Question

Raj is playing "Roll a dice-Number game" with his friend Rahul. If he gets a $5$ on the dice, he is supposed to collect an Irrational number from Rahul. Rahul has the set of numbers: $\{3, 3.14, \sqrt{2}, \frac{1}{2}\}$ with him. Raj takes the turn and rolls the dice. He gets a $5$ on the dice, so he picks up the number $\sqrt{2}$ from Rahul. Simran, who was just observing the game, said that the number collected by Raj is not an Irrational number. Help Raj to prove that the number collected by him from Rahul is an Irrational number.

Accepted Answer

To help Rahul prove that $\sqrt{2}$ is an irrational number,

Let us first assume that $\sqrt{2}$ is a rational number.

From the definition of Rational Numbers, we can write that

$\sqrt{2}$ can be expressed in the form $\frac{p}{q}$ , where $q \neq 0$ , and $p, q$ are both co-prime integers.

$∴ \sqrt{2} = \frac{p}{q}$

$\Rightarrow p = \sqrt{2} q$

$\Rightarrow p^{2} = 2 q^{2}$

$\Rightarrow p^{2}$ is an even number

$\Rightarrow p$ is an even number.

So, let $p = 2 m$ , where $m$ is an integer.

$\Rightarrow p^{2} = {(2 m)}^{2}$

$\Rightarrow 2 q^{2} = 4 m^{2} [∵ p^{2} = 2 q^{2}]$

$\Rightarrow q^{2} = 2 m^{2}$

$\Rightarrow q^{2}$ is an even number

$\Rightarrow q$ is an even number.

Thus, $p$ and $q$ are both even numbers.

So, there exists $2$ as the common factor of both $p$ and $q$ .

But, as per our assumption, there are no common factor in $p$ and $q$ , except $1$ .

This is a contradiction, and our assumption must be wrong.

$∴ \sqrt{2}$ is not a rational number.

Hence, $\sqrt{2}$ is an irrational number.

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