Simplify each of the following expressions :
$(i)$ $(3+\sqrt{3})(2+\sqrt{2})$
$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$
$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$
$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
$(i)$ $(3+\sqrt{3})(2+\sqrt{2})=2(3+\sqrt{3})+\sqrt{2}(3+\sqrt{3})$
$=(2 \times 3+2 \sqrt{3})+(3 \sqrt{2}+\sqrt{2} \times \sqrt{3})$
$=6+2 \sqrt{3}+3 \sqrt{2}+\sqrt{6}$
Thus, $(3+\sqrt{3})(2+\sqrt{2})=6+2 \sqrt{3}+3 \sqrt{2}+\sqrt{6}$
$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})=(3)^{2}-(\sqrt{3})^{2}$ $=3^{2}-3=9-3=6$
$\therefore $ $(3+\sqrt{3})(3-\sqrt{3})=6$
$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}=(\sqrt{5})^{2}+(\sqrt{2})^{2}+2(\sqrt{5})(\sqrt{2})$ $=5+2+2 \sqrt{10}=7+2 \sqrt{10}$
$\therefore $ $(\sqrt{5}+\sqrt{2})^{2}=7+2 \sqrt{10}$
$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^{2}-(\sqrt{2})^{2}$ $=5-2=3$
$\therefore $ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=3$
Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0),$ where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representations (expansions). Can you guess what property $q$ must satisfy ?
Write the following in decimal form and say what kind of decimal expansion each has :
$(i)$ $\frac{36}{100}$
$(ii)$ $\frac{1}{11}$
$(iii)$ $4 \frac{1}{8}$
$(iv)$ $\frac{3}{13}$
$(v)$ $\frac{2}{11}$
$(vi)$ $\frac{329}{400}$
Find :
$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
$(ii)$ $\left(\frac{1}{3^{3}}\right)^{7}$
$(iii)$ $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
$(iv)$ $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Visualise $4. \overline{26}$ . on the number line, up to $4$ decimal places.
Locate $\sqrt 2$ on the number line.