Question

Prove that √2 is an irrational number ?

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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

Answer:

Let us assume on the contrary that 2 is a rational number. Then, there exist positive integers a and b such that

2=ba where, a and b, are co-prime i.e. their HCF is 1

⇒(2)2=(ba)2 

⇒2=b2a2 

⇒2b2=a2 

⇒2∣a2[∵2∣2b2 and 2b2=a2] 

⇒2∣a...(i) 

⇒a=2c for some integer c

⇒a2=4c2 

⇒2b2=4c2[∵2b2=a2] 

⇒b2=2c2 

⇒2∣b2[∵2∣2c2] 

⇒2∣b...(ii)

From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.

Hence, 2 is an irrational number.

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