Question

Prove that √2 is irrational number ​

Answer 1

hope it helpful.

thanx.

Answer 2

$\star\;\LARGE{\underline{\underline{\sf{\purple{To\;ProvE:-}}}}}$

• $\sf \sqrt{2} \;is\;an\;irrational\;number.$

$\star\;\LARGE{\underline{\underline{\sf{\purple{Soluti0n:-}}}}}$

$\normalsize\sf Let\;us\;assume\;that\; \sqrt{2}\;is\;a\;rational\;number. \\\ \normalsize\sf Such\;that,\;a\;and\;b\;are\;integers\;where\;q\; doesn't\;equal\;to\;0. \\ \normalsize\sf and\;are\;coprime\;numbers.$

$:\implies\sf \sqrt{2} = \dfrac{a}{b}$

$\;\;\;\;\;\small{\underline{\underline{\sf{\red{\dag\;squaring\;both\;sides:-}}}}}$

$:\implies\normalsize\sf ( \sqrt{2})^2 = \bigg( \dfrac{a}{b} \bigg)^2 \\\\ :\implies\normalsize\sf 2 = \dfrac{a^2}{b^2} \\\\ :\implies\normalsize\sf 2b^2 = a^2 \\\\ :\implies\normalsize\sf 2 = \dfrac{a^2}{b^2} \\\\ \;\;\bullet\;\;\normalsize\sf 2\;divides\;a^2 \\\\ \;\;\bullet\;\;\normalsize\sf Therefore,\; 2\;also\;divides\;a\;\;\;\;\;[1] \\\\ :\implies\normalsize\sf Lets\;a = 2c\;for\;some\;integer\; \bf{c}. \\\\\ :\implies\normalsize\sf a^2 = 4c^2 \\\\ :\implies\normalsize\sf 2b^2 = 4c^2\;\;\;\;\;[\because\;2b^2 = a^2] \\\\\ :\implies\normalsize\sf b^2 = 2c^2 \\\\ \;\;\bullet\;\;\normalsize\sf 2\;divides\;b^2 \\\\ \;\;\bullet\;\;\normalsize\sf Therefore,\;2\;also\;divides\;b\;\;\;\;\;[2]$

$\;\;\;\;\normalsize{\underline{\underline{\sf{\red{\dag\;From\;eq[1]\;and\;eq[2]:-}}}}}$

2 divides both a and b. Therefore, 2 is a common factor of a and b.

Therefore, it becomes a contradiction the fact that a and b have no common factor other than 1.

This means that our supposition is wrong.

$\normalsize{\underline{\underline{\sf{\purple{\dag\;Hence,\; \sqrt{2}\;is\;an\;irrational\;number.}}}}}$

$\rule{200}3$