Exercise 1.1

Solution :

We know that a number is said to be a rational number if it can be represented in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \ne 0.$

Now, we say that zero is a rational number if it can be represented in the $\frac{p}{q}$ form.

Further, we see that zero can be represented as $\frac{0}{1},\frac{0}{2},\frac{0}{3}$ etc.

Therefore, zero is a rational number.

Solution :

An infinite number of rational numbers are possible between 3 and 4.

$3 = \frac{3}{1} \times \frac{{10}}{{10}} = \frac{{30}}{{10}}$

$4 = \frac{4}{1} \times \frac{{10}}{{10}} = \frac{{40}}{{10}}$

The required rational numbers are $\frac{{31}}{{10}},\frac{{32}}{{10}},\frac{{33}}{{10}},\frac{{34}}{{10}},\frac{{35}}{{10}},\frac{{36}}{{10}}.$

Solution :

(i) True, because we can say that whole numbers are nothing but natural numbers including zero. Therefore, every natural number is a whole number but every whole number is not a natural number. As 0 is not a natural number.

(ii) False, because integers include both positive and negative numbers. The whole numbers include only positive numbers and negative numbers are not whole numbers. Therefore, every integer is a not a whole number.

(iii) False, because rational numbers can also be in the form of fractions and these fractional numbers are not whole numbers. For example, $\frac{2}{3},\frac{3}{4},\frac{4}{5}$ are not whole numbers but they are rational numbers.

Exercise 1.2

Solution :

(i) True, since the real numbers are nothing but a combination of rational and irrational numbers.

(ii) N False, because the negative numbers on the number line cannot be expressed in the form $\sqrt m$

(iii) False, because real numbers contain both rational and irrational numbers. Therefore, every real number cannot be irrational.

Solution :

No, the square root of all positive numbers need not be irrational.

For example,. $\sqrt {\rm{4}} = {\rm{2}}\,$ and $\sqrt {\rm{9}} = {\rm{3}}$

Here, 2 and 3 are rational.

Therefore, the square roots of all positive integers are not irrational.

Solution :

To represent $\sqrt 5$ on the number line,

By Pythagoras theorem, we get

${\rm{O}}{{\rm{A}}^{\rm{2}}} = {\rm{O}}{{\rm{B}}^{\rm{2}}} + {\rm{A}}{{\rm{B}}^{\rm{2}}}$

${\rm{O}}{{\rm{A}}^{\rm{2}}} = {\left( 1 \right)^{\rm{2}}} + {\left( 2 \right)^{\rm{2}}}$

Take OB=2 units and make a perpendicular at B so that AB=1 unit.

Now, taking O as center and OB as radius, draw an arc intersecting number line at C.

OC is the required distance that represents $\sqrt 5$ .

${\rm{O}}{{\rm{A}}^{\rm{2}}} = 1 + 4$

${\rm{O}}{{\rm{A}}^{\rm{2}}} = 5$

${\rm{OA}} = \sqrt 5$

Exercise 1.3

Solution :

$\frac{{{\rm{36}}}}{{{\rm{1}}00}} = 0.{\rm{36}} \ldots \ldots ;$ Terminating.

Solution :

$\frac{{\rm{1}}}{{{\rm{11}}}} = 0.{\rm{9}}0{\rm{9}}0{\rm{9}} \ldots \ldots$ ; Non terminating and recurring decimal.

Solution :

$4\frac{1}{8} = \frac{{{\rm{33}}}}{{\rm{8}}} = {\rm{4}}.{\rm{125}} \ldots \ldots$ ; Terminating.

Solution :

$\frac{{\rm{3}}}{{{\rm{13}}}} = 0.{\rm{23}}0{\rm{76923}}0 \ldots \ldots$ ; Non terminating and recurring decimal.

Solution :

$\frac{{\rm{2}}}{{{\rm{11}}}} = 0.{\rm{1818181818}} \ldots \ldots$ ; Non terminating and recurring decimal.

Solution :

$\frac{{{\rm{329}}}}{{{\rm{4}}00}} = 0.{\rm{8225}}$ ; Terminating.

Solution :

Given,

$\frac{{\rm{1}}}{{\rm{7}}} = 0.\,\overline {{\rm{142857}}}$

$\frac{{\rm{2}}}{{\rm{7}}} = {\rm{2}} \times \frac{{\rm{1}}}{{\rm{7}}} = {\rm{2}} \times 0.\,\,\overline {{\rm{142857}}} = 0.\,\overline {{\rm{285714}}}$

$\frac{{\rm{3}}}{{\rm{7}}} = {\rm{3}} \times \frac{{\rm{1}}}{{\rm{7}}} = {\rm{3}} \times 0.\,\,\overline {{\rm{142857}}} = 0.\,\overline {{\rm{428571}}}$

$\frac{{\rm{4}}}{{\rm{7}}} = {\rm{4}} \times \frac{{\rm{1}}}{{\rm{7}}} = {\rm{4}} \times 0.\,\overline {\,{\rm{142857}}} = 0.\,\overline {{\rm{571428}}}$

$\frac{{\rm{5}}}{{\rm{7}}} = {\rm{5}} \times \frac{{\rm{1}}}{{\rm{7}}} = {\rm{5}} \times 0.\,\,\overline {{\rm{142857}}} = 0.\,\overline {{\rm{714285}}}$

$\frac{{\rm{6}}}{{\rm{7}}} = {\rm{6}} \times \frac{{\rm{1}}}{{\rm{7}}} = {\rm{6}} \times 0.\,\,\overline {{\rm{142857}}} = 0.\,\overline {{\rm{857142}}}$

Solution :

Let $x = 0.\,\overline 6 \,\,$

$x = 0.\,6666$ $....\left( 1 \right)$

Multiplying 10 on both the sides of equation (1), we get,

$10x = 6.6666\,\,\,\,\,\,\,\,\,....\left( 2 \right)$

Subtracting equation (2) from (1) we get,

$\,\,\,\,10x = 6.6666....\,\,\,\,\,\,\,$

$\underline {\,\,\,\,\,\,\,\,\,x = 0.\,6666....\,\,\,\,} \,\,$

$\,\,\,\,\,\,\,9x = 6$

$\Rightarrow \,\,\,\,x = \frac{6}{9}$

$\therefore \,\,\,\,\,\,x = \frac{2}{3}$

Solution :

Let $x = 0.\,4\overline 7 \,\,$

$x = 0.\,4777......$

Multiplying 10 on both the sides, we get,

$10x = 4.7777\,\,\,\,\,\,\,\,....\left( 1 \right)$

Multiplying 10 on both the sides of equation (1), we get,

$100x = 47.7777\,\,\,\,\,\,\,\,\,....\left( 2 \right)$

Subtracting equation (2) from (1) we get,

$\,\,\,\,100x = 47.7777....\,\,\,\,\,\,\,$

$\underline {\,\,\,\,\,\,\,10x = \,\,\,4.\,7777....\,\,\,\,} \,\,$

$\,\,\,\,\,\,\,90x = 43$

$\therefore \,\,\,\,\,\,x = \frac{{43}}{{90}}$

Solution :

Let $x = 0.\,\overline {001} \,\,$

$x = 0.\,001001\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Multiplying 1000 on both the sides of equation (1), we get,

$1000\,x = 1.001001\,\,\,\,\,\,\,\,\,\,\,\,....\left( 2 \right)$

Subtracting equation (2) from (1) we get,

$1000\,x = 1.001001....\,\,\,\,$

$\underline {\,\,\,\,\,\,\,\,\,\,\,x = 0.\,001001....\,\,} \,$

$\,\,\,999x = 1$

$\therefore \,\,\,\,\,\,\,\,x = \frac{1}{{999}}$

Solution :

Let

$x = 0.\,9999\,\,\,\,\,\,\,\,\,....\left( 1 \right)$

Multiplying 10 on both the sides of equation (1), we get,

$10x = 9.9999\,\,\,\,\,\,\,\,\,....\left(2 \right)$

Subtracting equation (2) from (1) we get,

$\,\,\,\,10x = 9.9999....\,\,\,\,\,$

$\underline {\,\,\,\,\,\,\,\,\,x = 0.\,9999....\,\,\,\,} \,$

$\,\,\,\,\,\,\,9x = 9$

$\Rightarrow \,\,\,\,x = \frac{9}{9}$

$\therefore \,\,\,\,\,\,x = 1$

Therefore 0.9999…. is very close to 1. Therefore, 0.99999 can be approximated to 1 and hence they are equal.

Solution :

$\frac{1}{{17}}$

Therefore, the maximum number of digits that can be in the repeating block of digits in the above expansion are 16.

Solution :

$\frac{3}{{20}} = 0.15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {20 = \,2 \times 2 \times 5} \right)$

$\frac{7}{{25}} = 0.28\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {25 = 5 \times 5} \right)$

$\frac{{21}}{8} = 2.375\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {8 = 2 \times 2 \times 2} \right)$

$\frac{{14}}{{75}} = 0.186666....\,\,\,\,\,\,\,\,\,\,\,\left( {75 = 3 \times 5 \times 5} \right)$

Terminating decimal expansion will occur when denominator of rational number are in the powers of 2 or 5 or both.

Solution :

There are infinite numbers having non-terminating and non-recurring expansions.

All irrational numbers are non-terminating non-recurring. Some examples are,

(i) $0.3756294....$

(ii) $\sqrt 3$

(iii) $11.47421......$

Solution :

By performing long division, the numbers can be represented as

$\frac{5}{7} = 0.{\rm{714285}} \ldots$ and $\frac{9}{{11}} = 0.{\rm{81818181}} \ldots$

Therefore, the required numbers can be:

(i) 0.720720072000...

(ii) 0.751751175111…

(iii) 0.770770077000777…

Solution :

$\sqrt {23} = {\rm{4}}.{\rm{79583152331}}......$

In the above number, the expansion is non-terminating and non-recurring; therefore, it is an irrational number.

Solution :

$\sqrt {225} = 15$

As the above number can be represented in $\frac{p}{q}$ form, it is a rational number.

Solution :

$0.{\rm{3796}}$

As the above number has terminating decimal expansion, it is a rational number.

Solution :

${\rm{7}}.{\rm{478478}}...... = {\rm{7}}.\,\overline {{\rm{478}}}$

From the decimals it is a recurring number, but it is non-terminating. Therefore, it is a rational number.

Solution :

$1.{\rm{1}}0{\rm{1}}00{\rm{1}}000{\rm{1}}0000{\rm{1}}......$

We see that the decimal expansion of this number is non-terminating and non-recurring. Therefore, it is an irrational number.

Exercise 1.4

Solution :

$2 - \sqrt 5 = 2 - 2.2360679... = - \,0.2360679.....$

We see that the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Solution :

$\left( {3 + \sqrt {23} } \right) - \sqrt {23} = 3 + \sqrt {23} - \sqrt {23} = 3$

We see that the number can be represented in $\frac{p}{q}$ form. Therefore, it is a rational number.

Solution :

$\frac{{2\sqrt 7 }}{{7\sqrt 7 }} = \frac{2}{7}$

The number can be represented in $\frac{p}{q}$ form. Therefore, it is a rational number.

Solution :

$\frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{\sqrt 2 }}{2} = {\rm{0}}{\rm{.7071067811}}.....$

We see that the decimal expansion of the above expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Solution :

$2\pi$

The decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Solution :

Given,

$\,\,\,\,\left( {3 + \sqrt 3 } \right)\left( {2 + \sqrt 2 } \right)$

$= \,6 + 3\sqrt 2 + 2\sqrt 3 + \sqrt 6$

Solution :

Given,

$\,\,\,\,\,\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)$

$= \,{\left( 3 \right)^2} - {\left( {\sqrt 3 } \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= 9 - 3$

$= 6$

Solution :

Given,

$\,\,\,\,\,{\left( {\sqrt 5 + \sqrt 2 } \right)^2}$

$= {\left( {\sqrt 5 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2\left( {\sqrt 5 } \right)\left( {\sqrt 2 } \right)\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab} \right]$

$= 5 + 2 + 2\sqrt {10}$

$= 7 + 2\sqrt {10}$

Solution :

Given,

$\,\,\,\,\,\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)$

$= \,\,{\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 2 } \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= \,5 - 2$

$= 3$

Solution :

There is no contradiction at all. When we measure a length with scale or any other instrument, we only obtain an approximate rational value which is rational. We never obtain an exact value. For this reason, we may not realize, that either c or d is irrational. Therefore, the fraction $\frac{c}{d}$ is irrational. Hence, $\pi$ is irrational.

Solution :

Steps:

We first need to mark a line segment OB=9.3 on number line.
Now, take BC of 1 unit.
Find the mid-point D of OC and draw a semi-circle on OC while taking 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 $\sqrt {{\rm{9}}.{\rm{3}}}$

Solution :

$\frac{1}{{\sqrt 7 }} \times \frac{{\sqrt 7 }}{{\sqrt 7 }} = \frac{{\sqrt 7 }}{7}$

Solution :

$\,\,\,\,\frac{1}{{\sqrt 7 - \sqrt 6 }} \times \frac{{\sqrt 7 + \sqrt 6 }}{{\sqrt 7 + \sqrt 6 }}$

$= \,\frac{{\sqrt 7 + \sqrt 6 }}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= \,\frac{{\sqrt 7 + \sqrt 6 }}{{7 - 6}}$

$= \frac{{\sqrt 7 + \sqrt 6 }}{1}$

$= \sqrt 7 + \sqrt 6$

Solution :

$\,\,\,\,\,\frac{1}{{\sqrt 5 + \sqrt 2 }} \times \frac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}$

$= \frac{{\sqrt 5 - \sqrt 2 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= \,\frac{{\sqrt 5 - \sqrt 2 }}{{5 - 2}}$

$= \,\frac{{\sqrt 5 - \sqrt 2 }}{3}$

Solution :

$\,\,\,\,\,\frac{1}{{\sqrt 7 - 2}} \times \frac{{\sqrt 7 + \,\,2}}{{\sqrt 7 + \,\,2}}$

$= \,\,\frac{{\sqrt 7 + \,\,2}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( 2 \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= \,\frac{{\sqrt 7 + \,\,2}}{{7 - 4}}$

$= \,\,\frac{{\sqrt 7 + \,\,2}}{3}$

Exercise 1.5

Solution :

$\,\,\,\,{\left( {64} \right)^{\frac{1}{2}}}$

$= \,{\left( {{8^2}} \right)^{\frac{1}{2}}}$

$= \,\,{8^{2 \times \frac{1}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{8^1}$

$= \,\,8$

Solution :

$\,\,\,\,{\left( {32} \right)^{\frac{1}{5}}}$

$= \,{\left( {{2^5}} \right)^{\frac{1}{5}}}$

$= \,\,{2^{5 \times \frac{1}{5}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{2^1}$

$= \,\,2$

Solution :

$\,\,\,\,{\left( {125} \right)^{\frac{1}{3}}}$

$= \,{\left( {{5^3}} \right)^{\frac{1}{3}}}$

$= \,\,{5^{3 \times \frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{5^1}$

$= \,\,5$

Solution :

$\,\,\,\,{\left( 9 \right)^{\frac{3}{2}}}$

$= \,{\left( {{3^2}} \right)^{\frac{3}{2}}}$

$= \,\,{3^{2 \times \frac{3}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{3^3}$

$= \,\,27$

Solution :

$\,\,\,\,{\left( {32} \right)^{\frac{2}{5}}}$

$= \,{\left( {{2^5}} \right)^{\frac{2}{5}}}$

$= \,\,{2^{5 \times \frac{2}{5}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{2^2}$

$= \,\,4$

Solution :

$\,\,\,\,{\left( {16} \right)^{\frac{3}{4}}}$

$= \,{\left( {{2^4}} \right)^{\frac{3}{4}}}$

$= \,\,{2^{4\, \times \,\frac{3}{4}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{2^3}$

$= \,\,8$

Solution :

$\,\,\,\,{\left( {125} \right)^{\frac{{ - 1}}{3}}}$

$= \,{\left( {{5^3}} \right)^{\frac{{ - 1}}{3}}}$

$= \,\,{5^{3 \times \left( { - \,\,\frac{1}{3}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,{5^{ - \,1}}$

$= \,\,\frac{1}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{a^{ - m}} = \frac{1}{{{a^m}}}} \right]$

Solution :

$\,\,\,\,\,{2^{\frac{2}{3}}}{.2^{\frac{1}{5}}}$

$= \,\,{2^{\frac{2}{3} + \frac{1}{5}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{a^m} \times {a^n} = {a^{m + n}}} \right]$

$= \,{2^{\frac{{10 + 3}}{{15}}}}$

$= {2^{\frac{{13}}{{15}}}}$

Solution :

$\,\,\,\,\,{\left( {\frac{1}{{{3^3}}}} \right)^7}$

$= \,{\left( {{3^{ - 3}}} \right)^7}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{a^{ - m}} = \frac{1}{{{a^m}}}} \right]$

$= {3^{\left( { - 3} \right) \times 7}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,{3^{ - 21}}$

Solution :

$\,\,\,\,\frac{{{{11}^{\frac{1}{2}}}}}{{{{11}^{\frac{1}{4}}}}}$

$= \,{11^{\frac{1}{2} - \frac{1}{4}}}\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right]$

$= {11^{\frac{1}{4}}}$

Solution :

$\,\,\,\,\,{7^{\frac{1}{2}}} \times {8^{\frac{1}{2}}}$

$= \,{\left( {7 \times 8} \right)^{\frac{1}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{since}}\,\,{a^m}\, \times \,\,{b^m} = {{\left( {ab} \right)}^m}} \right]$

$= \,{56^{\frac{1}{2}}}$

Exercise 1.1

Solution :

There are infinite number of rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ .

$\frac{3}{5} \times \frac{{10}}{{10}} = \frac{{30}}{{50}}$

$\frac{4}{5} \times \frac{{10}}{{10}} = \frac{{40}}{{50}}$

Therefore, the required numbers are $\frac{{31}}{{50}},\frac{{32}}{{50}},\frac{{33}}{{50}},\frac{{34}}{{50}},\frac{{35}}{{50}}$

Class 9th Ncert Solutions

Chapter List

Number Systems

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.5

Exercise 1.1

Class 9th Ncert Solutions

Chapter List

Number Systems

Exercise 1.1

Q 1. : Is zero a rational number? Can you write it in the form pq,\frac{p}{q}, where pp and qq are integers and pp

Q 2. : Find six rational numbers between 3 and 4.

Q 4. : State whether the following statements are true or false. Give reasons for your answers. (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number.

Exercise 1.2

Q 1. : 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 √m\sqrt m , where mm is a natural number. (iii) Every real number is an irrational number.

Q 2. : 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.

Q 3. : Show how √5\sqrt 5 can be represented on the number line.

Exercise 1.3

Q 1.(i) : Write the following in decimal form and say what kind of decimal expansion each has: 36100\frac{{36}}{{100}}

Q 1.(ii) : Write the following in decimal form and say what kind of decimal expansion each has: 111\frac{1}{{11}}

Q 1.(iii) : Write the following in decimal form and say what kind of decimal expansion each has : 4184\frac{1}{8}

Q 1.(iv) : Write the following in decimal form and say what kind of decimal expansion each has : 313\frac{3}{{13}}

Q 1.(v) : Write the following in decimal form and say what kind of decimal expansion each has : 211\frac{2}{{11}}

Q 1.(vi) : Write the following in decimal form and say what kind of decimal expansion each has : 329400\frac{{329}}{{400}}

Q 3.(i) : Express the following in the form pq\frac{p}{q}, where pp and qq are integers and q≠0.q \ne 0. 0.¯60.\,\overline 6

Q 3.(ii) : Express the following in the form \frac{p}{q}\frac{p}{q}, where pp and qq are integers and q \ne 0.q \ne 0. 0.\,\,4\overline 7 0.\,\,4\overline 7

Q 3.(iii) : Express the following in the form \frac{p}{q}\frac{p}{q}, where pp and qq are integers and q \ne 0.q \ne 0. 0.\,\overline {001} 0.\,\overline {001}

Q 4. : Express 0.99999…… in the form \frac{p}{q}\frac{p}{q}. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Q 5. : What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \frac{1}{{17}}\frac{1}{{17}}? Perform the division to check your answer.

Q 6. : Look at several examples of rational numbers in the form \frac{p}{q}\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?

Q 7. : Write three numbers whose decimal expansions are non-terminating non-recurring.

Q 8. : Find three different irrational numbers between the rational numbers \frac{5}{7}\frac{5}{7} and \frac{9}{{11}}\frac{9}{{11}}.

Q 9.(i) : Classify the following numbers as rational or irrational : \sqrt {23} \sqrt {23}

Q 9.(ii) : Classify the following numbers as rational or irrational : \sqrt {225} \sqrt {225}

Q 9.(iii) : Classify the following numbers as rational or irrational : 0.3796

Q 9.(iv) : Classify the following numbers as rational or irrational : 7.478478

Q 9.(v) : Classify the following numbers as rational or irrational : 1.101001000100001…

Exercise 1.4

Q 1.(i) : Classify the following numbers as rational or irrational : 2 - \sqrt 5 2 - \sqrt 5

Q 1.(ii) : Classify the following numbers as rational or irrational : \left( {3 + \sqrt {23} } \right) - \sqrt {23} \left( {3 + \sqrt {23} } \right) - \sqrt {23}

Q 1.(iii) : Classify the following numbers as rational or irrational : \frac{{2\sqrt 7 }}{{7\sqrt 7 }}\frac{{2\sqrt 7 }}{{7\sqrt 7 }}

Q 1.(iv) : Classify the following numbers as rational or irrational : \frac{1}{{\sqrt 2 }}\frac{1}{{\sqrt 2 }}

Q 1.(v) : Classify the following numbers as rational or irrational : 2\pi 2\pi

Q 2.(i) : Simplify each of the following expressions : \left( {3 + \sqrt 3 } \right)\left( {2 + \sqrt 2 } \right)\left( {3 + \sqrt 3 } \right)\left( {2 + \sqrt 2 } \right)

Q 2.(ii) : Simplify each of the following expressions : \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)

Q 2.(iii) : Simplify each of the following expressions : {\left( {\sqrt 5 + \sqrt 2 } \right)^2}{\left( {\sqrt 5 + \sqrt 2 } \right)^2}

Q 2.(iv) : Simplify each of the following expressions : \left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)

Q 3. : 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}\pi = \frac{c}{d}. This seems to contradict the fact that \pi \pi is irrational. How will you resolve this contradiction?

Q 4. : Represent \sqrt {{\rm{9}}.{\rm{3}}} \sqrt {{\rm{9}}.{\rm{3}}} on the number line.

Q 5.(i) : Rationalise the denominators of the following: \frac{1}{{\sqrt 7 }}\frac{1}{{\sqrt 7 }}

Q 5.(ii) : Rationalise the denominators of the following: \frac{1}{{\sqrt 7 - \sqrt 6 }}\frac{1}{{\sqrt 7 - \sqrt 6 }}

Q 5.(iii) : Rationalise the denominators of the following: \frac{1}{{\sqrt 5 + \sqrt 2 }}\frac{1}{{\sqrt 5 + \sqrt 2 }}

Q 5.(iv) : Rationalise the denominators of the following: \frac{1}{{\sqrt 7 - 2}}\frac{1}{{\sqrt 7 - 2}}

Exercise 1.5

Q 1.(I) : Find: {\left( {64} \right)^{\frac{1}{2}}}{\left( {64} \right)^{\frac{1}{2}}}

Q 1.(ii) : Find : {\left( {32} \right)^{\frac{1}{5}}}{\left( {32} \right)^{\frac{1}{5}}}

Q 1.(iii) : Find : {\left( {125} \right)^{\frac{1}{3}}}{\left( {125} \right)^{\frac{1}{3}}}

Q 2.(i) : Find : {\left( 9 \right)^{\frac{3}{2}}}{\left( 9 \right)^{\frac{3}{2}}}

Q 2.(ii) : Find : {\left( {32} \right)^{\frac{2}{5}}}{\left( {32} \right)^{\frac{2}{5}}}

Q 2.(iii) : Find : {\left( {16} \right)^{\frac{3}{4}}}{\left( {16} \right)^{\frac{3}{4}}}

Q 2.(iv) : Find : {\left( {125} \right)^{\frac{{ - 1}}{3}}}{\left( {125} \right)^{\frac{{ - 1}}{3}}}

Q 3.(i) : Find : {2^{\frac{2}{3}}}{.2^{\frac{1}{5}}}{2^{\frac{2}{3}}}{.2^{\frac{1}{5}}}

Q 3.(ii) : Find : {\left( {\frac{1}{{{3^3}}}} \right)^7}{\left( {\frac{1}{{{3^3}}}} \right)^7}

Q 3.(iii) : Find : \frac{{{{11}^{\frac{1}{2}}}}}{{{{11}^{\frac{1}{4}}}}}\frac{{{{11}^{\frac{1}{2}}}}}{{{{11}^{\frac{1}{4}}}}}

Q 3.(iv) : Find : {7^{\frac{1}{2}}} \times {8^{\frac{1}{2}}}{7^{\frac{1}{2}}} \times {8^{\frac{1}{2}}}

Exercise 1.1

Q 3. : Find five rational numbers between \frac{3}{5}\frac{3}{5} and \frac{4}{5}\frac{4}{5}.