Question

Prove that V2 is an irrational number​

Answer 1

Answer:

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

∴ ,Our supposition is wrong

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

Answer 2

To Prove :-

• √2 is an irrational number.

SoluTion :-

Let's assume on the contrary that √2 is a rational number.

Then, there exists two rational numbers a and b

such that √2 = a/b where, a and b are co primes.

(√2)² = (a/b)²

→ 2 = a²/b²

→ 2b² = a²

2 divides a²

So, 2 divides a.

a = 2k , (for some integer)

a² = 4k²

2b² = 4k²

b² = 2k²

2 divides b²

2 divides b

Now, 2 divides both a and b but this contradicts that a and b are co primes.

It happened due to our wrong assumption.

Hence, √2 is irrational.