Express $0.99999 \ldots$ in the form $\frac{p}{q}$. Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense.
Let $x=0.9999 \ldots$
Multiply both sides by $10,$ we have $[\because$ There is only one repeating digit.]
$10 \times x =10 \times(0.99999 \ldots)$
or $10 x=9.9999 \ldots$
Subtracting $(1)$ from $(2)$, we get
$10 x-x=(9.9999 \ldots)-(0.9999 \ldots)$
or $9 x=9$
or $x=\frac{9}{9}=1$
Thus, $\quad 0.9999 \ldots=1$
As $0.9999 \ldots$ goes on forever, there is no gap between $1$ and $0.9999 \ldots$
Hence both are equal.
Find :
$(i)$ $9^{\frac{3}{2}}$
$(ii)$ $32^{\frac{2}{5}}$
$(iii)$ $16^{\frac{3}{4}}$
$(iv)$ $125^{\frac{-1}{3}}$
Find an irrational number between $\frac {1}{7}$ and $\frac {2}{7}$
Express the following in the form $\frac {p}{q}$, where $p$ and $q$ are integers and $q \ne 0$.
$(i)$ $0 . \overline{6}$
$(ii)$ $0 . 4\overline{7}$
$(iii)$ $0 . \overline{001}$
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.
Represent $ \sqrt{9.3}$ on the number line.