Class 09 | Ganita Manjari 01 | Ch. 03 The World of Numbers

Class 09 | Ganita Manjari | Ch. 03 The World of Numbers | Solutions

1. Natural numbers: - When we begin to count 1, 2, 3, 4, 5, 6, 7, ....to infinity etc numbers comes naturally before us hence, mathematicians call the Counting Numbers as Natural Numbers. The counting numbers {1, 2, 3, … } that emerged at least tens of thousands of years ago due to humanity’s need to count.

Collection of Natural Numbers is denoted by Capital N. 

N = {1, 2, 3, 4, 5, 6, 7, 8}

1. प्राकृत संख्याएँ :- 1, 2, 3, 4, 5, 6, 7,से अनंत तक आदि संख्याएं हमारे सम्मुख प्राकृतिक रूप से आती हैं। इसीलिए, गणितज्ञ इन गणन (गिनती गिनने वाली) संख्याओं (Counting Numbers) को प्राकृत संख्याएँ (Natural Numbers) कहते हैं। गणन संख्याएँ {1, 2, 3, …} जो हज़ारों साल पहले इंसानों की गिनती करने की ज़रूरत के कारण सामने आईं।

प्राकृत संख्याओं के समूह को अंग्रेजी के बडे अक्षर एन० से दर्शाते हैं। N = {1, 2, 3, 4, 5, 6, 7, 8}

2. Whole Numbers: - The natural numbers along with zero form the collection of whole numbers i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8 Collection of Whole Numbers is denoted by Capital W. W = { 0, 1, 2, 3, 4, 5, 6, 7, 8}

2. पूर्ण संख्याएँ :- प्राकृत संख्याएँ शून्य के साथ मिलकर पूर्ण संख्याओं (Whole numbers) का संग्रह बनाती हैं। जैसे 0, 1, 2, 3, 4, 5, 6, 7, ....00 प्राकृत संख्याओं के समूह को अंग्रेजी के बडे अक्षर डब्ल्यु० से दर्शाते हैं। W={0, 1, 2, 3, 4, 5, 6, 7, ...8}

The Concept of Zero (Śhūnya) –> The Concept of Zero (Śhūnya) was formalised in India by philosophers and then brought into mathematics formally by Brahmagupta (629 CE), who transformed the philosophical state 

of ‘nothingness’ (Śhūnyatā) into an actual number on which one could perform arithmetic operations. Brahmagupta also introduced the negative numbers, i.e., numbers less than zero.

शून्य (Śhūnya) की अवधारणा –> भारत में दार्शनिकों ने शून्य (Śhūnya) की अवधारणा को औपचारिक रूप दिया और फिर ब्रह्मगुप्त (629 ईस्वी) ने इसे गणित में औपचारिक रूप से शामिल किया। उन्होंने 'शून्यता' (कुछ न होने) की दार्शनिक स्थिति को एक वास्तविक संख्या में बदल दिया, जिस पर गणितीय क्रियाएं की जा सकती थीं। ब्रह्मगुप्त ने ऋणात्मक संख्याओं (यानी शून्य से कम संख्याओं) की भी शुरुआत की।

From Philosophy to Mathematics: The Concept of Śhūnyatā

In the Upanishads and in the vast Buddhist literature starting well before the 7th century BCE, the concept of Shūnyatā (emptiness or nothingness)

was a profound state that was the goal of yoga and meditation. The word shūnya means zero, and the word shūnyatā (zeroness) was used extensively in these works to describe the state that one is trying to reach during meditation that of emptying one's mind of all vṛttis (fluctuations of the mind) to achieve the state of perfect stillness and tranquility. Patanjali in the Yoga Sutras around the 3rd century BCE also describes how shūnyatā can help lead to control over one's mind, body, and senses.

Because Indian thinkers revered this state of 'emptiness', they possessed the conceptual framework necessary to welcome 'nothingness' as a concept in philosophy. This concept of zeroness then found its way into many fields, such as architecture, literature, linguistics, and eventually made its way into mathematics in the works of Āryabhața and finally Brahmagupta.

Thus, the philosophical concept of emptiness crystallised into the mathematical zero.

3.2.2 The Bakhshālī Manuscript and Brahmagupta's Rules

In the Hindu Number System that we use today, the physical transition from a blank space to a symbol can be seen in the Bakhshālī Manuscript. Dated to the early centuries CE, it features a bold dot (bindu) used to represent zero.

However, a symbol is just a mark on a page until it has rules. The transformation of zero into a fully operational number occurred in the work of Brahmagupta. In his seminal work, the Brāhmasphuțasiddhānta (628 CE), he explicitly defined zero as the result of subtracting a number from itself (a - a = 0).

Brahmagupta then laid down the fundamental laws of arithmetic with shūnya:

Brahmagupta's Rules for Zero

When zero is added to a number, the number remains unchanged: a+0=a

When zero is subtracted from a number, the number remains unchanged: a-0=a.

When any number is multiplied by zero, the result is zero: a × 0 = 0.

दर्शन से गणित तक: शून्यता की अवधारणा

उपनिषदों और 7वीं सदी ईसा पूर्व से पहले के विशाल बौद्ध साहित्य में, शून्यता (खालीपन या कुछ न होने) की अवधारणा

एक गहरी अवस्था थी जो योग और ध्यान का लक्ष्य थी। 'शून्य' शब्द का अर्थ है ज़ीरो (zero), और 'शून्यता' (ज़ीरो होने का भाव) शब्द का इस्तेमाल इन रचनाओं में उस अवस्था का वर्णन करने के लिए बड़े पैमाने पर किया गया, जिसे पाने की कोशिश ध्यान के दौरान की जाती है—यानी मन को सभी 'वृत्तियों' (मन के उतार-चढ़ाव) से खाली करके पूरी तरह शांत और स्थिर अवस्था प्राप्त करना। लगभग तीसरी सदी ईसा पूर्व में पतंजलि ने योग सूत्रों में यह भी बताया है कि कैसे शून्यता मन, शरीर और इंद्रियों पर नियंत्रण पाने में मदद कर सकती है।

चूंकि भारतीय विचारकों ने 'खालीपन' की इस अवस्था का सम्मान किया, इसलिए उनके पास दर्शन में 'कुछ न होने' (nothingness) की अवधारणा को अपनाने के लिए ज़रूरी वैचारिक आधार मौजूद था। ज़ीरो होने की यह अवधारणा फिर कई क्षेत्रों में पहुँची, जैसे कि वास्तुकला, साहित्य, भाषा-विज्ञान, और आखिरकार आर्यभट्ट और फिर ब्रह्मगुप्त के कामों के ज़रिए गणित में शामिल हो गई।

इस तरह, खालीपन की दार्शनिक अवधारणा गणितीय शून्य के रूप में विकसित हुई।

3.2.2 बख्शाली पांडुलिपि और ब्रह्मगुप्त के नियम

आज हम जिस हिंदू संख्या प्रणाली का इस्तेमाल करते हैं, उसमें खाली जगह से एक प्रतीक (सिंबल) तक का बदलाव बख्शाली पांडुलिपि में देखा जा सकता है। ईस्वी सन् की शुरुआती सदियों की इस पांडुलिपि में शून्य को दिखाने के लिए एक मोटे बिंदु का इस्तेमाल किया गया है।

हालाँकि, कोई प्रतीक तब तक सिर्फ़ पन्ने पर बना एक निशान होता है जब तक उसके लिए नियम न हों। शून्य का पूरी तरह से काम करने वाली संख्या में बदलना ब्रह्मगुप्त के काम में हुआ। अपनी अहम रचना 'ब्राह्मस्फुटसिद्धांत' (628 ईस्वी) में, उन्होंने साफ़ तौर पर शून्य को किसी संख्या में से उसी संख्या को घटाने के नतीजे (a - a = 0) के रूप में परिभाषित किया। इसके बाद ब्रह्मगुप्त ने शून्य के साथ अंकगणित के बुनियादी नियम बताए:

शून्य के लिए ब्रह्मगुप्त के नियम

जब किसी संख्या में शून्य जोड़ा जाता है, तो संख्या में कोई बदलाव नहीं होता: a+0=a

जब किसी संख्या में से शून्य घटाया जाता है, तो संख्या में कोई बदलाव नहीं होता: a-0=a.

जब किसी संख्या को शून्य से गुणा किया जाता है, तो परिणाम शून्य होता है: a × 0 = 0.

3. Integer –>  The set of Negative and Positive whole numbers including zero is called Integer. Collection of Integers is denoted by Capital Z. Z = {–∞, .......,–3, –2, –1, 0, +1, +2 +3,....... +∞} 

3. पूर्काणांक –> नकारत्मक, शून्य तथा धनात्मक पूर्णांकों के समूह को इन्टीजर कहते हैं। इसे अंग्रेजी के बडे अक्षर जेड० से दर्शाते हैं। 

Z = {–∞, .......,–3, –2, –1, 0, +1, +2 +3,....... +∞} 

There are two types of integer number.

पूर्णांक संख्याएँ दो प्रकार की होती हैं।

01. Negative integer: - The integer with Negative sign and started from -tolis called Negative Integers i.e. –3, –2, –1.

01. ऋणात्मक पूर्णांक (Negative integer): - ऋणात्मक चिह्न वाली संख्याएँ जो ऋणात्मक दिशा में बढ़ती हैं, उन्हें ऋणात्मक पूर्णांक कहा जाता है, जैसे –3, –2, –1।

02. Zero: Integer 0 is neither negative nor positive.

02. शून्य (Zero): पूर्णांक 0 न तो ऋणात्मक होता है और न ही धनात्मक।

03. Positive integer: - The integer with positive number started from 1 to ∞is called Positive Integers i.e. +1, +2, +3, +8

03. धनात्मक पूर्णांक (Positive integer): - धनात्मक चिह्न वाली संख्याएँ जो 1 से शुरू होकर अनंत (∞) तक जाती हैं, उन्हें धनात्मक पूर्णांक कहा जाता है, जैसे +1, +2, +3, +8।

Rational Number: - The number in the form of p/q is called Rational number where q ≠ 0 and p and q are integer.

There are two types of rational number.

01. Negative Rational Number: A rational number is said to be Negative if either its Numerator or Denominator is Negative.  –2, –9, – 4/5, 4/–5, –2, –3/7, 3/–7, - 4/5  are Negative Rational Numbrs.

02. Zero Rational Number –> 0 is a rational Number because it can be written in the form of p/q with non zero q ≠ 0.

03. Positive Rational Number –> A rational number is said to be positive if its Numerator and Denominator both are Negative or both are Positive. 2, 9, 3/7, –3/–7, –4/–5, 4/5 are Positive Rational Number.

Note that: - p/0 = ∞ (Division by 0 is not define) and 0/q = 0

There are two types of Rational Numbers.

01. Terminating an

02. Non terminating number

01. Terminating Rational Numbers –> Which number divided completely by denominator is called Terminating Rational Number. As 1/2 = 0.5, 4/5 = 0.8, 1/25 = 0.04, 1/8 = 0.125 etc. 

02. Non-Terminating Number–> In which rational numbers, the division process never does not comes to end are called Non-Terminating Number.

These are of two types

a) Non-Terminating Repeating Numbers and

b) Non-Terminating Repeating Non Repeating

a) Non-Terminating Repeating Numbers or Non Terminating Recurring Numbers –> In which Non terminating numbers in the division process reminder starts repeating after a certain numbers are called Non Terminating Repeating Numbers or Non Terminating Recurring Numbers.

These Repeating Numbers are represented by putting a bar over the first block or group or pair and other repeating parts of number will omit after putting a Bar.

                                  

1/3 = 0.3333 or 0.3 

                                               

2/11 = 0.181818 .... or 0.18

                                              

 0.123123123...... or 0.123 

When reminder repeats itself then you should understand that the quotient repeats itself.

(b) Non-Terminating and Non Repeating Numbers –> In which Non terminating numbers, in the division process reminder starts repeating after a certain numbers are called Non Terminating and Non Repeating Numbers. 0.123456789101112......

Irrational Number –> The real numbers that are not expressible in the form of p/q is called Irrational number where q ≠ 0 and p and q are integer. π is an irrational number.

(NCERT Textbook Page No. 43)

EXERCISE SET 3.1

Q. 1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Sol. 

Given:

2 bags of spices = 15 copper ingots

12 bags of spices = ? copper ingots

By Unitary Method

First, find the number of copper ingots for 1 bag.

2 bags of spices = 15 copper ingots

1 bags of spices = 15/2 copper ingots

                               = 7.5 copper ingots

For 12 bags of      = 12 × 7.5

                               = 90 copper ingots

The merchant will receive 90 copper ingots for 12 bags.

Q. 2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Sol. 

What do these numbers have in common?

The numbers 11, 13, 17, and 19 are all prime numbers.

A prime number is a number that has only two factors: 1 and  The number itself

For example:

11 → Factors: 1, 11

13 → Factors: 1, 13

17 → Factors: 1, 17

19 → Factors: 1, 19

The next three prime numbers after 19 are:

23, 29, 31

23 → Factors: 1, 23

29 → Factors: 1, 29

31 → Factors: 1, 31

Common property: They are all prime numbers.

Next three numbers: 23, 29, 31.

Q. 3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Sol. 

Closer Property 

a + b = c

The sum of any two natural numbers is always a natural number

Explanation

A set is closed under subtraction if the subtraction of any two natural numbers always gives another natural number.

Let's check with examples.

Example 1:

8 –3 = 5 is a natural number. ✔

5 –3 = 2 is a natural number. ✔

Example 2:

2 − 5 = –3 is not a natural number. ✘

Since subtraction does not always give a natural number, thus natural numbers are not closed under subtraction.

No, Natural numbers are not closed under subtraction.

Q. *4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Sol.

Explanation

Excluding the thumb, one hand has 4 fingers.

Each finger has 3 joints.

The thumb is used to count these joints.

So, 

4 finger × 3 Joints = 13 Joints 

the total number of joints = 12

Therefore, using one hand, a person can count from 1 to 12.

For example:

Little finger → Joints 1, 2, 3

Ring finger → Joints 4, 5, 6

Middle finger → Joints 7, 8, 9

Index finger → Joints 10, 11, 12

Thus, one complete cycle of counting is 12.

This means we can count from 1 to 12 on one hand.

Relation to the Base-12 System

Since one hand can count up to 12, ancient people used 12 as one complete counting unit. This is why some ancient civilizations developed the base-12 (duodecimal) counting system.

1. Dozen

A dozen is a group of 12 items or units.

Examples:

1 dozen bananas means = 12 bananas

1 dozen eggs means = 12 eggs

1 dozen pencils means = 12 pencils

2. Gross

A gross is equal to 12 dozens.

Since 1 dozen = 12 items,

Sol 12 dozen means = 12 × 12 

                                     = 144 

Therefore,

1 gross = 144 items

3. Great Gross

A great gross is equal to 12 gross.

Since 1 gross = 144 items,

Since 1 Great gross = 12 × 144 

                                    = 1728 

Therefore,

1 great gross = 1,728 items lo

Think and Reflect (NCERT Textbook Page No. 46)

Q. Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt. 

(Hint: If someone takes away (–) four of your debts that are each worth ₹ 3 (that is, –3), you are effectively ₹ 12 richer! Therefore, (–3) × (–4) = (+12.)

Sol.

Think of a negative number as a debt and multiplication as an action being repeated.

Suppose you have a debt of ₹ 3. 

This is written as –3. 

Now, if someone removes this debt, what happens? You are no longer required to pay ₹ 3, so your position improves. Mathematically, you go from –3 to 0, which is an increase of ₹ 3. That is why removing a debt is the same as gaining money.

Now apply this idea to:

(–3) × (–4)

Here, –3 represents a debt of ₹ 3, and –4 represents the action of removing this debt 4 times. So the expression means: “Remove four debts of ₹ 3 each.”

Each time you remove a debt of ₹ 3, you gain ₹ 3. Repeating this action four times

After removing one debt → gain ₹ 3

After removing two debts → gain ₹ 6

After removing three debts → gain ₹ 9

After removing four debts → gain ₹ 12

So, after all debts are removed, you are ₹ 12 better off.

Therefore,

(—3) × (—4) = +12

In this way, a negative times a negative becomes positive.

Class 9 Maths Ganita Manjari Chapter 3 Exercise Set 3.2 Solution

Exercise Set 3.2 Class 9 Maths Solutions

Q. 1. The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature?

Sol.

Given:

Temperature at noon = 4°C

It drops by 15°C by midnight.

ATQ

A drop in temperature means we subtract 15 from 4.

4 – 15 = –11

So, the midnight temperature is –11 °C.

Q. 2. A spice trader takes a loan (debt) of ₹ 850. The next day, he makes a profit (fortune) of ₹ 1,200. The following week, he incurs a loss of ₹ 450. Write this sequence as an equation using integers and calculate his final financial standing.

Sol.

The trader first takes a loan of ₹ 850, which is represented as –850.

The next day, he earns a profit of ₹ 1200, represented as +1200.

Then he incurs a loss of ₹ 450, represented as –450.

So, the situation can be written as:

= – 850 + 1200 – 450

= + 350 – 450

= –100

So, his final financial standing is – ₹100, which means he is still in debt of ?100.

Q. 3. Calculate the following using Brahmagupta’s laws:

(i) (–12) × 5

(ii) (–8) × (–7)

(iii) 0 –(–14)

(iv) (– 20) ÷ 4

Sol.

(i) (–12) × 5

The product of a debt and a fortune is a debt(A negative multiplied by a positive is negative):

–12 × 5 = –60

(ii) (–8) × (–7)

Sol.

(–8) × (–7)

The product of two debts is a fortune (A negative multiplied by a negative is positive):

–8 × –7 = 56

(iii) 0 – (–14)

Sol.

0 – (–14)

Zero minus debt is a fortune (Subtracting a negative becomes addition):

0 + 14 = 14

(iv) (–20) ÷ 4

Sol.

(–20) ÷ 4

A debt divided by fortune is debt (A negative divided by a positive is negative):

–20 ÷ 4 = –5

Q. 4  Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 – (–5) = 15).

Sol.

Suppose you have ₹10.

Your friend owes you ₹5 (this debt is represented by – ₹5).

Now imagine the debt is cancelled or removed.

Mathematically,

= 10 – (–5)

= 10 + 5

= 15

Thus, subtracting a negative number is the same as adding a positive number because removing a debt increases your total.

Think and Reflect (NCERT Textbook Page No. 47)

Q. Can you explain why we need q ≠ 0 in the definition of a rational number?

Sol.

A rational number is written as p/q, where p and q are integers and q ≠ 0, because the denominator represents division, and division by zero is not defined. 

If q = 0, then p/0 would mean dividing a number into zero equal parts, which has no meaning.

Think and Reflect (NCERT Textbook Page No. 49)

Q. 1 While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Sol.

While adding or subtracting two rational numbers with different denominators, we make the denominators equal by finding a common denominator, usually the LCM (Least Common Multiple) of the two denominators. 

a/b ± c/d

LCM of b and d = m

we convert each fraction into an equivalent fraction with this common denominator by multiplying the numerator and denominator by the required number.

a/b(m/m) ± c/d(m/m)

Once the denominators are the same, we can easily add or subtract the numerators while keeping the denominator unchanged.

Q. 2 Verify the distributive law for rational numbers.

Sol.

To verify the distributive law for rational numbers, we check whether

a × (b + c) = a × b + a × c

Let us take three rational numbers:

a = 2, b = 3, c = 5

Now, take LHS 

a × (b + c) 

= 2 × (3+5) 

Next, take RHS

find sum of a × b and a × c

= a × b + a × c

= 2 × 3 + 2 × 5

= 6 + 10

= 16

Since both sides LHS and RHS are equal

a × (b + c) = a × b + a × c

Hence, the distributive law is verified for rational numbers.

Exercise Set 3.3 Class 9 Maths Solutions

EXERCISE SET 3.3

1. Prove that the following rational numbers are equal:

(i) 2/3 and 4/6                 (ii) 5/4 and 10/8

(iii) –3/5 and  –6/10         (iv) 9/3 and 3 

2. Find the sum:

(i) 2/5 + 3/10            (ii) 7/12 + 5/8

(iii)  –4/7 + 3/14       

3. Find the difference:

(i) 5/6 – 1/4               (ii) 11/8 – 3/4

(Iii) –7/9 –(–2/3)

4. Find the product:

(i) 2/3 × 3/10

(ii) 7/11 × 5/8

(iii)  – 4/7 × 5/14

5. Find the quotient:

(i) 2/3 ÷ (3/10)

(ii) 7/11 ÷  (5/8)

(iii) (– 4/7) / (5/14)

6. Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3

7. Simplify the following using the distributive property: 7/9(6/7 –3/4)

8. Find the rational number x such that: 5/6 × (x + 3/5) = 5/6 × x + 1/2

Sol. 

Q. 1 Prove that the following rational numbers are equal.

(i) 2/3 and 4/6

Sol.

Two rational numbers a/b and c/d are said to be equal if a.d = b.c.

2/3 and 4/6

a = 2, b = 3 and c = 4, d = 6

Here, 2 × 6 = 3 × 4

⇒ 12 = 12 [a × d = b × c]

So, both are equal.

(ii) 5/4 and 10/8

Sol. 

5/4 and 10/8;

a = 5 , b = 4 and c = 10, d = 8

5 × 8 = 4 × 10

⇒ 40 = 40 [a × d = b × c]

So, both are equal.

(iii) −3/5 and −6/10

Sol. 

–3/5 and −6/10;

a = –3, b = 5 and c = –6, d = 10

⇒ –30 = –30 [a × d = b × c]

So, both are equal.

(iv) 93 and 3

Sol. 

93 and 3;

a = 9, b = 3 and c = 3 d = 1

9 × 1 = 3 × 3

⇒ 9 = 9 [a × d = b × c]

So, both are equal.

Question 2.

Find the sum:

(i) 2/5+3/10

Sol. 

To find the sum of rational numbers, first make the denominators the same (LCM of denominators), then add the numerators.

2/5 + 3/10

LCM of 5 and 10 = 10

2/5 = 4/10

4/10 + 3/10 =7/10

(ii) 7/12+5/8

Sol. 

7/12+5/8

LCM of 12 and 8 = 24

7/12 = 14/24 and 5/8=15/24

14/24 +15/24 = 29/24

(iii) −4/7 + 3/14

Sol. 

LCM of 7 and 14 = 14

= –4/7 =–8/14

= –8/14 + 3/14 

=–5/14

Question 3.

Find the difference:

(i) 5/6 −1/4

Sol.

5/6 −1/4

LCM of 6 and 4 = 12

5/6=10/12, and 1/4=3/12

= 10/12−3/12

=712

(ii) 11/8−3/4

Sol. 

LCM of 8 and 4 = 8

3/4 = 6/8

11/8–6/8

=5/8

(iii) −7/9−(−2/3)

Sol. 

LCM of 9 and 3 = 9

2/3=6/9

= −7/9 + 6/9

=−1/9

Question 4.

Find the product:

(i) 2/3×3/10

Sol. 

To find the product of rational numbers, multiply the numerators and multiply the denominators, then simplify if possible.

2/3×3/10

= 2×3/3×10=6/30

=1/5

(ii) 7/11×5/8

Sol. 

7/11×5/8

7×5/11×8=35/88

(iii) −4/7×5/14

Sol. 

−4/7×5/14

−4×5/7×14=−20/98=−10/49

Q. 5.

Find the quotient:

(i) 2/5÷3/10

Sol. 

2/5÷3/10

=2/5×10/3

=20/15

=4/3

(ii) 7/11÷5/8

Sol. 

7/11÷5/8

=7/11×8/5

=56/55

(iii) −4/7÷5/14

Sol.

−4/7÷5/14

=−4/7×14/5

=−8/5

Q. 6. Show that: (1/2+3/4)×8/3=1/2×8/3+3/4×8/3

Sol. To show:

Q. 7. Simplify the following using the distributive property:

7/9(6/7−3/4)

Sol.

Apply distributive property:

7/9(6/7−3/4)=7/9×6/7−7/9×3/4

Simplify each term:

7/9×6/7=2/3,          7/9×3/4=7/12

Now subtract:

2/3−7/12

=8/12−7/12

=1/12

Therefore,

7/9(6/7−3/4)=1/12

Q. 8. Find the rational number x such that:

5/6(x+3/5)=(5/6)x+1/2.

Sol.

Apply the distributive property on the LHS:

5/6(x+3/5)=(5/6)x+5/6×3/5

Simplify the product:

5/6×3/5

=3/6

=12

So, LHS becomes:

5/6(x+3/5)=(5/6)x+1/2

So, the equation becomes:

(5/6)x+1/2=(5/6)x+1/2

Subtract 56 x from both sides: 

12=12

This is always true, so the equation holds true for all rational numbers.So, x can be any rational number.

Think and Reflect (NCERT Textbook Page No. 51)

Try and represent 8/5 and –7/4 on a number line.

Sol.

8/5 = 1.6

–7/4 = – 1.7

Example 1: |5/3| = 5/3 , |– 5/3| is also equal to 5/3 and |0|= 0.

Sol.

Thus, the absolute value of a positive number is the number itself. 

The absolute value of a negative number is its positive value. Therefore, the absolute value of any rational number is always non-negative, that is, |x| ≥ 0.

For two rational numbers a and b, the distance between them on the number line is given by |a – b|. Fig. 3.8 represents the distance between the integers –4 and 3.

Exercise Set 3.4 Class 9 Maths Solutions 

(page 52 –53)

Q. 1. Represent the rational numbers 2/3, –5/4 and 1½ on a single number line.

Sol.

To represent the rational numbers 2/3, –54 and 1 12on a single number line, first identify their positions.

Identify the values

2/3 lies between 0 and 1.

–5/4 = –1½, 

which lies between –2 and –1.

1½ = 3/2, 

which lies between 1 and 2.

Use suitable divisions

Mark integers -2, -1, 0, 1 and 2 on the number line.

For –5/4 : 

Divide the interval from –2 to –1 into 4 equal parts and mark one part to the left of –1 (–114)

For 2/3: 

Divide the interval from 0 to 1 into 3 equal parts and mark the second part.

For 1½: 

Divide the interval from 1 to 2 into 2 equal parts and mark the midpoint.

Thus, the number line:

Q. 2. Find three distinct rational numbers that lie strictly between –1/2 and 1/4

Sol.

First, take the average of –1/2 and 1/4:

(−1/2+1/4)/2

=(−2/4+1/4)/2

=(−1/4)/2

=−1

So, –18 lies between –12 and 14

Now find the average of –12 and –18:

(−1/2−1/8)/2

=(−4/8−1/8)/2

=(−5/8)/2

=−5/16

Next, find the average of –18 and 14:

(−1/8+1/4)/2

=(−1/8+2/8)2

=(1/8)/2

=1/16

Thus, three distinct rational numbers strictly between –12 and 14 are:

−5/16,−1/8,1/16

(Answer may vary)

Q. 3. Simplify the expression

Sol.

Find the LCM of 4 and 12, which is 12.

Convert – 14 to denominator 12:

−14=−312

Now add:

−312+512=212

Simplify

212=16

Therefore,

−14+512=16

Q. 4. A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can 4 he make?

Sol.

Convert mixed fractions to improper fractions:

15¾ =63/4,

2¼ = 9/4

Now divide:

63/4 ÷ 9/4

=63/4 × 4/9

= 63/9 

= 7

Therefore, he can make 7 kurtas.

Q. 5.Find three rational numbers between 3.1415 and 3.1416.

Sol.

To find rational numbers between 3.1415 and 3.1416, 

Write them with more decimal places:

3.1415 = 3.14150 and  for 3.1416 = 3.14160

Now choose any three numbers between 

3.14150 and 3.14160

these:

3.14151, 3.14152, 3.14153

These are all rational numbers and lie strictly between the given numbers.

Therefore, the required rational numbers are: 3.14151, 3.14152, 3.14153 (Answers may vary).

Q. 6. Can you think of other way(s) to find a rational number between any two rational numbers?

Sol.

One simple way is to make the denominators the same. Suppose the two rational numbers are a.b and c.d with ab<cd.

Convert them to equivalent fractions with a common denominator, then choose any fraction whose numerator lies between the two numerators. Another way is to multiply both numbers by the same positive number to create more space, then choose a number in between and divide back.

Think and Reflect (NCERT Textbook Page No. 53)

Can √2 be written as a rational number p/q ?

Sol.

No, cannot be written as a rational number .

Proof (by contradiction):

Assume that is a rational number. Then it can be written in the lowest terms as

√{2} = {p}/{q}

where p and q are coprime integers with no common factor (q ≠ 0). 

Squaring both sides,

2 = {p²} / {q²}

p² =2q²  ...........(1)

This shows that p² is divisible by 2, so p must also be divisible by 2. 

Then Let p = 2k

Substituting p = 2k in equation (1)

(2k)² = 2q²

(4k² = 2q²)÷2

q²=2k²

So, q² is also divisible by 2, which means q is divisible by 2.

But now both p and q are divisible by 2, which contradicts our assumption that p/q is in lowest terms. 

Hence, 

our original assumption is false, and thus √2 is irrational.

Think and Reflect (NCERT Textbook Page No. 55)

Q. 1. Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7 or √10?

Sol.

To prove that √3 is irrational using contradiction, assume the opposite of what we want to prove. Suppose √3 is rational. Then it can be written as a fraction in lowest terms:

√3 = p/q

where p and q are integers with no common factor other than 1, that means p and q are co-prime and q ≠ 0.

Now square both sides:

3 = p²/q² 

=> p² = 3q²

This shows that p² is divisible by 3, so p must also be divisible by 3 (since 3 is prime). 

Let p = 3k.

Substitute back:

(3 k)² = 3q²

=>  9 k² = 3q²

=> q² = 3k²

So, q² is also divisible by 3, which means q is divisible by 3.

But now both p and q are divisible by 3, which contradicts our assumption that p/q is in lowest terms. 

Hence, our original assumption is false, and thus √3 is irrational.

Next, for √5 or √7, the exact same method works. Both are prime numbers, and the argument depends on the fact that if a prime divides p², it must divide p. So 45 and 4l can be proved irrational in the same way.

For √10, the situation is slightly different because 10 is not prime (it is 2 ≠ 5). Still, the method can be adapted. If you

assume √10 = p/q, you get p² = 10q², which implies both 2 and 5 divide p², hence they divide p. This leads again to p and q sharing common factors, giving a contradiction. So, the same contradiction approach works for all these numbers—in fact, it works for the square root of any number that is not a perfect square.

Q. 2. We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?

Sol.

We obtain lines whose lengths are irrational by using geometric constructions based on the Pythagoras Theorem. 

For example, by constructing a right-angled triangle with both perpendicular sides of length 1 unit, the hypotenuse has length √2, which is an irrational number. Similarly, other irrational lengths can also be constructed.

For example:

Draw a line segment AB = 1 unit.

At point B, draw a line BC = 1 unit perpendicular to AB.

Join A and C.

In the right-angled triangle ABC

AC² = AB² + BC² 

        = 1² + 1²

 AC²  = 2,

AC  = √2,

Therefore, AC = √2,

Thus, the line segment AC has length √2, which is an irrational number.

Similarly, by using right triangles, we can construct line segments of lengths √3, √5, √6, and many other irrational numbers.

Think and Reflect (NCERT Textbook Page No. 56)

Q. Try to extend this method for constructing line segments of lengths 43 and 45 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form 4n , where n is a positive integer.

Sol.

To extend the same idea, we keep using right triangles and the Baudhayana-Pythagoras Theorem step by step.

For √3 : 

Start exactly like the construction of √2. We already have a right triangle with legs 1 and 1, so the hypotenuse is √2. Now, from the endpoint of this √2, draw a perpendicular and mark a length of 1 unit again. Join this new point to the origin. 

By the Baudhayana- Pythagoras theorem:

(√2)² + 1² = 2 + 1 = 3

So, the new hypotenuse is √3. Then, with centre at O and this length as radius, cut the number line to locate √3 .

For √5 : 

We can repeat the process, but there’s a simpler way. First construct √4 = 2 on the number line. At the point 2, draw a perpendicular of length 1 unit. Join this point to O. Then:

2² + 1² = 4 + 1 = 5

So, the hypotenuse gives √5. Transfer it onto the number line using a compass.

Generalisation for √n : This method works by building a chain of right triangles. Each time:

★ Take the previously obtained length n−1−−−−−√

★ Draw a perpendicular of length 1 unit

★ Join the new point to the origin

Then,

(n−1−−−−−√)² + 1² 

= (n – 1) + 1 

= n

So, by stacking right triangles, you can construct √2 , √3, √4, √5, ……… and hence any √n where n is a positive integer.

Example 2: is 3/8 a terminating or non terminating decimal ?

Sol.

3/8 

= 0.375. is terminating decimal.

Example 3: Is 5 /11 a terminating decimal or non terminating decimal ?

Sol. 

5 /11 

= 0.454545 … 

= 0.45 is non terminating decimal.

Think and Reflect (NCERT Textbook Page No. 57)

Q. Try to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?

Sol.

1. = 10/3

{10}/{3} = 3.333333.... 

                       _

               = 3. 3

Here, the digit 3 repeats forever

2. 11/12

{11}/{12} = 0.916666...... 

                             _

                 = 0.91 6

Here, after the digits 91, the digit 6 repeats forever.

Observation

★ has a non-terminating recurring (repeating) decimal expansion: .

★ also has a non-terminating recurring decimal expansion: .

★ In both cases, the digits after the decimal point eventually repeat in a fixed pattern.

Think and Reflect (NCERT Textbook Page No. 58)

Q. The decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?

Sol.:

Suppose p/q is in lowest terms, i.e., p and q have no common factor.

A decimal number terminates only if we can convert the denominator into a power of 10.

Since

10 = 2 × 5

every power of 10 is of the form

10ⁿ = 2ⁿ × 5ⁿ

Therefore, if the denominator q divides some power of 10, then q can contain only the prime factors 2 and 5.

For example:

3/8 = 375/1000 = 0.375 because 8 = 2³

7/20 = 35/100 = 0.35 because 20 = 2² × 5

But if q has any other prime factor, such as 3, 7, 11, etc., it can never divide any power of 10.

For example:

1/3 = 0.333.....

2/7 = 0.285714285713.......

These decimals never end; they repeat forever.

Hence,

The decimal expansion of p/q is terminating if and only if, after reducing the fraction to its lowest terms, the denominator is of the form

where  2ⁿ × 5ⁿ and  5ⁿ × 2ⁿ are whole numbers (possibly 0). Otherwise, the decimal expansion is non-terminating recurring.

Converting Rational Decimals into the Form p/q

Case 1: Terminating decimals

We have already seen in earlier grades how to convert terminating decimals into the form p/q.

Example 4: Convert 0.35 into the form p/q .

Sol. 

We know that 

0.35 

= 35 /100 

= 7/20. 

Now, let us understand how to convert non-terminating repeating decimals into the form p/q .

Case 2: Pure repeating decimals

A pure repeating decimal is one in which a digit or a sequence of digits begins repeating immediately after the decimal point.

Example 5: Convert 0.6 into the form p/q .

Sol.

                             _

Step 1: Let x = 0.6  ...............(1)

Step 2: Since one-digit repeats, multiply both sides by 10.

              _

10x = 6.6.      ..............(2)

Step 3: Subtract the first equation from the second:

                     _        _

10x – x = 6.6 – 0.6 

9x = 6.

Step 4: Solve for x: 

9x = 6. 

Thus, x = 6/9

x = 2/3

Example 6: Convert 0.45 into the form p/q. 

Sol.

                ___

Let x = 0.45       ..............(1)

Since two digits repeat, multiply by 10² = 100. We get 

                  ___

100x = 45.45.       ..............(2)

Subtracting the two equations, we get 

99x = 45.

So, x = 45/99 

x =  5 / 11.

Case 3: General repeating decimals

A general repeating decimal has some non-repeating digits just after the decimal point, followed by a repeating block.

Example 7: Convert 0.16 into the form p/q .

Sol.

                               __

Step 1: Let x = 0.1 6.

Step 2: First, shift the decimal to place the repeating part immediately after the decimal point. 

Since one digit is non-repeating, we multiply by 10¹ = 10 and obtain 

              __

10x = 1. 6.       ..............(1)

Step 3: Now, to move one full repeating cycle (1 digit repeating), we multiply by 10¹ = 10 again and get 

                  __

100x = 16.6..       ..............(2)

Step 4: Subtracting the first shifted equation from the second, we get, 

    100x – 10x = 16.6 – 1.6, 

=> 90x = 15. 

Thus, x = 15/90

x = 1/6

Example 8: Convert 2.357 into the form p/q . 

Sol.

                                 __

Step 1: Let x = 2.357.       ..............(1)

Step 2: Here, ‘35’ is non-repeating (2 digits), ‘7’ repeats (1 digit). 

Firstly, we multiply by 10² = 100 to move the non-repeating digits before the decimal and we obtain 

                    __

100x = 235.7        ..............(2)

Step 3: Now, we multiply by 10¹ = 10 to move the full repeating cycle and get 

                        __

1000x = 2357.7       ..............(3)

Step 4: Subtracting, we obtain 

                                     __           __

1000x – 100x = 2357.7 – 235.7.

Hence, 

900x = 2122  

x = 2122/900

x = 1061/450 .

                                          __

Example 9: Convert 2.4537 into the form p/q .

Sol.

                                  __

Step 1: Let x = 2.4537.

Step 2: Here, ‘45’ is non-repeating (2 digits), ‘37’ repeats (2 digits). 

Firstly, we multiply by 10² = 100 to move the non-repeating digits before the decimal and we obtain 

100x = 245.37.

Step 3: Now, we multiply by 10² = 100 to move the full repeating cycle and get 

10000x = 24537.37.

Step 4: Subtracting, we obtain 

10000x – 100x = 24537.37 – 245.37.

Hence, 

9900x = 24292 

x = 24292/9900

x = 6073/2475 .

Class 9 Maths Ganita Manjari Chapter 3 Exercise Set 3.5 Solutions

Exercise Set 3.5 Class 9 Maths Solutions

Q. 1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250

Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Sol.

For 7/20 :

20 = 2² × 5

Only 2 and 5 appear in the denominator, so the decimal terminates.

Now check by long division:

       0.35

20) 7.00

   – 6 0

       100

     – 100

            ×

So 7/20 = 0.35

For 4/15 :

15 = 3 × 5

Since 3 appears in the denominator, the decimal will be repeating.

Long division:

       0.266

15) 4.000

    – 3 0

       100

          90

            100

       –  100

               10

4 ÷ 15= 0.26̄

So, 13/250

250 = 2 × 5³

Only 2 and 5 appear in the denominator, so the decimal terminates.

Long division:

         0.052   

250) 13.000

       –1250 

            500

        – 500

               0

13 ÷ 250 = 0.052

So, 13/250 = 0.052

Q. 2. Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13etc. What do you notice?

Sol.

Long division for 1/13:

So, the repeating block is: 076923

Now,

      0.076924

13) 1.000000

      –91          

           90

         –78. 

            120

          –117

                  30

                –26

                    40

                 –39

                       1 (Repeat)

                              

1/13 = 0.074924...

                              

2/13 = 0.153846...

                              

3/13 = 0.230769...

                              

4/13 = 0.307692...

                              

5/13 = 0.384615...

                

repeating digits are the product of 2, 3, 4, 5, … to the repeating digit of 113.

Q. 3. Classify the following numbers as rational or irrational:

(i) √81

(ii) √12

(iii) 0.33333 .......

(iv) 0.123451234512345.........(v) 1.01001000100001 ..........

 (Notice the pattern: Is it repeating a single block?)

(vi) 23.560185612239874790120

Find the explicit fractions in case they are rational.

Sol.

(i) √81

√81 = 9

So it is rational.

Explicit fraction: 9 = 9, 1

(ii) √12

√12 = √4 × √3 = 2√3😀

Since √3 is irrational, 2√3 is also irrational.

So √12 is irrational.ñ(iii) 0.33333.........

This is a repeating decimal.

Let x = 0.33333.........        ..............(1)

Multiply by 10 both sides

Then 10x = 3.33333…       ..............(2)

Subtract equation (1) from equation (2):

10x – x = 3.33333…– 0.33333…

9x = 3

x = 3/9

x =1/3

So it is rational. 

Explicit fraction:

0.33333…= 1/3

(iv) 0.123451234512345…

The block 12345 repeats again and again. So, this is a repeating decimal, hence rational.

Let x = 0.1234512345…       ..............(1)

Since 5 digits repeat, multiply by 10⁵ = 100000

We get 

100000 x = 12345.12345…       ..............(2)

Subtract equation (1) from equation (2):

100000x – x = 12345.12346 ... – 0.12345 ....

99999 x = 12345

x = 12345/9999 

This can be simplified as 3:

x = 12345/99999 

x = 411533333

So, the number is rational.

Explicit fraction:

M0.123451234512345… = 411533333

(v) 1.01001000100001…

Here, the digits do not repeat in one fixed block. The pattern keeps changing because the zeros keep increasing.

So it is non-terminating, non-repeating.

Therefore, it is irrational

(vi) 23.560185612239874790120

This is a terminating decimal because it ends after a finite number of digits. So, it is rational.

As a fraction:

= 235601856122398747901201021

This may also be simplified to:58900464030599686975325000000000000000000

Q. 4. The number 0.9 (which means 0.99999…) is a rational number. Using algebra (let x = 0.9, multiply by 10, and subtract), explain why 0.9 is exactly equal to 1.

Sol.

Let x = 0.9999….       ..............(1)

Multiply both sides by 10:

10x = 9.9999….       ..............(2)

Subtract equation (1) from equation (2):

10x – x = 9.9999…-0.9999…

9x = 9

x = 9/9

x = 1

But we assumed x = 0.9999…,

So 0.9999… = 1

Hence, 0.9 (i.e., 0.9999…) is exactly equal to 1.

Q. 5.  We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1n) produce decimals with repeating blocks that are cyclic.

Sol. 

A cyclic number is one where its repeating digits rotate when multiplied by numbers.

We already know:

1/7 = 0.142857

The block 142857 is cyclic because multiplying it by 2,3, 4, 5, or 6 gives rotations of the same digits.

Now, to find more such numbers, we look for values of n where:

• The decimal expansion of 1n is repeating.

• The repeating block has special rotation properties.

These usually occur when:

• n is a prime number.

• The length of the repeating cycle is n -1 (called a full reptend prime).

Example,

1. 1/17 =0.0588235294117647 17

The repeating block has 16 digits and shows cyclic-like properties.

2. 1/19 = 0.052631578947368421 19

A number n (more precisely, a prime p) has this cyclic property when the decimal expansion of 1p repeats with the maximum possible length, which is P – 1. Such primes are called full reptend primes. These primes generate repeating decimals where the digit block cycles under multiplication, forming cyclic numbers.

Think and Reflect (Page 64)

Consider this puzzle: What is the square root of –1? We know that 1 x 1 = 1. We also know that (–1) × (–1) = 1. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, √–1 cannot exist on number line.

Sol.

We know that 

 1 × 1 = 1

And 

–1 × –1 = 1

But no real number multiplied by itself gives a negative number.

Therefore,

√–1 

cannot be represented on the real number line.

To solve this, mathematicians introduced a new number called the Imaginary Unit (i).

It is defined as:

i = √–1 

Therefore,

i²  = –1 

Final Answer: 

The square root of –1 is i, called the imaginary unit. It is not a real number, so it cannot be shown on the ordinary real number line.

Class 9 Maths Chapter 3 

End of Chapter Exercises Solutions

The World of Numbers End of Chapter Exercises Solutions

Q. 1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) 3/50

Sol.

3/50 = 0.06

       0.06

  50)3.00

      –300

             0

This is a terminating decimal.

(ii) 2/9

Sol.

       0.22

   9)2.000

    –18. 

           20

         –18

               2R

.                                   __

2/9 = 0.2222……… = 0.2̄

This is a non-terminating repeating decimal.

Q. 2. Prove that √5 is an irrational number.

Sol.

We will use proof by contradiction.

Assume that √5 is rational.

Then it can be written in the form:

√5 = p/q.         ...... (1)

where p and q are integers having no common factor (i.e., in lowest form), and q ≠ 0.

Squaring both sides:

5 = p² / q²

p² = 5q²

This means p² is divisible by 5, so √p² or p must also be divisible by 5.

Let p = 5k for some integer k.

Substitute p = 5k in equation (1) :

(5k)² = 5q²

25k² = 5q²

Divide both sides by 5:

5k² = q²

This shows q² is divisible by 5, so √q² or q is also divisible by 5. So, both p and q are divisible by 5, which contradicts our assumption that p/q is in lowest form.

Thus, our assumption is wrong.

Hence, √5 is irrational.

Q. 3. Convert the following decimal numbers into the form p/q.

(i) 12.6

Sol.

Let x = 12.6

          = 126/10

          = 63/5

(ii) 0.0120

Sol.

0.0120 

= 120/10000. 

= 12/1000        [Devide by 4 to D & N]

= 3/250

            __

(iii) 3.052.....

Sol.

Let x = 3.0525252…       ...... (1)

Multiply by 10¹ = 10 both sides of equation (1)

10x = 30.5252…       ...... (2)

Multiply by 10³ = 1000 both sides of equation (1)

1000x = 3052.5252…       ...... (3)

Subtract equation (2) from equation (3)

1000x = 3052.5252…

    10x =      30.5252…

⇒ 990x = 3022

⇒ x = 3022/990

     x =1511/495

            __

(iv) 1.235

Sol.

Let x = 1.2353535….   .......(1)

Multiply by 10¹ = 10 both sides of equation (1)

10x = 12.3535…   .......(2)

Multiply by 10³ = 1000 both sides of equation (1)

1000x = 1235.3535…   .......(3)

Subtract equation (2) from equation (3)

1000x = 1235.3535…

     10x =     12.3535…

⇒ 990x = 1223

⇒ x = 1223/990

          __

(v) 0.23

Sol.

Let x = 0.232323…. .......(1)

Multiply by 10² = 100 both sides of equation (1)

100x = 23.2323….   .......(2)

Subtract equation (1) from equation (2)

100x = 23.232323….

       x =   0.232323…. 

99x = 23

x = 23/99

             _

(vi) 2.05

Sol.

Let x = 2.05555… .......(1)

Multiply by 10¹ = 10 both sides of equation (1)

10 x = 20.5555… .......(2)

Multiply by 10² = 100 both sides of equation (1)

100x = 205.5555… .......(3)

Subtract equation (2) from equation (3)

100x = 205.5555…    

  10x =    20.5555…

90x = 185

     x = 185/90

     x = 37/18

               _

(vii) 2.125

Sol.

Let x = 2.125555… .......(1)

Multiply by 10² = 100 both sides of equation (1)

100.x = 212.5555… .......(2)

Multiply by 10¹ = 10 both sides of equation (1)

1000x = 2125.5555… .......(3)

Subtract equation (2) from equation (3)

1000x = 2125.5555…

   100x =   212.5555…

900x = 1913

⇒ x = 1913/900

                _

(viii) 3.125

Sol.

Let x = 3.125555… .......(1)

Multiply by 10² = 100 both sides of equation (1)

100x = 312.5555… .......(2)

Multiply by 10¹ = 10 both sides of equation (1)

1000x = 3125.5555… .......(3)

Subtract equation (2) from equation (3)

1000x = 3125.5555…

  100x =    312.5555… 

900x = 2813

x = 2813/900

           ____

(ix) 2.1625

Sol.

Let x = 2.162516251625… .......(1)

Multiply by 10⁴ = 10000 both sides of equation (1)

10000x= 21625.162516251625… .......(2)

Subtract equation (1) from equation (2)

10000x= 21625.162516251625…

           x =         2.162516251625…

9999x = 21623

x = 21623/9999

Q. 4 Locate the following rational numbers on the number line,
(i) 0.532
(ii) 1.15
Solution:
(i) 0.532 lies between 0.530 and 0.540. Divide the segment from 0.530 to 0.540 into 10 equal parts; the 2nd mark after 0.530 gives 0.532

(ii) Convert the repeating decimal to a fraction first: Let x — 1.15555…
Then 10x= 11.5555…
100x = 115.555…
Subtract:
100x – 10x= 115.5555… — 11.555…
= 10.4
90x = 104
x = 10490=5245
So,
1.15555….. = 5245
5245 = 1 + 745
It lies between 1 and 2. Divide the segment from 1 to 2 into 45 equal parts. Count 7 parts after 1.

Question 5.
Find 6 rational numbers between 3 and 4.
Solution:
Write 3 and 4 with a common denominator:
3 = 3010, 4 = 4010

Now choose any 6 rational numbers between them:
3110,3210,3310,3410,3510,3610

So, 6 rational numbers between 3 and 4 are:
3110,3210,3310,3410,3510,3610
(Answer may vary)

Question 6.
Find 5 rational numbers between 23 and 35.
Solution:
Make denominators larger to create more:
25=2050,35=3050

Now take any 5 rational numbers between them:
2150,2250,2350,2450,2550

So the required rational numbers are:
2150,2250,2350,2450,2550
(Answer may vary)

Question 7.
Find 5 rational numbers between 16 and 25.
Solution:
Take LCM of denominators 6 and 5, which is 30:
16=530,25=1230

Now pick any 5 rational numbers between them:
630,730,830,930,1030

So the required rational numbers numbers are:
630,730,830,930,1030
(Answer may vary)

Question 8.
If 12, find the rational number x.
Solution:
Take LCM of 3 and 5, which is 15.
x3=5x15,x5=3x15

So, the equation becomes:
5x+3x15=1615
8x15=1615

Multiply both sides by 15:
8x = 16
x = 168 = 2
x = 2

Q. 9.  Let a and b be two non-zero rational numbers such that a + 1b = 0
Without assigning numerical values, determine whether ab is positive or negative. Justify your answer.
Sol.
Given a + 1/ b = 0
Rearrange: a = –1/b
Now multiply both sides by b:

a × b = –1/b × b
ab = –1
Since –1 is a negative number, so, ab is negative.

Q. 10.  A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place.
Show that such a number can be written in the form, p/10⁴ where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in lowest form, is divisible by 2⁴ or 5⁴? Give reasons.
Sol.
Let the number be x.
Since the decimal expansion terminates and the last non¬zero digit is at the 4th decimal place, the number must be of the form:
x= a.bcde
where e 0 and there are no non-zero digits beyond the 4th decimal place.

Multiply by 10⁴ = 10000:
1000x = an integer = p
So, x = p10⁴

Now, since the last non-zero digit is exactly at the 4th decimal place, the decimal does not terminate earlier. This means p cannot be divisible by 10, because if p were divisible by 10, then:
p/10⁴ = p/10×1103
which would shift the decimal termination to an earlier place — a contradiction.

Hence, x = p10⁴, where p is not divisible by 10.
Now consider the second part.
10⁴ = (2 × 5)⁴ = 2⁴. 5⁴
So initially, the denominator contains both 2⁴ and 5⁴.

When reducing p10⁴ to lowest terms:

• Since p is not divisible by 10, it cannot be divisible by both 2 and 5 together.
• However, p may be divisible by 2 or by 5 individually.

So, some powers of 2 or 5 may cancel, but not both completely at the same time.
Therefore, it is not necessary that the denominator in lowest form is divisible by 2⁴ or 5⁴.
After simplification, the denominator will be of the form 2^m5ⁿ, where m ≤ 4, n ≤ 4, and possibly smaller depending on common factors with p.

Q. 11.  Without performing division, determine whether the decimal expansion of 18125 is terminating or non-terminating. If it terminates, state the number of decimal places.
Sol.
A rational number pq (in lowest form) has a terminating decimal expansion if the prime factorisation of the denominator q contains only powers of 2 and/or 5.

First, check if the fraction is in lowest form:
HCF (18, 125)= 1
So it is already in lowest term.
Now factorise the denominator:
125 = 5³

Since the denominator contains only the prime factor 5, the decimal expansion is terminating.
To find the number of decimal places, we need to convert the denominator into a power of 10.
10³ = 2³ × 5³

Multiply numerator and denominator by 2³ = 8:
18125=18×8125×8=1441000
Thus, the decimal has 3 decimal places.

Q. 12.  A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Sol.

Given denominator:
2³ × 5 = 8 × 5 = 40
A rational number has a terminating decimal if the denominator (in lowest form) is of the type 2^m × 5ⁿ.
To find the number of decimal places, we convert the denominator into a power of 10.
10 = 2 × 5

We need equal powers of 2 and 5. Here we have:
2³ × 5¹

To balance, multiply by 5²:
2³ × 5³ = 10³ = 1000
So, 140=251000
This shows the decimal will have 3 decimal places.

Q. 13  Let a = 712 and b = 56. Express both a and b in the form k₁1/m and k₂/m where k_1, k₂ and m are integers and k₂ – k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ – k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Sol.
First, write both fractions with a common denominator.
a = 7/12, 

b = 5/6

    =10/12
Here,
k₁ = 7, k₂ = 10, m = 12 => k₂ – k₁ = 3

But 3 < 6, so we need a larger common denominator.
Multiply both by 4:
a = 7/12

    =28/48

b = 10/12

     =40/48

Now,
k₁ = 28, k₂ = 40, m = 48
⇒ k₂ – k₁ = 12 > 6

Now write five rational numbers between them:

28/48, < 29/48, < 30/48, < 31/48, < 32/48, < 33/48 ...........40/48

These all lie strictly between ∀ and ∀.

Why the condition k₂ – k₁ > n + 1 is necessary:
Between k1/m and k2/m, the numbers with denominator m are:
(k₁+1)/m, (k₁+2)/m, ..........…,(k₂ −1)/m

Total numbers between them:
k₂ – k₁ – 1 > n
To find at least numbers, we need:
k₂ – k₁ – 1 > n
k₂ – k₁ > n + 1
Hence, the condition:
k₂ – k₁ > n + 1
ensures enough integers exist between k₁ and k₂ to form rational numbers.

Q. 14 Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
Sol.
Given:
x + y + z = 0 ............ (1)

xy + yz + zx = 0. ............(2)

Now,

x + y + z = 0 ............ (1)

z = – x – y 

z = – (x + y) .................(3)

Substitute (Place) this in the second equation
xy + yz + zx = 0
xy + y (–x – y) + x (–x – y) = 0

Simplify:
xy – xy – y² – x² – xy = 0
– (x² + y² + xy) = 0

x² + y² + xy = 0/–1   

x² + y² + xy = 0       …..............(4)

Adding and subtracting xy in equation (4)

x² + y² + xy +xy – xy = 0 

(x² + y² + 2xy) – xy = 0 

(x + y)² – xy = 0 

(x + y)²  = xy

So,

 (x + y)² = xy …....... (5)

But from equation (3), 

z = –(x +y)

Squaring both sides 
z² = (x + y)²
So from equation (3) and equation (5)
z² = xy ..........…(6)

Now similarly, by symmetry:
x² = yz, 

y² = zx
Add all three:
x² + y² + z² = xy + yz + zx

But given:
xy + yz + zx = 0
So,
x² + y² + z² = 0

Since squares of rational numbers are non-negative, the only way their sum is zero is:
x² = y² = z² = 0
⇒ x = y = z = 0
z = y = z = 0

Q. 15 Show that the rational number lies between the rational numbers a and b.
Sol.
Assume a < b (the case b < a is similar).
We need to prove:
a < (a+b)/2 < b

First part: 

Prove a < (a+b)/2
Multiply both sides by 2 (positive number, so inequality stays same):
2a < (a + b)
a < b (which is true)
Hence,
a < (a+b)/2

Second part: 

Prove (a+b)/2 < b
Multiply both sides by 2:
(a + b) < 2b
a < b (which is true)
Hence,
(a+b)/2 < b
Conclusion
a < (a+b)/2 < b
So, (a+b)/2 lies between a and b

Q. 16 Find the lengths of the hypotenuses of all the right triangles in Figure which isjeferred to as the square root spiral.

Square root spiral
Sol.
Each new triangle is formed by taking the previous hypotenuse as one side and 1 as the other side. We can find the hypotenuse of each triangle using the Baudhayana Pythagoras Theorem:
(hypotenuse)² = (base)² + (perpendicular)²

For triangle 1: 

Sides: 1, 1
Hypotenuse = √[(1)² + (1)²] 

= √(1 + 1) = √2

For triangle 2: 

Sides: √2, 1
Hypotenuse = √[(√2)² + (1)²]

= √(2+1) = √3

Similarly:

Triangle 3: 

Sides: √3 , 1; Hypotenuse: √4 = 2

Triangle 4: 

Sides: 2, 1; Hypotenuse: √5

Triangle 5: 

Sides: √5 , 1;  Hypotenuse: √6

Triangle 6: 

Sides: √6, 1; Hypotenuse: √7

Triangle 7: 

Sides: √7, 1; Hypotenuse: √8

Triangle 8: 

Sides: √8, 1; Hypotenuse: √9 = 3

Triangle 9: 

Sides: 3, 1; Hypotenuse: √10

Triangle 10: 

Sides: √10, 1; Hypotenuse: √11

Natural Numbers

Natural Numbers are the counting numbers 

{1, 2, 3, … } that emerged at least tens of thousands of years ago due to humanity’s need to count.

The Concept of Zero (Śhūnya)

The Concept of Zero (Śhūnya) was formalised in India by philosophers and then brought into mathematics formally by Brahmagupta (629 CE), who transformed the philosophical state 

of ‘nothingness’ (Śhūnyatā) into an actual number on which one could perform arithmetic operations. Brahmagupta also introduced the negative numbers, i.e., numbers less than zero.

Integers (Z) 

Integers (Z) extend the number line to the left of 1 to include zero as well as the negative numbers, which Brahmagupta historically 

categorised as ‘debts’ (ṛiṇa) in contrast with the positive number ‘fortunes’ (dhana).

Chapter 3.indd 66 4/15/2026 2:55:10 PM The World of Numbers 67

Brahmagupta’s Laws

Brahmagupta’s Laws provided the first rigorous framework for arithmetic with signed numbers, establishing rules such as ‛the product of two debts is a fortune’ (– × – = +), as well as the rules for zero.

Rational Numbers (Q)

Rational Numbers (Q) are defined as any number that can be expressed as a ratio p

q (where p and q are integers and q ≠ 0). 

Brahmagupta also gave the formal rules for addition, subtraction, multiplication, and division of rational numbers.

Rational numbers are dense, meaning a rational number always exists between any two other rational numbers.

Irrational Numbers are values like √2 and π that cannot be written as fractions. Their existence proves that the number line contains gaps that rational ratios alone cannot fill. The 

first proof of the irrationality of a number was that of √2, by Hippasus (c. 400 BCE). It was a proof by contradiction. A proof that π is irrational (as it was suspected to be by Āryabhaṭa in 499 CE) was given by the Swiss mathematician Johann Lambert in 1761.

Real Numbers (R)

Real Numbers (R) represent the total union of all rational and irrational numbers, forming a perfectly continuous and unbroken line where every real physical measurement has a 

corresponding point.

Decimal expansions serve as a mathematical signature for rational vs. irrational: rational numbers always result in terminating or repeating decimals, while irrational numbers 

produce non-repeating decimals that continue infinitely.

• Cyclic Numbers, such as the digits found in the repeating block of 1/7 (i.e., 142857), reveal the elegant and symmetrical internal patterns hidden within the rational numbers.

• Imaginary Numbers are introduced as a final conceptual frontier to handle operations like −1, which cannot be solved on the real number line and require a new dimension of mathematics.

Example: To find a rational number between 1/3 and 1/2, write them with denominator 6:

13=26,12=36

Now a number between them is:

512

since

26<512<36

We can also express the rational numbers in decimal form and then find the rational numbers between them.

So, besides the average method, we can use common denominators, scaling, by expressing in decimal form or simply pick a fraction with a suitable numerator after rewriting.