Question

prove that√3 is an irrational no.​

Answer 1

Answer:

Prove root 3 is an irrational number

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Step-by-step explanation:

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Answer 2

$\huge\underline\bold\red{QUESTION??}$

Prove that√3 is an irrational number.

$\sf\underline{\underline{\green{To \: prove:- \: }}}$

√3 is an irrational number.

$\huge\underline\mathbb{\red S\pink {0} \purple {L} \blue {UT} \orange {1}\green {ON :}}$

Let us assume that,

√3 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

         ∴√3 = $\frac{a}{b}$

On squaring both sides, we get ;

3 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 3b²

Clearly, a² is divisible by 3.

So, a is also divisible by 3.

$\sf\underline{\underline{\green{Now, \: }}}$

let some integer be c.

⇒ a = 3c

Substituting for a, we get ;

⇒ 3b² = 3c

Squaring both sides,

⇒ 3b² = 9c²

⇒ b² = 3c²

This means that, 3 divides b², and so 2 divides b.

$\sf\underline{\underline{\green{Therefore, \: }}}$

a and b have at least 3 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √3 is rational number.

So, we conclude that √3 is irrational number.

$\sf\underline{\underline{\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{</strong><strong>H</strong><strong>E</strong><strong>N</strong><strong>C</strong><strong>E</strong><strong> \: </strong><strong>PROVED</strong><strong> \: }}}$