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

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

zeros of a polynomial

Question

Show that there is no positive integer n for which

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

is rational.

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Suppose there exists a positive integer

n

for which is a rational number.
Where p and q positive integers and

q \neq 0

\Rightarrow \frac{q}{p} = \frac{1}{\sqrt{n - 1} + \sqrt{n + 1}}

\Rightarrow \frac{q}{p} = \frac{\sqrt{n - 1} - \sqrt{n + 1}}{(n - 1) - (n + 1)} = \frac{\sqrt{n - 1} - \sqrt{n + 1}}{- 2}

\Rightarrow \frac{2 q}{p} = \sqrt{n + 1} - \sqrt{n - 1}

(\sqrt{n - 1} + \sqrt{n + 1}) + (\sqrt{n + 1} - \sqrt{n - 1}) = \frac{p}{q} + \frac{2 q}{p}

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

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

.....(1)

(\sqrt{n - 1} + \sqrt{n + 1}) - (\sqrt{n + 1} - \sqrt{n - 1}) = \frac{p}{q} - \frac{2 q}{p}

\Rightarrow 2 \sqrt{n - 1} = \frac{p^{2} - 2 q^{2}}{p q}

\Rightarrow \sqrt{n - 1} = \frac{p^{2} - 2 q^{2}}{2 p q}

.....(2)

From eq1 and eq 2

\sqrt{n + 1}, a n d, \sqrt{n - 1}, a r e r a t i o n a l

∵, p a n d, q a r e, i n t e g e r s, \Rightarrow \frac{p^{2} + 2 q^{2}}{2 p q}, a n d, \frac{p^{2} - 2 q^{2}}{2 p q}

\Rightarrow, n + 1 a n d n - 1, p e r f e c t s q u a r e o f p o s i t i v e i n t e g e r s

Now

\sqrt{(} n + 1) - \sqrt{(} n - 1) = 2

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

3

Hence there is no positive integer

n

for which is a rational number.