Square Root of Sum as Sum of Square Roots/Proof 1

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 $d$.