Question

Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$.
(i) $0.\overline{47}$
(ii) $0.\overline{001}$
(iii) $0.5\overline{7}$
(iv) $0.2\overline{45}$
(v) $0.\overline{6}$
(vi) $1.\overline{5}$.

Answer 1

(i) Let $x=0.\overline{47}$. Then $x=0.474747$....
Since two digits are repeating, multiplying both sides by $100$, we get
$100x=\mathrm{47.474747...}=47+\mathrm{0.474747...}=47+x$
$99x=47$
$x=\frac{47}{99}$
$\therefore 0.\overline{47}=\frac{47}{99}$

(ii) Let $x=0.\overline{001}$. Then $x=\mathrm{0.001001001...}$
Since three digits are repeating, multiplying both sides by $1000$, we get
$1000x=\mathrm{1.001001001...}=1+\mathrm{0.001001001...}=1+x$
$1000x-x=1$
$999x=1$
$x=\frac{1}{999}$
$\therefore 0.\overline{001}=\frac{1}{999}$

(iii) Let $x=0.5\overline{7}$. Then $x=0.57777$....
Multiplying both sides by $10,$ we get
$10x=\mathrm{5.7777....}=5.2+\mathrm{0.57777...}=5.2+x$
$9x=5.2$
$x=\frac{5.2}{9}$
$x=\frac{52}{90}$
$\therefore 0.5\overline{7}=\frac{52}{90}=\frac{26}{45}$

(iv) Let $x=0.2\overline{45}$. Then $x=\mathrm{0.2454545...}$
Multiplying both sides by $100$, we get
$100x=\mathrm{24.545454...}=24.3+\mathrm{0.2454545...}=24.3+x$
$99x=24.3$
$x=\frac{24.3}{99}$
$0.2\overline{45}=\frac{243}{990}=\frac{27}{110}$

(v) Let $x=0.\overline{6}$. Then $x=0.66666$...
Multiplying both sides by $10$, we get
$10x=\mathrm{6.66666....}=6+\mathrm{0.6666...}=6+x$
$9x=6$
$x=\frac{6}{9}=\frac{2}{3}$
$\therefore 0.\overline{6}=\frac{2}{3}$

(vi) Let $x=1.\overline{5}$. Then $x=\mathrm{1.55555...}$
Multiplying both sides by $10$, we get
$10x=\mathrm{15.5555...}=14+\mathrm{1.5555...}=114+x$
$9x=14$
$x=\frac{14}{9}$
$\therefore 1.\overline{5}=1\frac{5}{9}$
So, every number with a non-terminating and recurring decimal expansion can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ not equal to zero.