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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.
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