NCERT Solutions for Class 10 Maths Exercise 1.2 Chapter 1 Real Numbers
1. Prove that √5 is irrational.
Step 1: Assume to the contrary that √5 is rational.
Then, by definition of rational numbers, √5 can be expressed as a fraction a/b, where a and b are integers that have no common factors other than 1, and b is not zero.
Step 2: Express the assumption in equation form.
√5 = a/b
5 = a²/b²
Step 3: Cross multiply to get rid of the denominator.
5b² = a²
Step 4: Since a² is divisible by 5, a must also be divisible by 5.
Let a = 5k, where k is an integer.
Step 5: Substitute the value of a in the equation from Step 3.
5b² = (5k)²
5b² = 25k²
b² = 5k²
Step 6: Since b² is divisible by 5, b must also be divisible by 5.
This is a contradiction because a and b have a common factor of 5, which contradicts our initial statement that a and b have no common factors other than 1.
Conclusion: Our assumption that √5 is rational is incorrect. Therefore, √5 is irrational.
2. Prove that 3 + 2√5 is irrational.
Step 1: Assume to the contrary that 3 + 2√5 is rational.
Step 2: Then 3 + 2√5 can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Express the assumption in equation form.
3 + 2√5 = a/b
Step 4: Rearrange the equation to isolate √5.
√5 = (a/b – 3)/2
Step 5: Multiply both sides by 2b to clear the fraction.
2b√5 = a – 3b
Step 6: Square both sides to eliminate the square root.
(2b√5)² = (a – 3b)²
20b² = a² – 6ab + 9b²
Step 7: Rearrange the terms to bring them all to one side.
a² – 6ab – 11b² = 0
Step 8: This equation suggests that 11b² is a perfect square since a² and 6ab are perfect squares. However, 11 is not a perfect square, and b² is an integer. Thus, the assumption that 11b² is a perfect square leads to a contradiction.
Conclusion: Our assumption that 3 + 2√5 is rational is incorrect. Therefore, 3 + 2√5 is irrational.
3. Prove that the following are irrationals:
(i) 1/√2
Step 1: Assume to the contrary that 1/√2 is rational.
Step 2: Then 1/√2 can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Express the assumption in equation form.
1/√2 = a/b
Step 4: Square both sides to eliminate the square root.
(1/√2)² = (a/b)²
1/2 = a²/b²
Step 5: Cross multiply to get rid of the denominator.
b² = 2a²
Step 6: This suggests that b² is even, hence b must be even.
Step 7: If b is even, then a² must also be even, and therefore a must be even.
Step 8: Having both a and b even contradicts the assumption that they are co-prime.
Conclusion: Our assumption that 1/√2 is rational is incorrect. Therefore, 1/√2 is irrational.
(ii) 7√5
Step 1: We already proved that √5 is irrational in question 1.
Step 2: Multiplying an irrational number by a rational number (7 in this case) does not result in a rational number.
Conclusion: Therefore, 7√5 is irrational.
(iii) 6 + √2
Step 1: Assume to the contrary that 6 + √2 is rational.
Step 2: Then 6 + √2 can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Rearrange to isolate √2.
√2 = (a/b) – 6
Step 4: Square both sides to eliminate the square root.
(√2)² = [(a/b) – 6]²
2 = (a²/b²) – 12a/b + 36
Step 5: Rearrange to form a quadratic equation in a/b.
a²/b² – 12a/b + 34 = 0
Step 6: This equation has no solution in integers for a and b because 2 cannot be expressed as the sum of an integer and the square of a rational number unless that rational number is an integer itself. But √2 is not an integer.
Conclusion: Our assumption that 6 + √2 is rational is incorrect. Therefore, 6 + √2 is irrational.
Additional Questions for Ex. 1.2 Real Numbers for Class 10 NCERT Book
1. Prove that √3 is irrational.
Step 1: Assume to the contrary that √3 is rational. Then, by definition of rational numbers, √3 can be expressed as a fraction a/b, where a and b are co-prime integers and b ≠ 0.
Step 2: Express the assumption in equation form.
√3 = a/b
Step 3: Square both sides to eliminate the square root.
3 = a²/b²
Step 4: Cross multiply to get rid of the denominator.
3b² = a²
Step 5: Since a² is divisible by 3, a must also be divisible by 3. Let a = 3k, where k is an integer.
Step 6: Substitute the value of a in the equation from Step 3.
3b² = (3k)²
3b² = 9k²
b² = 3k²
Step 7: Since b² is divisible by 3, b must also be divisible by 3. This is a contradiction because a and b have a common factor of 3, which contradicts our initial statement that a and b are co-prime.
Conclusion: Our assumption that √3 is rational is incorrect. Therefore, √3 is irrational.
2. Prove that √2 + √3 is irrational.
Step 1: Assume to the contrary that √2 + √3 is rational.
Step 2: Then √2 + √3 can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Rearrange the equation to isolate √3.
√3 = (a/b) – √2
Step 4: Square both sides to eliminate the square roots.
3 = (a²/b²) – 2(a/b)√2 + 2
Step 5: Rearrange the equation to solve for √2.
(a/b)√2 = (a² – 2b² + 3b²)/2b²
√2 = (a² + b²)/2ab
Step 6: Since √2 is irrational, the right side of the equation cannot be rational, which contradicts the assumption that √2 + √3 is rational.
Conclusion: Our assumption that √2 + √3 is rational is incorrect. Therefore, √2 + √3 is irrational.
3. Prove that √(2) + √(1/2) is irrational.
Step 1: Assume to the contrary that √(2) + √(1/2) is rational.
Step 2: Then √(2) + √(1/2) can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Rearrange the equation to isolate √(1/2).
√(1/2) = (a/b) – √(2)
Step 4: Square both sides to eliminate the square roots.
1/2 = (a²/b²) – 2(a/b)√(2) + 2
Step 5: Rearrange to solve for √(2).
√(2) = (a² – 2b² + b²)/2ab
√(2) = (a² – b²)/2ab
Step 6: Since √(2) is irrational, the right side of the equation cannot be rational, leading to a contradiction.
Conclusion: Our assumption that √(2) + √(1/2) is rational is incorrect. Therefore, √(2) + √(1/2) is irrational.
4. Prove that 1 + 1/√2 is irrational.
Step 1: Assume to the contrary that 1 + 1/√2 is rational.
Step 2: Then 1 + 1/√2 can be written as a rational number a/b, where a and b are co-prime integers and b ≠ 0.
Step 3: Rearrange the equation to isolate 1/√2.
1/√2 = (a/b) – 1
Step 4: Multiply both sides by √2 to clear the square root on the left side.
1 = √2(a/b) – √2
Step 5: Square both sides to eliminate the square root.
1 = 2(a²/b²) – 2√2(a/b) + 2
Step 6: Rearrange to bring all terms to one side and solve for √2.
√2(a/b) = (a² + b² – 2b²)/2ab
Step 7: Since √2 is irrational, the right side of the equation cannot be rational, which contradicts the assumption that 1 + 1/√2 is rational.
Conclusion: Our assumption that 1 + 1/√2 is rational is incorrect. Therefore, 1 + 1/√2 is irrational.