Question

Show that there is no positive integer n for which √n-1 + √n+1 is rational​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar\:{\underline{\pmb{\sf{\purple{\underline{Let's \: show \: the \: given \: question \: says...}}}}}}}$

${\sf{:\implies Given = \sqrt{n-1} + \sqrt{n+1}}}$

We have to show that there is no positive integer ${\sf{n}}$ for which ${\sf{\sqrt{n-1} + \sqrt{n+1}}}$ is a rational number.

~ First let us assume that there is positive integer ${\sf{n}}$ for which ${\sf{\sqrt{n-1} + \sqrt{n+1}}}$ is a rational number.

• Here, p and q are positive integers.

• q is not equal to zero.

Now let us solve this question!

$:\implies \sf \dfrac{q}{p} \: = \dfrac{1}{\sqrt{n-1} + \sqrt{n+1}} \\ \\ :\implies \sf \dfrac{q}{p} \: = \dfrac{\sqrt{n-1} -\sqrt{n+1}}{(\sqrt{n-1} + \sqrt{n+1}) (\sqrt{n-1} - \sqrt{n+1})} \\ \\ :\implies \sf \dfrac{q}{p} \: = \dfrac{\sqrt{n-1} -\sqrt{n+1}}{(n-1) - (n+1)} \\ \\ :\implies \sf \dfrac{q}{p} \: = \dfrac{\sqrt{n-1} -\sqrt{n+1}}{-2} \\ \\ :\implies \sf \dfrac{2q}{p} \: = \sqrt{n+1} - \sqrt{n-1} \\ \\ :\implies \sf (\sqrt{n-1} + \sqrt{n+1}) + \sqrt{n+1} - \sqrt{n-1} \: = \dfrac{p}{q} + \dfrac{2q}{p} \\ \\ :\implies \sf 2\sqrt{n+1} \: = \dfrac{p^2 + 2q^2}{pq} \\ \\ :\implies \sf \sqrt{n+1} \: = \dfrac{p^2 + 2q^2}{2pq} \dots 1 \\ \\ :\implies \sf (\sqrt{n-1} + \sqrt{n+1}) + \sqrt{n+1} - \sqrt{n-1} \: = \dfrac{p}{q} - \dfrac{2q}{p} \\ \\ :\implies \sf 2\sqrt{n-1} \: = \dfrac{p^2 - 2q^2}{pq} \\ \\ :\implies \sf \sqrt{n-1} \: = \dfrac{p^2 - 2q^2}{2pq} \dots 2$

Now from ...1 and ...2 As we assume numbers as rational so henceforth, firstly as

${\sf{:\implies \sqrt{n+1} \: and \sqrt{n-1} \: are \: rational}}$

${\sf{:\implies \dfrac{p^2 - 2q^2}{2pq} \: and \: \dfrac{p^2 + 2q^2}{2pq}}}$

Now as we know that,

${\sf{:\implies (n+1)-(n-1) \: are \: perfect \: sq. \: integers}}$

${\sf{:\implies (n+1)-(n-1) \: = 2 \: is \: not \: possible}}$

It is not possible because it must be squared by atleast 3

${\bigstar\:{\underline{\pmb{\sf{\purple{\underline{Using \: identity...}}}}}}}$

★ x²-y² = (x+y)(x-y)

• Don't be confused that how step sixth is came. It is came due to this identity.

Answer 2

Answer:

Question :-

$\dashrightarrow$ Show that there is no positive integer n for which $\sf\sqrt{n - 1}$ + $\sf\sqrt{n + 1}$ is rational.

Solution :-

Suppose, there exists a number for which $\sf\sqrt{n - 1}$ + $\sf\sqrt{n + 1}$ is rational.

Let,

$\longmapsto \sf\bold{\purple{\sqrt{n - 1} + \sqrt{n + 1} =\: \dfrac{p}{q}\: ------\: (Equation\: No\: 1)}}$

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

$\implies \sf \dfrac{1}{\sqrt{n - 1} + \sqrt{n + 1}} \times \dfrac{\sqrt{n - 1} - \sqrt{n + 1}}{\sqrt{n - 1} - \sqrt{n + 1}} =\: \dfrac{q}{p}$

$\implies \sf \dfrac{\sqrt{n - 1} - \sqrt{n + 1}}{\sqrt{n - 1} + \sqrt{n + 1} \times \sqrt{n - 1} - \sqrt{n + 1}} =\: \dfrac{q}{p}$

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

$\longmapsto \sf\bold{\purple{\sqrt{n + 1} - \sqrt{n - 1} =\: \dfrac{2q}{p}\: ------\: (Equation\: No\: 2)}}$

Now, by adding the equation no 1 and 2 we get,

$\implies \sf 2\sqrt{n + 1} =\: \dfrac{p}{q} + \dfrac{2q}{p}$

$\implies \sf 2\sqrt{n + 1} =\: \dfrac{{p}^{2} + 2{q}^{2}}{pq}$

$\longmapsto \sf\bold{\purple{\sqrt{n + 1} =\: \dfrac{{p}^{2} + 2{q}^{2}}{2pq}\: ------\: (Equation\: No\: 3)}}$

Again, by subtracting the equation no 1 and 2 we get,

$\implies \sf 2\sqrt{n - 1} =\: \dfrac{p}{q} - \dfrac{2q}{p}$

$\implies \sf 2\sqrt{n - 1} =\: \dfrac{{p}^{2} - 2{q}^{2}}{pq}$

$\longmapsto \sf\bold{\purple{\sqrt{n - 1} =\: \dfrac{{p}^{2} - 2{q}^{2}}{2pq}\: ------\: (Equation\: No\: 4)}}$

Now, from equation no 3 and 4 it is clear that √n - 1 and √n + 1 rational .

[ It is possible only if (n - 1) and (n + 1) are perfect square number.

But no two perfect square number differ by 2. ]

Hence, there is no number for which √n - 1 + √n + 1 are rational.