Note that $\sqrt{2}$ is a solution to the equation

1. Use the change of sign rule to show that

   $$1<\sqrt{2}<\frac{3}{2}$$

2. Show that the Newton-Raphson method, when applied to the equation $x^2=2,$ yields the recursive formula

   $$x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$$

3. Use a starting value of $x_0=\frac{3}{2}$ to find two further rational approximations, $x_1$ and $x_2$, for $\sqrt{2}.$

Given that $x_2$ is of the form $\frac{p}{q}$, where $p,q$ are coprime natural numbers, give the value of $pq$.