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Class 10 Maths Chapter 1 Exercise 1.2 Solutions
In this section of chapter 1 “Real Numbers“, you have studied the following points ⚡:
- Prove that √2, √3, √5 and, in general, √p is irrational, where p is a prime.
- One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
- Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.
In class 10 maths chapter 1 exercise 1.2 solutions, let’s try to find the answers to these questions —
EXERCISE 1.2
Question 1: Prove that √5 is irrational.
Answer 1: Let √5 be a rational number. Then we can find two integers a and b such that √5 = a/b, where b ≠ 0.
Let a and b have a common factor other than 1. Then we can divide a and b by the common factor, and assume that a and b are co-prime —
a = √5 b
⇒ a2 = 5 b2
Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5. Let a = 5k, where k is an integer.
(5k)2 = 5 b2
⇒ 5 k2 = b2
This means that b2 is divisible by 5 and hence b is divisible by 5. This implies that a and b have 5 as a common factor. This is a contradiction to the fact that a and b are co-prime. Hence, √5 cannot be expressed in the form of p/q or it can be said that √5 is irrational.
Question 2: Prove that 3 + 2√5 is irrational.
Answer 2: Let 3 + 2√5 be a rational number. Then we can find two co-prime integers a and b such that 3 + 2√5 = a/b, where b ≠ 0. So —
2√5 = a/b − 3
⇒ √5 = 1/2 (a/b − 3)
Since a and b are integers, 1/2 (a/b − 3) is rational. Therefore, √5 is also rational. This contradicts the fact that √5 is irrational. Hence, our assumption that 3 + 2√5 is rational, is false. This proves that 3 + 2√5 is irrational.
Question 3: Prove that the following are irrationals :
| (i) 1/√2 | (ii) 7√5 | (iii) 6 + √2 |
Answer 3: (i) Let 1/√2 be a rational number. Then we can find two co-prime integers a and b such that 1/√2 = a/b, where b ≠ 0. So —
√2 = b/a
b/a is a rational number as a and b are integers. Therefore, √2 is rational which contradicts the fact that √2 is irrational. Hence, our assumption that 1/√2 is rational, is false. This proves that 1/√2 is irrational.
(ii) Let 7√5 be a rational number. Then we can find two co-prime integers a and b such that 7√5 = a/b, where b ≠ 0. So —
√5 = a/7b
a/7b is a rational number as a and b are integers. Therefore, √5 should be rational. This contradicts the fact that √5 is irrational. Hence, our assumption that 7√5 is rational, is false. This proves that 7√5 is irrational.
(iii) Let 6 + √2 be a rational number. Then we can find two co-prime integers a and b such that 6 + √2 = a/b, where b ≠ 0. So —
√2 = a/b − 6
a/b − 6 is a rational number as a and b are integers. Therefore, √2 is rational which contradicts the fact that √2 is irrational. Hence, our assumption that 6 + √2 is rational, is false. This proves that 6 + √2 is irrational.