Question

prove that
$\sqrt{3}$

Answer 1

Hello dear ... ur prove is here .

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If possible , let √3 be rational and let it's simplest form be a/b.

Than , a and b are integers having no common factor other than 1 , and b ≠ 0.

Now ,
$\sqrt{3} = \frac{a}{b} = > 3 = \frac{ {a}^{2} }{ {b}^{2} } \: \: \:$
{ on squaring both side }

$= > 3 {b}^{2} = {a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: ........(1) \\ = > 3 \: \: divides \: {a}^{2}${3 divides 3b^2}

$= > 3 \: \: divides \: a$
{ 3 is prime and 3 divides a^2 => 3 divides a }

Let , a = 3c for some integer C.
putting a = 3c in (1 ) , we get

$3 {b}^{2} = 9 {c}^{2} \\ = > {b}^{2} = 3 {c}^{2}$
=> 3 divides b^2 [ 3 divides 3c^2]
=> 3 divides b

[ 3 is prime and 3 divides b^2 => 3 divides b ].

Thus , 3 is a common factor and a and b .
but , this contradict the fact that a and b have no common factor other than 1.

The contradicts arises by assuming that √3 is rational .

Hence , √3 is irrational .
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Hope it's helps you.
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