Identify which of the following are rational and irrational numbers:
(i) $\sqrt{3}$ (ii) $\frac{1}{6}$ (iii) $\pi$ (iv) $\frac{15}{2}$ (v) $7.25$ (vi)$\sqrt{29}$
Solution
Convert the following fraction into decimal fraction.
(i) $\frac{17}{25}$ (ii) $\frac{19}{4}$ (iii)$\frac{57}{8}$
(iv) $\frac{205}{18}$ (v) $\frac{5}{8}$ (vi) $\frac{25}{38}$
Soluation
Which of the statements are true and which of the false?
Soluaton
Represent the following numbers on the number line.
(i) $\frac{2}{3}$ (ii) $-\frac{4}{5}$ (iii) $\frac{3}{4}$
(iv) $-2\frac{5}{8}$ (v) $2\frac{3}{4}$ (vi) $\sqrt{5}$
Soluation
(i)
(ii)
(iii)
(iv)
(v)
(vi)
$$(m OB)^2 = (m OA)^2 + (m AB)^2 \quad (Pythagorean\,\, Theorem)$$ $$(m \overline{OB})^2 = 4 + 1 = 5 $$ therefore $$m \overline{OB} = \sqrt{5}$$ thus $$m \overline{OB} = m \overline{OM} = \sqrt{5}$$
Give a rational number between $\frac{3}{4}$ and $\frac{5}{9}$.
Soluation
The mean of the numbers is between given numbers. Therefore $\frac{3/4+5/9}{2}= \frac{47}{72}$ is a number between required numbers.
Express the following recurring decimals as the rational number $\frac{p}{q}$, where p , q are integers and $q \neq 0$.
(i) $0.\overline{5}$ (ii) $0.\overline{13}$ (iii) $0.\overline{67}$
Soluation
(i) Let $x = 0.\overline{5}$. That is $$x = 0.5555....\qquad (1)$$
Only one digit 5 is being repeated, multiply 10 on both sides
$$10x = (0.5555.....) \times 10$$ $$10x = 5.5555..... \qquad (2)$$ Subtract (1) from (2), we get
$$9x = 5$$ $$\therefore \,\, x = \frac{5}{9}$$ $$0.\overline{5} = \frac{5}{9}$$
(ii) Let $x = 0.\overline{13}$. That is $$x = 0.13131313.... \qquad (1)$$ Only one digit 13 is being repeated, multiply 100 on both sides $$100x = (0.13131313.....) \times 100$$ $$100x = 13.13131313..... \qquad (2)$$ Subtract (1) from (2) $$99x = 13$$ $$\therefore \,\, x = \frac{13}{99}$$ $$0.\overline{13} = \frac{13}{99}$$
(iii) Let $x = 0.\overline{67}$. That is $$x = 0.67676767.... \qquad (1)$$ Only one digit 67 is being repeated, multiply 100 on both sides $$100x = (0.67676767.....) \times 100$$ $$100x = 67.67676767..... \qquad (2)$$ Subtract (1) from (2) $$99x = 67$$ $$\therefore\,\, x = \frac{67}{99}$$ $$0.\overline{67} = \frac{67}{99}$$