NCERT Class 10 Maths

Let us assume, to the contrary, that $\sqrt{5}$ is rational.

Therefore, we can find two integers $a$ and $b$ ($b \neq 0$) such that:

Let $a$ and $b$ have a common factor other than 1. Then we can divide them by the common factor, and assume that $a$ and $b$ are co-prime.

Squaring both sides:

Therefore, $a^2$ is divisible by 5, and it can be said that $a$ is divisible by 5.

Let $a = 5k$, where $k$ is an integer. Substituting:

This means that $b^2$ is divisible by 5, and hence $b$ is divisible by 5.

Hence, $\sqrt{5}$ cannot be expressed as $\frac{a}{b}$.

Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.

Therefore, we can find two integers $a$ and $b$ ($b \neq 0$) such that:

Rearranging:

Since $a$ and $b$ are integers, $\frac{a – 3b}{2b}$ will also be rational.

Therefore, $\sqrt{5}$ is rational.

Hence, our assumption that $3 + 2\sqrt{5}$ is rational is false.

Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.

Therefore, we can find two integers $a$ and $b$ ($b \neq 0$) such that:

Rearranging:

Since $a$ and $b$ are integers, $\frac{b}{a}$ is rational.

Therefore, $\sqrt{2}$ should be rational.

Hence, our assumption is false.

Let us assume, to the contrary, that $7\sqrt{5}$ is rational.

Therefore, we can find two integers $a$ and $b$ ($b \neq 0$) such that:

Rearranging:

Since $a$ and $b$ are integers, $\frac{a}{7b}$ is also rational.

Therefore, $\sqrt{5}$ should be rational.

Hence, our assumption that $7\sqrt{5}$ is rational is false.

Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.

Therefore, we can find two integers $a$ and $b$ ($b \neq 0$) such that:

Rearranging:

Since $a$ and $b$ are integers, $\frac{a – 6b}{b}$ is also rational.

Hence, $\sqrt{2}$ should be rational.

Therefore, our assumption is false.