Visualize the representation of $5.3 \overline{7}$. on the number line upto $5$ decimal places, that is, up to $5.37777$.
Once again we proceed by successive magnification, and successively decrease the lengths of the portions of the number line in which $5.3 \overline{7}$ is located. First, we see that $5.3 \overline{7}$ is located between $5$ and $6 .$ In the next step, we locate $5.3 \overline{7}$ between $5.3$ and $5.4 .$ To get a more accurate visualization of the representation, we divide this portion of the number line into $10$ equal parts and use a magnifying glass to visualize that $5.3 \overline{7}$ lies between $5.3 \overline{7}$ and $5.38 .$ To visualize $5.3 \overline{7}$ more accurately, we again divide the portion between $5.3 \overline{7}$ and 5.38 into ten equal parts and use a magnifying glass to visualize that $5.3 \overline{7}$ lies between $5.377$ and $5.378 .$ Now to visualize $5.3 \overline{7}$ still more accurately, we divide the portion between $5.377 $ an $5.378$ into $10$ equal parts, and visualize the representation of $5.3 \overline{7}$ as in Fig. $(iv)$. Notice that $5.3 \overline{7}$ is located closer to $5.3778$ than to $5.3777$ [see Fig $(iv)$].
Find :
$(i)$ $9^{\frac{3}{2}}$
$(ii)$ $32^{\frac{2}{5}}$
$(iii)$ $16^{\frac{3}{4}}$
$(iv)$ $125^{\frac{-1}{3}}$
State whether the following statements are true or false. Justify your answers.
$(i)$ Every irrational number is a real number.
$(ii)$ Every point on the number line is of the form $\sqrt m$ , where $m$ is a natural number.
$(iii)$ Every real number is an irrational number.
Show that $3.142678$ is a rational number. In other words, express $3.142678$ in the form $\frac {p }{q }$, where $p$ and $q$ are integers and $q \ne 0$.
Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.
Rationalise the denominator of $\frac{1}{2+\sqrt{3}}$.