Question

prove that √3 is an irrational number​

Answer 1

$\huge\underline \mathfrak\purple {Question}$

★ Prove that √3 is an irrational number

$\huge\mathfrak\red{Answer}$

⇝Let us assume √3 is a Rational Number

⇝Now,

$\star√3 \: can \: be \: represented \: as \: \frac{a}{b}$

In which,

⇝a and b are Prime integers and there is no common factor between a and b

$Hence \: √3 = \frac{a}{b}$

$\sqrt{3} \: b = a$

⇝Now, squaring on both side

$( \sqrt{3} b) {}^{2} = {a}^{2}$

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

$\frac{ {a}^{2} }{3} = {b}^{2}$

$Hence \: \frac{3}{ {a}^{2} }$

[∵ 3 divides 3b square ]

3 divides a ........ eqn (1)

a = 3 c for some integer c

${a}^{2} = {9c}^{2}$

${3b}^{2} = {9c}^{2}$

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

$\frac{3}{ {b}^{2} }$

$\frac{3}{b} \: eqn \: (2)$

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 an irrational number.