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Answer 1

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Answer 2

Question 3.

By the method of contradiction..

Let √11 be rational , then there should exist √11=p/q ,where p & q are coprime and q≠0(by the definition of rational number). So,

√11= p/q

On squaring both side, we get,

11= p²/q² or,

11q² = p². …………….eqñ (i)

Since , 11q² = p² so ,11 divides p² & 11 divides p

Let 11 divides p for some integer c ,

so ,

p= 11c

On putting this value in eqñ(i) we get,

11q²= 121p²

or, q²= 11p²

So, 11 divides q² for p²

Therefore 11 divides q.

So we get 11 as a common factor of p & q but we assumpt that p & q are coprime so it contradicts our statement. Our supposition is wrong and √11 is irrational.

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