Math, asked by AmulyaA08032005, 11 months ago

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Answered by Nereida
3

\huge\star{\red{\underline{\mathfrak{Answer :-}}}}

Let us assume that for some positive integer n, \sqrt {n - 1}+ \sqrt {n + 1} is rational.

\therefore \sqrt{n - 1}+ \sqrt {n + 1} = \dfrac {p}{q}, where p and q are positive integers and q \neq 0.\\\\

\leadsto{\sqrt {n-1}+\sqrt {n+1}=\dfrac {p}{q}}

\leadsto{({\sqrt {n-1}+\sqrt {n+1}})^{2}=\dfrac{ {p}^{2}}{{q}^{2}}}

Applying identity :- (a+b)^2= a^2 + b^2 + 2ab

\leadsto{n-1+n+1+2 (\sqrt {{n}^{2}-1})=\dfrac{{p}^{2}}{{q}^{2}}}

\leadsto{2n + 2\sqrt {({n}^{2}-1)} = \dfrac {{p}^{2}}{{q}^{2}}}

\leadsto{2 (n+\sqrt {({n}^{2}-1)}) {q}^{2} = {p}^{2}}

Now, we see that p^2 can be divided by 2, then p can be divided by 2.

Let p=2a.

\leadsto{2 (n+\sqrt {({n}^{2}-1)}) {q}^{2} = ({2a})^{2}}

\leadsto{2 (n+\sqrt {({n}^{2}-1)} {q}^{2} = 4{a}^{2}}

\leadsto{{q}^{2} =\dfrac {2\:\:\cancel {4} {k}^{2}}{\cancel {2} (n+\sqrt {({n}^{2}-1)}}}

\leadsto{{q}^{2} =\dfrac {2 {k}^{2}}{ (n+\sqrt {({n}^{2}-1)}}}

Now, q^2 is divisible by 2, then q is also divisible by 2.

Therefore, are pre assumption is contradicted, and so there is no positive integer n, for which \sqrt {n - 1}+ \sqrt {n + 1} is rational.

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