Question

Prove√2 is irrational number.

Answer 1

Answer:

Let √2 be a rational number.

√2 = a/b (where a and b are co prime integers & b is not equal to 0)

squaring both sides

(√2)² = (a/b)²

=> 2 = a²/b²

=> 2b² = a²

=> a² is divisible by 2

=> a is divisible by 2

a = 2c (where c is an integer)

squaring both sides

a² = (2c)²

=> a² = 4c²

also, a² = 2b²

=> 2b² = 4c²

=> b² = 2c²

=> b² is divisible by 2

=> b is divisible by 2

Which is a contradiction as a & b are co prime integers.

=> Our assumption is wrong.

=> √2 is an irrational number.

HOPE IT HELPS U !!

Answer 2

Step-by-step explanation:

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.