Question

Prove
$prove \sqrt{2 \:} \: is \: irrational \: number$
​

Answer 1

$\mathfrak{\huge{\blue{\underline{\underline{★ Aditya ★}}}}}$

Let √2 be a rational number

Therefore, √2= p/q [ p and q are in their least 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.

Answer 2

Answer:

Let √2 be a rational number

Therefore, √2= p/q [ p and q are in their least 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

Hence √2 is not a rational number

i.e., irrational number.