Prove that √2 is irrational with steps
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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.
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