Question

7. Prove that
$\sqrt{n} \: is \: irrational \: if \: \: n \: is \: not \: a \: \\ perfect \: square.$
Standard:- 10

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Answer 1

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By the method of Contradiction,

let √n be a rational number

Then √n = p/q where p and q not equal to 0 and HCF(p,q) = 1

=> √n = p/q

=> p = √n(q)

Square both sides

=> p² = nq² -------(1)

=> p² is divisible by n

=> p is also divisible by n

Let p = nm

Square both sides

=> p² = n²m²

From (1)

=> nq² = n²m²

=> q² = nm²

=> q² is divisible by n

=> q is also divisible by n

Here HCF(p,q) = n but it should be 1

This is contradictory to our assumption that √n is rational

Thus √n is not rational

=> Thus √n is an Irrational number

HENCE PROVED

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