Try to prove the irrationality of $\sqrt{3}$ using the approach of proof by contradiction. Will the same approach work for $\sqrt{5}$, $\sqrt{7}$ or $\sqrt{10}$? [Page No. 55]

We have to prove $\sqrt{3}$ is irrational.

Let us assume the opposite  i.e., $\sqrt{3}$ is rational.

Hence, $\sqrt{3}$ can be written in the form $\frac{a}{b}$ where $a$ and $b$ ($b \neq 0$) are co-prime (no common factor other than 1)

Hence,
$$\sqrt{3} = \frac{a}{b}$$

$$\sqrt{3}b = a$$

Squaring both sides

$$(\sqrt{3}b)^2 = a^2$$

$$3b^2 = a^2$$

$$\frac{a^2}{3} = b^2$$

Hence, 3 divides $a^2$.

By theorem: If $p$ is a prime number, and $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive number.

So, 3 shall divide $a$ also. $\text{-(i)}$

Hence, we can say 
$$\frac{a}{3} = c \text{ where } c \text{ is some integer}$$

$$\text{So, } a = 3c$$

Now we know that

$$3b^2 = a^2$$

Putting $a = 3c$

$$3b^2 = (3c)^2$$

$$3b^2 = 9c^2$$

$$b^2 = \frac{1}{3} \times 9c^2$$

$$b = 3c^2$$

$$\frac{b^2}{3} = c^2$$

Hence, 3 divides $b^2$.

By theorem: If $p$ is a prime number, and $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive number.

So, 3 divides $b$ also ...(ii)

By (i) and (ii)

3 divides both $a$ & $b$

Hence, 3 is factor of $a$ and $b$.

So, $a$ & $b$ have a factor 3.

Therefore, $a$ & $b$ are not co-prime.

Hence, our assumption is wrong.

$\therefore$ By contradiction, $\sqrt{3}$ is irrational.

Now, we are asked

Will the same approach work for $\sqrt{5}, \sqrt{7}$, or $\sqrt{10}$?

Yes, we use the same approach for $\sqrt{5}, \sqrt{7}$, or $\sqrt{10}$ and indeed any square root of a number that isn't a perfect square.

If we swap the '3' for a '5', the logic holds perfectly.