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By Rohit RP
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Updated on 12 Aug 2026, 12:44 IST
The NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers provide clear and step-by-step solutions to the questions given in the current NCERT textbook. Chapter 1 introduces students to important concepts such as the Fundamental Theorem of Arithmetic, prime factorisation, HCF and LCM, rational and irrational numbers, and proof of irrationality.
The current Class 10 Maths Chapter 1 Real Numbers contains Exercise 1.1 and Exercise 1.2. Exercise 1.1 mainly focuses on prime factorisation, HCF, LCM, composite numbers, and applications of the Fundamental Theorem of Arithmetic. Exercise 1.2 focuses on irrational numbers and teaches students how to prove that particular numbers and expressions are irrational.
These Real Numbers Class 10 NCERT Solutions explain every question using simple and systematic methods. Students can use the solutions to understand concepts, check their answers, complete homework, revise the chapter, and prepare effectively for examinations.
Students can download the NCERT Solutions Class 10 Maths Chapter 1 Real Numbers PDF for offline practice and quick revision. The PDF provides exercise-wise solutions along with the important concepts and mathematical results required to understand the chapter.
Students can use the NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF for homework, self-study, revision, and exam preparation.
The current NCERT Class 10 Maths Chapter 1 Real Numbers contains two exercises.
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| Exercise | Number of Main Questions | Main Topics |
| Class 10 Maths Chapter 1 Exercise 1.1 | 7 Questions | Prime factorisation, HCF, LCM, Fundamental Theorem of Arithmetic, composite numbers |
| Class 10 Maths Chapter 1 Exercise 1.2 | 3 Questions | Irrational numbers and proof of irrationality |
The exercise-wise solutions below provide the question, complete working, and final answer in an easy-to-follow format.
Real numbers include all numbers that can be represented on a number line. They consist of both rational numbers and irrational numbers.
In Class 10 Maths Chapter 1 Real Numbers, students learn how composite numbers can be expressed as products of prime numbers using the Fundamental Theorem of Arithmetic. Prime factorisation is then used to find the HCF and LCM of numbers and solve different numerical problems.
Students also learn an important relationship between HCF and LCM for two positive integers:
HCF(a, b) × LCM(a, b) = a × b
Another major part of the chapter is understanding irrational numbers. Students learn how mathematical reasoning can be used to prove that numbers such as:
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√2
√3
√5
are irrational.
The chapter also develops students' understanding of proof by contradiction, which is used to establish the irrationality of different numbers and expressions.
A good understanding of Real Numbers strengthens number sense, logical reasoning, factorisation skills, and mathematical proof-writing.
The important topics covered in Real Numbers Class 10 include:
Understanding these concepts helps students solve questions from Class 10 Maths Chapter 1 Exercise 1.1 and Class 10 Maths Chapter 1 Exercise 1.2 more confidently.
All rational numbers and irrational numbers together are called real numbers.
In simple terms, every number that can be represented on a number line is a real number.
Examples of real numbers include:
2
-5
3/4
0.25
√2
√5
Real numbers can broadly be classified into two groups:
A number is called a rational number if it can be written in the form:
p/q
where p and q are integers and:
q ≠ 0
Examples of rational numbers include:
1/2
-3/4
5
0
0.75
A whole number such as 5 is also rational because it can be written as:
5 = 5/1
Similarly:
0 = 0/1
Therefore, integers and whole numbers are also rational numbers.
A number that cannot be expressed in the form:
p/q
where p and q are integers and:
q ≠ 0
is called an irrational number.
Examples include:
√2
√3
√5
These numbers cannot be expressed as ratios of two integers.
The concept of irrational numbers is especially important in Class 10 Maths Chapter 1 Exercise 1.2.
A prime number is a natural number greater than 1 that has exactly two positive factors:
Examples:
2, 3, 5, 7, 11, 13, 17
For example, the factors of 7 are:
1 and 7
Therefore, 7 is a prime number.
A composite number is a natural number greater than 1 that has more than two factors.
Examples:
4, 6, 8, 9, 10, 12
For example:
12 = 2 × 2 × 3
The factors of 12 are:
1, 2, 3, 4, 6, 12
Since 12 has more than two factors, it is a composite number.
Two positive integers are called coprime numbers if their HCF is 1.
For example, consider 8 and 15.
Factors of 8:
1, 2, 4, 8
Factors of 15:
1, 3, 5, 15
Their only common factor is 1.
Therefore:
HCF(8, 15) = 1
Hence, 8 and 15 are coprime numbers.
Two coprime numbers do not need to be prime numbers themselves. For example, both 8 and 15 are composite numbers, but they are coprime because their HCF is 1.
The idea of coprime integers is particularly important while proving that a number is irrational.
The Fundamental Theorem of Arithmetic Class 10 is one of the most important concepts in Chapter 1 Real Numbers.
It states:
Every composite number can be expressed as a product of prime numbers, and this prime factorisation is unique except for the order of the prime factors.
For example:
140 = 2 × 2 × 5 × 7
Therefore:
140 = 2² × 5 × 7
Even if we begin factorising 140 in a different way, its final prime factors remain the same.
Another example:
60 = 2 × 2 × 3 × 5
Therefore:
60 = 2² × 3 × 5
The Fundamental Theorem of Arithmetic is useful for:
Prime factorisation means expressing a composite number as a product of prime numbers.
For example:
60 = 2 × 2 × 3 × 5
Therefore:
60 = 2² × 3 × 5
Another example:
72 = 2 × 2 × 2 × 3 × 3
Therefore:
72 = 2³ × 3²
A general prime factorisation may be written as:
N = p₁ᵃ × p₂ᵇ × p₃ᶜ ...
where p₁, p₂, p₃, and so on are prime numbers.
Prime factorisation is used extensively in Class 10 Maths Chapter 1 Exercise 1.1 to find HCF and LCM and solve different number-based problems.
The Highest Common Factor, or HCF, is the greatest number that divides each of the given numbers exactly.
The Least Common Multiple, or LCM, is the smallest positive number that is exactly divisible by each of the given numbers.
To find the HCF:
Take the smallest powers of the common prime factors.
To find the LCM:
Take the greatest powers of all the prime factors involved.
Prime factorisation:
12 = 2² × 3
18 = 2 × 3²
For HCF, take the smallest powers of the common prime factors:
HCF = 2 × 3
HCF = 6
For LCM, take the greatest powers:
LCM = 2² × 3²
LCM = 4 × 9
LCM = 36
Therefore:
HCF = 6
LCM = 36
For two positive integers a and b, the relationship between HCF and LCM is:
HCF(a, b) × LCM(a, b) = a × b
This formula can also be used to find LCM when HCF is known:
LCM(a, b) = (a × b) / HCF(a, b)
Similarly, HCF can be found when LCM is known:
HCF(a, b) = (a × b) / LCM(a, b)
For 12 and 18:
HCF = 6
LCM = 36
Now:
HCF × LCM = 6 × 36
HCF × LCM = 216
Also:
12 × 18 = 216
Therefore:
HCF × LCM = Product of the two numbers
The simple relationship:
HCF(a, b) × LCM(a, b) = a × b
is directly applicable to two positive integers.
For three numbers, the following statement is generally not true:
HCF(a, b, c) × LCM(a, b, c) = a × b × c
Students should therefore avoid applying the two-number formula directly to three numbers.
Express each of the following numbers as a product of its prime factors.
Start dividing 140 by prime numbers:
140 = 2 × 70
70 = 2 × 35
35 = 5 × 7
Therefore:
140 = 2 × 2 × 5 × 7
Using powers:
140 = 2² × 5 × 7
140 = 2² × 5 × 7
156 = 2 × 78
78 = 2 × 39
39 = 3 × 13
Therefore:
156 = 2 × 2 × 3 × 13
Using powers:
156 = 2² × 3 × 13
156 = 2² × 3 × 13
3825 = 3 × 1275
1275 = 3 × 425
425 = 5 × 85
85 = 5 × 17
Therefore:
3825 = 3 × 3 × 5 × 5 × 17
Using powers:
3825 = 3² × 5² × 17
3825 = 3² × 5² × 17
5005 = 5 × 1001
1001 = 7 × 143
143 = 11 × 13
Therefore:
5005 = 5 × 7 × 11 × 13
5005 = 5 × 7 × 11 × 13
7429 = 17 × 437
Now:
437 = 19 × 23
Therefore:
7429 = 17 × 19 × 23
7429 = 17 × 19 × 23
Find the LCM and HCF of the following pairs of integers and verify that:
HCF × LCM = Product of the two numbers
Prime factorisation:
26 = 2 × 13
91 = 7 × 13
The common prime factor is 13.
Therefore:
HCF = 13
For LCM, take all prime factors with their greatest powers:
LCM = 2 × 7 × 13
LCM = 182
Verification:
HCF × LCM = 13 × 182
HCF × LCM = 2366
Product of the two numbers:
26 × 91 = 2366
Therefore:
HCF × LCM = 26 × 91
HCF = 13
LCM = 182
The relationship is verified.
Prime factorisation:
510 = 2 × 3 × 5 × 17
92 = 2² × 23
The common prime factor is 2.
Therefore:
HCF = 2
For LCM:
LCM = 2² × 3 × 5 × 17 × 23
LCM = 4 × 3 × 5 × 17 × 23
LCM = 23460
Verification:
HCF × LCM = 2 × 23460
HCF × LCM = 46920
Product of the numbers:
510 × 92 = 46920
Therefore:
HCF × LCM = 510 × 92
HCF = 2
LCM = 23460
The relationship is verified.
Prime factorisation:
336 = 2⁴ × 3 × 7
54 = 2 × 3³
For HCF, take the smallest powers of the common prime factors:
HCF = 2 × 3
HCF = 6
For LCM, take the greatest powers:
LCM = 2⁴ × 3³ × 7
LCM = 16 × 27 × 7
LCM = 3024
Verification:
HCF × LCM = 6 × 3024
HCF × LCM = 18144
Product of the numbers:
336 × 54 = 18144
Therefore:
HCF × LCM = 336 × 54
HCF = 6
LCM = 3024
The relationship is verified.
Find the HCF and LCM of the following integers by applying the prime factorisation method.
Prime factorisation:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
The common prime factor in all three numbers is 3.
Therefore:
HCF = 3
For LCM, take the greatest powers of all prime factors:
LCM = 2² × 3 × 5 × 7
LCM = 4 × 3 × 5 × 7
LCM = 420
HCF = 3
LCM = 420
17, 23 and 29 are different prime numbers.
They have no common prime factor other than 1.
Therefore:
HCF = 1
For LCM:
LCM = 17 × 23 × 29
First:
17 × 23 = 391
Then:
391 × 29 = 11339
Therefore:
LCM = 11339
HCF = 1
LCM = 11339
Prime factorisation:
8 = 2³
9 = 3²
25 = 5²
There is no common prime factor among all three numbers.
Therefore:
HCF = 1
For LCM:
LCM = 2³ × 3² × 5²
LCM = 8 × 9 × 25
LCM = 1800
HCF = 1
LCM = 1800
Given:
HCF(306, 657) = 9
Find:
LCM(306, 657)
For two positive integers:
HCF × LCM = Product of the two numbers
Therefore:
9 × LCM = 306 × 657
So:
LCM = (306 × 657) / 9
Since:
657 / 9 = 73
Therefore:
LCM = 306 × 73
LCM = 22338
LCM(306, 657) = 22338
Check whether 6ⁿ can end with the digit 0 for any natural number n.
A number ending in 0 must be divisible by 10.
Prime factorisation of 10 is:
10 = 2 × 5
Therefore, a number that ends in 0 must contain both 2 and 5 as prime factors.
Now:
6 = 2 × 3
Therefore:
6ⁿ = 2ⁿ × 3ⁿ
The prime factorisation of 6ⁿ contains only the prime factors 2 and 3.
It does not contain 5 as a prime factor.
Therefore, 6ⁿ cannot be divisible by 10.
Hence, it cannot end with the digit 0.
6ⁿ cannot end with the digit 0 for any natural number n.
Explain why the following numbers are composite.
Given:
7 × 11 × 13 + 13
Take 13 as a common factor:
7 × 11 × 13 + 13 = 13(7 × 11 + 1)
= 13(77 + 1)
= 13 × 78
The given number can be expressed as a product of two integers greater than 1.
Therefore, it is a composite number.
7 × 11 × 13 + 13 is composite because it can be written as 13 × 78.
Given:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Take 5 as a common factor:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
Now:
7 × 6 × 4 × 3 × 2 × 1 = 1008
Therefore:
= 5(1008 + 1)
= 5 × 1009
The number can be expressed as a product of two integers greater than 1.
Therefore, it is composite.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite because it can be written as 5 × 1009.
Sonia takes 18 minutes to complete one round of a circular path, while Ravi takes 12 minutes. They start from the same point at the same time and move in the same direction. After how many minutes will they meet again at the starting point?
To find when both will return to the starting point together, calculate the LCM of 18 and 12.
Prime factorisation:
18 = 2 × 3²
12 = 2² × 3
For LCM, take the greatest powers:
LCM = 2² × 3²
LCM = 4 × 9
LCM = 36
Therefore, both will return to the starting point together after 36 minutes.
Sonia and Ravi will meet again at the starting point after 36 minutes.
An irrational number is a real number that cannot be expressed in the form:
p/q
where p and q are integers and:
q ≠ 0
Examples include:
√2
√3
√5
In Class 10 Maths Chapter 1 Exercise 1.2, students learn how to prove that particular numbers and expressions are irrational.
The main method used for these questions is proof by contradiction.
Proof by contradiction begins by assuming that the opposite of what we want to prove is true.
To prove that a number is irrational, students can follow these general steps:
This method is particularly useful for proving the irrationality of square roots.
Prove that √5 is irrational.
Assume that √5 is rational.
Then it can be written as:
√5 = a/b
where a and b are coprime integers and:
b ≠ 0
Squaring both sides:
5 = a²/b²
Therefore:
a² = 5b²
This shows that a² is divisible by 5.
Therefore, a is also divisible by 5.
Let:
a = 5k
for some integer k.
Substitute this value of a:
a² = 5b²
becomes:
(5k)² = 5b²
Therefore:
25k² = 5b²
Dividing both sides by 5:
5k² = b²
Therefore:
b² = 5k²
This shows that b² is divisible by 5.
Therefore, b is also divisible by 5.
So both a and b are divisible by 5.
This means that a and b have a common factor of 5.
But we assumed that a and b are coprime.
This is a contradiction.
Therefore, our original assumption that √5 is rational is false.
√5 is irrational.
Prove that:
3 + 2√5
is irrational.
Assume that:
3 + 2√5
is rational.
Let:
3 + 2√5 = r
where r is a rational number.
Subtract 3 from both sides:
2√5 = r - 3
Divide both sides by 2:
√5 = (r - 3)/2
Since r is rational, r - 3 is also rational.
Dividing a rational number by 2 also gives a rational number.
Therefore:
(r - 3)/2
is rational.
This would mean that √5 is rational.
But √5 is irrational.
This is a contradiction.
Therefore, the assumption that 3 + 2√5 is rational is false.
3 + 2√5 is irrational.
Prove that:
1/√2
is irrational.
Assume that:
1/√2
is rational.
Let:
1/√2 = r
where r is a non-zero rational number.
Then:
√2 = 1/r
Since r is rational and non-zero, 1/r is also rational.
This would mean that √2 is rational.
But √2 is irrational.
This is a contradiction.
Therefore, the original assumption is false.
1/√2 is irrational.
Prove that:
7√5
is irrational.
Assume that:
7√5
is rational.
Let:
7√5 = r
where r is a rational number.
Divide both sides by 7:
√5 = r/7
Since r is rational, r/7 is also rational.
This would mean that √5 is rational.
But √5 is irrational.
This is a contradiction.
Therefore, the original assumption is false.
7√5 is irrational.
Prove that:
6 + √2
is irrational.
Assume that:
6 + √2
is rational.
Let:
6 + √2 = r
where r is a rational number.
Subtract 6 from both sides:
√2 = r - 6
Since both r and 6 are rational numbers:
r - 6
is also rational.
This would mean that √2 is rational.
But √2 is irrational.
This is a contradiction.
Therefore, the original assumption is false.
6 + √2 is irrational.
Students should remember the following important mathematical results while solving questions from NCERT Solutions Class 10 Maths Chapter 1 Real Numbers.
A rational number can be expressed as:
p/q
where:
q ≠ 0
A composite number may be expressed as:
N = p₁ᵃ × p₂ᵇ × p₃ᶜ ...
where p₁, p₂, p₃, and so on are prime numbers.
To find the HCF using prime factorisation:
Take the smallest powers of the common prime factors.
To find the LCM using prime factorisation:
Take the greatest powers of all the prime factors involved.
For two positive integers a and b:
HCF(a, b) × LCM(a, b) = a × b
LCM(a, b) = (a × b) / HCF(a, b)
HCF(a, b) = (a × b) / LCM(a, b)
The following relationship is generally not true for three numbers:
HCF(a, b, c) × LCM(a, b, c) = a × b × c
The simple HCF-LCM product relationship should be applied directly to two positive integers.
Students often make small mistakes while solving Real Numbers questions. Avoiding these errors can improve both accuracy and presentation.
For HCF, always take the smallest powers of the common prime factors.
For example:
12 = 2² × 3
18 = 2 × 3²
Therefore:
HCF = 2 × 3 = 6
Do not take:
2² × 3²
because those are the greatest powers and are used for LCM.
For LCM, take the greatest powers of all prime factors involved.
For:
12 = 2² × 3
and:
18 = 2 × 3²
we get:
LCM = 2² × 3²
LCM = 36
Prime factorisation must contain only prime numbers.
For example:
140 = 4 × 35
is not the complete prime factorisation because both 4 and 35 are composite.
The complete prime factorisation is:
140 = 2² × 5 × 7
The relationship:
HCF(a, b) × LCM(a, b) = a × b
is directly applicable to two positive integers.
Students should not automatically write:
HCF(a, b, c) × LCM(a, b, c) = a × b × c
because this is generally not true.
While proving irrationality, students may begin with:
√5 = a/b
but forget to mention that a and b are coprime.
The correct assumption is:
√5 = a/b
where a and b are coprime integers and:
b ≠ 0
This condition is important because the contradiction depends on proving that a and b actually have a common factor.
At the end of an irrationality proof, clearly explain why the result contradicts the original assumption.
For example:
Both a and b are divisible by 5, which contradicts the assumption that a and b are coprime.
Therefore, √5 is irrational.
When finding LCM, include every prime factor that appears in any of the given numbers.
For example:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
Therefore:
LCM = 2² × 3 × 5 × 7
LCM = 420
Leaving out 5 or 7 would give an incorrect answer.
Questions involving prime factorisation, HCF, LCM, and irrationality proofs should include proper working.
Instead of writing only:
HCF = 6
show how the HCF was obtained from the prime factorisation.
This makes the solution easier to understand and verify.
Students can read each chapter online or use the chapter-wise PDF links below for offline revision.
| Chapter | Download PDF |
| 1. Real Numbers | Real Numbers Class 10 notes PDF |
| 2. Polynomials | Polynomials Class 10 notes PDF |
| 3. Pair of Linear Equations in Two Variables | Pair of Linear Equations notes PDF |
| 4. Quadratic Equations | Quadratic Equations Class 10 notes PDF |
| 5. Arithmetic Progressions | Arithmetic Progressions notes PDF |
| 6. Triangles | Triangles Class 10 notes PDF |
| 7. Coordinate Geometry | Coordinate Geometry notes PDF |
| 8. Introduction to Trigonometry | Introduction to Trigonometry notes PDF |
| 9. Applications of Trigonometry | Applications of Trigonometry notes PDF |
| 10. Circles | Circles Class 10 notes PDF |
| 11. Areas Related to Circles | Areas Related to Circles notes PDF |
| 12. Surface Areas and Volumes | Surface Areas and Volumes notes PDF |
| 13. Statistics | Statistics Class 10 notes PDF |
| 14. Probability | Probability Class 10 notes PDF |
The NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers can help students understand the chapter systematically and practise textbook questions effectively.
Each question is solved using a clear sequence of steps so students can understand how the final answer is obtained.
The solutions show how composite numbers can be expressed as prime factors and how these factors are used in HCF and LCM questions.
Students learn how to select the correct powers of prime factors and apply the relationship:
HCF × LCM = Product of the two numbers
where appropriate.
Questions based on irrationality encourage students to think logically and understand how a contradiction can be used to prove a mathematical statement.
The Class 10 Maths Chapter 1 Exercise 1.2 solutions show students how to organise an irrationality proof in a clear and logical sequence.
Students can compare their own solutions with the step-by-step methods to identify mistakes independently.
The solutions make it easier to revise Exercise 1.1 and Exercise 1.2 before school tests and examinations.
Regular practice of NCERT textbook questions helps students strengthen the concepts covered in the chapter and become more confident while solving similar problems.
Students can use chapter-wise NCERT Solutions to understand textbook questions, revise important concepts, and prepare systematically for Class 10 Maths examinations.
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The NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers provide step-by-step solutions to the questions given in the NCERT textbook. They cover concepts such as the Fundamental Theorem of Arithmetic, prime factorisation, HCF and LCM, irrational numbers, and proof of irrationality.
The main topics include:
Fundamental Theorem of Arithmetic
Prime factorisation
HCF and LCM
Relationship between HCF and LCM
Prime and composite numbers
Coprime numbers
Rational numbers
Irrational numbers
Proof by contradiction
Proof of irrationality
The current chapter contains:
Exercise 1.1
Exercise 1.2
Class 10 Maths Chapter 1 Exercise 1.1 contains 7 main questions.
These questions cover prime factorisation, HCF, LCM, the relationship between HCF and LCM, composite numbers, and applications of LCM.
The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers and that this prime factorisation is unique except for the order of the prime factors.
For example:
60 = 2² × 3 × 5
Prime factorisation is the process of expressing a composite number as a product of prime numbers.
For example:
72 = 2 × 2 × 2 × 3 × 3
Therefore:
72 = 2³ × 3²
First write each number as a product of prime factors.
Then take the smallest powers of the common prime factors.
For example:
12 = 2² × 3
18 = 2 × 3²
Therefore:
HCF = 2 × 3
HCF = 6
Write all numbers in prime factor form and take the greatest powers of all prime factors involved.
For example:
12 = 2² × 3
18 = 2 × 3²
Therefore:
LCM = 2² × 3²
LCM = 36
For two positive integers a and b:
HCF(a, b) × LCM(a, b) = a × b
This relationship can be used to calculate LCM if HCF is known or HCF if LCM is known.
Use the formula:
LCM(a, b) = (a × b) / HCF(a, b)
For example, if:
HCF(306, 657) = 9
then:
LCM = (306 × 657) / 9
LCM = 22338
The formula:
HCF(a, b) × LCM(a, b) = a × b
is directly applicable to two positive integers.
The following statement is generally not true for three numbers:
HCF(a, b, c) × LCM(a, b, c) = a × b × c
A rational number is a number that can be expressed in the form:
p/q
where p and q are integers and:
q ≠ 0
Examples include:
1/2
-3/4
5
0.75
An irrational number is a real number that cannot be expressed in the form:
p/q
where p and q are integers and:
q ≠ 0
Examples include:
√2
√3
√5
Assume that √5 is rational and write:
√5 = a/b
where a and b are coprime integers and:
b ≠ 0
After squaring and simplifying, it can be shown that both a and b are divisible by 5.
This contradicts the assumption that a and b are coprime.
Therefore:
√5 is irrational.
Proof by contradiction is a mathematical method in which we first assume that the opposite of the statement we want to prove is true.
If that assumption leads to a contradiction, the assumption must be false.
Therefore, the original statement is true.
This method is commonly used in proof of irrationality Class 10 questions.
Yes.
If:
3 + 2√5
were rational, then rearranging the expression would give:
√5 = (r - 3)/2
for some rational number r.
This would mean that √5 is rational, which contradicts the fact that √5 is irrational.
Therefore:
3 + 2√5 is irrational.
Yes.
If:
1/√2
were rational, then its reciprocal √2 would also be rational.
But √2 is irrational.
Therefore:
1/√2 is irrational.
A number ending in 0 must contain both 2 and 5 as prime factors because:
10 = 2 × 5
But:
6 = 2 × 3
Therefore:
6ⁿ = 2ⁿ × 3ⁿ
The prime factorisation does not contain 5.
Hence:
6ⁿ cannot end with the digit 0 for any natural number n.
When proving a square root irrational, we usually assume:
√n = a/b
where a and b are coprime integers.
If the proof later shows that both a and b have a common factor, it creates a contradiction.
That contradiction proves that the original assumption was false and that the number is irrational.
Students can use the NCERT Solutions Class 10 Maths Chapter 1 Real Numbers PDF available on Infinity Learn website for offline practice and revision. It includes solutions for Exercise 1.1 and Exercise 1.2 along with important concepts, formulas, mathematical results, and revision points from the chapter.