n+1-1)}}\frac{(x_1^2 - a)^2}{x_1^2} \end{split} $$ Therefore the estimates holds true for every $n\geq2$ and this, by the sandwich theorem, implies $x_n^2\to a$ for any choice of the positive rational $x_1$.

A few notes:

1. The proof uses only elementary inequalities and the result of every step is in $\mathbb{Q}$, since this field is closed respect to the four basic arithmetical operations. If we take $a=2$ we have the direct answer to the question of Vishal.
2. However, despite being needed by the didactic aims of Vishal, the hypothesis $a, x_1\in\mathbb{Q}$ is not needed by the logical structure of the reasoning: all works perfectly for any $a,x_1\in\mathbb{R}_+$. This is exactly what Emanuel Fisher asks to do in his beautiful (even if flawed by many typos) "Intermediate Real Analysis" (1983, Springer Verlag, exercise III.8.8 page 139). However, since in that textbook the reals have been introduced earlier as Dedekind cuts, the formal development needed to solve the exercise is slightly simpler due to the fact that it is not needed to square the terms of the sequence in order to work inside $\mathbb{Q}$.
3. An observation from the "approximation theorist" point of view: the approximation error halves at every iteration whatever the value of the initial value $x_1>0$ is.

Answer 3

Look at $x_{n+1}^2-2$ compared with $x_n^2-2$. We have $$x_{n+1}=\frac {x_n}2+\frac 1{x_n}\\ x_{n+1}^2-2=\left(\frac {x_n}2+\frac 1{x_n}\right)-2\\ =\frac 14(x_n^2-2)-\frac 12-\frac 1{(x_n^2-2)+2}\\ \approx\frac 14(x_n^2-2)-\frac 12-\frac 12(1-\frac 12(x_n^2-2)+\frac 14(x_n^2-2)^2)\\ \approx \frac 18(x_n^2-2)^2$$ And the error becomes as small as we want.

Answer 4

The usual proof of convergence using Newton's method also includes an error estimate.

Answer 5

Let $d_n=|x_n^2-2|$.

Then $$d_{n+1}=\left|\frac{x_n^2}4-1+\frac1{x_n^2}\right|=\left|\frac{x_n^4-4x_n^2+4}{4x_n^2}\right|=\left(\frac{d_n}{2x_n}\right)^2$$

So it suffices to prove that $x_n>1/2$ for some $n$, because then $d_n\to 0$.

But if $0<x_1\le1/2$ then $$x_2=\frac{x_n^2+2}{2x_n}\ge\frac21$$