(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that [tex]\sqrt{n - 1} + \sqrt{n + 1}[/tex] is irrational for every positive integer n.

2. Relevant equations

[tex]\sqrt{n - 1} + \sqrt{n + 1}[/tex]

3. The attempt at a solution

[tex]\exists p,q \in Z[/tex] s.t. [tex]\sqrt{n - 1} + \sqrt{n + 1} = \frac{p}{q}[/tex]

Then [tex]2n + \sqrt{n^2 -1} = \frac{p^2}{q^2}[/tex]

So "all" I have to do is to show that [tex]\sqrt{n^2 - 1}[/tex] is irrational.

What's the easiest way? I could show that the quantity under the square root is not a perfect square, but since we have not learned that the square root of a non-perfectsquare is irrational, I'd appreciate a proof of that also.

What I did was

[tex]\exists p,q \in Z[/tex] s.t. [tex]\sqrt{n^2 - 1} = \frac{p}{q}[/tex] and gcd(p,q) = 1.

Then [tex]n^2 -1 = \frac{p^2}{q^2}[/tex] or [tex]n^2 = \frac{p^2 + q^2}{q^2}[/tex]

Any rational solution is of the form [tex]\frac{m}{n}[/tex], where m divides [tex]\frac{p^2 + q^2}{q^2}[/tex]. That means m divides both p^2 and q^2, therefore they cannot be relatively prime.

Is that valid?

Or is there a more elegant way?

Also can anybody provide me with the proof that only square roots of perfect squares are rationals?

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# Homework Help: Problem involving irrationals

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