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I do not understand how to do this? I have tried to prove it by contradiction and I proved, assuming $\sqrt{n+1}+\sqrt{n-1} = \frac{a}{b}$, that $2n + \sqrt{n^2 - 1}$ divides $a$ and $b$ and so it is not rational but that does not prove it (both $a$ and $b$ are co-primes and $b$ is not equal to $0$)

there are proofs on the internet but I do not understand why they did what they did in the very first step: taking the reciprocal

Potato
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    It might help to link some of the proofs to explain it to you. – Cameron Williams Dec 19 '21 at 23:12
  • https://www.toppr.com/ask/question/show-that-there-is-no-positive-integer-n-for-which-sqrt-n1sqrt-n1-is-rational/ here is one but all the others follow the same method – Potato Dec 19 '21 at 23:15
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    They're multiplying by the conjugate. Look up that technique with. You will want to specify square roots most likely. – Cameron Williams Dec 19 '21 at 23:17
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    They're not reciprocating, they're rationalising. Essentially it boils down to the fact that$$(\sqrt{n + 1} + \sqrt{n - 1})(\sqrt{n + 1} - \sqrt{n - 1}) = 2 \in \Bbb{Q}.$$If one of the factors is rational, then both are rational, and if both are rational, then $\sqrt{n + 1}$ and $\sqrt{n - 1}$ are both rational. – Theo Bendit Dec 19 '21 at 23:18

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If $n = 1 \implies a =\sqrt{2} $ is irrational. If $n \ge 2$, then if $a = \sqrt{n+1} + \sqrt{n-1}$ is a rational number, then $a^2 = 2n+2\sqrt{n^2-1}$ is also a rational number. But this shows that $\sqrt{n^2-1}$ is a rational as well. And because $n$ is an integer, this shows that $n^2-1$ must be a perfect square, and this is not possible. Thus there is no $n$ such that $\sqrt{n+1}+\sqrt{n-1}$ is a rational number.

Wang YeFei
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  • Wait, doesn't it only need to be the square of a rational number? Not necessarily perfect square? – user_194421 Dec 19 '21 at 23:08
  • wait but doesn't that prove that the expression itself is irrational despite n being an integer or rational or irrational – Potato Dec 19 '21 at 23:15
  • Well, if $n=1$ then $n^2 - 1 = 0^2$ is a perfect square; so for that case, you need to go back to $a = \sqrt{2}$ being irrational. – Daniel Schepler Dec 19 '21 at 23:26
  • @user_194421 Let $\sqrt{n^2-1}=\frac pq$, with $q>1$ and $\gcd(p,q)=1$. Then $n^2=\frac{p^2+q^2}{q^2}$. But $\gcd(p^2+q^2,q^2)=\gcd(p^2,q^2)=1$, so $n^2$ isn't an integer. – PM 2Ring Dec 21 '21 at 04:20