Question

prove that √3 is an irrational​

Answer 1

$\huge\tt\underline{Question}$:

prove that √3 is an irrational

$\huge\tt\underline{to\:proof}$:

➜ √3 Is an irrational

$\huge\tt\underline{solution}$:

✰let us assume on the contrary that √3 is a rational number.

✰then,there exist positive integers a and b such that:

$\sqrt{3} = \frac{a}{b}$where,a and b,are co - prime,i.e:

their HCF Is 1

now,$\sqrt{3} = \frac{a}{b}$

$= > 3 = \frac{ {a}^{2} }{b^{2} } </p><p>$

$= > {3b}^{2} = {a}^{2}$

$= > 3 \: divides \: {a}^{2} ( 3\: divides \: {3b}^{2)}$

$= > 3 \: divides \: a \: ...(1)$

$= > a = 3c \: for \: some \: integer \: c$

$= > {a}^{2} = {9}^{ {c}^{2} } ( {a}^{2} = {3b}^{2} )$

$= > {b}^{2} = {3c}^{2}$

$= > 3 \: divides \: {b}^{2} (3 \: divides \: {3c}^{2} )$

$= > 3 \: divides \: b....(2)$

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

hence,√3 is irrational number