Question

prove that √2 is irrational​

Answer 1

$\huge{\underline{\mathfrak{\textcolor{blue}{Answer :}}}}$
$\huge\boxed{\fcolorbox{red}{pink}{irrational}}$
$\huge{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}$

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$
⚪ root over 2 [ √2 ]

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}$
⚪ $\sqrt{2}$ is an irrational number.

$\LARGE{\underline{\mathtt{\textcolor{teal}{Concept \: \: used :-}}}}$
⚪ Real numbers
⚪ Irrational numbers

$\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}$
✒ If possible, $\sqrt{2}$ be a rational number.

let $\sqrt{2} = \Large{ \frac{a}{b} }$, where a and b are co-primes and b ≠ 0.

Then,
✒ $\sqrt{2} = \Large{ \frac{a}{b} }$

On squaring both the sides :

$\Rightarrow { \left( \sqrt{2} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }$

$\Rightarrow 2 = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }$

$\Rightarrow 2 {b}^{2} = {a}^{2}$

$\Rightarrow {a}^{2} = 2 {b}^{2}$ ——————— ( 1 )

Therefore, ${a}^{2}$ is divisible by 2.
Therefore, a is also divisible by 2.

let a = 2c, for some integer c. ——————— ( 2 )

On substituting ( 2 ) in ( 1 ) , we get :

${(2c)}^{2} = 2 {b}^{2}$

$\Rightarrow 4 {c}^{2} = 2 {b}^{2}$

$\Rightarrow {}^{2} \: \cancel{4} {c}^{2} = {}^{1} \: \cancel{2} {b}^{2}$

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

Therefore, ${b}^{2}$ is divisible by 2.
Therefore, b is also divisible by 2.

Therefore, a and b have a common factor 2.
This contradicts the fact that a and b are co-primes.
This contradiction arises on assuming $\sqrt{2}$ to be a rational number.
So, our assumption is wrong.
Hence, $\sqrt{2}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$
⚪$\sqrt{5}$ is an $\sf\green{\underline{\blue{irrational \: number}}}[\tex] . [tex]\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

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