Square root of 6

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

${\displaystyle [2;2,4,2,4,2,\ldots ]=2+{\cfrac {1}{2+{\cfrac {1}{4+{\cfrac {1}{2+{\cfrac {1}{4+\dots }}}}}}}}.}$ (sequence A040003 in the OEIS)

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

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

Their numerators are 2, 5, 22, 49, 218, 485, 2158, 4801, 21362, 47525, 211462, …(sequence A041006 in the OEIS), and their denominators are 1, 2, 9, 20, 89, 198, 881, 1960, 8721, 19402, 86329, …(sequence A041007 in the OEIS).[5]

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

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

The convergents, expressed as x/y, satisfy alternately the Pell's equations[5]

${\displaystyle x^{2}-6y^{2}=-2\quad \mathrm {and} \quad x^{2}-6y^{2}=1}$

When ${\displaystyle {\sqrt {6}}}$ is approximated with the Babylonian method, starting with x₀ = 2 and using x_n+1 = 1/2(x_n + 6/x_n), the nth approximant x_n is equal to the 2ⁿth convergent of the continued fraction:

${\displaystyle x_{0}=2,\quad x_{1}={\frac {5}{2}}=2.5,\quad x_{2}={\frac {49}{20}}=2.45,\quad x_{3}={\frac {4801}{1960}}=2.449489796...,\quad x_{4}={\frac {46099201}{18819920}}=2.449489742783179...,\quad \dots }$

A Logarex system Darmstadt slide rule 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 ${\displaystyle x^{2}-6}$. The Newton's method update, ${\displaystyle x_{n+1}=x_{n}-f(x_{n})/f'(x_{n}),}$ is equal to ${\displaystyle (x_{n}+6/x_{n})/2}$ when ${\displaystyle f(x)=x^{2}-6}$. The method therefore converges quadratically.

A regular octahedron with an inscribed sphere, illustrating the square root of 6 ratio between edge length and radius

Root rectangles illustrate a construction of the square root of 6

An equilateral triangle with circumscribed rectangle and square; the side of the square is ${\displaystyle ({\sqrt {6}}+{\sqrt {2}})/2}$, and the diagonal of the rectangle is the square root of 7.

In plane geometry, the square root of 6 can be constructed via a sequence of dynamic rectangles, as illustrated here.[6][7][8]

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).[8]

The edge length of a cube with total surface area of 1 is ${\displaystyle {\frac {\sqrt {6}}{6}}}$ or the reciprocal square root of 6. The edge lengths of a regular tetrahedron (t), a regular octahedron (o), and a cube (c) of equal total surface areas satisfy ${\displaystyle {\frac {t\cdot o}{c^{2}}}={\sqrt {6}}}$.[3][9]

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).[10]

The square root of 6 appears in various other geometry contexts, such as the side length ${\displaystyle {\frac {{\sqrt {6}}+{\sqrt {2}}}{2}}}$ 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 (${\displaystyle \pi /12}$ radians).[11]

:{| class="wikitable" style="text-align: center;"

!Radians!!Degrees!!sin!!cos!!tan!!cot!!sec!!csc |- ! ${\displaystyle {\frac {\pi }{12}}}$!! ${\displaystyle 15^{\circ }}$ |${\displaystyle {\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}}$|| ${\displaystyle {\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}}$|| ${\displaystyle 2-{\sqrt {3}}}$|| ${\displaystyle 2+{\sqrt {3}}}$|| ${\displaystyle {\sqrt {6}}-{\sqrt {2}}}$|| ${\displaystyle {\sqrt {6}}+{\sqrt {2}}}$ |- ! ${\displaystyle {\frac {5\pi }{12}}}$!! ${\displaystyle 75^{\circ }}$ |${\displaystyle {\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}}$|| ${\displaystyle {\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}}$|| ${\displaystyle 2+{\sqrt {3}}}$|| ${\displaystyle 2-{\sqrt {3}}}$|| ${\displaystyle {\sqrt {6}}+{\sqrt {2}}}$|| ${\displaystyle {\sqrt {6}}-{\sqrt {2}}}$ |}

13th-century fifth-point arch shape, according to Branner's 1960 interpretation (Paris, Bibliothèque nationale de France, MS Fr 19093) of the 13th-century Picard artist Villard de Honnecourt

The square root of six (actually its reciprocal, "the square root of six over six") appears in Star Wars dialogue.[12]

The question of "whether the square root of six is three" has been posited as a question that might be answered by economic methods, if social issues can be so addressed.[13][14][15][16]

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.[17][18]

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