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Class 9th Ncert Solutions

Number Systems


Exercise 1.1

Solution :

We know that a number is said to be a rational number if it can be represented in the form pq where p and q are integers and q0.

Now, we say that zero is a rational number if it can be represented in the pq form.

Further, we see that zero can be represented as 01,02,03 etc.

Therefore, zero is a rational number.

Solution :

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

3=31×1010=3010

4=41×1010=4010

 The required rational numbers are 3110,3210,3310,3410,3510,3610.

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, 23,34,45 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 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,.4=2 and 9=3

Here, 2 and 3 are rational.

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

Solution :

To represent 5 on the number line,

By Pythagoras theorem, we get

OA2=OB2+AB2

OA2=(1)2+(2)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 5.

OA2=1+4

OA2=5

OA=5

Exercise 1.3

Solution :

36100=0.36; Terminating.

Solution :

111=0.90909; Non terminating and recurring decimal.

Solution :

418=338=4.125; Terminating.

Solution :

313=0.230769230; Non terminating and recurring decimal.

Solution :

211=0.1818181818; Non terminating and recurring decimal.

Solution :

329400=0.8225; Terminating.

Solution :

Given,

17=0.¯142857

27=2×17=2×0.¯142857=0.¯285714

37=3×17=3×0.¯142857=0.¯428571

47=4×17=4×0.¯142857=0.¯571428

57=5×17=5×0.¯142857=0.¯714285

67=6×17=6×0.¯142857=0.¯857142

Solution :

Let x=0.¯6

       x=0.6666   ....(1)

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

  10x=6.6666....(2)

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

10x=6.6666....

x=0.6666...._

9x=6

x=69

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:

  1. We first need to mark a line segment OB=9.3 on number line.
  2. Now, take BC of 1 unit.
  3. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre.
  4. Draw a perpendicular to line OC passing through point B.
  5. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F.
  6. 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}}