Square root of 5

The diagonal of a2 × 1 rectangle has length √5.
Rationality	Irrational
Representations
Decimal	2.23606797749978969...
Algebraic form	${\displaystyle \sqrt{5}}$
Continued fraction	${\displaystyle 2+\frac{1}{4+\frac{1}{4+\frac{1}{4+\genfrac{}{}{0ex}{}{\phantom{x}}{{\displaystyle \ddots }}}}}}$

Thesquare root of 5, denoted⁠${\displaystyle \sqrt{5}}$⁠, is the positivereal number that, when multiplied by itself, gives the natural number5. Along with itsconjugate⁠${\displaystyle -\sqrt{5}}$⁠, it solves thequadratic equation⁠${\displaystyle x^2-5=0}$⁠, making it aquadratic integer, a type ofalgebraic number.⁠${\displaystyle \sqrt{5}}$⁠ is anirrational number, meaning it cannot be written as a fraction ofintegers.^[1] The first forty significant digits of itsdecimal expansion are:

2.236067977499789696409173668731276235440... (sequenceA002163 in theOEIS).

A length of⁠${\displaystyle \sqrt{5}}$⁠ can be constructed as thediagonal of a⁠${\displaystyle 2\times 1}$⁠ unitrectangle.⁠${\displaystyle \sqrt{5}}$⁠ also appears throughout in the metrical geometry of shapes with fivefold symmetry; the ratio between diagonal and side of aregular pentagon is thegolden ratio⁠${\displaystyle \phi =\frac{1}{2}(1+\sqrt{5})}$⁠.

The square root of 5 is anirrational number, meaning it can not be exactly represented as a fraction⁠${\displaystyle x/y}$⁠ where⁠${\displaystyle x}$⁠ and⁠${\displaystyle y}$⁠ areintegers. However, it can be approximated arbitrarily closely by suchrational numbers.

Particularly good approximations are the integer solutions ofPell's equations,

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

which can be algebraically rearranged into the form

${\displaystyle \frac{x}{y}=\sqrt{5\pm \frac{1}{y^2}}.}$

For example, the approximation⁠${\displaystyle 2=\sqrt{5-1}}$⁠, which is accurate to about 10%, satisfies the negative Pell's equation,⁠${\displaystyle 2^2-5\cdot 1^2=4-5=-1}$⁠; likewise, the approximation⁠${\displaystyle \frac{9}{4}=\sqrt{5+\frac{1}{16}}=2.25}$⁠, which is accurate within 1%, satisfies the positive equation,⁠${\displaystyle 9^2-5\cdot 4^2=81-80=1}$⁠. These two approximations are the respective fundamental solutions of each Pell's equation, to whichadditional solutions are algebraically related.

Solutions to both Pell's equations can also be found systematically by following theEuclidean algorithm, resulting in thesimple continued fraction for⁠${\displaystyle \sqrt{5}}$⁠,^[2]

${\displaystyle \sqrt{5}=[2;4,4,4,\dots ]=2+\frac{1}{4+\frac{1}{4+\frac{1}{4+\genfrac{}{}{0ex}{}{\phantom{x}}{{\displaystyle \ddots }}}}}.}$

Each step⁠${\displaystyle n}$⁠ of the algorithm produces a better approximation⁠${\displaystyle x_n/y_n}$⁠, one of theconvergents (partial evaluations) of this continued fraction. These are a sequence ofbest rational approximations to⁠${\displaystyle \sqrt{5}}$⁠, each more accurate than any other rational approximation with the same or smaller denominator. They give all of the solutions to Pell's equations, satisfying⁠${\displaystyle x_n^2-5y_n^2=(-1{)}^n}$⁠.^[3] The first several convergents to the continued fraction are:^[4]

⁠${\displaystyle \mathit{n}}$⁠	⁠${\displaystyle 0}$⁠	⁠${\displaystyle 1}$⁠	⁠${\displaystyle 2}$⁠	⁠${\displaystyle 3}$⁠	⁠${\displaystyle 4}$⁠	⁠${\displaystyle 5}$⁠	⁠${\displaystyle 6}$⁠	⁠${\displaystyle 7}$⁠	⁠${\displaystyle 8}$⁠	⁠${\displaystyle 9}$⁠	⁠${\displaystyle \dots }$⁠
⁠${\displaystyle \frac{{\mathit{x}}_{\mathit{n}}\phantom{t}}{{\mathit{y}}_{\mathit{n}}}}$⁠	${\displaystyle \frac{1}{0}}$	${\displaystyle \frac{2}{1}}$	${\displaystyle \frac{9}{4}}$	${\displaystyle \frac{38}{17}}$	${\displaystyle \frac{161}{72}}$	${\displaystyle \frac{682}{305}}$	${\displaystyle \frac{2889}{1292}}$	${\displaystyle \frac{12238}{5473}}$	${\displaystyle \frac{51841}{23184}}$	${\displaystyle \frac{219602}{98209}}$	${\displaystyle \dots }$

In thelimit, these approximations converge to⁠${\displaystyle \sqrt{5}}$⁠. That is,⁠${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{x}_n/y_n=\sqrt{5}}$⁠.

One of the oldest methods of calculating a square root of a number⁠${\displaystyle d}$⁠, theBabylonian method,^[5] starts with an initial guess⁠${\displaystyle x_0}$⁠, and at each step finds a new approximation by averaging the previous approximation and⁠${\displaystyle d}$⁠ times itsreciprocal,⁠${\displaystyle x_{n+1}=\frac{1}{2}(x_n+d/x_n)}$⁠. This is the special case, for the function⁠${\displaystyle f(x)=x^2-d}$⁠, ofNewton's method for finding the root of an arbitrary function. For a typical guess, the approximationconverges quadratically (roughly doubles the number of correct digits at each step).^[6]

The initial guess is somewhat arbitrary, but when approximating⁠${\displaystyle \sqrt{5}}$⁠ by this method, usually⁠${\displaystyle x_0=2}$⁠ is chosen.^[7] With this choice, the⁠${\displaystyle n}$⁠th approximation is equal to the⁠${\displaystyle 2^n}$⁠th convergent of the continued fraction for⁠${\displaystyle \sqrt{5}}$⁠.^[8]

with digits that differ from the decimal expansion of⁠${\displaystyle \sqrt{5}}$⁠ highlighted in red.

Thegolden ratio⁠${\displaystyle \phi }$⁠ is thearithmetic mean of 1 and${\displaystyle \sqrt{5}}$.^[9]${\displaystyle \sqrt{5}}$ has a relationship to the golden ratio andits algebraic conjugate⁠${\displaystyle \overline{\phi }}$⁠ as is expressed in the following formulae:

${\displaystyle \sqrt{5}}$ then figures in the closed form expression for theFibonacci numbers:^[10]

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

The quotient⁠${\displaystyle \sqrt{5}/\phi }$⁠ provides an interesting pattern of continued fractions and are related to the ratios between the Fibonacci numbers and theLucas numbers:^[11]

The convergents feature the Lucas numbers as numerators and the Fibonacci numbers as denominators:

$${\displaystyle \frac{1}{1},\frac{3}{2},\frac{4}{3},\frac{7}{5},\frac{11}{8},\frac{18}{13},\frac{29}{21},\dots ,\frac{L_n}{F_{n+1}},\dots }$$

In thelimit,$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{F_{n+1}}{F_n}=\frac{L_{n+1}}{L_n}=\phi ,\underset{n\to \mathrm{\infty }}{lim}\frac{L_n}{F_n}=\sqrt{5}.}$$

More precisely, the convergents to the continued fraction for⁠${\displaystyle \sqrt{5}}$⁠ (see§ Rational approximations above) are:

${\displaystyle \frac{2}{1},\frac{9}{4},\frac{38}{17},\frac{161}{72},\frac{682}{305},\dots ,\frac{\frac{1}{2}{L}_{3n}}{\frac{1}{2}{F}_{3n}},\dots .}$

Geometrically,${\displaystyle \sqrt{5}}$ corresponds to thediagonal of a rectangle whose sides are of length1 and2, as is evident from thePythagorean 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 asubdivision surface.^[12] Together with the algebraic relationship between${\displaystyle \sqrt{5}}$ and⁠${\displaystyle \phi }$⁠, this forms the basis for the geometrical construction of agolden rectangle from a square, and for the construction of aregularpentagon given its side (since the side-to-diagonal ratio in a regular pentagon is⁠${\displaystyle \phi }$⁠).

Since two adjacent faces of acube would unfold into a⁠${\displaystyle 1:2}$⁠ rectangle, the ratio between the length of the cube'sedge and the shortest distance from one of itsvertices to the opposite one, when traversing the cubesurface, is${\displaystyle \sqrt{5}}$. By contrast, the shortest distance when traversing through theinside of the cube corresponds to the length of the cube diagonal, which is thesquare root of three times the edge.^[13]

A rectangle with side proportions${\displaystyle 1:\sqrt{5}}$ is part of the series ofdynamic rectangles, which are based on proportions⁠${\displaystyle \sqrt{1}}$⁠,⁠${\displaystyle \sqrt{2}}$⁠,⁠${\displaystyle \sqrt{3}}$⁠,⁠${\displaystyle \sqrt{4}}$⁠,⁠${\displaystyle \sqrt{5}}$⁠, ... and successively constructed using the diagonal of the previous root rectangle, starting from a square.^[14] A root-5 rectangle is particularly notable in that it can be split into a square and two equal golden rectangles or into two golden rectangles of different sizes.^[15] It can also be decomposed as the union of two equal golden rectangles whose intersection forms a square. These shapes pictorially represent the algebraic relationships between${\displaystyle \sqrt{5}}$,⁠${\displaystyle \phi }$⁠ and⁠${\displaystyle {\phi }^{-1}}$⁠ mentioned above.

The square root of 5 appears intrigonometric constants related to the angles in a regular pentagon and decagon, which when combined which can be combined with other angles involving${\displaystyle \sqrt{2}}$ and${\displaystyle \sqrt{3}}$ to describesines and cosines of everyangle whose measure indegrees isdivisible by 3 but not by 15.^[16] The simplest of these are

Computing its value was therefore historically important for generatingtrigonometric 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 adodecahedron.^[13]

Hurwitz's theorem inDiophantine approximations states that everyirrational numberx can be approximated byinfinitely manyrational numbers⁠m/n⁠ inlowest terms in such a way that

and that${\displaystyle \sqrt{5}}$ is best possible, in the sense that for any larger constant than${\displaystyle \sqrt{5}}$, there are some irrational numbersx for which only finitely many such approximations exist.^[17]

Closely related to this is the theorem^[18] that of any three consecutiveconvergents⁠p_i/q_i⁠,⁠p_i+1/q_i+1⁠,⁠p_i+2/q_i+2⁠, of a numberα, at least one of the three inequalities holds:

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.^[18]

The twoquadratic fields⁠${\displaystyle \mathbb{Q}(\sqrt{5})}$⁠ and⁠${\displaystyle \mathbb{Q}(\sqrt{-5})}$⁠,field extensions of therational numbers, and their associated rings of integers,⁠${\displaystyle \mathbb{Z}[\frac{1}{2}+\frac{1}{2}\sqrt{5}]}$⁠ and⁠${\displaystyle \mathbb{Z}[\sqrt{-5}]}$⁠, respectively, are basic examples and have been studied extensively.

Thering${\displaystyle \mathbb{Z}[\sqrt{-5}]}$ contains numbers of the form${\displaystyle a+b\sqrt{-5}}$, wherea andb areintegers and${\displaystyle \sqrt{-5}}$ is theimaginary number${\displaystyle i\sqrt{5}}$. This ring is a frequently cited example of anintegral domain that is not aunique factorization domain.^[19] For example, the number 6 has two inequivalent factorizations within this ring:${\displaystyle 6=2\cdot 3=(1-\sqrt{-5})(1+\sqrt{-5}).}$

On the other hand, the realquadratic integer ring ofgolden integers${\displaystyle \mathbb{Z}[\phi ]}$, adjoining thegolden ratio${\displaystyle \phi =\frac{1}{2}(1+\sqrt{5})}$, was shown to beEuclidean, and hence a unique factorization domain, by Dedekind. This is the ring of integers in thegolden field⁠${\displaystyle \mathbb{Q}(\sqrt{5})}$⁠.^[20]

Thefield${\displaystyle \mathbb{Q}[\sqrt{-5}],}$ like any otherquadratic field, is anabelian extension of the rational numbers. TheKronecker–Weber theorem therefore guarantees that the square root of five can be written as a rationallinear combination ofroots of unity:

${\displaystyle \sqrt{5}=e^{2\pi i/5}-e^{4\pi i/5}-e^{6\pi i/5}+e^{8\pi i/5}.}$^{[citation needed]}

As of January 2022, the numerical value in decimal of the square root of 5 has been computed to at least 2.25 trillion digits.^[21]

• Golden ratio
• Square root
• Square root of 2
• Square root of 3
• Square root of 6
• Square root of 7
• Square root of 10

• Quadratic irrational square roots

• Articles with short description
• Short description is different from Wikidata
• All articles with unsourced statements
• Articles with unsourced statements from July 2025