Class 10 Maths Ch 1 Real Numbers Ex 1.2 is based on the Euclidβs Division Lemma and Algorithm, which are important tools to find the HCF (Highest Common Factor) of two numbers. In this exercise, students learn how to apply the division algorithm step-by-step to solve problems accurately. In this section, you will find clear, step-by-step detailed answers to all questions of Exercise 1.2, explained in a simple and easy-to-understand manner.
In this exercise, students learn how to:
- Apply Euclidβs Division Algorithm to find the HCF of two numbers.
- Use prime factorization to find the HCF (Highest Common Factor) and LCM (Least Common Multiple) of two or more numbers.
- Verify the relation:
HCF Γ LCM = Product of the two numbers
The questions in this exercise help build a strong foundation in understanding how numbers are broken down into their basic prime factors and how this can be used to solve real-life problems involving divisibility and multiples.
Class 10 Maths Ch 1 Real Numbers Ex 1.2-Textbook Solutions
1. Prove that β5 is irrational
Solution:
Proof (by contradiction):
Assume β5 is rational.
So, β5 = p/q, where p and q are integers and in lowest form (HCF = 1).
Squaring both sides:
5 = pΒ²/qΒ²
β pΒ² = 5qΒ²
So, pΒ² is divisible by 5 β p is divisible by 5
Let p = 5k
Substitute:
(5k)Β² = 5qΒ²
β 25kΒ² = 5qΒ²
β qΒ² = 5kΒ²
So, q is also divisible by 5
This means p and q have a common factor 5
But we assumed they are coprime.
Contradiction β β5 is irrational.
2. Prove that 3 + β5 is irrational
Solution:
Proof (by contradiction):
Assume 3 + β5 is rational.
Then,
β5 = (3 + β5) β 3
Right side = rational β rational = rational
So β5 becomes rational β
But we already proved β5 is irrational.
Contradiction β 3 + β5 is irrational.
3. Prove the following are irrational
(i) 1/β2
Solution:
Assume 1/β2 is rational.
Then,
β2 = 1 Γ· (1/β2)
Right side = rational Γ· rational = rational β
But β2 is irrational.
Contradiction β 1/β2 is irrational.
(ii) 7β5
Assume 7β5 is rational.
Then,
β5 = (7β5) Γ· 7
Right side = rational Γ· rational = rational β
But β5 is irrational.
Contradiction β 7β5 is irrational.
(iii) 6 + β2
Assume 6 + β2 is rational.
Then,
β2 = (6 + β2) β 6
Right side = rational β rational = rational β
But β2 is irrational.
Contradiction β 6 + β2 is irrational.
Class-wise Solutions – Class 10 Maths Ch 1 Real Numbers Ex 1.2
Class 12:
Class 12 Physics – NCERT Solutions
Class 12 Chemistry – NCERT Solutions
Class 11:
- Class 11 Physics – NCERT Solutions
- Class 11 Chemistry – NCERT Solutions
- Class 11 Biology – NCERT Solutions
- Class 11 Math – NCERT Solutions
Class 10:
Class 9:
Class 8:
Class 7:
Class 6:
Subject-wise Solutions – Class 10 Maths Ch 1 Real Numbers Ex 1.2
Physics:
Chemistry:
Biology:
Math:
- Class 11 Math – NCERT Solutions
- Class 10 Math – NCERT Solutions
- Class 9 Math – NCERT Solutions
- Class 8 Math – NCERT Solutions
Science:
- Class 10 Science – NCERT Solutions
- Class 9 Science – NCERT Solutions
- Class 8 Science – Oxford Solutions
- Class 7 Science – Oxford Solutions
- Class 6 Science – Oxford Solutions
NEET BIOLOGY
- Evolution
- Breathing and Exchange of Gases
- Anatomy of Flowering Plants
- Body Fluids and Circulation
- Human Health and Disease
- Microbes in Human Welfare
- Cell Cycle and Cell Division
- Biotechnology and Its Applications
- Biodiversity and Conservation
- Morphology of Flowering Plants
For the official Class 8 Mathematics Solutions, you can visit:
- NCERT Textbooks (for Class 8):
