Question

show that there is no positive integer n for which √n-1+√n+1 is rational​

Answer 1

Let √(n - 1) + √(n + 1) be a rational number p/q.

$\sqrt{n - 1} + \sqrt{n + 1} = \frac{a}{b} ...(1)$

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On doing reciprocal of equation

$\frac{1}{ \sqrt{n - 1} + \sqrt{n + 1} } = \frac{b}{a}$

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On rationalisation we get,

$\frac{ \sqrt{n - 1} - \sqrt{n + 1} }{ { \sqrt{(n - 1)} }^{2} - \sqrt{ {(n + 1)}^{2} } } = \frac{b}{a} \\ \\ \frac{ \sqrt{n - 1} - \sqrt{n + 1} }{n - 1 - n - 1} = \frac{b}{a} \\ \\ \sqrt{n - 1} - \sqrt{n + 1} = \frac{ - 2b}{a} ..(2)$

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On adding both equation be get,

• 2√(n - 1) = - 2b/a + a/b

• √(n - 1) = - b/a + a/2b

This means that n - 1 is a perfect square.

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On subtrafting hoth equations we get,

• 2√(n + 1) = a/b + 2b/a

• √(n + 1) = a/2b + b/a

This means that n + 1 is also a perfect square.

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But smallest perfect squares i.e., 4 and 1 differ by 3.

While n + 1 and n - 1 differ by 2.

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Therefore due to this contradiction they aren't perfect squares.

Hence √(n - 1) + √(n + 1) is an irrational number.

Q.E.D