Find: <br/><br/>(i)$64^{\frac{1}{2}}$ <br/><br/> (ii) $32^{\frac{1}{5}}$ <br/><br/>(iii)$125^{\frac{1}{3}}$
Solution: (i) $64^{\frac{1}{2}}=\left(2^{6}\right)^{\frac{1}{2}}$ $=2^{6 x \frac{1}{2}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$ $=2^{3}=8$ (ii) $32^{\frac{1}{5}}=\left(2^{5}\right)^{\frac{1}{5}}$ $=(2)^{5 \times \frac{1}{5}}$ $\left[\left(a^{m}\right)^{n}=a^{m n n}\right]$ $=2^{1}=2$ (iii) $(125)^{\frac{1}{3}}=\left(5^{3}\right)^{\frac{1}{3}}$ $=5^{3 \times \frac{1}{3}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$ $=5^{1}=5$...
Read More →Rationalise the denominators of the following: <br/> <br/> (i)$\frac{1}{\sqrt{7}}$<br/> <br/> (ii)$\frac{1}{\sqrt{7}-\sqrt{6}}$ <br/> <br/>(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$ <br/> <br/> (iv) $\frac{1}{\sqrt{7}-2}$
Solution: (i) $\frac{1}{\sqrt{7}}=\frac{1 \times \sqrt{7}}{1 \times \sqrt{7}}=\frac{\sqrt{7}}{7}$ (ii) $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1 \quad(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$ $=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$ $=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\frac{\sqrt{7}+\sqrt{6}}{1}=\sqrt{7}+\sqrt{6}$ (iii) $\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$ $=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\frac{\sqrt{...
Read More →Represent √9.3 on the number line.
Solution: Mark a line segment $\mathrm{OB}=9.3$ on number line. Further, take $\mathrm{BC}$ of 1 unit. Find the mid-point $\mathrm{D}$ of $\mathrm{OC}$ and draw a semi-circle on $\mathrm{OC}$ while taking $\mathrm{D}$ as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at $F$. BF is $\sqrt{9.3}$....
Read More →Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi=\frac{c}{d}$. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution: There is no contradiction. When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value. For this reason, we may not realise that either $c$ or $d$ is irrational. Therefore, the fraction $\frac{c}{d}$ is irrational. Hence, $\pi$ is irrational....
Read More →Simplify each of the following expressions: <br/> <br/> (i) $(3+\sqrt{3})(2+\sqrt{2})$ <br/> <br/>(ii) $(3+\sqrt{3})(3-\sqrt{3})$ <br/> <br/>(iii) $(\sqrt{5}+\sqrt{2})^{2}$ <br/> <br/>(iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
Solution: (i) $(3+\sqrt{3})(2+\sqrt{2})=3(2+\sqrt{2})+\sqrt{3}(2+\sqrt{2})$ $=6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}$ (ii) $(3+\sqrt{3})(3-\sqrt{3})=(3)^{2}-(\sqrt{3})^{2}$ $=9-3=6$ (iii) $(\sqrt{5}+\sqrt{2})^{2}=(\sqrt{5})^{2}+(\sqrt{2})^{2}+2(\sqrt{5})(\sqrt{2})$ $=5+2+2 \sqrt{10}=7+2 \sqrt{10}$ (iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^{2}-(\sqrt{2})^{2}$ $=5-2=3$...
Read More →Classify the following numbers as rational or irrational: <br/> <br/>(i)$2-\sqrt{5}$<br/> <br/>(ii) $(3+\sqrt{23})-\sqrt{23}$ <br/> <br/> (iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}$ <br/> <br/> (iv) $\frac{1}{\sqrt{2}}$ <br/> <br/> (v) $2 \pi$
Solution: (i) $2-\sqrt{5}=2-2.2360679 \ldots$ $=-0.2360679 \ldots$ As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. (ii) $(3+\sqrt{23})-\sqrt{23}=3=\frac{3}{1}$ As it can be represented in $\frac{p}{q}$ form, therefore, it is a rational number. (iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}$ As it can be represented in $\frac{p}{q}$ form, therefore, it is a rational number. (iv) $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}=0.7071067...
Read More →Visualise $4 . \overline{26}$ on the number line, up to 4 decimal places.
Solution: $4 . \overline{26}=4.2626 \ldots$ 4.2626 can be visualised as in the following steps....
Read More →Visualise 3.765 on the number line using successive magnification.
Solution: 3.765 can be visualised as in the following steps....
Read More →Classify the following numbers as rational or irrational: <br/><br/>(i) √23<br/><br/> (ii) √225 <br/><br/>(iii) 0.3196 <br/><br/>(iv) 7.478478 <br/><br/>(v) 1.101001000100001…
Solution: (i) $\sqrt{23}=4.79583152331 \ldots$ As the decimal expansion of this number is non-terminating non-recurring, therefore, it is an irrational number. (ii) $\sqrt{225}=15=\frac{15}{1}$ It is a rational number as it can be represented in $\frac{p}{q}$ form. (iii) $0.3796$ As the decimal expansion of this number is terminating, therefore, it is a rational number. (iv) $7.478478 \ldots=7 . \overline{478}$ As the decimal expansion of this number is non-terminating recurring, therefore, it i...
Read More →Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$
Solution: $\frac{5}{7}=0 . \overline{714285}$ $\frac{9}{11}=0.81$ 3 irrational numbers are as follows. $0.73073007300073000073 \ldots$ $0.75075007500075000075 \ldots$ $0.79079007900079000079 \ldots$...
Read More →Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution: 3 numbers whose decimal expansions are non-terminating non-recurring are as follows. $0.505005000500005000005 \ldots$ $0.7207200720007200007200000 \ldots$ $0.080080008000080000080000008 \ldots$...
Read More →Look at several examples of rational numbers in the form $\frac{p}{q}$ (q ≠ 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?
Solution: Terminating decimal expansion will occur when denominator $q$ of rational number $\frac{p}{q}$ is either of $2,4,5,8,10$, and so on... $\frac{9}{4}=2.25$ $\frac{11}{8}=1.375$ $\frac{27}{5}=5.4$ It can be observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both....
Read More →What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.
Solution: It can be observed that, $\frac{1}{17}=0 . \overline{0588235294117647}$ There are 16 digits in the repeating block of the decimal expansion of $\frac{1}{17}$....
Read More →Express 0.99999…in the form $\frac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution: Let $x=0.9999 \ldots$ $10 x=9.9999 \ldots$ $10 x=9+x$ $9 x=9$ $x=1$...
Read More →Express the following in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \overline{6}$<br/> <br/> (ii) $0.4 \overline{7}$ <br/> <br/>(iii) $0 . \overline{001}$
Solution: (i) $0 . \overline{6}=0.666 \ldots$ Let $x=0.666 \ldots$ $10 x=6.666 \ldots$ $10 x=6+x$ $9 x=6$ $x=\frac{2}{3}$ (ii) $0 . \overline{47}=0.4777 \ldots . .$ $=\frac{4}{10}+\frac{0.777}{10}$ $10 x=7.777 \ldots$ $10 x=7+x$ $x=\frac{7}{9}$ $\frac{4}{10}+\frac{0.777 \ldots}{10}=\frac{4}{10}+\frac{7}{90}$ $=\frac{36+7}{90}=\frac{43}{90}$ (iii) $0 . \overline{001}=0.001001 \ldots$ Let $x=0.001001 \ldots$ $1000 x=1.001001 \ldots$ $1000 x=1+x$ $999 x=1$ $x=\frac{1}{999}$...
Read More →You know that $\frac{1}{7}=0 . \overline{142857}$ Can you predict what the decimal expansion of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ are, without actually doing the long division? If so, how?
Solution: Yes. It can be done as follows. $\frac{2}{7}=2 \times \frac{1}{7}=2 \times 0 . \overline{142857}=0 . \overline{285714}$ $\frac{3}{7}=3 \times \frac{1}{7}=3 \times 0 . \overline{142857}=0 . \overline{428571}$ $\frac{4}{7}=4 \times \frac{1}{7}=4 \times 0 . \overline{142857}=0 . \overline{571428}$ $\frac{5}{7}=5 \times \frac{1}{7}=5 \times 0 . \overline{142857}=0 . \overline{714285}$ $\frac{6}{7}=6 \times \frac{1}{7}=6 \times 0 . \overline{142857}=0 . \overline{857142}$...
Read More →Show how √5 can be represented on the number line.
Solution: We know that, $\sqrt{4}=2$ And, $\sqrt{5}=\sqrt{(2)^{2}+(1)^{2}}$ Mark a point ‘A’ representing 2 on number line. Now, construct AB of unit length perpendicular to OA. Then, taking O as centre and OB as radius, draw an arc intersecting number line at C. $C$ is representing $\sqrt{5}$....
Read More →Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution: If numbers such as $\sqrt{4}=2, \sqrt{9}=3$ are considered, Then here, 2 and 3 are rational numbers. Thus, the square roots of all positive integers are not irrational....
Read More →State whether the following statements are true or false. Give reasons for your answers. <br/><br/>(i) Every natural number is a whole number. <br/><br/> (ii) Every integer is a whole number. <br/><br/> (iii) Every rational number is a whole number.
Solution: (i) True; since the collection of whole numbers contains all natural numbers. (ii) False; as integers may be negative but whole numbers are positive. For example: −3 is an integer but not a whole number. (iii) False; as rational numbers may be fractional but whole numbers may not be. For example: $\frac{1}{5}$ is a rational number but not a whole number....
Read More →Find five rational numbers between 3/5 and 4/5
Solution: There are infinite rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ $\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}$ $\frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}$ Therefore, rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ are $\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}$...
Read More →Find six rational numbers between 3 and 4.
Solution: There are infinite rational numbers in between 3 and 4. 3 and 4 can be represented as $\frac{24}{8}$ and $\frac{32}{8}$ respectively. Therefore, rational numbers between 3 and 4 are $\frac{25}{8}, \frac{26}{8}, \frac{27}{8}, \frac{28}{8}, \frac{29}{8}, \frac{30}{8}$...
Read More →Is zero a rational number? Can you write it in the form, where p and q are integers and q ≠ 0?
Yes. Zero is a rational number as it can be represented as $\frac{0}{1}$ or $\frac{0}{2}$ or $\frac{0}{3}$ etc....
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