Show that there is no positive integer n for which \sqrt{n

Foundation

Mathematics Foundation

Inequalities

Properties of Real Number

Irrational Number

Question

Show that there is no positive integer

n

for which

\sqrt{n - 1} + \sqrt{n + 1}

is rational.

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If possible, let there be a positive integer

n

for which

\sqrt{n - 1} + \sqrt{n + 1}

is rational, which is equal to

\frac{a}{b}

(say), where

a, b

are positive integers.

Then,

\frac{a}{b} = \sqrt{n - 1} + \sqrt{n + 1}

..... (1)

\Rightarrow

\frac{b}{a} = \frac{1}{< b r / > \sqrt{n - 1} + \sqrt{n + 1},}

\Rightarrow

\frac{b}{a} = \frac{\sqrt{} < b r / > n + 1 - \sqrt{n - 1},}{{\sqrt{n + 1} + \sqrt{< b r / > n - 1},} {\sqrt{n + 1} - \sqrt{n - 1}, < b r / >},}

\Rightarrow \frac{b}{a} = \frac{\sqrt{n} + 1 - \sqrt{n - 1},}{< b r / > (n + 1) - (n - 1),}

\Rightarrow \frac{b}{a} = \frac{\sqrt{< b r / > n + 1} - \sqrt{n - 1},}{2}

\Rightarrow

\frac{2 b}{a} = \sqrt{n + 1} < b r / > - \sqrt{n - 1}

..... (2)

Adding (1) and (2), we get

2 \sqrt{n + 1} = \frac{a}{b} + \frac{2 b}{a}

Subtracting (2) from (1), we get

2 \sqrt{n - 1} = \frac{a}{b} - \frac{2 b}{a}

\Rightarrow

\sqrt{n + 1} = \frac{a^{2 < b r / >} + 2 b^{2}}{2 a b}

and

\sqrt{n - 1} = \frac{{< b r / > a}^{2} - 2 b^{2}}{2 a b}

\Rightarrow \sqrt{n + 1}

and

\sqrt{n - 1}

are rationals

[

∵ a, b

are integers

∴ \frac{a^{2} + b^{2}}{2 a b}

and

\frac{a^{2} - 2 b^{2}}{2 a b}

are rationals

]

\Rightarrow (n + 1)

and

(n - 1)

are perfect square of positive integers.

Now,

(n + 1) - (n - 1) = 2

which is not possible as any two perfect squares differ at least by

3

Hence, there is no positive integer

n

for which

(\sqrt{n - 1} + \sqrt{n + 1})

is rational.