Square Root of Number Minus Square Root/Proof 1

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

Then:

Proof
We are given that $a^2 - b > 0$.

Then:
 * $a > \sqrt b$

and so $\ds \sqrt {a - \sqrt b}$ is defined on the real numbers.

Let $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are (strictly) positive real numbers.

Observe that:
 * $\ds 0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$

Squaring both sides gives:

Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$

From $\sqrt b = 2 \sqrt {x y}$ we get:

By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:


 * $z^2 - a z + \dfrac b 4 = 0$

From Solution to Quadratic Equation:


 * $z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$

where $a^2 - b > 0$ (which is a given)

Because we have that $x > y$:

Subsituting into $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$: