Square Root of Sum as Sum of Square Roots

Theorem
Let $a, b \in \R, a \ge b$.

Then:
 * $\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$

Proof
Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$.

From Square of Sum:
 * $a + b = c + d + 2 \sqrt {c d}$

We now need to solve the simultaneous equations:
 * $a = c + d$
 * $b = 2 \sqrt {c d}$

First:

Solving for $c$:

Solving for $d$:

From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$.