Square Root of Number Plus or Minus Square Root

Theorem

Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.

Then:

$\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$

$\ds \sqrt {a - \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$

$\ds \sqrt {2 - \sqrt 3}$

$=$

$\ds \sqrt {\dfrac {2 + \sqrt {2^2 - 3} } 2} - \sqrt {\dfrac {2 - \sqrt {2^2 - 3} } 2}$

$\ds $

$=$

$\ds \sqrt {\dfrac 3 2} - \sqrt {\dfrac 1 2}$

$\ds $

$=$

$\ds \dfrac {\sqrt 3 - 1} {\sqrt 2}$

$\ds \sqrt {3 - \sqrt 8}$

$=$

$\ds \sqrt {\dfrac {3 + \sqrt {3^2 - 8} } 2} - \sqrt {\dfrac {3 - \sqrt {3^2 - 8} } 2}$

$\ds $

$=$

$\ds \sqrt {\dfrac 4 2} - \sqrt {\dfrac 2 2}$

$\ds $

$=$

$\ds \sqrt 2 - 1$