Question

Show that there is no positive integer 'n' for which $\tt{( \sqrt{n - 1} + \sqrt{n + 1} )}$ is rational.

$\bf{Class\:X}$
$\bf{No\:spam\:answers\:please!}$​

Answer 1

For √(n-1) to be rational

n-1 = t^2 (t is an integer)

n= t^2 + 1

Now for √n+1 to be rational

n + 1 = L^2 ( L is integer)

t^2 +1 +1 = L^2

t^2 +2 = L^2

2 = L^2 - t^2

2 = (L-T)( L+T)

Now L And T both are integers

So L-T and L+T are also integers

If L- T = 1 or 2

then L+T can't be 2 or 1

SO IT'S ALWAYS IRRATIONAL

Answer 2

$\mathfrak{\huge{Hi!}}$

Let's assume $\sf{( \sqrt{n - 1} + \sqrt{n + 1} )}$ to be a rational number. Thus :-

$\sf{(\sqrt{n - 1} + \sqrt{n + 1} )= \frac{p}{q}}$, where p and q are co-primes and q ><0 ...(1)

Do the reciprocal of this equation :

=》 $\sf{\frac{1}{( \sqrt{n - 1} + \sqrt{n + 1} )} = \frac{q}{p}}$

Rationalise the Left hand side of this equation :

=》 $\sf{\tiny{\frac{1}{ \sqrt{n + 1} + \sqrt{n - 1} } \times \frac{\sqrt{n + 1} - \sqrt{n - 1}}{\sqrt{ n + 1} - \sqrt{n -1}} = \frac{q}{p}}}$

=》 $\sf{\frac{\sqrt{n + 1} - \sqrt{ n - 1}}{n + 1 - n + 1} = \frac{q}{p}}$

=》 $\sf{( \sqrt{n + 1} - \sqrt{n - 1} ) = \frac{2q}{p}}$ ...(2)

Adding and subtracting (1) and (2) :

Adding first :

=》 $\sf{\tiny{\sqrt{n - 1} + \sqrt{n + 1} + \sqrt{n + 1} - \sqrt{n - 1}= \frac{p}{q} + \frac{2q}{p}}}$

=》 $\sf{ \sqrt{n + 1} = \frac{p^{2} + 2q^{2}}{qp}}$

Subtracting now :

=》 $\sf{\tiny{\sqrt{n - 1} + \sqrt{n + 1} - \sqrt{n + 1} + \sqrt{n - 1}= \frac{p}{q} - \frac{2q}{p}}}$

=》 $\sf{ \sqrt{n - 1} = \frac{p^{2} - 2q^{2}}{qp}}$

$\tt{\therefore Both \: \sqrt{n + 1}\: and \: \sqrt{n - 1}\: are\:rationals.}$

$\tt{\therefore Both\: (n + 1)\: and \: (n - 1)\: should \: be \: perfect\: squares.}$

Difference between ( n + 1 ) and ( n - 1 ) = n + 1 - n + 1 = 2

$\tt{\therefore These\:numbers\:aren't \:perfect\:squares.}$ This is because for any two perfect squares, the minimum difference is 3.

$\tt{\therefore These \:numbers\:aren't \:rationals.}$

$\bf{\huge{Hence\:Proved!}}$