Add $2 \sqrt{2} + 5 \sqrt{3}$ and $\sqrt{2} - 3 \sqrt{3}$.

Recall,$\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is,$\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

Check whether $7 \sqrt{5}$,$\frac{7}{\sqrt{5}}$,$\sqrt{2}+21$,and $\pi-2$ are irrational numbers or not.

Simplify each of the following expressions:
$(i)$ $(3+\sqrt{3})(2+\sqrt{2})$
$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$
$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$
$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Find five rational numbers between $1$ and $2$.

Show that $0.3333... = 0.\overline{3}$ can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \ne 0$.