Prove irrational

[Phlembac Adib Hasan]

Prove that $\sqrt{\sqrt{37}+6}+\sqrt{\sqrt{37}-6}$ is irrational.

(বুড়ো খ্যোকারা, পোস্ট দেওয়ার আগে উপরে ক্যাটাগরি কোনটা খেয়াল কর)

[Prosenjit Basak]

Let us assume that the given expression is rational. Then
$\sqrt{\sqrt{37}+6}+\sqrt{\sqrt{37}-6} = \frac{a}{b}$ where $a,b$ are integer and $b$ is not equal to zero.

We can substitute $\sqrt{\sqrt{37}+6} = m$ and
$\sqrt{\sqrt{37}-6} = n$
Now $\sqrt{m} + \sqrt{n} = \frac{a}{b}$
$\Rightarrow (\sqrt{m} + \sqrt{n})^2 = \frac{a^2}{b^2}$
$\Rightarrow m + 2\sqrt{mn} + n = \frac{a^2}{b^2}$
Now
$mn = (\sqrt{\sqrt{37}+6})(\sqrt{\sqrt{37}-6})
= 37 - 36
= 1$ and
$ m + n = \sqrt{37} + 6 + \sqrt{37} - 6
= 2\sqrt{37}$
So
$2\sqrt{37} + 1 = \frac{a^2}{b^2}$
or, $\displaystyle 2\sqrt{37} = \frac{a^2 - b^2}{b^2}$
or, $\displaystyle \sqrt{37} = \frac{a^2 - b^2}{2b^2}$
Now it is seen that a irrational number is equal to a rational number which is impossible. So that expression has to be irrational.
( Latex use করতে এত কষ্ট জানতাম নাহ )