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end3r7
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Homework Statement
Prove that [tex]\sqrt{n - 1} + \sqrt{n + 1}[/tex] is irrational for every positive integer n.
Homework Equations
[tex]\sqrt{n - 1} + \sqrt{n + 1}[/tex]
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?