Question

prove that
$prove that \sqrt{2} is irrational$
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Answer 1

Hope it helps you Nd I think it was correct

Answer 2

Solution :

Let us Find the Square root of 2 by Long Division Method. Refer to the Attachment.

$\sf \: \sqrt{2} = 1.4142135$

Clearly,The Decimal Representation of √2 is neither Terminating nor Repeating.

$\sf \red{Hence \: \sqrt{2} \: is \: an \: irrational \: number \: }$

Aliter :

We shall prove it by the process of Contradiction. Let us assume √2 is a rational Number. Then,

$\sf \: \sqrt{2} = \dfrac{p}{q}$

Where p and q are integers having no common factor and q is not equal to 0

Squaring Both The Sides :

$\rightarrow \sf \: (\sqrt{2} )^{2} =( \dfrac{p}{q} )^{2}$

$\rightarrow \sf \: 2 =( \dfrac{ {p}^{2} }{ {q}^{2} } )$

$\rightarrow \sf \: {p}^{2} = 2 {q}^{2} ....(i)$

$\sf \bullet \: {p}^{2} \: is \: an \: even \: integer$

$\sf \: \bullet \: {p} \: \: is \: an \: even \: integer$

$\rightarrow \sf \: p = 2m , \: where\: m \: is \: an \: integer$

Squaring both the sides :

$\rightarrow \sf \: {p}^{2} = 4 {m}^{2}$

$\rightarrow \sf \: 2{q}^{2} = 4 {m}^{2} ..... Using \: i)$

$\rightarrow \sf \: {q}^{2} = 2 {m}^{2}$

$\sf \bullet \: {q}^{2} \: is \: an \: even \: integer$

$\bullet \: \sf \: {q} \: \: is \: an \: even \: integer$

So,Both p and q are even integers and therefore have a common factor 2. But,this contradicts that p and q have no common factor.

Hence,Our Assumption is Wrong.

$\sf \red{ \: \sqrt{2} \: is \: an \: irrational \: number \: }$