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Problem involving irrationals

  1. Jan 16, 2008 #1
    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?
  2. jcsd
  3. Jan 16, 2008 #2
    Since [tex]y=\sqrt{x^2-1}[/tex] is the upper half of a hyperbola with foci on the x-axis, for positive values of x, it approaches the line [tex]y=x[/tex].

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

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

    Last edited: Jan 16, 2008
  4. Jan 16, 2008 #3
    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.
  5. Jan 16, 2008 #4
    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. =)
  6. Jan 16, 2008 #5


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    Homework Helper

    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
  7. Jan 16, 2008 #6
    Doesn't my post just above yours take care of it?
  8. Jan 16, 2008 #7


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    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.
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