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

### 2 Answers

- kbLv 71 decade agoFavorite 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!

- Login to reply the answers

- 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.

- Login to reply the answers