# NCERT Solutions for Class 9 Maths Chapter 1 | NCERT class 9th maths

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# NCERT Solutions for Class 9 Maths Chapter 1 Number System

### class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.1**

**1. Is zero a rational number? Can you write it in the form \(\displaystyle \frac{p}{q}\) , where p and q are integers and q ≠ 0?**

**Solutions :**

Yes, zero is a rational number and it can be written as \(\displaystyle \frac{p}{q}\)

Example : \(\displaystyle \frac{0}{5},\,\,\frac{0}{{10}},\,\,\frac{0}{1}\) etc.

**2. Find six rational numbers between 3 and 4.**

**Solutions :**

On multiplying and dividing the numbers 3 and 4 by the numbers 7 (3 + 4) –

\(\displaystyle 3\times \frac{8}{8}=\frac{{24}}{8}\)

\(\displaystyle 4\times \frac{8}{8}=\frac{{32}}{8}\)

Hence six rational numbers between 3 and 4 are \(\displaystyle \frac{{25}}{8},\,\,\frac{{26}}{8},\,\,\frac{{27}}{8},\,\,\frac{{28}}{8},\,\,\frac{{29}}{8},\,\,\frac{{30}}{8}\)

**3. Find five rational numbers between \(\displaystyle \frac{3}{5}\) and \(\displaystyle \frac{4}{5}\) .**

**Solutions :**

Both the rational numbers have the same denominator, so on multiplying and dividing both by the number 6 (5 + 1) –

\(\displaystyle \frac{3}{5}\times \frac{6}{6}=\frac{{18}}{{24}}\)

\(\displaystyle \frac{4}{5}\times \frac{6}{6}=\frac{{24}}{{30}}\)

So, five rational number between \(\displaystyle \frac{3}{5}\) and \(\displaystyle \frac{4}{5}\) are \(\displaystyle \frac{{19}}{{30}},\,\,\frac{{20}}{{30}},\,\,\frac{{21}}{{30}},\,\,\frac{{22}}{{30}},\,\,\frac{{23}}{{30}}\) .

**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. |

**Solutions : **

**(i) Every natural number is a whole number.**

This statement is true because natural numbers start from 1 and go on till infinity whereas whole numbers start from 0 which go on till infinity. Therefore, only whole numbers are natural numbers.

**(ii) Every integer is a whole number.**

This statement is false because integer numbers start from 0 and go to infinity on both sides of the number line whereas whole numbers start from 0 which go till infinity. So not all whole numbers are integer numbers.

**(iii) Every rational number is a whole number.**

This statement is false, because whole numbers do not have fractions, whereas rational numbers have fractions as well as decimal numbers. Ex. \(\displaystyle \frac{1}{2}\) is a rational number, but not a whole number.

## NCERT Solutions for class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.2**

**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 \(\displaystyle \sqrt{m}\) , where m is a natural number. | |

(iii) | Every real number is an irrational number. |

**Solutions :**

**(i) Every irrational number is a real number.**

This statement is true, because the collection of real numbers consists of rational and irrational numbers, so every irrational number is real.

**(ii) Every point on the number line is of the form \(\displaystyle \sqrt{m}\) , where m is a natural number.**

This statement is false, If m is a natural number then there should be only \(\displaystyle \sqrt{1}\) , \(\displaystyle \sqrt{2}\) , \(\displaystyle \sqrt{3}\) . . . . . points on the number line while there are infinite number of numbers between two consecutive numbers on the number line.

**(iii) Every real number is an irrational number.**

This statement is false, because the collection of real numbers consists of both rational and irrational numbers. So every real number need not be irrational.

**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.**

**Solutions :** No, the square roots of all positive integers are not irrational.

Ex \(\displaystyle \sqrt{25}\) = 5, which is a rational number.

**3. Show how \(\displaystyle \sqrt{5}\) can be represented on the number line.**

**Solutions :** We know that –

5 = 4 + 1

\(\displaystyle {{\left( {\sqrt{5}} \right)}^{2}}\) = 22 + 12

\(\displaystyle \sqrt{5}=\sqrt{{{{2}^{2}}+{{1}^{2}}}}\)

First we will draw a number line.

Let us take OA = 2 units on the number line.

Let ∠OAM = 90° and cut AB = 1 unit from AM.

Join OB.

After this by Pythagoras theorem in right angle OAB –

OB^{2} = OA^{2} + AB^{2}

OB^{2} = (2)^{2} + (1)^{2}

OB^{2} = 4 + 1

OB^{2} = 5

OB = \(\displaystyle \sqrt{5}\) unit

With O as center, cut an arc of radius OB, which cuts the number line at P.

So OB = OP = \(\displaystyle \sqrt{5}\) unit.

## NCERT Solutions for class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.3**

1. Write the following in decimal form and say what kind of decimal expansion each has :

(i) \(\displaystyle \frac{36}{100}\) | (ii) \(\displaystyle \frac{1}{11}\) | (iii) \(\displaystyle 4\frac{1}{8}\) |

(iv) \(\displaystyle \frac{3}{13}\) | (v) \(\displaystyle \frac{2}{11}\) | (vi) \(\displaystyle \frac{329}{400}\) |

Solutions :

**(i) \(\displaystyle \frac{36}{100}\)**

\(\displaystyle 100\overset{{0.36}}{\overline{\left){\begin{array}{l}360\\\underline{{300}}\\\,\,\,600\\\,\,\,\underline{{600}}\\\,\,\,\,\,\,0\end{array}}\right.}}\)

On dividing 100 by 36, the quotient is 0.36 and the remainder is 0.

So, the decimal expansion of \(\displaystyle \frac{36}{100}\) is 0.36 terminating.

**(ii) \(\displaystyle \frac{1}{11}\)**

\(\displaystyle 11\overset{{0.090909…}}{\overline{\left){\begin{array}{l}1.0000\,\,\,\,\,\\\underline{{\,\,99}}\\\,\,\,\,100\\\,\,\,\,\underline{{\,\,99}}\\\,\,\,\,\,\,\,\,100\\\,\,\,\,\,\,\,\underline{{\,\,\,99}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,1\end{array}}\right.}}\)
When 1 is divided by 11, the division does not end and the remainder is 1 repeatedly.

So the decimal expansion of \(\displaystyle \frac{1}{11}\) \(\displaystyle 0.\overline{{09}}\) is non-terminating and recurring.

**(iii) \(\displaystyle 4\frac{1}{8}\)**

Converting the compound fraction \(\displaystyle 4\frac{1}{8}\) to a simple fraction gives \(\displaystyle \frac{{33}}{8}\) .

\(\displaystyle 8\overset{{4.125}}{\overline{\left){\begin{array}{l}33\,\,\,\,\\\underline{{32}}\\\,\,\,10\\\,\,\,\underline{{\,\,8}}\\\,\,\,\,\,20\\\,\,\,\,\underline{{\,16}}\\\,\,\,\,\,\,\,40\\\,\,\,\,\,\,\,\underline{{40}}\\\,\,\,\,\,\,\,\,\,0\end{array}}\right.}}\)

On dividing 33 by 8, the remainder is zero.

So the decimal expansion of \(\displaystyle 4\frac{1}{8}\) is 4.125 terminating.

**(iv) \(\displaystyle \frac{3}{13}\)**

\(\displaystyle 13\overset{{0.2307692…}}{\overline{\left){\begin{array}{l}3.0\\\underline{{26}}\\\,\,\,40\\\,\,\underline{{\,39}}\\\,\,\,\,\,\,100\\\,\,\,\,\,\,\underline{{\,\,91}}\\\,\,\,\,\,\,\,\,\,\,90\\\,\,\,\,\,\,\,\,\,\underline{{\,78}}\\\,\,\,\,\,\,\,\,\,\,120\\\,\,\,\,\,\,\,\,\,\,\underline{{117}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,30\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{26}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\end{array}}\right.}}\)

On dividing 13 by 3, the division does not end and the quotient is repeated.

So the decimal expansion of \(\displaystyle \frac{3}{13}\) \(\displaystyle 0.\overline{{230769}}\) is non-terminating and recurring.

**(v) \(\displaystyle \frac{2}{11}\)**

\(\displaystyle 11\overset{{0.1818…}}{\overline{\left){\begin{array}{l}2.0\,\,\,\,\,\,\\\underline{{\,11}}\\\,\,90\\\,\,\underline{{88}}\\\,\,\,\,20\\\,\,\,\,\underline{{11}}\\\,\,\,\,\,90\\\,\,\,\,\,\underline{{88}}\\\,\,\,\,\,\,\,20\\\,\,\,\,\,\,\underline{{\,11}}\\\,\,\,\,\,\,\,\,\,9\\\,\,\,\,\,\,\,\end{array}}\right.}}\)

On dividing 11 by 2, the division does not end and the quotient is repeated.

So the decimal expansion of \(\displaystyle \frac{2}{11}\) \(\displaystyle 0.\overline{{18}}\) is non-terminating and recurring.

**(vi) \(\displaystyle \frac{329}{400}\)**

\(\displaystyle 400\overset{{0.8225}}{\overline{\left){\begin{array}{l}329.0\\\underline{{3200}}\\\,\,\,900\\\,\,\,\underline{{800}}\\\,\,\,1000\\\,\,\,\,\,\underline{{800}}\\\,\,\,\,\,2000\\\,\,\,\,\,\underline{{2000}}\\\,\,\,\,\,\,\,\,\,0\end{array}}\right.}}\)

On dividing 400 by 329, the remainder is zero.

So the decimal expansion of \(\displaystyle \frac{329}{400}\) is 0.8225 terminating.

**2. You know that \(\displaystyle \frac{1}{7}=0.\overline{{142857}}\) . Can you predict what the decimal expansions of \(\displaystyle \frac{2}{7}\) , \(\displaystyle \frac{3}{7}\) , \(\displaystyle \frac{4}{7}\) , \(\displaystyle \frac{5}{7}\) , \(\displaystyle \frac{6}{7}\) are, without actually doing the long division? If so, how?**

[**Hint :** Study the remainders while finding the value of \(\displaystyle \frac{1}{7}\) carefully.]

Solutions :

Yes, we can find the decimal expansion of the above number without division as follows.

because, we know that \(\displaystyle \frac{1}{7}=0.\overline{{142857}}\)

So,

\(\displaystyle \frac{2}{7}=2\times \frac{1}{7}=2\times 0.\overline{{142857}}=0.\overline{{285714}}\)

\(\displaystyle \frac{3}{7}=3\times \frac{1}{7}=3\times 0.\overline{{142857}}=0.\overline{{428571}}\)

\(\displaystyle \frac{4}{7}=4\times \frac{1}{7}=4\times 0.\overline{{142857}}=0.\overline{{571428}}\)

\(\displaystyle \frac{5}{7}=5\times \frac{1}{7}=5\times 0.\overline{{142857}}=0.\overline{{714285}}\)

\(\displaystyle \frac{6}{7}=6\times \frac{1}{7}=6\times 0.\overline{{142857}}=0.\overline{{857142}}\)

**3. Express the following in the form \(\displaystyle \frac{p}{q}\) where p and q are integers and q ≠ 0.**

(i) \(\displaystyle 0.\overline{6}\) | (ii) 0.47 | (iii) 0.001 |

**Solutions :
**

**(i) \(\displaystyle 0.\overline{6}\)**

suppose \(\displaystyle 0.\overline{6}\) =

*x*

*x*= \(\displaystyle 0.\overline{6}\)

*x*= 0.666 . . . _____ (1)

On multiplying both the sides by 10 –

10

*x*= 6.666 . . . _____ (2)

On subtracting equation (1) from equation (2) –

9

*x*= 6

\(\displaystyle x=\frac{6}{9}\)

\(\displaystyle x=\frac{2}{3}\)

So, \(\displaystyle 0.\overline{6}=\frac{2}{3}\)

**(ii) \(\displaystyle 0.4\overline{7}\)**

suppose \(\displaystyle 0.4\overline{7}\) = *x*

*x* = \(\displaystyle 0.4\overline{7}\)

On multiplying both the sides by 10 –

10*x* = 4.777 . . . ________ (1)

On multiplying both the sides by 10 again –

100*x* = 47.777 . . . _______ (2)

On subtracting equation (1) from equation (2) –

90*x* = 43

\(\displaystyle x=\frac{43}{90}\)

So, \(\displaystyle 0.4\overline{7}=\frac{43}{90}\)

**(iii) \(\displaystyle 0.\overline{001}\)**

suppose \(\displaystyle 0.\overline{001}\) = *x*

*x* = 0.001001001 . . . _______ (1)

On multiplying both the sides by 1000 –

1000*x* = 1.001001001 _____ (2)

On subtracting equation (1) from equation (2) –

999*x* = 1

\(\displaystyle x=\frac{1}{999}\)

So, \(\displaystyle 0.\overline{{001}}=\frac{1}{{999}}\)

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

**Solutions :**

suppose x = 0.99999 . . . ______ (1)

On multiplying both the sides by 1000 –

10x = 9.9999 . . . _________ (2)

On subtracting equation (1) from equation (2) –

9x = 9

x = 1

0.9999 . . . = 1

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

**Solutions :**

On dividing 1 by 17, any number from 1 to 16 can be obtained as a remainder. After this the repetition of numbers will definitely happen.

\(\displaystyle \begin{array}{l}17\overset{{0.0588235294117647}}{\overline{\left){{1.0000000000000000}}\right.}}\\\,\,\,\,\,\,\,\,\,\underline{{\,\,85}}\\\,\,\,\,\,\,\,\,\,\,150\\\,\,\,\,\,\,\,\,\,\,\underline{{136}}\\\,\,\,\,\,\,\,\,\,\,\,\,140\\\,\,\,\,\,\,\,\,\,\,\,\underline{{\,136}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,40\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,34}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,60\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,51}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,90\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{85}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,50\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,34}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,160\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,153}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,70\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,68}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,20\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,17}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,30\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,17}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,130\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,119}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,110\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,102}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,80\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,68}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,120\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,119}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\end{array}\)
So the maximum number of digits in the repetition section of the decimal expansion of \(\displaystyle \frac{1}{{17}}\) will be 16.

∴ \(\displaystyle \frac{1}{{17}}\) = 0.0588235294117647 . . .

So, \(\displaystyle \frac{1}{{17}}\) = \(\displaystyle 0.\overline{{0588235294117647}}\)

**6. Look at several examples of rational numbers in the form \(\displaystyle \frac{1}{17}\) (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?**

**Solutions :**

The decimal expansion of rational numbers in the form of \(\displaystyle \frac{1}{{17}}\) will terminate only when q divided by p leaves the remainder zero while p and q have no common factors other than 1. where p and q are integers and q ≠ 0.

On dividing the divisor in a dividend, the remainder will be zero only if –

- The divisor is 2 or any power of 2.
- The divisor is 5 or any power of 5.
- The divisor is the product of any power of 2 and any power of 5.

So q must be equal to 2 or 5 or any power thereof or must be equal to the multiplication of any power of 2 and any power of 5.

i.e. *q* = 2* ^{m}* × 5

^{n}where m and n are whole numbers.

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

Solutions :

The decimal expansions of all irrational numbers are non-recurring.

Eg: √2, √3, √5, √7 etc.

**8. Find three different irrational numbers between the rational numbers \(\displaystyle \frac{5}{{7}}\) and \(\displaystyle \frac{9}{{11}}\) .**

**Solutions :**

Decimal Expansion of \(\displaystyle \frac{5}{{7}}\) = \(\displaystyle 0.\overline{{714285}}\)

Decimal Expansion of \(\displaystyle \frac{9}{{11}}\) = \(\displaystyle 0.\overline{{81}}\)

So three irrational numbers between the rational numbers \(\displaystyle \frac{5}{{7}}\) and \(\displaystyle \frac{9}{{11}}\) are –

0.73073007300073000 . . . , 0.750750075000750000 . . ., 0.790790079000790000 . . .

**9. Classify the following numbers as rational or irrational :**

(i) \(\displaystyle \sqrt{{23}}\) | (ii) \(\displaystyle \sqrt{{225}}\) | (iii) 0.3796 |

(iv) 7.478478 . . . | (v) 1.101001000100001 . . . |

**Solutions :**

**(i) \(\displaystyle \sqrt{{23}}\)**

\(\displaystyle \sqrt{{23}}\) = 4.79583

The decimal expansion of \(\displaystyle \sqrt{{23}}\) is non-terminating and non-recurring.

So, \(\displaystyle \sqrt{{23}}\) is an irrational number.

**(ii) \(\displaystyle \sqrt{{225}}\)**

\(\displaystyle \sqrt{{225}}\) = 15

Number 15 is a natural number.

So, \(\displaystyle \sqrt{{225}}\) is an rational number.

**(iii) 0.3796**

The decimal expansion of 0.3796 is terminating.

So, 0.3796 is an rational number.

**(iv) 1.101001000100001 . . .**

The decimal expansion of 1.101001000100001 . . . is non-terminating and non-recurring.

So, 1.101001000100001 . . . is an irrational number.

## NCERT Solutions for class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.4**

1. Visualise 3.765 on the number line, using successive magnification.

Solutions :

Solution : First we magnify the interval between 3 and 4 and divide it into 10 equal parts.

The given number is 3.765 which lies between 3.7 and 3.8.

Therefore, divide the interval between 3.7 and 3.8 also into 10 equal parts and magnify it.

The given number is 3.765 which lies between 3.76 and 3.78.

Therefore, magnify the interval between 3.76 and 3.77 and divide it into 10 equal parts.

Its magnification has 3.765 fifth part.

## NCERT Solutions for class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.5**

**1. Classify the following numbers as rational or irrational:**

(i) \(\displaystyle 2-\sqrt{5}\) | (ii) \(\displaystyle \left( {3+\sqrt{{23}}} \right)-\sqrt{{23}}\) |

(iv) \(\displaystyle \frac{1}{{\sqrt{2}}}\) | (v) 2π |

**Solutions :**

**(i) \(\displaystyle 2-\sqrt{5}\)**

Because we know that 2 is a rational number and \(\displaystyle \sqrt{5}\) is a irrational number.

So we know that the difference between a rational number and an irrational number is an irrational number.

So, \(\displaystyle 2-\sqrt{5}\) is a irrational number.

**(ii) \(\displaystyle \left( {3+\sqrt{{23}}} \right)-\sqrt{{23}}\)**

= \(\displaystyle 3+\sqrt{{23}}-\sqrt{{23}}\)

= 3

So, \(\displaystyle \left( {3+\sqrt{{23}}} \right)-\sqrt{{23}}\) is a rational number.

**(iii) \(\displaystyle \frac{{2\sqrt{7}}}{{2\sqrt{7}}}\)**

\(\displaystyle \frac{{2\sqrt{7}}}{{2\sqrt{7}}}\) = \(\displaystyle \frac{2}{7}\)

So, \(\displaystyle \frac{{2\sqrt{7}}}{{2\sqrt{7}}}\) rational number.

**(iv) \(\displaystyle \frac{1}{{\sqrt{2}}}\)**

Because we know that 2 is a rational number and \(\displaystyle \sqrt{2}\) is a rational number.

We also know that dividing a rational number by an irrational number gives an irrational number.

So, \(\displaystyle \frac{1}{{\sqrt{2}}}\) is a irrational number.

**(v) 2 π**

We know that 2 is a rational number and is an irrational number.

We also know that multiplying a rational number by an irrational number gives an irrational number.

Hence 2π is an irrational number.

**2. Simplify each of the following expressions:**

(i) \(\displaystyle \left( {3+\sqrt{3}} \right)\left( {2+\sqrt{2}} \right)\) | (ii) \(\displaystyle \left( {3+\sqrt{3}} \right)\left( {3-\sqrt{3}} \right)\) |

(iii) \(\displaystyle {{\left( {\sqrt{5}+\sqrt{2}} \right)}^{2}}\) | (iv) \(\displaystyle \left( {\sqrt{5}-\sqrt{2}} \right)\left( {\sqrt{5}+\sqrt{2}} \right)\) |

**Solutions :**

**(i) \(\displaystyle \left( {3+\sqrt{3}} \right)\left( {2+\sqrt{2}} \right)\)**

= \(\displaystyle 3\times \left( {2+\sqrt{2}} \right)+\sqrt{3}\times \left( {2+\sqrt{2}} \right)\)

= \(\displaystyle 3\times 2+3\times \sqrt{2}+\sqrt{3}\times 2+\sqrt{3}\times \sqrt{2}\)

= \(\displaystyle 6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\)

**(ii) \(\displaystyle \left( {3+\sqrt{3}} \right)\left( {3-\sqrt{3}} \right)\)**

= \(\displaystyle 3\times 3-3\times \sqrt{3}+\sqrt{3}\times 3-\sqrt{3}\times \sqrt{3}\)

= \(\displaystyle 9-3\sqrt{3}+3\sqrt{3}-3\)

= 9 − 3

= 6

**(iii) \(\displaystyle {{\left( {\sqrt{5}+\sqrt{2}} \right)}^{2}}\)**

= \(\displaystyle {{\left( {\sqrt{5}} \right)}^{2}}+2\times \sqrt{5}\times \sqrt{2}+{{\left( {\sqrt{2}} \right)}^{2}}\)

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= \(\displaystyle 5+2\sqrt{{10}}+2\)

= \(\displaystyle 7+2\sqrt{{10}}\)

**(iv) \(\displaystyle \left( {\sqrt{5}-\sqrt{2}} \right)\left( {\sqrt{5}+\sqrt{2}} \right)\)**

= \(\displaystyle {{\left( {\sqrt{5}} \right)}^{2}}-{{\left( {\sqrt{2}} \right)}^{2}}\)

[∵ (a)^{2} − (b)^{2} = (a + b) (a − b)]

= 5 − 2

= 3

**3. Recall, p is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, \(\displaystyle \pi =\frac{c}{d}\) This seems to contradict the fact that p is irrational. How will you resolve this contradiction?**

**Solutions :**

\(\displaystyle \pi =\frac{c}{d}\)

Measuring c and d with a scale, we only get an approximate measure from which it is not known whether c or d are rational numbers or irrational numbers. For this reason, we get the illusion of thinking c and d as rational numbers and we move on to understand the ratio of c and d as rational numbers, which creates a contradiction. In fact, there is no contradiction in being irrational.

**5. Rationalise the denominators of the following:**

(i) \(\displaystyle \frac{1}{{\sqrt{7}}}\) | (ii) \(\displaystyle \frac{1}{{\sqrt{7}-\sqrt{6}}}\) | (iii) \(\displaystyle \frac{1}{{\sqrt{5}+\sqrt{2}}}\) | (iv) \(\displaystyle \frac{1}{{\sqrt{7}-2}}\) |

**Solutions : **

**(i) \(\displaystyle \frac{1}{{\sqrt{7}}}\)**

Here the irrational number \(\displaystyle \sqrt{7}\) is given in the denominator.

So, On multiplying the numerator by \(\displaystyle \sqrt{7}\) and the denominator by –

= \(\displaystyle \frac{{1\times \sqrt{7}}}{{\sqrt{7}\times \sqrt{7}}}=\frac{{\sqrt{7}}}{7}\)

**(ii) \(\displaystyle \frac{1}{{\sqrt{7}-\sqrt{6}}}\)**

Here the irrational number \(\displaystyle \left( {\sqrt{7}-\sqrt{6}} \right)\) is given in the denominator.

Hence its rational factor is \(\displaystyle \left( {\sqrt{7}+\sqrt{6}} \right)\) .

So, On multiplying \(\displaystyle \left( {\sqrt{7}+\sqrt{6}} \right)\) by numerator and denominator –

= \(\displaystyle \frac{1}{{\left( {\sqrt{7}-\sqrt{6}} \right)}}\times \frac{{\sqrt{7}+\sqrt{6}}}{{\sqrt{7}+\sqrt{6}}}\)

= \(\displaystyle \frac{{\sqrt{7}+\sqrt{6}}}{{{{{\left( {\sqrt{7}} \right)}}^{2}}-{{{\left( {\sqrt{6}} \right)}}^{2}}}}\)

= \(\displaystyle \frac{{\sqrt{7}+\sqrt{6}}}{{7-6}}\)

= \(\displaystyle \frac{{\sqrt{7}+\sqrt{6}}}{1}\)

= \(\displaystyle \sqrt{7}+\sqrt{6}\)

**(iii) \(\displaystyle \frac{1}{{\sqrt{5}+\sqrt{2}}}\)**

Here the irrational number \(\displaystyle \sqrt{5}+\sqrt{2}\) is given in the denominator.

Hence its rational factor is \(\displaystyle \sqrt{5}-\sqrt{2}\) .

So, \(\displaystyle \sqrt{5}-\sqrt{2}\) by numerator and denominator –

= \(\displaystyle \frac{1}{{\sqrt{5}+\sqrt{2}}}\times \frac{{\sqrt{5}-\sqrt{2}}}{{\sqrt{5}-\sqrt{2}}}\)

= \(\displaystyle \frac{{\sqrt{5}-\sqrt{2}}}{{{{{\left( {\sqrt{5}} \right)}}^{2}}-{{{\left( {\sqrt{2}} \right)}}^{2}}}}\)

= \(\displaystyle \frac{{\sqrt{5}-\sqrt{2}}}{{5-2}}\)

= \(\displaystyle \frac{{\sqrt{5}-\sqrt{2}}}{3}\)

**(iv) \(\displaystyle \frac{1}{{\sqrt{7}-2}}\)**

Here the irrational number \(\displaystyle \frac{1}{{\sqrt{7}-2}}\) is given in the denominator.

Hence its rational factor is \(\displaystyle \frac{1}{{\sqrt{7}+2}}\) .

So, \(\displaystyle \frac{1}{{\sqrt{7}+2}}\) by numerator and denominator –

= \(\displaystyle \frac{1}{{\sqrt{7}-2}}\times \frac{{\sqrt{7}+2}}{{\sqrt{7}+2}}\)

= \(\displaystyle \frac{{\sqrt{7}+2}}{{{{{\left( {\sqrt{7}} \right)}}^{2}}-{{{\left( 2 \right)}}^{2}}}}\)

= \(\displaystyle \frac{{\sqrt{7}+2}}{{7-4}}\)

= \(\displaystyle \frac{{\sqrt{7}+2}}{3}\)

## NCERT Solutions for class 9th maths Exercise 1 Number System

NCERT Solutions for Class 9 Maths Chapter 1 Number System

**Exercise 1.5**

1. Find :

(i) \(\displaystyle {{64}^{{\frac{1}{2}}}}\) | (ii) \(\displaystyle {{32}^{{\frac{1}{5}}}}\) | (iii) \(\displaystyle {{125}^{{\frac{1}{3}}}}\) |

Solutions :

**(i) \(\displaystyle {{64}^{{\frac{1}{2}}}}\)**

prime factors of 64

= \(\displaystyle {{(2\times 2\times 2\times 2\times 2\times 2)}^{{\frac{1}{2}}}}\)

= \(\displaystyle {{({{2}^{6}})}^{{\frac{1}{2}}}}\)

= \(\displaystyle {{(2)}^{{6\times \frac{1}{2}}}}\)

= (2)^{3}

= 8

**(ii) \(\displaystyle {{32}^{{\frac{1}{5}}}}\)**

prime factors of 32

= \(\displaystyle {{(2\times 2\times 2\times 2\times 2)}^{{\frac{1}{5}}}}\)

= \(\displaystyle {{({{2}^{5}})}^{{\frac{1}{5}}}}\)

= \(\displaystyle {{(2)}^{{5\times \frac{1}{5}}}}\)

= (2)^{1}

= 2

**(iii) \(\displaystyle {{125}^{{\frac{1}{3}}}}\)**

prime factors of 125

= \(\displaystyle {{(5\times 5\times 5)}^{{\frac{1}{3}}}}\)

= \(\displaystyle {{({{5}^{3}})}^{{\frac{1}{3}}}}\)

= \(\displaystyle {{(5)}^{{3\times \frac{1}{3}}}}\)

= 5^{1}

= 5

**2. Find :**

(i) \(\displaystyle {{9}^{{\frac{3}{2}}}}\) | (ii) \(\displaystyle {{32}^{{\frac{2}{5}}}}\) | (iii) \(\displaystyle {{16}^{{\frac{3}{4}}}}\) | (iv) \(\displaystyle {{125}^{{\frac{{-1}}{3}}}}\) |

**हल : **

**(i) \(\displaystyle {{9}^{{\frac{3}{2}}}}\)**

= \(\displaystyle {{(3\times 3)}^{{\frac{3}{2}}}}\)

= \(\displaystyle {{({{3}^{2}})}^{{\frac{3}{2}}}}\)

= \(\displaystyle {{(3)}^{{2\times \frac{3}{2}}}}\)

= 3^{3}

= 27