Question

prove that root 2 is irrational step by step.

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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.

Answer 2

Suppose that root 2is a rational number.Let root 2=m/n, where m and n are positive integers with no common factor greater than 1.Then, we have m square = 2n square, which implies that m square is even and hence m is even.

Let m=2k. then, we have 2n square=4k square which gives n square = 2k square.

Thus, n is also even.

It follows, that m and n are even numbers having a common factor 2.

Thus we arrived at a contradiction.

Hence, root 2 is irrational.

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