Math, asked by maskeen171, 9 months ago

power that √5 is a in irrational number ​

Answers

Answered by Anonymous
1

Step-by-step explanation:

It is an irrational algebraic number. 4.3×10−5). ... , its numerical value in decimal has been computed to at least 2,000,000,000,000 digits.

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Answered by Mihir1001
35
\huge{\underline{\mathfrak{\textcolor{blue}{Answer :}}}} \large{\fcolorbox{pink}{pink}{irrational}}
\huge{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}

\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}
⚪ root over 5 [ √5 ]

\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}
 \sqrt{5} is an irrational number.

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

\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}
✒ If possible,  \sqrt{5} be a rational number.

let  \sqrt{5} = \Large{ \frac{a}{b} }, where a and b are co-primes and b ≠ 0.

Then,
 \sqrt{5} = \Large{ \frac{a}{b} }

On squaring both the sides :

 \Rightarrow { \left( \sqrt{5} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }

 \Rightarrow 5 = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }

 \Rightarrow 5 {b}^{2} = {a}^{2}

 \Rightarrow {a}^{2} = 5 {b}^{2} ——————— ( 1 )

Therefore,  {a}^{2} is divisible by 5.
Therefore, a is also divisible by 5.

let a = 5c, for some integer c. ——————— ( 2 )

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

 {(5c)}^{2} = 5 {b}^{2}

 \Rightarrow 25 {c}^{2} = 5 {b}^{2}

 \Rightarrow {}^{5} \: \cancel{25} {c}^{2} = {}^{1} \: \cancel{5} {b}^{2}

 \Rightarrow {b}^{2} = 5 {c}^{2}

Therefore,  {b}^{2} is divisible by 5.
Therefore, b is also divisible by 5.

Therefore, a and b have a common factor 5.
This contradicts the fact that a and b are co-primes.
This contradiction arises on assuming  \sqrt{5} to be a rational number.
So, our assumption is wrong.
Hence,  \sqrt{5} is an \sf\green{\underline{\blue{irrational \: number}}}.

\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}
 \sqrt{5} is an \sf\green{\underline{\blue{irrational \: number}}}.

\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid

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