NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2

1. Suppose \(\sqrt{5}\) is rational. Then \(\sqrt{5} = \dfrac{a}{b}\) with \(a, b\) integers, \(b \ne 0\), and \(a, b\) coprime (no common factor other than 1).
2. Squaring: \(5b^2 = a^2\). So \(5\) divides \(a^2\), and since \(5\) is prime, \(5\) divides \(a\). Write \(a = 5c\).
3. Then \(5b^2 = 25c^2\), so \(b^2 = 5c^2\). So \(5\) divides \(b^2\), and therefore \(5\) divides \(b\).
4. Now \(5\) divides both \(a\) and \(b\), contradicting that they are coprime. So the assumption is false.

Where marks slip: State the coprime assumption at the start; examiners give a step mark for it and for naming the contradiction at the end.

1. Suppose \(3 + 2\sqrt{5} = r\), where \(r\) is rational.
2. Rearrange: \(\sqrt{5} = \dfrac{r - 3}{2}\).
3. The right-hand side is rational (rationals are closed under subtraction and division by a non-zero number), so \(\sqrt{5}\) would be rational.
4. This contradicts Q1 (\(\sqrt{5}\) is irrational). So \(3 + 2\sqrt{5}\) is irrational.

1. Suppose \(\dfrac{1}{\sqrt{2}} = r\), rational and non-zero. Then \(\sqrt{2} = \dfrac{1}{r}\), which is rational.
2. But \(\sqrt{2}\) is irrational (same proof as for \(\sqrt{5}\), with the prime \(2\)). Contradiction.

1. Suppose \(7\sqrt{5} = r\), rational. Then \(\sqrt{5} = \dfrac{r}{7}\), which is rational.
2. This contradicts \(\sqrt{5}\) being irrational.

1. Suppose \(6 + \sqrt{2} = r\), rational. Then \(\sqrt{2} = r - 6\), which is rational.
2. This contradicts \(\sqrt{2}\) being irrational.