# proof of square root of square root binomial

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

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

Since the squaring operation undoes the square roots, we obtain 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 root of the product, we have the following:

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

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

 $\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^% {2}-b}}{2}}.$
Title proof of square root of square root binomial ProofOfSquareRootOfSquareRootBinomial 2013-03-22 17:42:45 2013-03-22 17:42:45 rspuzio (6075) rspuzio (6075) 5 rspuzio (6075) Proof msc 11A25