Question

1) Prove that √2 is an irrational number.​

Answer 1

$\huge{ \bold{ \boxed { \color {blue}{✿ɑղՏաҽɾ✿}}}}$

To ProvE:

• √2 is irrational number.

SolutioN:

$Let \: \sqrt{2} = \frac{a}{b} \: where \: a, \: b \: are \: integers \: b \: ≠ \: 0 \: we \: also \\ \: suppose \: that \: \frac{a}{b} \: is \: written \: in \: the \: simplest \: form$

$Now \: \sqrt{2} = \frac{a}{b} \\ ⇒ 2 = a2 / b2 \\ ⇒ 2b2 = a2 \\ ∴ 2 b2 \: \: is \: \: divisible \: \: by \: \: 2 \\ ⇒ a2 \: \: is \: \: divisible \: \: by \: \: 2 \\ ⇒ a \: \: is \: \: divisible \: \: by \: \: 2$

$let \: \: a = 2c \\ a2 = 4c2 \\ ⇒ 2b2 = 4c2 \\ ⇒ b2 = 2c2 \\ ∴ \: \: 2c2 \: \: is \: \: divisible \: \: by \: \: 2 \\ ∴ \: \: b2 \: \: is \: \: divisible \: \: by \: \: 2 \\ ∴ \: \: b \: \: is \: \: divisible \: \: by \: \: 2$

$a \: \: are \: \: b \: \: are \: \: divisible \: \: by \: \: 2 \: \: \\ this \: \: contradicts \: \: our \: \: supposition \: \: \\ \: \: that \: \: \frac{a}{b} \: \: is \: \: written \: \: in \: \: the \: \: simplest \: \: form \: \: \\ \\ \\ Hence \: \: our \: \: supposition \: \: is \: \: wrong$

$\huge{\bold{ ∴ √2 \: \: is \: \: a \: \: irrational \: \: number.}}$

Answer 2

Answer:

$\sf \: Let \: \sqrt{2} \: be \: a \: rational \: number$

$\sf \therefore \: \sqrt{2} = \frac{ p}{q}  [ p \: and \: q \: are \: in \: their \: least \\ \sf \: terms \: i.e., HCF \: of \: (p,q)=1 \: and \: q ≠ 0$

On squaring both sides, we get 

                   p²= 2q²                                                                                    ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p²                                                                    [since, 2q²=p²]

⇒ 2 is a factor of p

 Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get 

            p²= 4 m²                                                                                          ...(2)

From (1) and (2), we get 

           2q² = 4m²      ⇒      q²= 2m²

Clearly, 2 is a factor of 2m²

⇒       2 is a factor of q²                                                      [since, q² = 2m²]

⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.