Question

prove that √2 is irrational??

Answer 1

This created problem....
We earlier considered that p and q we're coming primes, but they have a common factor 2 , So this proves that our consideration was wrong and √2 is a irrational no.

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Answer 2

Answer :

To prove,

√2 is an irrational no.

Proof :

Let √2 be a rational number in the form of p / q where q is not equal to zero at p and q are co-prime integers.

√2 = p/q

Whole sqauring both sides of this equation :-

2 = p^2/q^2

p^2 = 2q^2 (I)

From (I),

2 divided p^2

So, p divides p. (a)

Now , let p= 2k where k is any integer.

Substituting the values , we get :-

(2k)^2 = 2q^2

4k^2 = 2q^2

q^2 = 2k^2 (ii)

From (ii),

2 divides q^2.

Therefore, 2 divides q also. (b)

From statements (a) and (b) , we can say that :-

p and q have a common factor namely 2.

Hence, our assumption that p and q are co-prime is wrong. Hence , √2 is an irrational no.

Hence proved.

This method is called contradiction method.