Square root of 5

src: i.ytimg.com

The square root of 5 is the positive real number that, when multiplied by itself, gives the prime number 5. It is more precisely called the principal square root of 5, to distinguish it from the negative number with the same property. This number appears in the fractional expression for the golden ratio. It can be denoted in surd form as:

${\displaystyle {\sqrt {5}}.\,}$

It is an irrational algebraic number. The first sixty significant digits of its decimal expansion are:

2.23606797749978969640917366873127623544061835961152572427089... (sequence A002163 in the OEIS).

which can be rounded down to 2.236 to within 99.99% accuracy. The approximation 161/72 (? 2.23611) for the square root of five can be used. Despite having a denominator of only 72, it differs from the correct value by less than 1/10,000 (approx. 4.3×10^-5). As of December 2013, its numerical value in decimal has been computed to at least ten billion digits.

Video Square root of 5

Proofs of irrationality

1. This irrationality proof for the square root of 5 uses Fermat's method of infinite descent:

Suppose that ?5 is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as m/n for natural numbers m and n. Then ?5 can be expressed in lower terms as 5n - 2m/m - 2n, which is a contradiction. (The two fractional expressions are equal because equating them, cross-multiplying, and canceling like additive terms gives 5n² = m² and m/n = ?5, which is true by the premise. The second fractional expression for ?5 is in lower terms since, comparing denominators, m - 2n < n since m < 3n since m/n < 3 since ?5 < 3. And both the numerator and the denominator of the second fractional expression are positive since 2 < ?5 < 5/2 and m/n = ?5.)

2. This irrationality proof is also a proof by contradiction:

Suppose that ?5 = a/b where a/b is in reduced form.

Thus 5 = a²/b² and 5b² = a². If b were even, b², a², and a would be even making the fraction a/b not in reduced form. Thus b is odd, and by following a similar process, a is odd.

Now, let a = 2m + 1 and b = 2n + 1 where m and n are integers.

Substituting into 5b² = a² we get:

${\displaystyle 5(2n+1)^{2}=(2m+1)^{2}}$

which simplifies to:

${\displaystyle 5\left(4n^{2}+4n+1\right)=4m^{2}+4m+1}$

making:

${\displaystyle 20n^{2}+20n+5=4m^{2}+4m+1}$

By subtracting 1 from both sides, we get:

${\displaystyle 20n^{2}+20n+4=4m^{2}+4m}$

which reduces to:

${\displaystyle 5n^{2}+5n+1=m^{2}+m}$

In other words:

${\displaystyle 5n(n+1)+1=m(m+1)}$

The expression x(x + 1) is even for any integer x (since either x or x + 1 is even). So this says that 5 × even + 1 = even, or odd = even. Since there is no integer that is both even and odd, we have reached a contradiction and ?5 is irrational.

Maps Square root of 5

Continued fraction

It can be expressed as the continued fraction

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

The convergents and semiconvergents of this continued fraction are as follows (the black terms are the semiconvergents):

${\displaystyle {\color {red}{\frac {2}{1}}},{\frac {7}{3}},{\color {red}{\frac {9}{4}}},{\frac {20}{9}},{\frac {29}{13}},{\color {red}{\frac {38}{17}}},{\frac {123}{55}},{\color {red}{\frac {161}{72}}},{\frac {360}{161}},{\frac {521}{233}},{\color {red}{\frac {682}{305}}},{\frac {2207}{987}},{\color {red}{\frac {2889}{1292}}},\dots }$

Convergents of the continued fraction are colored red; 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 the best rational approximation of ?5; in other words, it is closer to ?5 than any rational with a smaller denominator.

src: i.ytimg.com

Babylonian method

When ?5 is computed with the Babylonian method, starting with r₀ = 2 and using r_n+1 = 1/2(r_n + 5/r_n), the nth approximant r_n is equal to the 2ⁿth convergent of the convergent sequence:

${\displaystyle {\frac {2}{1}}=2.0,\quad {\frac {9}{4}}=2.25,\quad {\frac {161}{72}}=2.23611\dots ,\quad {\frac {51841}{23184}}=2.2360679779\ldots }$

src: saylordotorg.github.io

Nested square expansions

The following nested square expressions converge to ${\displaystyle {\sqrt {5}}}$:

${\displaystyle {\begin{aligned}{\sqrt {5}}&=3-10\left({\frac {1}{5}}+\left({\frac {1}{5}}+\left({\frac {1}{5}}+\left({\frac {1}{5}}+\cdots \right)^{2}\right)^{2}\right)^{2}\right)^{2}\\&={\frac {9}{4}}-4\left({\frac {1}{16}}-\left({\frac {1}{16}}-\left({\frac {1}{16}}-\left({\frac {1}{16}}-\cdots \right)^{2}\right)^{2}\right)^{2}\right)^{2}\\&={\frac {9}{4}}-5\left({\frac {1}{20}}+\left({\frac {1}{20}}+\left({\frac {1}{20}}+\left({\frac {1}{20}}+\cdots \right)^{2}\right)^{2}\right)^{2}\right)^{2}\end{aligned}}}$

src: i.ytimg.com

Relation to the golden ratio and Fibonacci numbers

The golden ratio ? is the arithmetic mean of 1 and ?5. The algebraic relationship between ?5, the golden ratio and the conjugate of the golden ratio (? = -1/? = 1 - ?) 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 ?5 rectangle.)

?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 ?5 and ? (or the product of ?5 and ?), and its reciprocal, provide an interesting pattern of continued fractions and are related to the ratios between the Fibonacci numbers and the Lucas numbers:

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

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:

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

src: slideplayer.com

Geometry

Geometrically, ?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. Together with the algebraic relationship between ?5 and ?, 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 ?).

Forming a dihedral right angle with the two equal squares that halve a 1:2 rectangle, it can be seen that ?5 corresponds also to the ratio between the length of a cube edge and the shortest distance from one of its vertices to the opposite one, when traversing the cube surface (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).

The number ?5 can be algebraically and geometrically related to ?2 and ?3, as it is the length of the hypotenuse of a right triangle with catheti measuring ?2 and ?3 (again, the Pythagorean theorem proves this). Right triangles of such proportions can be found inside a cube: the sides of any triangle defined by the centre point of a cube, one of its vertices, and the middle point of a side located on one the faces containing that vertex and opposite to it, are in the ratio ?2:?3:?5. This follows from the geometrical relationships between a cube and the quantities ?2 (edge-to-face-diagonal ratio, or distance between opposite edges), ?3 (edge-to-cube-diagonal ratio) and ?5 (the relationship just mentioned above).

A rectangle with side proportions 1:?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 ?1 (= 1), ?2, ?3, ?4 (= 2), ?5... and successively constructed using the diagonal of the previous root rectangle, starting from a square. A root-5 rectangle is particularly notable in that it can be split into a square and two equal golden rectangles (of dimensions ? × 1), or into two golden rectangles of different sizes (of dimensions ? × 1 and 1 × ?). It can also be decomposed as the union of two equal golden rectangles (of dimensions 1 × ?) whose intersection forms a square. All this is can be seen as the geometric interpretation of the algebraic relationships between ?5, ? and ? 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 ?5/2 to both sides.

src: i.ytimg.com

Trigonometry

Like ?2 and ?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. 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 ?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.

src: i.pinimg.com

Diophantine approximations

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 ?5 is best possible, in the sense that for any larger constant than ?5, there are some irrational numbers x for which only finitely many such approximations exist.

Closely related to this is the theorem 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 ?, at least one of the three inequalities holds:

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

And the ?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.

src: i.ytimg.com

Algebra

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

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

The field Q[?-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:

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

src: krhsmath.files.wordpress.com

Identities of Ramanujan

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

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

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

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

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

src: i.ytimg.com

See also

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

src: westernpsa.com

References

Source of article : Wikipedia