Match the following columns:

Solution:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\pi$ is an irrational number.

(c) $\frac{1}{7}=.142857142857 \ldots$

Hence, its period is 6.

(d)

$x^{2}+\frac{1}{x^{2}}$

$=(2-\sqrt{3})^{2}+\frac{1}{(2-\sqrt{3})^{2}}$

$=\left(2^{2}+(\sqrt{3})^{3}-2 \times 2 \times \sqrt{3}\right)+\frac{1}{\left(2^{2}+(\sqrt{3})^{3}-2 \times 2 \times \sqrt{3}\right)}$

$=(4+3-4 \times \sqrt{3})+\frac{1}{(4+3-4 \times \sqrt{3})}$

$=(7-4 \times \sqrt{3})+\frac{1}{(7-4 \times \sqrt{3})}$

$=\frac{(7-4 \times \sqrt{3})^{2}}{(7-4 \times \sqrt{3})}+\frac{1}{(7-4 \times \sqrt{3})}$

$=\frac{7^{2}+(4 \sqrt{3})^{2}-2 \times 7 \times 4 \sqrt{3}+1}{(7-4 \times \sqrt{3})}$

$=\frac{49+48-56 \sqrt{3}+1}{(7-4 \times \sqrt{3})}$

$=\frac{98-56 \sqrt{3}}{(7-4 \times \sqrt{3})}$

$=14 \times \frac{((7-4 \times \sqrt{3}))}{(7-4 \times \sqrt{3})}$

$=14$