Prove that is $\sqrt{2}$ an irrational number.


Given: Number:$\sqrt{2}$.

To do: To prove that the given number is irrational number.

Solution:

Let $\sqrt{2}$  be rational.

$\sqrt{2}=\frac{p}{q}$ where $p$ and $q$ are co-prime integers and, 

$q =0$

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

$\Rightarrow 2q^{2}=p^{2}$                                .............$( i)$

$\Rightarrow 2\ divides\ p^{2}$

$\Rightarrow 2\ divides\ $p$                               ...............$( A)$

Let $p =2c$ for some integer

$p^{2}= 4c^{2}$

$2q^{2}=4c^{2}$

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

$\Rightarrow 2\ divides\ q^{2}$

$\Rightarrow 2\ divides\ q$                            .................$( B)$

 From $( A)$ and $( B)$, we get 2 is common factor of both $p$ and $q$. 

But this contradicts the fact that p and q have no common factor other than 1 

Our supposition is wrong Hence, 

 $\sqrt{2}$ is an irrational number.

Updated on: 10-Oct-2022

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