Question

prove that
$\sqrt{2}$
irrational number ​

Answer 1

Answer:

Since,

$\sqrt{2} =1.41421..............$

Therefore it is Irrational.

Answer 2

$\underline \mathbb{SOLUTION}$

》Let us assume that √2 is rational no.

$\underline \mathbb{THEN}$

$= > \sqrt{2} = \frac{p}{q}$

Where p and q are integers and coprime and q is not equal to 0.

$= > \sqrt{2} p = p$

$\underline \mathbb{THEN}$

Squaring both sides,

$= > ( \sqrt{2} {p}^{2} ) = {p}^{2}$

$= > 2 {p}^{2} = {p}^{2}$

Hence,

$= > 2 \: divides \: {p}^{2}$

$= > 2 \: divide \: p$

》Hence, 2 is a factor of p.

$= > 2 {q}^{2} = {p}^{2}$

Let p = 2c

$= > 2 {q}^{2} = 2 {c}^{2}$

$= > {q}^{2} = 2 {c}^{2}$

$= > 2 \: divides \: {q}^{2}$

$= > 2 \: divides \: q$

》Hence, 2 is a factor of q also.

So,

》2 is a factor of both p and q.

$\underline \mathbb{</strong><strong>B</strong><strong>u</strong><strong>t</strong><strong>,</strong><strong>}$

This is in contradiction to the assumption that p and q are coprime.

HENCE PROVED.

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