Yahoo Answers: Answers and Comments for Prove that sqrt(n1) + sqrt(n+1) is irrational for every positive integer n.? [Mathematics]
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From Anonymous
enUS
Sun, 06 Sep 2009 13:30:08 +0000
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Yahoo Answers: Answers and Comments for Prove that sqrt(n1) + sqrt(n+1) is irrational for every positive integer n.? [Mathematics]
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From T D: This is how I'd do it
We know that sqrt...
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Sun, 06 Sep 2009 13:44:00 +0000
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(n1) + 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.

From kb: If n = 1, then sqrt(n1) + sqrt(n+1) = sqrt(2)...
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Sun, 06 Sep 2009 13:42:02 +0000
If n = 1, then sqrt(n1) + sqrt(n+1) = sqrt(2), which is easily shown irrational. (I'll leave that to you!)
Suppose that n > 1.
Suppose that sqrt(n1) + sqrt(n+1) is rational.
Then its square [sqrt(n1) + 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!