proof of square root of square root binomial

We square the expression on the right-hand-side and expandusing the binomial formula:

${\left(\sqrt{{\displaystyle \frac{a+\sqrt{a^2-b}}{2}}}\pm \sqrt{{\displaystyle \frac{a-\sqrt{a^2-b}}{2}}}\right)}^2$$={\left(\sqrt{{\displaystyle \frac{a+\sqrt{a^2-b}}{2}}}\right)}^2$

$+{\left(\sqrt{{\displaystyle \frac{a-\sqrt{a^2-b}}{2}}}\right)}^2\pm 2\sqrt{{\displaystyle \frac{a+\sqrt{a^2-b}}{2}}}\sqrt{{\displaystyle \frac{a-\sqrt{a^2-b}}{2}}}$

Since the squaring operation undoes the square roots, weobtain the following:

$${\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\right)}^2+{\left(\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)}^2=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}=a$$

Since the product of square roots equals the square rootof the product, we have the following:

$\sqrt{{\displaystyle \frac{a+\sqrt{a^2-b}}{2}}}\sqrt{{\displaystyle \frac{a-\sqrt{a^2-b}}{2}}}$$=\sqrt{{\displaystyle \frac{a+\sqrt{a^2-b}}{2}}\cdot {\displaystyle \frac{a-\sqrt{a^2-b}}{2}}}$

$=\sqrt{{\displaystyle \frac{a^2-{(\sqrt{a^2-b})}^2}{4}}}$

$=\sqrt{{\displaystyle \frac{a^2-(a^2-b)}{4}}}$

$=\sqrt{{\displaystyle \frac{b}{4}}}={\displaystyle \frac{\sqrt{b}}{2}}$

Combining what we have calculated above, we obtain

$${\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)}^2=a\pm \sqrt{b}.$$

Because the square of the asserted value of the square root equals the radicand ($a\pm \sqrt{b}$) of the square root, and the asserted value of the square root is clearly non-negative, we have justified the validity of the formulas

$$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}.$$