Question

root 2 is an irrational no. prove it.

Answer 1

let root 2 be rational no
root 2=p/q
take square both sides
(root 2)^2 = (p/q)^2
2= p^2/q^2
2q^2=p^2
so p^2is divisible by 2 and also pis divisible by 2
now let p=2k
then
2q^2=2k^2
2q^2= 4k^2
divide 4by 2
q^2= 2k^2
now we can say q^2 is divisible by 2 and q is also divisible by 2
here HCF is not equal to 1
so it makes contradiction that root2 is rational no. thus it is irrational no

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.