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proof of square root of square root binomial
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(Proof)
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We square the expression on the right-hand-side and expand using the binomial formula:
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:
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 $$ \sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}. $$
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"proof of square root of square root binomial" is owned by rspuzio. [ full author list (2) ]
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Cross-references: formulas, radicand, product, square roots, operation, binomial formula, expand, expression, square
This is version 2 of proof of square root of square root binomial, born on 2007-12-24, modified 2007-12-24.
Object id is 10157, canonical name is ProofOfSquareRootOfSquareRootBinomial.
Accessed 1267 times total.
Classification:
| AMS MSC: | 11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas) |
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Pending Errata and Addenda
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