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 $\sqrt c + \sqrt d = \sqrt {a + b}$.

It follows that $a + b = c + d + 2 \sqrt {c d}$.

Simultaneously equating $a = c + d$ and $b = 2 \sqrt {c d}$:

Solving for $d$:

Substituting $d$ and 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 $d$.