Math, asked by INNOCENTDEVIL006, 9 months ago

By division method prove that √23 is an irrational number

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Answered by reenapandey581

Answer:

(i) 23 is not a perfect square values so that, it is an irrational number. The decimal expansion of above number is terminating, so that it is a rational number. The decimal expansion of above number is non-terminating recurring, so that, it is a rational number.

Step-by-step explanation:

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Answered by Mihir1001

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

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$
⚪ root over 23 [ √23 ]

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}$
⚪ $\sqrt{23}$ is an irrational number.

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

$\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}$
✒ If possible, $\sqrt{23}$ be a rational number.

let $\sqrt{23} = \Large{ \frac{a}{b} }$ , where a and b are co-primes and b ≠ 0.

Then,
✒ $\sqrt{23} = \Large{ \frac{a}{b} }$

On squaring both the sides :

$\Rightarrow { \left( \sqrt{23} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }$

$\Rightarrow 23 = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }$

$\Rightarrow 23 {b}^{2} = {a}^{2}$

$\Rightarrow {a}^{2} = 23 {b}^{2}$ ——————— ( 1 )

Therefore, ${a}^{2}$ is divisible by 23
Therefore, a is also divisible by 23.

let a = 23c, for some integer c. ——————— ( 2 )

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

${(23c)}^{2} = 23 {b}^{2}$

$\Rightarrow 529 {c}^{2} = 23 {b}^{2}$

$\Rightarrow {}^{23} \: \cancel{529} {c}^{2} = {}^{1} \: \cancel{23} {b}^{2}$

$\Rightarrow {b}^{2} = 23 {c}^{2}$

Therefore, ${b}^{2}$ is divisible by 23.
Therefore, b is also divisible by 23.

Therefore, a and b have a common factor 23.
This contradicts the fact that a and b are co-primes.
This contradiction arises on assuming $\sqrt{23}$ to be a rational number.
So, our assumption is wrong.
Hence, $\sqrt{23}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$ .

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$
⚪ $\sqrt{23}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$ .

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By division method prove that √23 is an irrational number

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