Square root of 5

The square root of 5 can be expressed as the continued fraction

${\displaystyle [2;4,4,4,4,4,\ldots ]=2+{\cfrac {1}{4+{\cfrac {1}{4+{\cfrac {1}{4+{\cfrac {1}{4+{{} \atop \displaystyle \ddots }}}}}}}}}.}$ (sequence A040002 in the OEIS)

The successive partial evaluations of the continued fraction, which are called its convergents, approach ${\displaystyle {\sqrt {5}}}$:

${\displaystyle {\frac {2}{1}},{\frac {9}{4}},{\frac {38}{17}},{\frac {161}{72}},{\frac {682}{305}},{\frac {2889}{1292}},{\frac {12238}{5473}},{\frac {51841}{23184}},\dots }$

Their numerators are 2, 9, 38, 161, … (sequence A001077 in the OEIS), and their denominators are 1, 4, 17, 72, … (sequence A001076 in the OEIS).

Each of these is a best rational approximation of ${\displaystyle {\sqrt {5}}}$; in other words, it is closer to ${\displaystyle {\sqrt {5}}}$ than any rational number with a smaller denominator.

The convergents, expressed as x/y, satisfy alternately the Pell's equations[3]

${\displaystyle x^{2}-5y^{2}=-1\quad {\text{and}}\quad x^{2}-5y^{2}=1}$

When ${\displaystyle {\sqrt {5}}}$ is approximated with the Babylonian method, starting with x₀ = 2 and using x_n+1 = 1/2(x_n + 5/x_n), the nth approximant x_n is equal to the 2ⁿth convergent of the continued fraction:

${\displaystyle x_{0}=2.0,\quad x_{1}={\frac {9}{4}}=2.25,\quad x_{2}={\frac {161}{72}}=2.23611\dots ,\quad x_{3}={\frac {51841}{23184}}=2.2360679779\ldots ,\quad x_{4}={\frac {5374978561}{2403763488}}=2.23606797749979\ldots }$

The Babylonian method is equivalent to Newton's method for root finding applied to the polynomial ${\displaystyle x^{2}-5}$. The Newton's method update, ${\displaystyle x_{n+1}=x_{n}-f(x_{n})/f'(x_{n})}$, is equal to ${\displaystyle (x_{n}+5/x_{n})/2}$ when ${\displaystyle f(x)=x^{2}-5}$. The method therefore converges quadratically.

The ${\displaystyle {\sqrt {5}}/2}$ diagonal of a half square forms the basis for the geometrical construction of a golden rectangle.

The golden ratio φ is the arithmetic mean of 1 and ${\displaystyle {\sqrt {5}}}$.[4] The algebraic relationship between ${\displaystyle {\sqrt {5}}}$, the golden ratio and the conjugate of the golden ratio (Φ = −1/φ = 1 − φ) is expressed in the following formulae:

${\displaystyle {\begin{aligned}{\sqrt {5}}&=\varphi -\Phi =2\varphi -1=1-2\Phi \\[5pt]\varphi &={\frac {1+{\sqrt {5}}}{2}}\\[5pt]\Phi &={\frac {1-{\sqrt {5}}}{2}}.\end{aligned}}}$

(See the section below for their geometrical interpretation as decompositions of a ${\displaystyle {\sqrt {5}}}$ rectangle.)

${\displaystyle {\sqrt {5}}}$ then naturally figures in the closed form expression for the Fibonacci numbers, a formula which is usually written in terms of the golden ratio:

${\displaystyle F(n)={\frac {\varphi ^{n}-(1-\varphi )^{n}}{\sqrt {5}}}.}$

The quotient of ${\displaystyle {\sqrt {5}}}$ and φ (or the product of ${\displaystyle {\sqrt {5}}}$ and Φ), and its reciprocal, provide an interesting pattern of continued fractions and are related to the ratios between the Fibonacci numbers and the Lucas numbers:[5]

${\displaystyle {\begin{aligned}{\frac {\sqrt {5}}{\varphi }}=\Phi \cdot {\sqrt {5}}={\frac {5-{\sqrt {5}}}{2}}&=1.3819660112501051518\dots \\&=[1;2,1,1,1,1,1,1,1,\ldots ]\\[5pt]{\frac {\varphi }{\sqrt {5}}}={\frac {1}{\Phi \cdot {\sqrt {5}}}}={\frac {5+{\sqrt {5}}}{10}}&=0.72360679774997896964\ldots \\&=[0;1,2,1,1,1,1,1,1,\ldots ].\end{aligned}}}$

The series of convergents to these values feature the series of Fibonacci numbers and the series of Lucas numbers as numerators and denominators, and vice versa, respectively:

${\displaystyle {\begin{aligned}&{1,{\frac {3}{2}},{\frac {4}{3}},{\frac {7}{5}},{\frac {11}{8}},{\frac {18}{13}},{\frac {29}{21}},{\frac {47}{34}},{\frac {76}{55}},{\frac {123}{89}}},\ldots \ldots [1;2,1,1,1,1,1,1,1,\ldots ]\\[8pt]&{1,{\frac {2}{3}},{\frac {3}{4}},{\frac {5}{7}},{\frac {8}{11}},{\frac {13}{18}},{\frac {21}{29}},{\frac {34}{47}},{\frac {55}{76}},{\frac {89}{123}}},\dots \dots [0;1,2,1,1,1,1,1,1,\dots ].\end{aligned}}}$

In fact, the limit of the quotient of the ${\displaystyle n^{th}}$ Lucas number ${\displaystyle L_{n}}$ and the ${\displaystyle n^{th}}$ Fibonacci number ${\displaystyle F_{n}}$ is directly equal to the square root of ${\displaystyle 5}$:

${\displaystyle \lim _{n\to \infty }{\frac {L_{n}}{F_{n}}}={\sqrt {5}}.}$

Decomposition of a ${\displaystyle 1\times 2}$ right triangle into five similar triangles, the basis for the aperiodic pinwheel tiling.

Jay Hambidge's construction of "root rectangles"

Distances between vertices of a double unit cube are square roots of the first six natural numbers, including the square root of 5 (√7 is not possible due to Legendre's three-square theorem)

Geometrically, ${\displaystyle {\sqrt {5}}}$ corresponds to the diagonal of a rectangle whose sides are of length 1 and 2, as is evident from the Pythagorean theorem. Such a rectangle can be obtained by halving a square, or by placing two equal squares side by side. This can be used to subdivide a square grid into a tilted square grid with five times as many squares, forming the basis for a subdivision surface.[6] Together with the algebraic relationship between ${\displaystyle {\sqrt {5}}}$ and φ, this forms the basis for the geometrical construction of a golden rectangle from a square, and for the construction of a regular pentagon given its side (since the side-to-diagonal ratio in a regular pentagon is φ).

Since two adjacent faces of a cube would unfold into a 1:2 rectangle, the ratio between the length of the cube's edge and the shortest distance from one of its vertices to the opposite one, when traversing the cube surface, is ${\displaystyle {\sqrt {5}}}$. By contrast, the shortest distance when traversing through the inside of the cube corresponds to the length of the cube diagonal, which is the square root of three times the edge.

A rectangle with side proportions 1:${\displaystyle {\sqrt {5}}}$ is called a root-five rectangle and is part of the series of root rectangles, a subset of dynamic rectangles, which are based on ${\displaystyle {\sqrt {1}}}$ (= 1), ${\displaystyle {\sqrt {2}}}$, ${\displaystyle {\sqrt {3}}}$, ${\displaystyle {\sqrt {4}}}$ (= 2), ${\displaystyle {\sqrt {5}}}$... and successively constructed using the diagonal of the previous root rectangle, starting from a square.[7] A root-5 rectangle is particularly notable in that it can be split into a square and two equal golden rectangles (of dimensions Φ × 1), or into two golden rectangles of different sizes (of dimensions Φ × 1 and 1 × φ).[8] It can also be decomposed as the union of two equal golden rectangles (of dimensions 1 × φ) whose intersection forms a square. All this is can be seen as the geometric interpretation of the algebraic relationships between ${\displaystyle {\sqrt {5}}}$, φ and Φ mentioned above. The root-5 rectangle can be constructed from a 1:2 rectangle (the root-4 rectangle), or directly from a square in a manner similar to the one for the golden rectangle shown in the illustration, but extending the arc of length ${\displaystyle {\sqrt {5}}/2}$ to both sides.

Like ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle {\sqrt {3}}}$, the square root of 5 appears extensively in the formulae for exact trigonometric constants, including in the sines and cosines of every angle whose measure in degrees is divisible by 3 but not by 15.[9] The simplest of these are

${\displaystyle {\begin{aligned}\sin {\frac {\pi }{10}}=\sin 18^{\circ }&={\tfrac {1}{4}}({\sqrt {5}}-1)={\frac {1}{{\sqrt {5}}+1}},\\[5pt]\sin {\frac {\pi }{5}}=\sin 36^{\circ }&={\tfrac {1}{4}}{\sqrt {2(5-{\sqrt {5}})}},\\[5pt]\sin {\frac {3\pi }{10}}=\sin 54^{\circ }&={\tfrac {1}{4}}({\sqrt {5}}+1)={\frac {1}{{\sqrt {5}}-1}},\\[5pt]\sin {\frac {2\pi }{5}}=\sin 72^{\circ }&={\tfrac {1}{4}}{\sqrt {2(5+{\sqrt {5}})}}\,.\end{aligned}}}$

As such the computation of its value is important for generating trigonometric tables. Since ${\displaystyle {\sqrt {5}}}$ is geometrically linked to half-square rectangles and to pentagons, it also appears frequently in formulae for the geometric properties of figures derived from them, such as in the formula for the volume of a dodecahedron.

Hurwitz's theorem in Diophantine approximations states that every irrational number x can be approximated by infinitely many rational numbers m/n in lowest terms in such a way that

${\displaystyle \left|x-{\frac {m}{n}}\right|<{\frac {1}{{\sqrt {5}}\,n^{2}}}}$

and that ${\displaystyle {\sqrt {5}}}$ is best possible, in the sense that for any larger constant than ${\displaystyle {\sqrt {5}}}$, there are some irrational numbers x for which only finitely many such approximations exist.[10]

Closely related to this is the theorem[11] that of any three consecutive convergents p_i/q_i, p_i+1/q_i+1, p_i+2/q_i+2, of a number α, at least one of the three inequalities holds:

${\displaystyle \left|\alpha -{p_{i} \over q_{i}}\right|<{1 \over {\sqrt {5}}q_{i}^{2}},\quad \left|\alpha -{p_{i+1} \over q_{i+1}}\right|<{1 \over {\sqrt {5}}q_{i+1}^{2}},\quad \left|\alpha -{p_{i+2} \over q_{i+2}}\right|<{1 \over {\sqrt {5}}q_{i+2}^{2}}.}$

And the ${\displaystyle {\sqrt {5}}}$ in the denominator is the best bound possible since the convergents of the golden ratio make the difference on the left-hand side arbitrarily close to the value on the right-hand side. In particular, one cannot obtain a tighter bound by considering sequences of four or more consecutive convergents.[11]

The ring ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ contains numbers of the form ${\displaystyle a+b{\sqrt {-5}}}$, where a and b are integers and ${\displaystyle {\sqrt {-5}}}$ is the imaginary number ${\displaystyle i{\sqrt {5}}}$. This ring is a frequently cited example of an integral domain that is not a unique factorization domain.[12] The number 6 has two inequivalent factorizations within this ring:

${\displaystyle 6=2\cdot 3=(1-{\sqrt {-5}})(1+{\sqrt {-5}}).\,}$

The field ${\displaystyle \mathbb {Q} [{\sqrt {-5}}],}$ like any other quadratic field, is an abelian extension of the rational numbers. The Kronecker–Weber theorem therefore guarantees that the square root of five can be written as a rational linear combination of roots of unity:

${\displaystyle {\sqrt {5}}=e^{{\frac {2\pi }{5}}i}-e^{{\frac {4\pi }{5}}i}-e^{{\frac {6\pi }{5}}i}+e^{{\frac {8\pi }{5}}i}.\,}$

The square root of 5 appears in various identities discovered by Srinivasa Ramanujan involving continued fractions.[13][14]

For example, this case of the Rogers–Ramanujan continued fraction:

${\displaystyle {\cfrac {1}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-6\pi }}{1+{{} \atop \displaystyle \ddots }}}}}}}}}=\left({\sqrt {\frac {5+{\sqrt {5}}}{2}}}-{\frac {{\sqrt {5}}+1}{2}}\right)e^{\frac {2\pi }{5}}=e^{\frac {2\pi }{5}}\left({\sqrt {\varphi {\sqrt {5}}}}-\varphi \right).}$

${\displaystyle {\cfrac {1}{1+{\cfrac {e^{-2\pi {\sqrt {5}}}}{1+{\cfrac {e^{-4\pi {\sqrt {5}}}}{1+{\cfrac {e^{-6\pi {\sqrt {5}}}}{1+{{} \atop \displaystyle \ddots }}}}}}}}}=\left({{\sqrt {5}} \over 1+{\sqrt[{5}]{5^{\frac {3}{4}}(\varphi -1)^{\frac {5}{2}}-1}}}-\varphi \right)e^{\frac {2\pi }{\sqrt {5}}}.}$

${\displaystyle 4\int _{0}^{\infty }{\frac {xe^{-x{\sqrt {5}}}}{\cosh x}}\,dx={\cfrac {1}{1+{\cfrac {1^{2}}{1+{\cfrac {1^{2}}{1+{\cfrac {2^{2}}{1+{\cfrac {2^{2}}{1+{\cfrac {3^{2}}{1+{\cfrac {3^{2}}{1+{{} \atop \displaystyle \ddots }}}}}}}}}}}}}}}\,.}$

• Golden ratio
• Square root
• Square root of 2
• Square root of 3
• Square root of 6
• Square root of 7