Show that $0.2353535 \ldots=0.2 \overline{35}$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0$.
Show that $0.2353535 \ldots=0.2 \overline{35}$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0$.
Let $x=0.2 \overline{35}$. Over here, note that $2$ does not repeat, but the block $35$ repeats. since two digits are repeating, we multiply $x$ by $100$ to get
$100 x=23.53535 \ldots$
So, $100 x=23.3+0.23535 \ldots=23.3+x$
Therefore, $99 x=23.3$
i.e., $99 x=\frac{233}{10},$ which gives $x=\frac{233}{990}$
You can also check the reverse that $\frac{233}{990}=0.2 \overline{35}$.
Similar Questions
Find :
$(i)$ $9^{\frac{3}{2}}$
$(ii)$ $32^{\frac{2}{5}}$
$(iii)$ $16^{\frac{3}{4}}$
$(iv)$ $125^{\frac{-1}{3}}$
Find :
$(i)$ $9^{\frac{3}{2}}$
$(ii)$ $32^{\frac{2}{5}}$
$(iii)$ $16^{\frac{3}{4}}$
$(iv)$ $125^{\frac{-1}{3}}$
Find the decimal expansions of $\frac{10}{3},\, \frac{7}{8}$ and $\frac{1}{7}$.
Find the decimal expansions of $\frac{10}{3},\, \frac{7}{8}$ and $\frac{1}{7}$.
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 ?
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 ?
Classify the following numbers as rational or irrational :
$(i)$ $2-\sqrt{5}$
$(ii)$ $(3+\sqrt{23})-\sqrt{23}$
$(iii)$ $\frac{2 \sqrt{7}}{7 \sqrt{7}}$
$(iv)$ $\frac{1}{\sqrt{2}}$
$(v)$ $2 \pi$
Classify the following numbers as rational or irrational :
$(i)$ $2-\sqrt{5}$
$(ii)$ $(3+\sqrt{23})-\sqrt{23}$
$(iii)$ $\frac{2 \sqrt{7}}{7 \sqrt{7}}$
$(iv)$ $\frac{1}{\sqrt{2}}$
$(v)$ $2 \pi$
Simplify
$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}}$
$(ii)$ $\left(3^{\frac{1}{5}}\right)^{4}$
$(iii)$ $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}$
$(iv)$ $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}$
Simplify
$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}}$
$(ii)$ $\left(3^{\frac{1}{5}}\right)^{4}$
$(iii)$ $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}$
$(iv)$ $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}$