Show that there is no positive integer n for which \[\sqrt{n-1}+\sqrt{n+1}\]is rational.

Hint: Consider that the given expression is equal to \[{}^{p}/{}_{q}\]. Take the inverse of the expression and simplify it. By adding these 2 expressions formed you will get that (n+1) is a perfect square. By subtracting these expressions, you will get that \[\sqrt{(n-1)}\] is also a perfect square.

Complete step-by-step answer:

Let there be a positive integer n for which \[\sqrt{n-1}+\sqrt{n+1}\] is a rational number.
Let us consider that \[\sqrt{n-1}+\sqrt{n+1}=\dfrac{p}{q}......(1)\]

where p and q are integers and \[q\ne 0.\]

\[\sqrt{n-1}+\sqrt{n+1}=\dfrac{p}{q}\]

If we take inverse of the above expression,

\[\dfrac{1}{\sqrt{n-1}+\sqrt{n+1}}=\dfrac{q}{p}\]

Now we need to rationalize it by multiplying \[\left( \sqrt{n-1}+\sqrt{n+1} \right)\] in the

numerator and denominator of LHS.

\[\dfrac{\sqrt{n-1}-\sqrt{n+1}}{\left( \sqrt{n-1}+\sqrt{n+1} \right)\left( \sqrt{n-1}-\sqrt{n+1}

\right)}=\dfrac{q}{p}.........(2)\]

We know \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}.\]

\[\therefore \left( \sqrt{n-1}+\sqrt{n+1} \right)\left( \sqrt{n-1}-\sqrt{n+1} \right)={{\left(

 \sqrt{n-1} \right)}^{2}}-{{\left( \sqrt{n+1} \right)}^{2}}=n-1-(n+1)=n-1-n-1=-2.\]

Thus substitute the value of \[\left( \sqrt{n-1}+\sqrt{n+1} \right)\left( \sqrt{n-1}-\sqrt{n+1}

\right)\] in equation (2).

\[\begin{align}
  & \therefore \dfrac{\sqrt{n-1}-\sqrt{n+1}}{-2}=\dfrac{q}{p} \\

 & \Rightarrow \dfrac{-\left( \sqrt{n-1}+\sqrt{n+1} \right)}{-2}=\dfrac{q}{p} \\

 & \dfrac{\sqrt{n+1}-\sqrt{n-1}}{2}=\dfrac{q}{p}........(3) \\

\end{align}\]

Now lets us add equation (1) and (3).

\[\begin{align}

  & \left( \sqrt{n-1}+\sqrt{n+1} \right)+\left( \sqrt{n+1}-\sqrt{n-1}

\right)=\dfrac{p}{q}+\dfrac{2q}{p} \\

 & \Rightarrow 2\sqrt{n+1}=\dfrac{{{p}^{2}}+2{{q}^{2}}}{pq}. \\

\end{align}\]

Cancelling out the like terms and simplify it, we get the expression,

\[\begin{align}

  & 2\sqrt{n+1}=\dfrac{{{p}^{2}}+2{{q}^{2}}}{pq} \\

 & \therefore \sqrt{n+1}=\dfrac{{{p}^{2}}+2{{q}^{2}}}{2pq}......(4) \\

\end{align}\]

(n+1) is a perfect square of positive integers.

Again let us subtract equation (3) from (1).

\[\begin{align}

  & \sqrt{n-1}+\sqrt{n+1}-\sqrt{n+1}+\sqrt{n-1}=\dfrac{p}{q}-\dfrac{2q}{p} \\

 & \Rightarrow 2\sqrt{n-2}=\dfrac{{{p}^{2}}-2{{q}^{2}}}{pq}.......(5) \\

\end{align}\]

Cancelling out like terms and simplifying it, we get the above expression.

\[\sqrt{n-1}\] is a rational number, and \[\dfrac{{{p}^{2}}-2{{q}^{2}}}{2pq}\] is a rational number.

\[\because \sqrt{n-1}\] is also a perfect square of positive integers.

Thus we can say that \[\sqrt{n+1}\] and \[\sqrt{n-1}\] are perfect squares of positive integers.

It contradicts the fact that two perfect squares differ by 3.

Thus there is no positive integer n for which \[\sqrt{n-1}+\sqrt{n+1}\] is rational.

Note: It is important to consider the expression equal to \[{}^{p}/{}_{q}\], where p and q are integers and \[q\ne 0\]. Take their inverse to formulate another expression. Be careful while rationalizing so that you don’t mix up the signs.