# Problem involving irrationals

1. Jan 16, 2008

### end3r7

1. The problem statement, all variables and given/known data
Prove that $$\sqrt{n - 1} + \sqrt{n + 1}$$ is irrational for every positive integer n.

2. Relevant equations
$$\sqrt{n - 1} + \sqrt{n + 1}$$

3. 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?

2. Jan 16, 2008

### 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.

Last edited: Jan 16, 2008
3. Jan 16, 2008

### 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.

4. Jan 16, 2008

### 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. =)

5. Jan 16, 2008

### 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?

Last edited: Jan 16, 2008
6. Jan 16, 2008

### end3r7

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

7. Jan 16, 2008

### Dick

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.