Proving the Irrationality of Square Roots of Non-Perfect Squares

[end3r7]

Homework Statement

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

Homework Equations

\sqrt{n - 1} + \sqrt{n + 1}

The Attempt at a Solution

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

Then 2n + \sqrt{n^2 -1} = \frac{p^2}{q^2}

So "all" I have to do is to show that \sqrt{n^2 - 1} 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
\exists p,q \in Z s.t. \sqrt{n^2 - 1} = \frac{p}{q} and gcd(p,q) = 1.Then n^2 -1 = \frac{p^2}{q^2} or n^2 = \frac{p^2 + q^2}{q^2}

Any rational solution is of the form \frac{m}{n}, where m divides \frac{p^2 + q^2}{q^2}. 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?

 

[foxjwill]

Since y=\sqrt{x^2-1} is the upper half of a hyperbola with foci on the x-axis, for positive values of x, it approaches the line y=x.

Also, since \sqrt{x^2-1} is monotonically increasing for positive x, \forall x \in \mathbb{N}, 0<x-\sqrt{x^2-1}<1.

Therefore, \forall x \in \mathbb{N}, \sqrt{x^2-1} \notin \mathbb{Q}.

Q.E.D.

 

[end3r7]

That's a neat way to do it, but I was really wondering about the methods I proposed. I'd like to see if my reasoning is correct.

 

[end3r7]

Actually I think I might know how to prove that any square of a composite nonperfect is an irrational.

let's x = (p1^n1)(p2^n2)...(pN^nN), where pi's are prime numbers.
In order for x to not be a nonperfect square then at least one ni is odd.

for each p, if the corresponding n>=2, then we can factor that p out of the square root until we are just left with p^1 power inside the square root. Therefore the square root only has a product of primes inside it.

To show that the square root of a product of primes is irrational is trivial by a proof by contradiction.

I think I'll proceed with that.

But again, thanks for helping foxjwill, your really was very prompt. =)

 

[Dick]

I don't think either one of them are particularly valid. If sqrt(n)=p/q with p and q relatively prime, then q^2*n=p^2. If r is any prime divisor of n, then r must divide p^2. If r divides p^2 then r divides p, so r^2 divides p^2. r^2 doesn't divide q^2 since p and q are relatively prime. If r^2 divides p^2 then it must divide ___. Fill in the blank. Can you take it from there?

 

[end3r7]

I don't think either one of them are particularly valid. If sqrt(n)=p/q with p and q relatively prime, then q^2*n=p^2. If r is any prime divisor of n, then r must divide p^2. If r divides p^2 then r divides p, so r^2 divides p^2. r^2 doesn't divide q^2 since p and q are relatively prime. If r^2 divides p^2 then it must divide ___. Fill in the blank. Can you take it from there?

Doesn't my post just above yours take care of it?

 

[Dick]

Doesn't my post just above yours take care of it?

Sure. Once you prove the square root of a prime is irrational and that the square root of two or more different primes is also irrational.