Question

Show that there is no positive integer n so that root(n-1)+root(n+1) is rational

Answer 1

p/q = $\sqrt{n-1} + \sqrt{n+1}$                                               ------ 1

q/p=1/$\sqrt{n-1} + \sqrt{n+1}$  = $\sqrt{n-1} + \sqrt{n+1}$/-2

2q/p= $\sqrt{n-1} - \sqrt{n+1}$                                               ------- 2

adding 1 and 2 

$2\sqrt{n+1} = \frac{p}{q}+ \frac{2q}{p} = p^{2} + 2 \frac{ q^{2} }{pq}$

\sqrt{n+1} [/tex] = p/q + 2q/p = p$x^{2}$ + $2 q^{2}$ /2pq --3

subtracting 1 from 2
 
[tex] \sqrt{n-1} [/tex] = $\frac{p}{q} - 2 \frac{q}{p} = \frac{ p^{2} +2q^{2} }{2pq}$  ------- 4

from 3 and 4

$\sqrt{n+1}$ and $\sqrt{n-1}$ are rational numbers as $\frac{ p^{2} + 2 q^{2}}{2pq}$ and  $\frac{ p^{2} - 2q^{2} }{2pq}$

are rational for integer p and q .

here n+1 and n-1 are perfect square of positive integer.

now,(n+1)-(n-1)=2 which is not possible since any perfect square differ by
 
at least 3.

thus there is no positive integer n which $\sqrt{n+1}$ and $\sqrt{n-1}$ is rational. 

Answer 2

Question:

Show that there is no positive integer 'n' such that $\sf \sqrt{n - 1} + \sf \sqrt{n + 1}$ is rational.

Solution:

We'll assume that there is a positive integer 'n' such that $\sf \sqrt{n - 1} + \sf \sqrt{n + 1}$ is a rational number.

Therefore, $\sf \sqrt{n - 1} + \sf \sqrt{n + 1}$ can be expressed in the p/q form, where p and q are integers (positive) and q ≠ 0.

$\Longrightarrow \sf \ \sqrt{n - 1} + \sf \sqrt{n + 1} = \dfrac{p}{q}$

Taking the reciprocal we get,

$\Longrightarrow \sf \ \dfrac{1}{\sqrt{n - 1} + \sf \sqrt{n + 1}} = \dfrac{q}{p}$

Rationalizing the denominator;

$\Longrightarrow \sf \ \dfrac{1}{\sqrt{n - 1} + \sf \sqrt{n + 1}} \ \dfrac{\sqrt{n - 1} - \sf \sqrt{n + 1}}{\sqrt{n - 1} - \sf \sqrt{n + 1}} = \dfrac{q}{p}$

$\Longrightarrow \sf \ \dfrac{\sqrt{n-1} - \sf \sqrt{n+1}}{(\sqrt{n-1}+ \sf \sqrt{n+1})(\sqrt{n-1} - \sf \sqrt{n+1})} = \dfrac{q}{p}$

Using (a - b) (a + b) = a² - b²

$\Longrightarrow \sf \ \dfrac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n-1})^2-(\sqrt{n+1})^2} = \dfrac{q}{p}$

$\Longrightarrow \sf \ \dfrac{\sqrt{n-1}-\sqrt{n+1}}{ \ (n-1)-(n+1)} = \dfrac{q}{p}$

$\Longrightarrow \sf \ \dfrac{\sqrt{n-1}-\sqrt{n+1}}{n-1-n-1} = \dfrac{q}{p}$

$\Longrightarrow \sf \ \dfrac{\sqrt{n-1}-\sqrt{n+1}}{-2} = \dfrac{q}{p}$

$\Longrightarrow \sf \sqrt{n-1}-\sqrt{n+1} = \dfrac{-2q}{p}$

$\Longrightarrow \sf -(-\sqrt{n-1}+\sqrt{n+1}) = -\left(\dfrac{2q}{p}\right)$

$\Longrightarrow \sf \sqrt{n+1}-\sqrt{n-1} = \dfrac{2q}{p} \ \ \sf {\longmapsto \textcircled{\scriptsize1}}$

We also know that,

$\Longrightarrow \sf \ \sqrt{n + 1} + \sqrt{n - 1} = \dfrac{p}{q} \ \ \sf {\longmapsto \textcircled{\scriptsize2}}$

Adding equations 1 and 2 we get.

$\sf \sqrt{n+1}-\sqrt{n-1} = \dfrac{2q}{p} \\ \\ \\ \ \sqrt{n + 1} + \sqrt{n - 1} = \ \dfrac{p}{q}\\ \\\rule{109}{1}$

$\sf 2\sqrt{n+1} = \dfrac{2q}{p} + \dfrac{p}{q}\\ \\ \\ \sf 2\sqrt{n+1} = \dfrac{2q^2 + p^2}{pq} \ \ \longmapsto \sf \textcircled{\scriptsize3}$

Subtracting Equations 2 by 1 we get,

$\sf \sqrt{n+1}-\sqrt{n-1} = \dfrac{2q}{p} \\ \\ \ _{(-)} \ \ \ \ \ \ _{(-)} \ \ \ \ \ \ \ \ \ _{(-)} \\ \\ \ \sqrt{n + 1} + \sqrt{n - 1} = \ \dfrac{p}{q}\\ \\\rule{109}{1}$

$\sf -2\sqrt{n-1} = \dfrac{2q}{p} - \dfrac{p}{q}\\ \\ \\ \sf -2\sqrt{n-1} = \dfrac{2q^2 - p^2}{pq} \ \\ \\ \\ \sf -(2\sqrt{n-1}) = -\left(\dfrac{-2q^2+ p^2}{pq} \right)\ \\ \\ \\ 2\sqrt{n-1} = \dfrac{p^2 - 2q^2}{pq} \ \ \ \longmapsto \textcircled{\scriptsize4}$

From 3 and 4 both $\sf \sqrt{n - 1}$ and $\sf \sqrt{n + 1}$ are rational numbers.

$\sf This \ is \ because \ both \ \dfrac{p^2 + 2q^2}{pq} \ \ and \ \ \dfrac{p^2 - 2q^2}{pq} \ are \ rational.$

Since $\sf \sqrt{n - 1} + \sf \sqrt{n + 1}$ are rational, we can call it a perfect square having rational square roots.

But On subtracting n - 1 from n + 1 we get,

= n + 1 - (n - 1)

= n - 1 - n + 1

= 2

But perfect squares don't differ by 2, but differ atleast by 3.

(Example: 2² - 1² = 3)

Therefore there is no positive integer 'n' so that $\sf \sqrt{n - 1} + \sf \sqrt{n + 1}$ is rational.