Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Prove that sqrt(n-1) + sqrt(n+1) is irrational for every positive integer n.?

2 Answers

Relevance
  • kb
    Lv 7
    1 decade ago
    Favorite Answer

    If n = 1, then sqrt(n-1) + sqrt(n+1) = sqrt(2), which is easily shown irrational. (I'll leave that to you!)

    Suppose that n > 1.

    Suppose that sqrt(n-1) + sqrt(n+1) is rational.

    Then its square [sqrt(n-1) + sqrt(n+1)]^2 = 2n + 2 * sqrt(n^2 -1) is also rational.

    Next, since 2 and n are rational, by the closure laws of Q, we have that

    sqrt(n^2 - 1) is rational. This proof will be complete if we can prove the following fact.

    Claim: sqrt(n^2 - 1) is irrational.

    This follows from the claim that consecutive squares are spaced more than 1 unit apart as long as n^2 > 1. [(n+1)^2 - n^2 = 2n + 1.]

    More precisely, since (n - 1)^2 < n^2 - 1< n^2 for all integers n > 1, taking square roots shows that sqrt(n^2 - 1) is between two consecutive perfect squares.

    I hope that helps!

  • 1 decade ago

    This is how I'd do it

    We know that sqrt(n) is rational if and only if n is a perfect square.

    We assume X = sqrt(n-1) + sqrt(n+1) is rational. Hence X^2 is also rational. X^2 = n - 1 + n + 1 - 2sqrt(n^2 - 1). Therefore sqrt(n^2 - 1) must also be rational. But that means n^2 - 1 must be a perfect square. n^2 is already a perfect square, so this is clearly impossible.

    You can also show that n > sqrt(n^2 - 1) > n - 1, meaning no sqrt(n^2 - 1) is not an integer, hence n^2 - 1 is not perfect square.

Still have questions? Get your answers by asking now.