Question

Prove that root 5 is not a rational number using conduction method

Answer 1

$\star\;\;{\underline{\underline{\sf{\pink{To\;Prove:-}}}}}$

• $\sqrt{5}$ is an irrational number ( not a rational).

$\star\;\;{\underline{\underline{\sf{\pink{Solution:-}}}}}$

★ Lets $\sqrt{5}$ is a rational number.

࿋ Therefore, it can be written in form of $\sf \dfrac{a}{b}$ where b ≠ 0 and a and b are co prime number(no common factor other than 1).

Hence,

$:\implies\sf \sqrt{5} = \dfrac{a}{b} \\\\ :\implies\sf \sqrt{5} b = a \\\\ \;\;\;\;\;\small{\underline{\underline{\sf{\red{\dag\;Squaring\;both\;side:-}}}}} \\\\ :\implies\sf ( \sqrt{5} b)^2 = (a)^2 \\\\ :\implies\sf 5b^2 = a^2 \\\\ :\implies\sf \dfrac{a^2}{5} = b^2 \\\\ :\implies\sf Hence,\;5\;divides\; \bf{a^2} \\\\ :\implies\sf So,\;5\;divides\; \bf{a} \;also \;\;\;\;\;---[1]$

$\sf Hence\;we\;can\;say\;that: \\\\ \sf \dfrac{a}{5} = c, where\;c\;is\;some\;integer \\\\ \sf So,\; \bf{a = 5c}$

$\sf Now\;we\;know\;that \\\\ :\implies\sf 5b^2 = a^2 \\\\\ \;\;\;\;\;\small{\underline{\underline{\sf{\red{\dag\;Putting\;a = 5c}}}}}$

$:\implies\sf 5b^2 = (5c)^2 \\\\ :\implies\sf \cancel{5}b^2 = \cancel{25} c^2 \\\\\ :\implies\sf b^2 = 5c^2 \\\\ :\implies\sf \dfrac{b^2}{5} = c^2$

$:\implies\sf Hence,\;5\;divides\; \bf{b^2} \\\\ :\implies\sf So,\;5\;divides\; \bf{a} \;also \;\;\;\;---[2] \\\\ :\implies\sf By\;Eq[1]\;and\;Eq[2]:-$

$\normalsize\sf 5\;divides\;both\;a\;and\;b \\\\ \normalsize\sf Hence,\;5\;is\;a\;common\;factor\;of\;both\;a\;and\;b \\\\ \normalsize\sf Therefore,\;a\;and\;b\;are\;not\;co - prime$

$\normalsize\sf This\;become\;a\;contradiction. \\\\ \normalsize\sf Hence,\;our\;supposition\;is\;wrong. \\\\ {\underline{\underline{\sf{\purple{\dag\;Hence,\; \sqrt{5} \;is\;an\; irrational\;number.}}}}}$

$\rule{200}{3}$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Question}}}}}}}$

• Prove that root 5 is not a rational number using conduction method

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Explanation}}}}}}}$

Let √5 is a rational number

$\sqrt {5} = \dfrac{a}{b}$

• squaring both the sides

$5 = \dfrac{a^2}{b^2}$

$a^2 = 5b^2$

$a^2 \: is \: divisible \: by \: 5$

${\blue{a \: is \: also\ divisible \ by \ 5 }}$---------(i)

$Let \ a = 5c$

• squaring both the sides

$a^2 = 25c^2$

• $a^2 = 5b^2$

$5b^2 = 25c^2$

$b^2 = 5c^2$

${b^2 \: is \:divisible \:by \:5}$

${\blue{b \: is\: also \: divisible \: by \: 5}}$-----------(ii)

From (i) and (ii) , we get a and b both are divisible by 5 .

i.e. a and b have a common factor 5 .

This contradicts our assumptions that $\dfrac{a}{b}$ is a rational number i.e. a and b do not have any common factor other than unity (1).

therefore

$\sf\implies {\red{\dfrac{a}{b} \: is \: not\: rational}}$

$\sf\implies {\red{\sqrt 5 \: is \: not\: rational}}$

$\sf\implies{\red{\sqrt 5 \: is \: irrational}}$