Question

How to prove under root 2 is irrational number

Answer 1

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

    Therefore, Our supposition is wrong

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

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Answer 2

Qᴜᴇsᴛɪᴏɴ :-

Prove that √2 is an irrational number

Sᴏʟᴜᴛɪᴏɴ :-

Let us assume to the contrary that √2 is rational.

⛬ √2 = a/b (a & B are co-primes, b ≠ 0)

Squaring on both sides, we get

⟾ (√2)² = (a/b)²

⟾ 2 = a²/b²

⟾ 2b² = a²

So, 2 divides a², Then 2 divides a

⛬ 2c = a

Putting the value of a in eq1

⟾ 2b² = a²

⟾ 2b² = (2c)²

⟾ 2b² = 4c²

⟾ b² = 2c²

So, 2 divides b², Then 2 divides b

⛬ a and b have at least 2 as a common factor

This contradicts the fact that a and b have no common factor

This contradiction has arisen because of our own assumption.

So, √2 is an irrational number