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Let us assume that √2 is not an irrational
√2 = p/q
squaring on both sides
2 = p²/q²
2q² = p²  --------- (1)
q² = p²/2
2 divides p² ⇒ 2 divides p ------- (2)
let p = 2m ⇒p² = 4m²
sub p² = 4m² in ----(1)
2q² = 4m²
 q² = 2m²
m² = q²/2
2 divides q² ⇒ 2 divides q ------------ (3)
from (2) and (3) we can say that p and q had an other common factor other than 1. it is not true because p and q are co primes and doesnt have any common factor.
it is contradiction to our assumption.
√2 is an irrational number.
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ok its your wish but u have to give a correct answer for the given question
of course dere r mny methods na.. i done vth one of dose methods.
where u r from
If the number which is in the form of p/q ,where q is not equal to zero and also p and q must be an integer.but √2 is not an integer , even though √2 can be written in the form of p/q as √2/1.so √2 is an irrational number.
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