Square root of 6

The square root of 6 can be expressed as the simple continued fraction

: $[2;\; 2,\; 4,\; 2,\; 4,\; 2,\backslash ldots]\; =\; 2\; +\; \backslash cfrac\; 1\; \{2\; +\; \backslash cfrac\; 1\; \{4\; +\; \backslash cfrac\; 1\; \{2\; +\; \backslash cfrac\; 1\; \{4\; +\; \backslash dots\}\}\}\}.$ {{OEIS|id=A040003}}

The successive partial evaluations of the continued fraction, which are called its convergents, approach $\backslash sqrt\{6\}$:

:$\backslash frac\{2\}\{1\},\; \backslash frac\{5\}\{2\},\; \backslash frac\{22\}\{9\},\; \backslash frac\{49\}\{20\},\; \backslash frac\{218\}\{89\},\; \backslash frac\{485\}\{198\},\; \backslash frac\{2158\}\{881\},\; \backslash frac\{4801\}\{1960\},\; \backslash dots$

Their numerators are 2, 5, 22, 49, 218, 485, 2158, 4801, 21362, 47525, 211462, …{{OEIS|id=A041006}}, and their denominators are 1, 2, 9, 20, 89, 198, 881, 1960, 8721, 19402, 86329, …{{OEIS|id=A041007}}.{{Cite web |last=Conrad |first=Keith |title=Pell's Equation II |url=https://kconrad.math.uconn.edu/blurbs/ugradnumthy/pelleqn2.pdf |access-date=17 March 2022 |website=uconn.edu |quote=The continued fraction of √6 is [2; {{overline|2, 4}}], and the table of convergents below suggests (and it is true) that every other convergent provides a solution to {{math |x² − 6y² {{=}} 1}}.}}

Each convergent is a best rational approximation of $\backslash sqrt\{6\}$; in other words, it is closer to $\backslash sqrt\{6\}$ than any rational with a smaller denominator. Decimal equivalents improve linearly, at a rate of nearly one digit per convergent:

:$\backslash frac\{2\}\{1\}\; =\; 2.0,\backslash quad\; \backslash frac\{5\}\{2\}\; =\; 2.5,\backslash quad\; \backslash frac\{22\}\{9\}\; =\; 2.4444\backslash dots,\backslash quad\; \backslash frac\{49\}\{20\}\; =\; 2.45,\backslash quad\; \backslash frac\{218\}\{89\}\; =\; 2.44943...,\backslash quad\; \backslash frac\{485\}\{198\}\; =\; 2.449494...,\; \backslash quad\; \backslash ldots$

The convergents, expressed as {{math|{{sfrac|x|y}}}}, satisfy alternately the Pell's equations

:$x^2\; -\; 6y^2\; =\; -2\backslash quad\; \backslash mathrm\{and\}\; \backslash quad\; x^2\; -\; 6y^2\; =\; 1$

When $\backslash sqrt\{6\}$ is approximated with the Babylonian method, starting with {{math|x₀ {{=}} 2}} and using {{math|x_n+1 {{=}} {{sfrac|1|2}}{{big|{{big|(}}}}x_n + {{sfrac|6|x_n}}{{big|{{big|)}}}}}}, the {{math|n}}th approximant {{math|x_n}} is equal to the {{math|2ⁿ}}th convergent of the continued fraction:

:$x\_0\; =\; 2,\; \backslash quad\; x\_1\; =\; \backslash frac\{5\}\{2\}\; =\; 2.5,\; \backslash quad\; x\_2\; =\; \backslash frac\{49\}\{20\}\; =2.45,\; \backslash quad\; x\_3\; =\; \backslash frac\{4801\}\{1960\}\; =\; 2.449489796...,\; \backslash quad\; x\_4\; =\; \backslash frac\{46099201\}\{18819920\}\; =\; 2.449489742783179...,\; \backslash quad\; \backslash dots$

File:Slide rule with square roots of 6 and 7.jpg with 7 and 6 on A and B scales, and square roots of 6 and of 7 on C and D scales, which can be read as slightly less than 2.45 and somewhat more than 2.64, respectively]]

The Babylonian method is equivalent to Newton's method for root finding applied to the polynomial $x^2-6$. The Newton's method update, $x\_\{n+1\}\; =\; x\_n\; -\; f(x\_n)/f\text{'}(x\_n),$ is equal to $(x\_n\; +\; 6/x\_n)/2$ when $f(x)\; =\; x^2\; -\; 6$. The method therefore converges quadratically.

File:Regular octahedron with inscribed sphere annotated.png

File:Root rectangles up to 6.png

File:Equilateral triangle with circumscribed rectangle and square.png.]]

In plane geometry, the square root of 6 can be constructed via a sequence of dynamic rectangles, as illustrated here.{{Cite book |last=Jay Hambidge |url=https://archive.org/details/bub_gb_Qq4gAAAAMAAJ |title=Dynamic Symmetry: The Greek Vase |publisher=Kessinger Publishing |year=1920 |isbn=0-7661-7679-7 |edition=Reprint of original Yale University Press |location=Whitefish, MT |pages=[https://archive.org/details/bub_gb_Qq4gAAAAMAAJ/page/n29 19]–29 |quote=Dynamic Symmetry root rectangles. |orig-year=1920}}{{Cite book |last=Matila Ghyka |url=https://archive.org/details/geometryofartlif00mati |title=The Geometry of Art and Life |publisher=Courier Dover Publications |year=1977 |isbn=9780486235424 |pages=[https://archive.org/details/geometryofartlif00mati/page/126 126–127] |url-access=registration}}{{Cite book |last=Fletcher |first=Rachel |url=https://infinitemeasure.com/publications/infinite-measure/ |title=Infinite Measure: Learning to Design in Geometric Harmony with Art, Architecture, and Nature |date=2013 |publisher=George F Thompson Publishing |isbn=978-1-938086-02-1}}

In solid geometry, the square root of 6 appears as the longest distances between corners (vertices) of the double cube, as illustrated above. The square roots of all lower natural numbers appear as the distances between other vertex pairs in the double cube (including the vertices of the included two cubes).

The edge length of a cube with total surface area of 1 is $\backslash frac\{\backslash sqrt\{6\}\}\{6\}$ or the reciprocal square root of 6. The edge lengths of a regular tetrahedron ({{mvar|t}}), a regular octahedron ({{mvar|o}}), and a cube ({{mvar|c}}) of equal total surface areas satisfy $\backslash frac\{t\backslash cdot\; o\}\{c^2\}\; =\; \backslash sqrt\{6\}$.{{Cite web |last=Rechtman |first=Ana |title=Un défi par semaine Avril 2016, 3e défi (Solution du 2e défi d'Avril) |url=https://images.math.cnrs.fr/Avril-2016-3e-defi.html |access-date=23 March 2022 |website=Images des Mathématiques}}

The edge length of a regular octahedron is the square root of 6 times the radius of an inscribed sphere (that is, the distance from the center of the solid to the center of each face).{{Cite book |last=S. C. & L. M. Gould |url=https://books.google.com/books?id=WywAAAAAYAAJ&pg=PA342 |title=The Bizarre Notes and Queries in History, Folk-lore, Mathematics, Mysticism, Art, Science, Etc, Volumes 7-8 |year=1890 |location=Manchester, N. H. |page=342 |quote="In the octahedron whose diameter is 2, the linear edge equals the square root of 6." |access-date=19 March 2022}}

The square root of 6 appears in various other geometry contexts, such as the side length $\backslash frac\{\backslash sqrt\{6\}+\backslash sqrt\{2\}\}\{2\}$ for the square enclosing an equilateral triangle of side 2 (see figure).

The square root of 6, with the square root of 2 added or subtracted, appears in several exact trigonometric values for angles at multiples of 15 degrees ($\backslash pi/12$ radians).{{Cite book |url=https://archive.org/details/handbookofmathe000abra |title=Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables |publisher=Dover Publications |year=1972 |isbn=978-0-486-61272-0 |editor-last=Abramowitz |editor-first=Milton |editor-link=Milton Abramowitz |location=New York |page=74 |editor-last2=Stegun |editor-first2=Irene A. |editor-link2=Irene Stegun}}

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File:13th-century fifth-point arch shape.png artist Villard de Honnecourt]]

Villard de Honnecourt's 13th century construction of a Gothic "fifth-point arch" with circular arcs of radius 5 has a height of twice the square root of 6, as illustrated here.{{Cite journal |last=Branner |first=Robert |date=1960 |title=Villard de Honnecourt, Archimedes, and Chartres |url=https://www.jstor.org/stable/988023 |journal=Journal of the Society of Architectural Historians |volume=19 |issue=3 |pages=91–96 |doi=10.2307/988023 |jstor=988023 |access-date=25 March 2022|url-access=subscription }}{{Cite journal |last=Shelby |first=Lon R. |year=1969 |title=Setting Out the Keystones of Pointed Arches: A Note on Medieval 'Baugeometrie' |url=https://www.jstor.org/stable/3101574 |journal=Technology and Culture |volume=10 |issue=4 |pages=537–548 |doi=10.2307/3101574 |jstor=3101574 |access-date=25 March 2022|url-access=subscription }}

• Square root
• Square root of 2
• Square root of 3
• Square root of 5
• Square root of 7

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{{Algebraic numbers}}

{{Irrational number}}

Category:Mathematical constants

Category:Quadratic irrational numbers