Math, asked by ssanidhya75, 9 months ago

prove that root 5 is irrational

Answers

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 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 $\sf\green{\underline{\blue{irrational \: number}}}$ .

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Answered by Anonymous

Answer:

Step-by-step explanation:

Question : Prove that√5 is irrational.

Answer :

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

Hence proved

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