NCERT Solutions Class 10 Maths Chapter 1: Real Numbers

Let us assume, to the contrary, that √5 is rational.
Therefore, we can find two integers a and b (where b ≠ 0) such that:
√5 = a / b
Assume that a and b are coprime (they have no common factor other than 1).

5 = a² / b²
5b² = a² ... (Equation 1)
This means a² is divisible by 5. Therefore, a is also divisible by 5.

Since a is divisible by 5, we can write a = 5c for some integer c.
Substitute this into Equation 1:
5b² = (5c)²
5b² = 25c²
b² = 5c²
This means b² is divisible by 5, so b is also divisible by 5.

From Steps 2 and 3, both a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
Hence, √5 is irrational.

Let us assume that 3 + 2√5 is rational.
Then we can find coprime integers a and b (b ≠ 0) such that:
3 + 2√5 = a / b

2√5 = (a / b) - 3
√5 = (a - 3b) / 2b

Since a and b are integers, (a - 3b) / 2b is rational.
This implies that √5 is rational.
But we know that √5 is irrational.
This contradiction has arisen because of our incorrect assumption.
Hence, 3 + 2√5 is irrational.

Assume 1 / √2 is rational.
1 / √2 = a / b
Inverting both sides: √2 = b / a
Since a, b are integers, b / a is rational, implying √2 is rational.
But √2 is irrational. So, assumption is false.
Answer: Irrational.

Assume 7√5 is rational.
7√5 = a / b
√5 = a / 7b
Here, a / 7b is rational because a, b, 7 are integers.
This implies √5 is rational, which contradicts the fact that √5 is irrational.
Answer: Irrational.

Assume 6 + √2 is rational.
6 + √2 = a / b
√2 = (a / b) - 6
√2 = (a - 6b) / b
The RHS is rational, but LHS (√2) is irrational.
This contradiction proves the number is irrational.
Answer: Irrational.