Yahoo Answers: Answers and Comments for Prove that sqrt(n  1) + sqrt(n + 1) is irrational for every potitive integer n...? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
From Anonymous
enSG
Wed, 21 Jan 2009 15:03:19 +0000
3
Yahoo Answers: Answers and Comments for Prove that sqrt(n  1) + sqrt(n + 1) is irrational for every potitive integer n...? [Mathematics]
292
38
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
https://s.yimg.com/zz/combo?images/emaillogosg.png

From Anonymous: If
q = √(n+1) + √(n1)
is rational, then
r...
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
Wed, 21 Jan 2009 15:26:00 +0000
If
q = √(n+1) + √(n1)
is rational, then
r = 2/q = √(n+1)  √(n1)
is also rational.
Then
q²  r² = 4 √(n+1) √(n1) = 4 √(n²1)
must be also rational.
But (n²1) cannot be a perfect square, except when n = 1.
And in case n=1, we have q = √(n+1) + √(n1) = √2.

From JayJay: Let our number be N = √(n  1) + √(n + 1) for ...
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
Wed, 21 Jan 2009 15:35:25 +0000
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.

From galaz: it rather is how i could do all of it of us co...
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
https://sg.answers.yahoo.com/question/index?qid=20090121150319AAHn973
Tue, 13 Dec 2016 10:47:48 +0000
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(na 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..