Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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

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• JayJay
Lv 5

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.

• Anonymous

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.