Square root of 5

The square root of 5 (written ) is the positive, real number that, when multiplied by itself, results in the prime number 5 . From an algebraic point of view, it can therefore be defined as the positive solution of the quadratic equation . The square root of 5 is an irrational number . ${\displaystyle {\sqrt {5}}}$ ${\displaystyle x^{2}-5=0}$

Decimal places

The first 100 decimal places:

2.2360679774 9978969640 9173668731 2762354406 1835961152 5724270897 2454105209 2563780489 9414414408 3787822749 ...

Further decimal places can also be found under sequence A002163 in OEIS .

The current world record for calculating the decimal places (from July 4, 2019) is 2,000,000,000,000 and was achieved by Hiroyuki Oodaira (大平 寛 之).

Proof of irrationality

Similar to the proof of the irrationality of the square root of 2, the proof for the irrationality takes place indirectly, i.e. by refuting the opposite assumption. Suppose would be rational. Then you could write the number as a fraction of two natural numbers and : ${\displaystyle {\sqrt {5}}}$ ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle {\sqrt {5}}={\frac {a}{b}}}$.

Squaring the equation gives

${\displaystyle 5={\tfrac {a^{2}}{b^{2}}}}$

and it follows

${\displaystyle 5b^{2}=a^{2}}$.

The prime factor 5 occurs in or twice as often as in or , in any case with an even number, whereby 0 occurrences are of course also permitted. So the prime factor occurs odd-numbered on the left side of this equation, while on the right-hand side it occurs evenly, and we get a contradiction to the uniqueness of the prime factorization. Hence is irrational. ${\displaystyle a^{2}}$ ${\displaystyle b^{2}}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle {\sqrt {5}}}$

Continued fraction development

The continued fraction expansion of the square root of 5 is:

${\displaystyle [2;4,4,4,4,4,\ldots ]=2+{\cfrac {1}{4+{\cfrac {1}{4+{\cfrac {1}{4+{\cfrac {1}{4+\dots }}}}}}}}.}$(Follow A040002 in OEIS )

Golden ratio and Fibonacci sequence

The ratio of the golden section

${\displaystyle \Phi ={\frac {1}{2}}(1+{\sqrt {5}})\approx 1{,}6180339887}$

is the arithmetic mean of the number 1 and the square root of 5. The same applies to the limit value of the quotient of consecutive Fibonacci numbers

${\displaystyle \lim _{n\to \infty }{\frac {f_{n+1}}{f_{n}}}=\Phi ={\frac {1}{2}}(1+{\sqrt {5}}).}$

Also in the explicit formula for the Fibonacci numbers

${\displaystyle f_{n}={\frac {\Phi ^{n}-(-\Phi )^{-n}}{\sqrt {5}}}}$

the square root of 5 occurs.

geometry

Conway's decomposition of a triangle into smaller similar triangles

Geometric corresponds to the diagonal of a rectangle with the side lengths 1 and 2, which results directly from the Pythagorean theorem. Such a rectangle is obtained by halving a square or by joining two squares of the same size side by side. Together with the algebraic relationship between and , this is the basis for the geometric construction of a golden rectangle from a square and thus for the construction of a regular pentagon with a given side length. is namely the ratio of a pentagonal diagonal to the side length. ${\displaystyle {\sqrt {5}}}$ ${\displaystyle {\sqrt {5}}}$ ${\displaystyle \Phi }$ ${\displaystyle \Phi }$

trigonometry

Similar to and , the square root of 5 often occurs in the exact trigonometric values ​​of special angles, especially in the sine and cosine values ​​of the angles, the degrees of which are divisible by 3 but not by 15. Simple examples are: ${\displaystyle {\sqrt {2}}}$ ${\displaystyle {\sqrt {3}}}$

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

algebra

The ring contains the numbers of the form , where and are whole numbers and symbolize the imaginary number . This ring is a frequently cited example of an integrity ring that is not a factorial ring (ZPE ring). This can be seen, for example, from the fact that the number 6 has two non-equivalent factorizations within this ring: ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ ${\displaystyle a+b{\sqrt {-5}}}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle {\sqrt {-5}}}$ ${\displaystyle i{\sqrt {5}}}$

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

The body is like any quadratic fields an abelian extension of the rational numbers. The theorem of Kronecker and Weber therefore guarantees that the square root of 5 can be written as a rational linear combination of unit roots : ${\displaystyle \mathbb {Q} [{\sqrt {5}}]}$

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

Identities of Ramanujan

The square root of 5 appears in various identities discovered by Srinivasa Ramanujan that contain continued fractions.

An example is the following case of a Rogers-Ramanujan continued fraction :

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

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

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

Individual evidence

1. ^ Records set by y-cruncher. Retrieved March 5, 2020 . 
2. ^ Julian DA Wiseman, "Sin and cos in surds"
3. ^ KG Ramanathan: On the Rogers-Ramanujan continued fraction . In: Proceedings of the Indian Academy of Sciences - Section A . tape 93 , no. 2-3 , December 1984, ISSN  0370-0089 , p. 67-77 , doi : 10.1007 / BF02840651 ( springer.com [accessed March 12, 2020]).
4. ↑ Eric W. Weisstein: Ramanujan Continued Fractions. Retrieved March 12, 2020 .