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Solution:

If possible, suppose √n-1 +√n+1 is rational for some positive integer equal to a/b where a and b are positive integers.

Then, a/b = √n-1 + √n+1 ------(i)

=> b/a= 1/(√n+1 +√n-1)

=> b/a = (√n+1 -√n-1)/(√n+1 +√n-1)(√n+1 - √n-1)

=> b/a = (√n+1 - √n-1)/2

=> 2b/a = √n+1 - √n-1 -------(¡¡)

Adding eq(¡) and (¡¡) and Subtracting eq(ii) from (¡), we get,

2√n+1 = a/b +2b/a and 2√n-1 = a/b - 2b/a

=> √n+1 = a²+2b²/2ab and √n-1 = a²-2b²/2ab

since, RHS = a²+2b²/2ab and a²-2b²/2ab are rational as a, b are integers.

Then, LHS = √n+1 and √n-1 are also rationals.

=> (n+1) and(n-1) are perfect squares of positive integers.

But it is not possible because every positive perfect square differs at least by 3.

So, our supposition is wrong.

Hence, there is no positive integer n for which √n-1 + √n+1 is rational.

---------------------------xProvedx--------------------------------

Hope this helped you...