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

### 3 Answers

- JayJayLv 51 decade agoFavourite answer
Let our number be N = √(n - 1) + √(n + 1) for integers n ≥ 1

If a number p / q is rational then its square p² / q² must also be rational. Some irrational numbers square to give rational numbers: √2 / 2 squares to 1 /2, for example. Other irrational numbers square to give irrational numbers: √2 + 1 squares to 3 + 2√2, for example. Since only irrational numbers can square to give irrational numbers, to prove N is irrational it is sufficient to prove N² must be irrational:

N² = [√(n - 1) + √(n + 1)]²

N² = [√(n - 1)]² + 2 * √(n - 1) * √(n + 1) + [√(n + 1)]²

N² = (n - 1) + 2 * √[(n - 1) * (n + 1)] + (n + 1)

N² = 2n + 2 * √(n² - 1)

First, dispose of the trivial case n = 1 where N² = 2. We know N is irrational as N = √(1 - 1) + √(1 + 1) = √2

For integers n ≥ 2, it is clear that N² is rational only if n² - 1 is a perfect square. But n² is itself a perfect square, and so we would need two perfect squares that differ by exactly 1. This is impossible, as the smallest possible difference between perfect squares occurs when the perfect squares are consecutive, and this difference is

(k + 1)² - k² = 2k + 1 for integers k ≥ 1. In other words, the smallest difference between consecutive perfect squares is 2(1) + 1 = 3 ≠ 1 and so as n² is a perfect square, n² - 1 cannot be a perfect square, and thus √(n² - 1) is irrational.

Having shown that N is irrational for n = 1 and N² (and thus N) is irrational for integers n ≥ 2, it follows that N is irrational for all integers n ≥ 1.

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- Anonymous1 decade ago
If

q = √(n+1) + √(n-1)

is rational, then

r = 2/q = √(n+1) - √(n-1)

is also rational.

Then

q² - r² = 4 √(n+1) √(n-1) = 4 √(n²-1)

must be also rational.

But (n²-1) cannot be a perfect square, except when n = 1.

And in case n=1, we have q = √(n+1) + √(n-1) = √2.

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- Anonymous4 years ago
it rather is how i could do all of it of us comprehend that sqrt(n) is rational if and provided that n is a suited sq.. We assume X = sqrt(n-a million) + sqrt(n+a million) is rational. hence X^2 is likewise rational. X^2 = n - a million + n + a million - 2sqrt(n^2 - a million). hence sqrt(n^2 - a million) additionally must be rational. yet which skill n^2 - a million might desire to be a suited sq.. n^2 is already a suited sq., so it rather is obviously impossible. you additionally can tutor that n > sqrt(n^2 - a million) > n - a million, which skill no sqrt(n^2 - a million) isn't an integer, hence n^2 - a million isn't suited sq..

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