ProofOfSquareRootOfSquareRootBinomial 2007-12-24 14:42:29 2007-12-24 17:36:11 Proof square root of square root binomial 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here We square the expression on the right-hand-side and expand using the binomial formula: \begin{align*} \left( \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}} \right)^2 &= \left( \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \right)^2 \\ &+ \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}} \end{align*} 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: \begin{align*} \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \sqrt{\frac{a-\sqrt{a^2-b}}{2}} &= \sqrt{\frac{a+\sqrt{a^2-b}}{2} \cdot \frac{a-\sqrt{a^2-b}}{2}} \\ &= \sqrt{\frac{a^2 - (\sqrt{a^2-b})^2}{4}} \\ &= \sqrt{\frac{a^2 - (a^2-b)}{4}} \\ &= \sqrt{\frac{b}{4}} = \frac{\sqrt{b}}{2} \end{align*} 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 \[ \sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}. \]