Question

answer this .....................

Answer 1

Let 
√n+1+√n−1 is rational and can be expressed by pq
where p and q prime to each other and q≠0

So 
√n+1+√n−1=pq.........(1)

Inverting (1) we get

1√n+1+√n−1=qp

⇒√n+1−√n−1(√n+1+√n−1)(√n+1−√n−1)=qp

⇒√n+1−√n−12=qp

⇒(√n+1−√n−1)=2qp.....(2)

Adding (1) and (2) we get

2√n+1=pq+2qp

⇒√n+1=p2+2q22pq.....(3)

Similarly subtracting (2) from (1) we get

⇒√n−1=p2−2q22pq.....(4)

Since p and q are integers then eqution (3) and equation(4) reveal

that both √n+1and√n−1

are rational as their RHS rational

So both (n+1)and(n−1) will be perfect square

Their difference becomes (n+1)−(n−1)=2

But we know any two perfect square differ by at least by 3

Hence it can be inferred that there is no positive integer for which

√n+1+√n−1 is rational

Hope this answer will help you.

Answer 2

This one is not a technical proof, but using common sense and general properties of numbers.

So if √(n+1) +√(n-1) was to be rational, then √(n+1) and √(n-1) should be rational too. For these to be rational, n+1 and n-1 should be square numbers. If so, then the difference between them is 2.
Now, we know that there no square numbers that only differ by two. (Difference between the first square numbers 1 and 4 is 3 and the difference only grows from there). Hence n+1 and n-1 are not square numbers >> √(n+1) and √(n-1) are irrational >> √(n+1) + √(n-1) is irrational.

Hence, proved.