We are tasked with showing that there exists a sequence of rational numbers $r_n$ such that $|r_n^2 - 2| \to 0$, meaning that the sequence of rational numbers $r_n$ converges to the square root of 2. We are given a classical iterative formula:

$r_{n+1} = \frac{1}{2} \left( r_n + \frac{2}{r_n} \right)$

Step 1: Understanding the Iterative Formula

This is a classical method known as the Newton-Raphson method for finding square roots. The sequence defined by this iteration should converge to $\sqrt{2}$. To understand why this is the case, let's explore the iteration in more detail.

• Let $f(x) = x^2 - 2$. We are seeking the root of $f(x) = 0$, i.e., $x = \sqrt{2}$.
• The Newton-Raphson method gives the iterative formula:

$r_{n+1} = r_n - \frac{f(r_n)}{f'(r_n)} = r_n - \frac{r_n^2 - 2}{2r_n} = \frac{r_n}{2} + \frac{1}{r_n}$

This matches the formula we are given. Therefore, the iteration should converge to $\sqrt{2}$ if it starts with a rational $r_0$.

Step 2: Convergence of the Sequence

We want to show that $|r_n^2 - 2| \to 0$, i.e., $r_n \to \sqrt{2}$. To do this, we can examine the error in the approximation at each step. Define the error $e_n = r_n^2 - 2$. Our goal is to show that $e_n \to 0$.

1. Error expression: First, observe that:

   $r_n^2 - 2 = e_n$

2. Relation between successive terms: Let’s try to bound the ratio between successive terms of the sequence. We compute $r_{n+1}^2$ in terms of $r_n$:

   $r_{n+1} = \frac{1}{2} \left( r_n + \frac{2}{r_n} \right)$

   Squaring both sides:

   $r_{n+1}^2 = \frac{1}{4} \left( r_n + \frac{2}{r_n} \right)^2 = \frac{1}{4} \left( r_n^2 + 2 + \frac{4}{r_n^2} \right)$

   Therefore:

   $r_{n+1}^2 - 2 = \frac{1}{4} \left( r_n^2 + 2 + \frac{4}{r_n^2} \right) - 2 = \frac{1}{4} \left( r_n^2 + \frac{4}{r_n^2} - 6 \right)$

   This expression shows how the error at step $n+1$, $r_{n+1}^2 - 2$, depends on the error at step $n$, $r_n^2 - 2$.

3. Bounding the error: If $r_n$ is sufficiently close to $\sqrt{2}$, the term $r_n^2 - 2$ will be small. Since $r_n \to \sqrt{2}$, the term $r_n^2 \approx 2$, and the fraction $\frac{4}{r_n^2}$ approaches 2. This gives us:

   $r_{n+1}^2 - 2 \approx \frac{1}{4} \left( e_n + \frac{4}{r_n^2} - 6 \right)$

   We see that the error decreases quadratically as $r_n \to \sqrt{2}$, implying that $e_n \to 0$ as $n \to \infty$.

Step 3: Conclusion

The iteration defined by

$r_{n+1} = \frac{1}{2} \left( r_n + \frac{2}{r_n} \right)$

will converge to $\sqrt{2}$, and the error $|r_n^2 - 2|$ will tend to 0 as $n \to \infty$. Hence, we have shown that there exists a sequence of rational numbers $r_n$ such that $|r_n^2 - 2| \to 0$.

Next Steps and Further Investigation:

1. Can we establish a formal rate of convergence for this method?
2. What happens if we start with irrational initial conditions? Does it still converge to $\sqrt{2}$?
3. Can this method be generalized to approximate other square roots?
4. Can we find a simpler sequence that converges to $\sqrt{2}$?
5. How does this method compare with other methods for approximating square roots?

Tip: Newton’s method converges quadratically, meaning the number of correct digits approximately doubles with each step once the approximation is close enough. This makes it a very efficient method for numerical computation.