Irrational finding

Multiply both the numerator and denominator of each fraction by the conjugate of the denominator. Then, for example, the first term simplifies to

$\dfrac{1}{\sqrt{1} + \sqrt{2}} \cdot \dfrac{\sqrt{2} - \sqrt{1}}{\sqrt{2} - \sqrt{1}} = \sqrt{2} - \sqrt{1}.$

If you keep doing this for each successive fraction, by difference of squares you'll keep getting $1$ in the denominator to obtain $\left(\sqrt{2} - \sqrt{1}\right) + \left(\sqrt{3} - \sqrt{2}\right) + \left(\sqrt{4} - \sqrt{3}\right) + \cdots$

Notice that this is a telescoping series, and that $\sqrt{1} = 1$ does not "collapse" as the other terms do. Our partial sums are

$\displaystyle \begin{aligned} & S_1 = \sqrt{2} - 1 \\ & S_2 = \sqrt{3} - 1 \\ & S_3 = \sqrt{4} - 1 \\ & S_4 = \sqrt{5} - 1 \\ & \cdot \\ & \cdot \\ & \cdot \\ & S_k = \sqrt{k + 1} - 1 \end{aligned}$

Thus, $S_k$ can only be rational if $k + 1$ is a perfect square. There are only two perfect squares between $2$ and $11$ inclusive: $4$ and $9$ .

$\displaystyle \begin{aligned} k + 1 & = 4 \ \ \Longrightarrow \ \ a_3 = \sqrt{4} - 1 = 3 \\ k + 1 & = 9 \ \ \Longrightarrow \ \ a_8 = \sqrt{9} - 1 = 8 \end{aligned}$

Since $2$ of these partial sums are rational, $\boxed{8}$ of them are irrational.