Question

prove that root 2 is irretional​

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow \sqrt{2}$

$\large\underline\bold{TO\:PROVE, \sqrt{2}\:IS\:AN\: IRRATIONAL\: NUMBER.}$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore \green{ let \: us \:assume\: that\: \sqrt{2}\: is \:a \:rational\: number.}$

$\sf\therefore \sqrt{2} = \dfrac{a}{b}$$\sf\red{\:---[ where\:a\:and\:b\:are\:integers\:and\:coprime\:and\:where\: b \neq 0]}$

$\sf\implies 2b=a$

$\sf\large\star squaring\:both\:sides .$

$\sf\implies 2b^2=a^2\:------ \boxed{1}$

$\sf\therefore 2\:is\:a\:factor\:of\:a^2$

$\sf\therefore 2\:is\:a\:factor\:of\:a$

NOW,

$\sf\star TAKING\: a=2m$$\sf\red{\:----[for\:m\:is\:a\:positive\:integer.]}$

$\sf\large\star squaring\:both\:sides .$

$\sf\implies a^2=4m^2\:----\boxed{2}$

$\sf\therefore \purple{FROM \:EQUATION \:1\: AND \:2.\: WE\: GET,}$

$\sf\implies 2b^2=4m^2$

$\sf\implies b^2= \dfrac{4m^2}{2}$

$\sf\implies b^2= \cancel \dfrac{4m^2}{2}$

$\sf\implies b^2= 2m^2$

$\sf\therefore 2\:is\:a\:factor\:of\:b^2.$

$\sf\therefore 2\:is\:a\:factor\:of\:b.$

$\sf\therefore here,a \:and\:b\:are\:coprimes \:and\:2\:cannot\:divide\:both(a\:and\:b) .$

$\sf\therefore hence,\:H.C.F\:of\:a\:and\:b\:is\:1.$

$\sf\therefore these\:contradiction\:has\:occured\:due\:to\:our\:wrong\: assumption\:that\:\sqrt{2}\:is\:a\:rational\:no.$

WE CONCLUDE,

$\large\underline\bold\red{\sqrt{2}\:IS\:AN\: IRRATIONAL\: NUMBER}$

______________

Answer 2

$\huge\mathcal{\fcolorbox{lime}{black}{\pink{Answer}}}$

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow \sqrt{2}$

$\large\underline\bold{TO\:PROVE, \sqrt{2}\:IS\:AN\: IRRATIONAL\: NUMBER.}$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore \green{ let \: us \:assume\: that\: \sqrt{2}\: is \:a \:rational\: number.}$

$\sf\therefore \sqrt{2} = \dfrac{a}{b}$$\sf\red{\:---[ where\:a\:and\:b\:are\:integers\:and\:coprime\:and\:where\: b \neq 0]}$

$\sf\implies 2b=a$

$\sf\large\star squaring\:both\:sides .$

$\sf\implies 2b^2=a^2\:------ \boxed{1}$

$\sf\therefore 2\:is\:a\:factor\:of\:a^2$

$\sf\therefore 2\:is\:a\:factor\:of\:a$

NOW,

$\sf\star TAKING\: a=2m$$\sf\red{\:----[for\:m\:is\:a\:positive\:integer.]}$

$\sf\large\star squaring\:both\:sides .$

$\sf\implies a^2=4m^2\:----\boxed{2}$

$\sf\therefore \purple{FROM \:EQUATION \:1\: AND \:2.\: WE\: GET,}$

$\sf\implies 2b^2=4m^2$

$\sf\implies b^2= \dfrac{4m^2}{2}$

$\sf\implies b^2= \cancel \dfrac{4m^2}{2}$

$\sf\implies b^2= 2m^2$

$\sf\therefore 2\:is\:a\:factor\:of\:b^2.$

$\sf\therefore 2\:is\:a\:factor\:of\:b.$

$\sf\therefore here,a \:and\:b\:are\:coprimes \:and\:2\:cannot\:divide\:both(a\:and\:b) .$

$\sf\therefore hence,\:H.C.F\:of\:a\:and\:b\:is\:1.$

$\sf\therefore these\:contradiction\:has\:occured\:due\:to\:our\:wrong\: assumption\:that\:\sqrt{2}\:is\:a\:rational\:no.$

WE CONCLUDE,

$\large\underline\bold\red{\sqrt{2}\:IS\:AN\: IRRATIONAL\: NUMBER}$