Question

Show that there is no positive integer n for which root of n-1 + root of n+1 is rational .

Answer 1

So 
sqrt(n+1)+sqrt(n-1)=p/q.........(1)

Inverting (1) we get

1/(sqrt(n+1)+sqrt(n-1))=q/p

=>(sqrt(n+1)-sqrt(n-1))/((sqrt(n+1)+sqrt(n-1))(sqrt(n+1)-sqrt(n-1)))=q/p

=>(sqrt(n+1)-sqrt(n-1))/2=q/p

=>(sqrt(n+1)-sqrt(n-1))=(2q)/p.....(2)

Adding (1) and (2) we get

2sqrt(n+1)=p/q+(2q)/p

=>sqrt(n+1)=(p^2+2q^2)/(2pq).....(3)

Similarly subtracting (2) from (1) we get

=>sqrt(n-1)=(p^2-2q^2)/(2pq).....(4)

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

that both" "sqrt(n+1) and sqrt(n-1)

color (blue)(" 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

sqrt(n+1)+sqrt(n-1)" is rational"

Helpful??

Answer 2

Answer:

REFER TO THE ATTACHED FILE