Online Self Test
Learn & Practice Quizzes & Challenges Competitive Exams
 

Number Systems – CBSE NCERT Study Resources

Previous ChapterNext Chapter
Previous Chapter Next Chapter

Study Materials

9th

9th - Mathematics

Number Systems

Jump to Question Sets

Number Systems - Grade 9 Mathematics

Chapter Overview

This chapter introduces the concept of number systems, focusing on real numbers, rational and irrational numbers, and their representation on the number line. Students will learn about decimal expansions, operations on real numbers, and laws of exponents for real numbers.

Number System: A system for representing numbers using digits or symbols in a consistent manner.

Types of Numbers

Natural Numbers

Counting numbers starting from 1 (1, 2, 3, ...).

Whole Numbers

Natural numbers including zero (0, 1, 2, 3, ...).

Integers

All whole numbers and their negatives (..., -2, -1, 0, 1, 2, ...).

Rational Numbers

Numbers that can be expressed as p/q where p and q are integers and q ≠ 0.

Rational Number: A number is called rational if it can be expressed in the form p/q where p and q are integers and q ≠ 0.

Irrational Numbers

Numbers that cannot be expressed as p/q (where p and q are integers and q ≠ 0). Their decimal expansions are non-terminating and non-repeating.

Real Numbers

The collection of all rational and irrational numbers forms the set of real numbers.

Representation of Real Numbers on the Number Line

Every real number corresponds to a unique point on the number line, and every point on the number line represents a unique real number.

Decimal Expansions

Terminating Decimals

Decimal expansions that end after a finite number of digits (e.g., 0.5, 0.75).

Non-Terminating Repeating Decimals

Decimal expansions that continue infinitely with repeating patterns (e.g., 0.333..., 0.142857142857...).

Non-Terminating Non-Repeating Decimals

Decimal expansions that continue infinitely without any repeating pattern (e.g., √2 = 1.41421356..., π = 3.14159265...).

Operations on Real Numbers

Real numbers can be added, subtracted, multiplied, and divided (except division by zero). Some important properties:

  • Commutative Property
  • Associative Property
  • Distributive Property

Laws of Exponents for Real Numbers

For any non-zero real number a and integers m and n:

  1. am × an = am+n
  2. (am)n = amn
  3. am/an = am-n
  4. am × bm = (ab)m

Rationalizing the Denominator

The process of eliminating radicals from the denominator of a fraction by multiplying both numerator and denominator by a suitable expression.

Rationalization: The process of removing radicals from the denominator of a fraction.

Summary

This chapter covers the fundamental concepts of number systems, including classification of numbers, their representation on the number line, operations on real numbers, and laws of exponents. Understanding these concepts forms the foundation for more advanced mathematical topics.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the decimal expansion of 3/8?
Answer:
0.375
Question 2:
Is √2 a rational number?
Answer:

No, it is irrational.

Question 3:
Express 0.6̄ as a fraction.
Answer:
2/3
Question 4:
Find the value of √9 × √16.
Answer:
12
Question 5:
What is the sum of 0.3̄ and 0.4̄?
Answer:
0.7̄
Question 6:
Is π a rational number?
Answer:

No, it is irrational.

Question 7:
Write the decimal form of 1/5.
Answer:
0.2
Question 8:
Simplify: (√5 + √3)(√5 - √3).
Answer:
2
Question 9:
What is the product of 0.2 and 0.5?
Answer:
0.1
Question 10:
Classify -7 as natural, whole, or integer.
Answer:

Integer.

Question 11:
Find the square root of 144.
Answer:
12
Question 12:
Is 1.010010001... a rational number?
Answer:

No, it is irrational.

Question 13:
What is the decimal representation of the rational number 5/4?
Answer:

The decimal representation of 5/4 is 1.25.
Calculation:
5 ÷ 4 = 1.25

Question 14:
Identify whether √3 is a rational or irrational number.
Answer:

√3 is an irrational number because it cannot be expressed as a fraction p/q where p and q are integers and q ≠ 0.

Question 15:
Express 0.6̄ (repeating) as a fraction in simplest form.
Answer:

0.6̄ can be expressed as 2/3.
Let x = 0.666...
10x = 6.666...
Subtract: 10x - x = 6.666... - 0.666...
9x = 6
x = 6/9 = 2/3

Question 16:
Find three rational numbers between 1/2 and 3/4.
Answer:

Three rational numbers between 1/2 (0.5) and 3/4 (0.75) are:
0.6, 0.65, and 0.7.
These can also be written as 3/5, 13/20, and 7/10.

Question 17:
Is zero a rational number? Justify your answer.
Answer:

Yes, zero is a rational number because it can be expressed as 0/1, where both numerator and denominator are integers and the denominator is not zero.

Question 18:
Simplify: (√5 + √2)(√5 - √2).
Answer:

Using the identity (a + b)(a - b) = a² - b²:
(√5 + √2)(√5 - √2) = (√5)² - (√2)²
= 5 - 2
= 3

Question 19:
Convert the decimal 0.125 into a fraction in simplest form.
Answer:

0.125 can be written as 1/8.
Calculation:
0.125 = 125/1000
= 1/8 (simplified)

Question 20:
What is the value of √(9 × 16)?
Answer:

√(9 × 16) = √9 × √16
= 3 × 4
= 12

Question 21:
Write the decimal expansion of 1/7 up to 3 decimal places.
Answer:

The decimal expansion of 1/7 up to 3 decimal places is 0.142.
Calculation:
1 ÷ 7 ≈ 0.142857...

Question 22:
Find the value of (3² + 4²)^(1/2).
Answer:

(3² + 4²)^(1/2) = (9 + 16)^(1/2)
= √25
= 5

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Express 0.6 in the form p/q, where p and q are integers and q ≠ 0.
Answer:

Let x = 0.6.
Multiply both sides by 10: 10x = 6.6.
Subtract the original equation: 10x - x = 6.6 - 0.6.
Simplify: 9x = 6.
Divide by 9: x = 6/9.
Simplify the fraction: x = 2/3.

Question 2:
Find three rational numbers between 3/5 and 4/5.
Answer:

Convert the fractions to have a common denominator (e.g., 20):
3/5 = 12/20 and 4/5 = 16/20.
Three rational numbers between them are:
13/20, 14/20 (7/10), and 15/20 (3/4).

Question 3:
State whether √2 is a rational or irrational number. Justify your answer.
Answer:

√2 is an irrational number.
It cannot be expressed as p/q where p and q are integers and q ≠ 0.
Its decimal expansion is non-terminating and non-repeating.

Question 4:
Simplify: (√5 + √2)².
Answer:

Using the identity (a + b)² = a² + 2ab + b²:
(√5)² + 2(√5)(√2) + (√2)²
= 5 + 2√10 + 2
= 7 + 2√10.

Question 5:
Write the decimal expansion of 1/7 and state its type.
Answer:

The decimal expansion of 1/7 is 0.ᐨ57 (repeating).
It is a non-terminating repeating decimal, hence a rational number.

Question 6:
Find the value of √(9 × 16).
Answer:

√(9 × 16) = √9 × √16
= 3 × 4
= 12.

Question 7:
Is π a rational number? Give a reason.
Answer:

π is an irrational number.
It cannot be expressed as a fraction p/q (p, q integers, q ≠ 0).
Its decimal expansion is non-terminating and non-repeating.

Question 8:
Express 0.999... in the form p/q. Is it equal to 1?
Answer:

Let x = 0.999...
Multiply by 10: 10x = 9.999...
Subtract the original: 9x = 9.
Thus, x = 1.
Yes, 0.999... is exactly equal to 1.

Question 9:
Classify √25 as rational or irrational.
Answer:

√25 = 5, which is a rational number.
It can be expressed as 5/1, where 5 and 1 are integers and 1 ≠ 0.

Question 10:
Simplify: (3 + √3)(2 + √2).
Answer:

Use the distributive property:
3 × 2 + 3 × √2 + √3 × 2 + √3 × √2
= 6 + 3√2 + 2√3 + √6.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Express 0.6̄ in the form of p/q, where p and q are integers and q ≠ 0.
Answer:

Let x = 0.6̄, which means x = 0.6666...
Multiply both sides by 10: 10x = 6.6666...
Subtract the first equation from the second: 10x - x = 6.6666... - 0.6666...
This simplifies to: 9x = 6
Divide both sides by 9: x = 6/9
Simplify the fraction: x = 2/3
Therefore, 0.6̄ in p/q form is 2/3.

Question 2:
Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Answer:

First, convert 5/7 and 9/11 to decimal form:
5/7 ≈ 0.714285...
9/11 ≈ 0.818181...
Three irrational numbers between them can be:
1. 0.720720072000... (non-terminating and non-repeating)
2. 0.750750075000... (non-terminating and non-repeating)
3. 0.80800800080000... (non-terminating and non-repeating)
These numbers are irrational because their decimal expansions neither terminate nor repeat.

Question 3:
Show that √5 is an irrational number using the method of contradiction.
Answer:

Assume that √5 is a rational number, so it can be written as √5 = p/q, where p and q are co-prime integers and q ≠ 0.
Squaring both sides: 5 = p²/q² → p² = 5q².
This means p² is divisible by 5, so p must also be divisible by 5 (since 5 is a prime number).
Let p = 5k, then substitute: (5k)² = 5q² → 25k² = 5q² → q² = 5k².
This implies q² is divisible by 5, so q must also be divisible by 5.
But this contradicts our assumption that p and q are co-prime.
Hence, √5 is an irrational number.

Question 4:
Simplify: (3 + √3)(2 + √2) and state whether the result is rational or irrational.
Answer:

Use the distributive property:
(3 + √3)(2 + √2) = 3 × 2 + 3 × √2 + √3 × 2 + √3 × √2
Simplify each term: = 6 + 3√2 + 2√3 + √6
The simplified form is 6 + 3√2 + 2√3 + √6.
Since √2, √3, and √6 are irrational, their sum with rational numbers remains irrational.
Therefore, the result is an irrational number.

Question 5:
Represent √9.3 on the number line using a geometric method.
Answer:

To represent √9.3 on the number line:
1. Draw a horizontal line and mark point O (origin) and A at 9.3 units.
2. Extend the line by 1 unit to point B (total length = 10.3 units).
3. Find the midpoint M of AB (at 5.15 units).
4. Draw a semicircle with diameter AB.
5. At point A, draw a perpendicular line intersecting the semicircle at C.
6. The length AC is √9.3.
7. Using a compass, transfer this length to the number line from O.
The point where the arc meets the number line represents √9.3.

Question 6:
Express 0.6̄ as a fraction in simplest form.
Answer:

Let x = 0.6̄ (where the bar denotes repeating decimal).
Multiply both sides by 10: 10x = 6.6̄
Subtract the original equation: 10x - x = 6.6̄ - 0.6̄
Simplify: 9x = 6
Divide by 9: x = 6/9 = 2/3 (simplified form).

Question 7:
Find three irrational numbers between √2 and √3.
Answer:

Irrational numbers cannot be expressed as fractions. Since √2 ≈ 1.414 and √3 ≈ 1.732, examples include:
1. 1.5010010001... (non-repeating, non-terminating)
2. 1.6010010001...
3. 1.7010010001...
Note: The pattern ensures irrationality by avoiding repetition.

Question 8:
Simplify: (3 + √5)(3 - √5).
Answer:

This is a difference of squares problem: (a + b)(a - b) = a² - b².
Here, a = 3 and b = √5.
Apply the formula: 3² - (√5)²
Calculate: 9 - 5 = 4.

Question 9:
Prove that √7 is irrational.
Answer:

Assume √7 is rational, so √7 = p/q (where p, q are co-prime integers).
Square both sides: 7 = p²/q² → 7q² = p².
Thus, p² is divisible by 7 → p is divisible by 7.
Let p = 7k. Substitute: 7q² = (7k)² → q² = 7k².
This implies q is also divisible by 7, contradicting co-primality. Hence, √7 is irrational.

Question 10:
Represent √9.3 on the number line using successive magnification.
Answer:

Steps:
1. Draw a number line with integers 3 and 4 (since 3² = 9 < 9.3 < 16 = 4²).
2. Magnify the interval [3, 4] into 10 equal parts (3.0 to 3.1, 3.1 to 3.2, etc.).
3. Locate 3.0² = 9 and 3.1² = 9.61. Since 9.3 < 9.61, focus on [3.0, 3.1].
4. Further divide [3.0, 3.1] into 10 parts (e.g., 3.04² = 9.2416, 3.05² = 9.3025).
5. √9.3 lies between 3.04 and 3.05. Mark it approximately.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain why every rational number is either a terminating or a repeating decimal. Support your answer with an example from NCERT.
Answer:
Introduction

We studied that rational numbers can be expressed as fractions p/q where q ≠ 0. Their decimal forms are either terminating or repeating.


Argument 1
  • Terminating decimals occur when the denominator q (in simplest form) has only 2 or 5 as prime factors. Example: 3/8 = 0.375 (terminates).

Argument 2
  • Repeating decimals occur when q has other prime factors. Example (NCERT): 1/3 = 0.333... (repeats).

Conclusion

Thus, rational numbers always exhibit one of these two patterns, as shown in our textbook.

Question 2:
Prove that √3 is an irrational number using the method of contradiction, similar to the proof for √2 in NCERT.
Answer:
Introduction

Our textbook shows that √2 is irrational. We use the same method for √3.


Argument 1
  • Assume √3 is rational: √3 = p/q (simplest form). Squaring gives 3 = p²/q² → p² = 3q².

Argument 2
  • This implies p² is divisible by 3, so p is divisible by 3. Let p = 3k. Substituting: 9k² = 3q² → q² = 3k², making q also divisible by 3.

Conclusion

This contradicts p/q being in simplest form. Hence, √3 is irrational.

Question 3:
Describe how real numbers are used in measuring lengths. Include an example of finding the diagonal of a square with side 1 unit.
Answer:
Introduction

Real numbers include both rational and irrational numbers, essential for precise measurements.


Argument 1
  • Lengths often involve irrational numbers. Example: Diagonal of a unit square = √(1² + 1²) = √2 units.

Argument 2
  • √2 is irrational (proven in NCERT), showing real numbers are needed even for simple geometry.

Conclusion

Thus, real numbers bridge theory and practical applications like measurement.

Question 4:
Explain why √2 is an irrational number using proof by contradiction. Support your answer with steps from NCERT.
Answer:
Introduction

We studied that irrational numbers cannot be expressed as fractions. √2 is a classic example.


Argument 1
  • Assume √2 is rational, so it can be written as a/b where a and b are co-prime.
  • Then, 2 = a²/b², leading to a² = 2b².

Argument 2
  • This implies a² is even, so a must be even (let a = 2k).
  • Substituting gives 4k² = 2b², so b² = 2k², making b even.

Conclusion

But a and b cannot both be even (contradicting co-prime). Hence, √2 is irrational.

Question 5:
Compare rational and irrational numbers with examples from daily life and NCERT.
Answer:
Introduction

Rational and irrational numbers differ in their representation and properties.


Argument 1
  • Rational numbers like 3/4 or 0.5 can be written as fractions. Our textbook shows examples like measuring lengths.
  • Irrational numbers like π or √3 cannot be expressed as fractions.

Argument 2
  • In daily life, rational numbers appear in recipes (e.g., 1/2 cup sugar), while irrational numbers occur in circular objects (e.g., wheel circumference).

Conclusion

Both types are essential, but their applications differ based on precision needs.

Question 6:
Show how to represent √5 on the number line using NCERT methods.
Answer:
Introduction

We can locate √5 on the number line geometrically, as shown in NCERT.


Argument 1
  • Draw a number line and mark 0 and 2. Construct a perpendicular of 1 unit at 2.
  • Join the points to form a right triangle with hypotenuse √(2² + 1²) = √5.

Argument 2
  • Using a compass, draw an arc with radius √5 from 0 to intersect the number line.

Conclusion

The intersection point represents √5, demonstrating its exact position.

Question 7:
Explain why every rational number is either a terminating or a repeating decimal. Support your answer with NCERT examples.
Answer:
Introduction

We studied that rational numbers can be expressed as fractions p/q where q ≠ 0. Their decimal forms are either terminating or repeating.


Argument 1
  • Terminating decimals occur when the denominator q (in simplest form) has only 2 or 5 as prime factors. Example: 3/8 = 0.375 (NCERT Ex 1.4).

Argument 2
  • Repeating decimals arise when q has prime factors other than 2 or 5. Example: 1/3 = 0.333... (NCERT Ex 1.3).

Conclusion

This property helps us classify and operate with rational numbers efficiently in real-life measurements.

Question 8:
Prove that √3 is an irrational number using the method of contradiction, as shown in NCERT.
Answer:
Introduction

We use contradiction to prove √3 is irrational, assuming it is rational first.


Argument 1
  • Let √3 = p/q (simplest form). Squaring gives 3 = p²/q² ⇒ p² = 3q², so p² is divisible by 3 (NCERT Thm 1.3).

Argument 2
  • If p² is divisible by 3, then p is also divisible by 3. Let p = 3k. Substituting gives 9k² = 3q² ⇒ q² = 3k², making q also divisible by 3.

Conclusion

This contradicts our assumption that p/q is in simplest form. Hence, √3 is irrational.

Question 9:
How are real numbers used to represent quantities in daily life? Illustrate with NCERT-based examples.
Answer:
Introduction

Real numbers include both rational and irrational numbers, helping us measure continuous quantities.


Argument 1
  • Lengths often involve irrational numbers. Example: Diagonal of a 1m square is √2m (NCERT Ex 1.2).

Argument 2
  • Money transactions use terminating decimals. Example: ₹12.75 for 3 pens implies ₹4.25 per pen (NCERT Ex 1.1).

Conclusion

From construction to finance, real numbers provide precise measurements, proving their universal utility.

Question 10:
Explain why √2 is an irrational number. Use the method of contradiction as shown in NCERT.
Answer:
Introduction

We studied that irrational numbers cannot be expressed as fractions. √2 is a classic example.


Argument 1

Assume √2 is rational, so it can be written as a/b where a and b are co-prime integers.


Argument 2

Squaring both sides gives 2 = a²/b², so a² = 2b². This means a² is even, hence a is even. Let a = 2k. Substituting, we get b² = 2k², so b is also even. But this contradicts our assumption that a and b are co-prime.


Conclusion

Thus, √2 cannot be rational and must be irrational.

Question 11:
Show how to represent 3.765 on the number line using successive magnification, referencing NCERT methods.
Answer:
Introduction

Our textbook shows how to represent decimals on a number line using magnification.


Argument 1

First, locate 3.7 and 3.8 on the line. Divide this segment into 10 equal parts to represent tenths.


Argument 2

Next, focus on 3.76 to 3.77. Divide this smaller segment into 10 parts for hundredths. Finally, locate 3.765 between 3.76 and 3.77 by estimating the position.


Conclusion

This step-by-step magnification helps us accurately plot 3.765 on the number line.

Question 12:
Prove that the sum of a rational number and an irrational number is always irrational, with an example from NCERT.
Answer:
Introduction

We learned that rational and irrational numbers have distinct properties. Let’s prove their sum is irrational.


Argument 1

Let r be rational and i be irrational. Assume r + i is rational, say a/b.


Argument 2

Then, i = (a/b) - r. Since rational numbers are closed under subtraction, i must be rational. But this contradicts the given that i is irrational.


Conclusion

Hence, r + i must be irrational. For example, 2 + √3 is irrational.

Question 13:
Explain the difference between rational and irrational numbers with examples. Also, justify why the sum of a rational and an irrational number is always irrational.
Answer:

Rational numbers are numbers that can be expressed as a fraction p/q, where p and q are integers and q ≠ 0. Examples include 1/2, 0.75 (which is 3/4), and 5 (which is 5/1).

Irrational numbers cannot be expressed as simple fractions and have non-terminating, non-repeating decimal expansions. Examples include √2, π, and e.

Sum of a rational and irrational number is always irrational:
Let’s assume the sum is rational. Then, for a rational number r and irrational number i, r + i = s (where s is rational).
Rearranging, i = s - r.
Since the difference of two rational numbers is rational, i must be rational, which contradicts the definition of irrational numbers. Hence, the sum must be irrational.

Question 14:
Represent √5 on the number line using a step-by-step geometric construction. Explain each step clearly.
Answer:

Steps to represent √5 on the number line:
1. Draw a horizontal number line and mark points O (0) and A (2).
2. At point A, draw a perpendicular line AB of length 1 unit.
3. Join points O and B to form a right-angled triangle OAB.
4. Using the Pythagorean theorem, OB = √(OA² + AB²) = √(2² + 1²) = √5.
5. With O as the center and OB as the radius, draw an arc intersecting the number line at point C.
6. The point C represents √5 on the number line.

Explanation:
The construction uses the property of right triangles to locate irrational lengths geometrically. The hypotenuse OB gives the value of √5, which is then transferred to the number line.

Question 15:
Prove that 3 + 2√5 is an irrational number, given that √5 is irrational.
Answer:

Proof by contradiction:
Assume 3 + 2√5 is rational. Then, it can be written as p/q, where p and q are integers with no common factors (other than 1) and q ≠ 0.
So, 3 + 2√5 = p/q.
Rearranging, 2√5 = (p/q) - 3.
Simplifying, √5 = (p - 3q)/(2q).
Since p and q are integers, (p - 3q)/(2q) is a rational number.
But this contradicts the given fact that √5 is irrational. Hence, our assumption is false, and 3 + 2√5 must be irrational.

Question 16:
Prove that 3 + 2√5 is an irrational number. Assume that √5 is irrational.
Answer:

Proof by contradiction:
Assume 3 + 2√5 is rational. Then, it can be written as p/q, where p and q are integers with no common factors (other than 1) and q ≠ 0.

Let 3 + 2√5 = p/q.
Rearranging: 2√5 = (p/q) - 3.
Simplify: √5 = (p - 3q)/(2q).

Since p and q are integers, (p - 3q)/(2q) is a rational number. But √5 is irrational, leading to a contradiction. Hence, 3 + 2√5 must be irrational.

Key takeaway:
The sum or product of a rational and irrational number is always irrational, as demonstrated here.

Question 17:
Prove that √5 is an irrational number using the method of contradiction.
Answer:

To prove that √5 is an irrational number, we will use the method of contradiction. We assume the opposite, i.e., √5 is a rational number, and then show that this leads to a contradiction.


Step 1: Assume √5 is rational. Then, it can be expressed as a fraction in its simplest form:


√5 = a/b, where a and b are coprime integers (i.e., their greatest common divisor is 1) and b ≠ 0.


Step 2: Square both sides of the equation:


(√5)² = (a/b)²
5 = a²/b²
5b² = a²


Step 3: This implies that a² is divisible by 5, so a must also be divisible by 5 (since 5 is a prime number). Let a = 5k, where k is an integer.


Step 4: Substitute a = 5k back into the equation:


5b² = (5k)²
5b² = 25k²
b² = 5k²


Step 5: This shows that b² is also divisible by 5, so b must be divisible by 5.


Step 6: However, this contradicts our initial assumption that a and b are coprime (since both are divisible by 5).


Conclusion: Since our assumption leads to a contradiction, √5 cannot be a rational number. Therefore, √5 is an irrational number.


Value-added note: This proof technique is widely used in mathematics to establish the irrationality of numbers like √2, √3, etc. It highlights the importance of prime factorization and the properties of coprime integers.

Question 18:
Explain how to represent √3 on the number line using a geometric construction. Include steps and a diagram description.
Answer:

To represent √3 on the number line, follow these steps geometrically:


Step 1: Draw a horizontal number line with points O (origin) and A at 1 unit distance.
Step 2: At point A, draw a perpendicular line AB of 1 unit length.
Step 3: Join O to B to form a right triangle OAB.
Step 4: By the Pythagoras theorem, OB = √(OA² + AB²) = √(1² + 1²) = √2.
Step 5: Now, from point B, draw another perpendicular BC of 1 unit length.
Step 6: Join O to C to form a new right triangle OBC.
Step 7: Now, OC = √(OB² + BC²) = √((√2)² + 1²) = √(2 + 1) = √3.
Step 8: Using a compass, measure the length OC and mark an arc on the number line from O. The intersection point represents √3.


Diagram Description: A number line with points O, A (1 unit), B (vertical from A), and C (vertical from B). The hypotenuse OC equals √3.

Question 19:
Prove that √3 is an irrational number using the method of contradiction.
Answer:

To prove that √3 is an irrational number, we will use the method of contradiction. We assume the opposite, i.e., √3 is a rational number, and then show that this leads to a contradiction.


Assumption: Let √3 be a rational number. This means it can be expressed in the form p/q, where p and q are coprime integers (i.e., their HCF is 1) and q ≠ 0.


Step 1: Write √3 = p/q.
Step 2: Squaring both sides, we get 3 = p²/q².
Step 3: Rearranging, 3q² = p².


This implies that p² is divisible by 3, and hence p must also be divisible by 3 (since 3 is a prime number). Let p = 3k, where k is an integer.


Step 4: Substitute p = 3k into the equation 3q² = p².
Step 5: This gives 3q² = (3k)², which simplifies to 3q² = 9k².
Step 6: Dividing both sides by 3, we get q² = 3k².


This shows that q² is also divisible by 3, and hence q must be divisible by 3. But this contradicts our initial assumption that p and q are coprime (since both are divisible by 3).


Conclusion: Our assumption that √3 is rational is false. Therefore, √3 is an irrational number.

Question 20:
Prove that √5 is an irrational number using the method of contradiction.
Answer:

To prove that √5 is an irrational number, we will use the method of contradiction. We assume the opposite, i.e., √5 is a rational number, and show that this leads to a contradiction.


Assumption: Let √5 be a rational number. Then, it can be expressed in the form p/q, where p and q are coprime integers (i.e., their HCF is 1) and q ≠ 0.


Step 1: Write √5 = p/q.
Squaring both sides, we get: 5 = p²/q².
This implies: p² = 5q².


Step 2: From p² = 5q², we deduce that p² is divisible by 5. Therefore, p must also be divisible by 5 (since 5 is a prime number).
Let p = 5k, where k is an integer.


Step 3: Substitute p = 5k into p² = 5q²:
(5k)² = 5q²
25k² = 5q²
Simplifying, we get: q² = 5k².


Step 4: From q² = 5k², we conclude that q² is divisible by 5, and hence q must also be divisible by 5.


Contradiction: Both p and q are divisible by 5, which contradicts our initial assumption that p and q are coprime. Therefore, our assumption that √5 is rational is false.


Conclusion: √5 is an irrational number.


Additional Insight: This proof method can be generalized to show that the square root of any prime number is irrational.

Question 21:
Explain the difference between rational and irrational numbers with examples. Also, prove that √5 is an irrational number.
Answer:

Rational numbers are numbers that can be expressed as a fraction p/q, where p and q are integers and q ≠ 0. Examples include 3/4, -2/5, and 0.6 (which is 3/5).

Irrational numbers cannot be expressed as simple fractions and have non-terminating, non-repeating decimal expansions. Examples include √2, π, and √5.

Proof that √5 is irrational:
Assume the opposite, that √5 is rational. Then, it can be written as √5 = a/b, where a and b are co-prime integers (no common factors other than 1).
Squaring both sides: 5 = a²/b² → 5b² = a².
This means a² is divisible by 5, so a must also be divisible by 5 (since 5 is prime).
Let a = 5k, then substituting: 5b² = (5k)² → 5b² = 25k² → b² = 5k².
This implies b² is divisible by 5, so b must also be divisible by 5.
But this contradicts our assumption that a and b are co-prime. Hence, √5 is irrational.

Question 22:
Represent √9.3 on the number line using successive magnification. Explain each step clearly.
Answer:

To represent √9.3 on the number line using successive magnification, follow these steps:

  • Step 1: Locate 9.3 between the integers 9 and 10 on the number line.
  • Step 2: Draw a perpendicular line at 9.3 of length 1 unit to form a right-angled triangle.
  • Step 3: The hypotenuse of this triangle will be √(9.3² + 1²) = √(86.49 + 1) = √87.49.
  • Step 4: Using a compass, mark an arc from 0 to the hypotenuse length (√87.49 ≈ 9.35).
  • Step 5: Now, refine the process by magnifying the region between 9.2 and 9.4 to locate √9.3 more precisely.

This method ensures accurate representation of irrational numbers on the number line through geometric construction.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A shopkeeper has 72 apples and 108 oranges. He wants to pack them into boxes such that each box contains the same number of fruits of each type with no leftovers. What is the maximum number of boxes he can use?
Answer:
Problem Interpretation

We need to find the maximum number of boxes where each box has equal apples and oranges without leftovers.


Mathematical Modeling

This is a problem of finding the HCF of 72 and 108.


Solution
  • Prime factors of 72: 23 × 32
  • Prime factors of 108: 22 × 33
  • HCF = 22 × 32 = 36

Thus, the shopkeeper can use a maximum of 36 boxes.

Question 2:
A student measures the length of a room as 5.6 meters and width as 4.2 meters. Express the area in the form of p/q where p and q are integers.
Answer:
Problem Interpretation

We need to find the area of the room and express it as a fraction.


Mathematical Modeling

Area = Length × Width = 5.6 × 4.2.


Solution
  • Convert decimals to fractions: 5.6 = 56/10, 4.2 = 42/10
  • Multiply: (56/10) × (42/10) = 2352/100
  • Simplify: 2352 ÷ 16 / 100 ÷ 4 = 588/25

The area is 588/25 m2.

Question 3:
A shopkeeper has 72 apples and 108 oranges. He wants to pack them into boxes such that each box contains the same number of fruits and no fruit is left unpacked. What is the maximum number of boxes he can use?
Answer:
Problem Interpretation

We need to find the maximum number of boxes such that apples and oranges are equally distributed without leftovers.

Mathematical Modeling

This is a problem of finding the HCF of 72 and 108.

Solution
  • Prime factors of 72: 23 × 32
  • Prime factors of 108: 22 × 33
  • HCF = 22 × 32 = 36

The shopkeeper can use a maximum of 36 boxes.

Question 4:
A student measures the length of a classroom as 8.25 meters using a scale. If the actual length is 8.20 meters, find the absolute error and relative error in the measurement.
Answer:
Problem Interpretation

We need to calculate the difference between measured and actual values (absolute error) and its ratio to the actual value (relative error).

Mathematical Modeling

Absolute Error = |Measured Value − Actual Value|

Relative Error = Absolute Error / Actual Value

Solution
  • Absolute Error = |8.25 − 8.20| = 0.05 meters
  • Relative Error = 0.05 / 8.20 ≈ 0.0061 (or 0.61%)

The errors help assess measurement accuracy.

Question 5:
A farmer has a rectangular field of length √50 meters and width √18 meters. He wants to divide it into equal square plots. What is the maximum side length of each square plot? Justify your answer.
Answer:
Problem Interpretation

We need to find the largest possible square plot size that can divide the field evenly.

Mathematical Modeling

The side length must be a common factor of √50 and √18. We simplify these:

  • √50 = 5√2
  • √18 = 3√2
Solution

The greatest common divisor is √2. Thus, the maximum side length is √2 meters.

Question 6:
A student writes 0.999... as a fraction. Our textbook shows recurring decimals can be converted. Prove whether 0.999... equals 1 using algebraic methods.
Answer:
Problem Interpretation

We need to verify if the recurring decimal 0.999... equals 1.

Mathematical Modeling

Let x = 0.999...
Then 10x = 9.999...

Solution

Subtracting the equations: 10x - x = 9.999... - 0.999... → 9x = 9 → x = 1. Thus, 0.999... = 1.

Question 7:
A shopkeeper sells rational numbers as prices of items. He marks a shirt for ₹(√16 × 3) and a book for ₹(5 + 0.333...).

(i) Convert these prices to simplest fractional form.
(ii) Which number system do these prices belong to? Justify.
Answer:
Problem Interpretation

We need to simplify the given prices and identify their number system.


Mathematical Modeling
  • Shirt price: √16 × 3 = 4 × 3 = ₹12 (integer)
  • Book price: 5 + 0.333... = 5 + ⅓ = 16/3 (fraction)

Solution

Both values are rational numbers as they can be expressed as p/q (q≠0). Our textbook shows all integers and fractions are rational.

Question 8:
A student measures a square's side as 2√5 cm.

(i) Find its area in exact form.
(ii) Convert the area to decimal representation (up to 2 places). What type of number results?
Answer:
Problem Interpretation

We must calculate area using irrational numbers and analyze the output.


Mathematical Modeling
  • Area = (2√5)² = 4×5 = 20 cm² (exact)
  • Decimal form: 20.00 (terminating)

Solution

The exact form shows whole number area. Decimal form confirms it's rational as it terminates. We studied that √5 is irrational but its square becomes rational.

Question 9:
A student measures the length of a room as 5.6 meters and width as 4.2 meters. Express the area in the form of a/b where a and b are coprime integers.
Answer:
Problem Interpretation

We need to find the area of the room and express it as a simplified fraction.


Mathematical Modeling

Area = Length × Width = 5.6 × 4.2.


Solution
  • Convert decimals to fractions: 5.6 = 28/5, 4.2 = 21/5.
  • Area = (28/5) × (21/5) = 588/25.
  • 588 and 25 are coprime.

The area is 588/25 m².

Question 10:
A farmer has a rectangular field of length √50 meters and width √18 meters. He wants to divide it into equal square plots. What is the maximum side length of each square plot?
Answer:
Problem Interpretation

We need to find the largest possible square plot that can divide the field without leftovers.


Mathematical Modeling

We studied that the side length must be the HCF of √50 and √18. Simplifying, √50 = 5√2 and √18 = 3√2.


Solution

The HCF of coefficients (5 and 3) is 1. Thus, the maximum side length is √2 meters.

Question 11:
A pizza is divided into 8 equal slices. If 3 slices are eaten, express the remaining portion as a decimal and fraction in simplest form.
Answer:
Problem Interpretation

We must represent the leftover pizza as a fraction and decimal.


Mathematical Modeling

Total slices = 8, eaten = 3. Remaining = 8 - 3 = 5 slices.


Solution

Fraction: 5/8 (simplest form). Decimal: 5 ÷ 8 = 0.625. Our textbook shows similar conversions.

Question 12:

Rahul was solving a problem involving rational numbers. He needed to find three rational numbers between 1/4 and 1/2. Help him by providing the correct numbers and explaining the method step-by-step.

Answer:

To find three rational numbers between 1/4 and 1/2, we can use the mean method or convert them to equivalent fractions with a common denominator.


Method 1: Mean Method
1. Find the mean of 1/4 and 1/2: (1/4 + 1/2)/2 = (3/4)/2 = 3/8.
2. Now, find the mean of 1/4 and 3/8: (1/4 + 3/8)/2 = (5/8)/2 = 5/16.
3. Next, find the mean of 3/8 and 1/2: (3/8 + 1/2)/2 = (7/8)/2 = 7/16.
Three rational numbers: 5/16, 3/8, 7/16.

Method 2: Common Denominator
1. Convert 1/4 and 1/2 to fractions with denominator 8: 2/8 and 4/8.
2. Now, three rational numbers between them are: 3/8, 5/16, and 7/16 (as above).

Both methods are valid, and the numbers lie strictly between 1/4 and 1/2.

Question 13:

Priya was learning about irrational numbers. Her teacher asked her to prove that √5 is irrational. Help her by writing a step-by-step proof using the contradiction method.

Answer:

To prove that √5 is irrational, we assume the opposite (that it is rational) and arrive at a contradiction.


Step 1: Assume √5 is rational.
Let √5 = a/b, where a and b are coprime integers (no common factors other than 1) and b ≠ 0.

Step 2: Square both sides.
5 = a²/b² ⇒ a² = 5b².
This means a² is divisible by 5, so a must also be divisible by 5 (since 5 is prime).

Step 3: Let a = 5k for some integer k.
Substitute into a² = 5b²: (5k)² = 5b² ⇒ 25k² = 5b² ⇒ b² = 5k².
This implies b² is divisible by 5, so b must also be divisible by 5.

Step 4: Contradiction.
Both a and b are divisible by 5, which contradicts our assumption that they are coprime.

Hence, √5 cannot be expressed as a fraction a/b, proving it is irrational.

Question 14:

Rahul was given a problem to find two irrational numbers between 0.5 and 0.6. He proposed √0.26 and √0.35 as his answers.

(i) Are his answers correct? Justify.

(ii) Find two more irrational numbers in the same range using a different method.

Answer:

(i) Verification of Rahul's answers:

√0.26 ≈ 0.5099 (which is > 0.5)


√0.35 ≈ 0.5916 (which is < 0.6)


Since both numbers are non-terminating, non-repeating decimals, they are irrational and lie between 0.5 and 0.6. Thus, Rahul's answers are correct.


(ii) Alternate method:

We can use the property that the sum of a rational and an irrational number is irrational.


Example 1: 0.5 + (√2/10) ≈ 0.5 + 0.1414 = 0.6414 (Adjust denominator to fit range)


Example 2: 0.5 + (π/20) ≈ 0.5 + 0.157 = 0.657 (Use π for guaranteed irrationality)


Note: The exact values need adjustment to ensure they fall strictly between 0.5 and 0.6.

Question 15:

In a school math fair, students were asked to represent real numbers on a number line. Priya chose to represent 2√3, while Rohan selected (1 + √5)/2.

(i) Explain how to geometrically construct these numbers on a number line using ruler and compass.

(ii) Between 2√3 and (1 + √5)/2, which is greater? Show your work.

Answer:

(i) Geometric construction steps:

For 2√3:


1. Draw base line of 2 units (0 to 2).


2. At point 1, construct perpendicular of √3 units using right triangle method (1² + (√3)² = 2²).


3. Use compass to transfer this length to main number line.


For (1 + √5)/2:


1. Construct √5 as hypotenuse of 1×2 rectangle.


2. Add 1 unit to √5, then bisect the total length geometrically.


(ii) Comparison:


Calculate approximate decimal values:


2√3 ≈ 3.464


(1 + √5)/2 ≈ (1 + 2.236)/2 ≈ 1.618


Thus, 2√3 > (1 + √5)/2


Note: (1 + √5)/2 is the famous golden ratio φ, while 2√3 appears in regular hexagon geometry.

Question 16:

Rahul was solving a problem involving irrational numbers. He came across the number √5 and wondered how to represent it on the number line. Help him by explaining the step-by-step process to locate √5 on the number line using a geometrical method.

Answer:

To locate √5 on the number line, we can use the Pythagoras theorem and follow these steps:


  • Draw a horizontal number line and mark points O (0) and A (2) on it.
  • At point A, draw a perpendicular line AB of length 1 unit.
  • Join points O and B to form a right-angled triangle OAB.
  • Using the Pythagoras theorem: OB² = OA² + AB² = 2² + 1² = 5.
  • Thus, OB = √5 units.
  • With O as the center and OB as the radius, draw an arc intersecting the number line at point C.
  • Point C represents √5 on the number line.

This method ensures accurate representation of irrational numbers geometrically.

Question 17:

Priya was given two numbers: 3.142857... and 3.1416. She needs to classify them as rational or irrational and justify her answer. Help her by explaining the nature of these numbers with proper reasoning.

Answer:

Let's analyze both numbers one by one:


1. 3.142857...

  • This number is a repeating decimal expansion, as the sequence "142857" repeats indefinitely.
  • Any number with a repeating or terminating decimal expansion is a rational number.
  • Thus, 3.142857... is a rational number.

2. 3.1416

  • This number is a terminating decimal, as it ends after four decimal places.
  • Terminating decimals are always rational numbers because they can be expressed as a fraction (e.g., 31416/10000).
  • Thus, 3.1416 is also a rational number.

Note: If the number were π (3.14159265...), it would be irrational due to its non-terminating and non-repeating nature.

Question 18:

Rahul was solving a problem involving rational numbers. He had to find three rational numbers between 1/4 and 1/2. Help him by providing the correct numbers and explaining the method step-by-step.

Answer:

To find three rational numbers between 1/4 and 1/2, we can use the method of equidistant rational numbers.


Step 1: Convert the fractions to have a common denominator.
1/4 = 2/8
1/2 = 4/8

Step 2: Since we need three numbers, we divide the interval into four equal parts.
The numbers will be:
3/8 (since 2/8 + 1/8 = 3/8)
4/8 (which is 1/2, but we exclude it as it's the upper bound)
Instead, we can find numbers between 2/8 and 4/8 by averaging:
(2/8 + 3/8)/2 = 5/16
(3/8 + 4/8)/2 = 7/16

Final Answer: Three rational numbers between 1/4 and 1/2 are 3/8, 5/16, and 7/16.
Question 19:

Priya was learning about irrational numbers. Her teacher asked her to prove that √5 is irrational. Help her by providing a step-by-step proof using the method of contradiction.

Answer:

To prove that √5 is irrational, we use the method of contradiction.


Step 1: Assume √5 is rational, so it can be written as a fraction a/b in simplest form, where a and b are co-prime integers.
√5 = a/b

Step 2: Square both sides to eliminate the square root.
5 = a²/b²
=> a² = 5b²

Step 3: This implies a² is divisible by 5, so a must also be divisible by 5 (since 5 is prime).
Let a = 5k for some integer k.

Step 4: Substitute a = 5k into the equation.
(5k)² = 5b²
25k² = 5b²
=> b² = 5k²

Step 5: This implies b² is divisible by 5, so b must also be divisible by 5.
But this contradicts our assumption that a and b are co-prime (no common factors other than 1).

Conclusion: Our initial assumption is false, so √5 must be irrational.
Question 20:

Rahul was solving a problem involving rational numbers. He came across two numbers, 5/7 and 12/13. He needs to find a rational number between them. Help him by providing a step-by-step solution.

Answer:

To find a rational number between 5/7 and 12/13, we can use the average method.


Step 1: Convert both fractions to have a common denominator.
LCM of 7 and 13 is 91.
5/7 = (5 × 13)/91 = 65/91
12/13 = (12 × 7)/91 = 84/91
Step 2: Find the average of the two fractions.
(65/91 + 84/91) ÷ 2 = (149/91) ÷ 2 = 149/182

The rational number 149/182 lies between 5/7 and 12/13.

Note: There are infinitely many rational numbers between any two rational numbers.

Question 21:

Priya was exploring irrational numbers and wondered if the sum of √2 and √3 is rational or irrational. Explain with reasoning.

Answer:

The sum of √2 and √3 is an irrational number. Here's why:


Step 1: Assume √2 + √3 is rational.
Let √2 + √3 = p/q, where p and q are integers with no common factors.
Step 2: Square both sides.
(√2 + √3)² = (p/q)²
2 + 2√6 + 3 = p²/q²
5 + 2√6 = p²/q²
Step 3: Rearrange to isolate the irrational part.
2√6 = (p²/q²) - 5
√6 = (p² - 5q²)/(2q²)

Since p and q are integers, the right side is rational, but √6 is irrational. This is a contradiction.

Thus, our assumption is false, and √2 + √3 must be irrational.

Chat on WhatsApp