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Square root of 2

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From Wikipedia, the free encyclopedia

Unique positive real number which when multiplied by itself gives 2

"Pythagoras's constant" redirects here; not to be confused withPythagoras number.

Square root of 2
Representations
The square root of 2 is equal to the length of thehypotenuse of anisosceles right triangle with legs of length 1.
Rationality	Irrational
Decimal	1.4142135623730950488...
Continued fraction	$1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+\ddots }}}}}}}}$

Thesquare root of 2 (approximately 1.4142) is the positivereal number that, when multiplied by itself or squared, equals thenumber 2. It may be written as ${\sqrt {2}}$ or $2^{1/2}$ . It is analgebraic number, and therefore not atranscendental number. Technically, it should be called theprincipalsquare root of 2, to distinguish it from the negative number with the same property.

Geometrically, the square root of 2 is the length of a diagonal across asquare with sides of one unit of length; this follows from thePythagorean theorem. It was probably the first number known to beirrational.^[1] The fraction⁠99/70⁠ (≈1.4142857) is sometimes used as a goodrational approximation with a reasonably smalldenominator.

SequenceA002193 in theOn-Line Encyclopedia of Integer Sequences consists of the digits in thedecimal expansion of the square root of 2, here truncated to 30 decimal places:^[2]

1.414213562373095048801688724209

History

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Babylonian clay tabletYBC 7289 with annotations. Besides showing the square root of 2 insexagesimal (1 24 51 10), the tablet also gives an example where one side of the square is 30 and the diagonal then is42 25 35. The sexagesimal digit 30 can also stand for0 30 =⁠1/2⁠, in which case0 42 25 35 is approximately 0.7071065.

TheBabylonian clay tabletYBC 7289 (c. 1800–1600 BC) gives an approximation of ${\sqrt {2}}$ in four sexagesimal figures,1 24 51 10, which is accurate to about sixdecimal digits, and is the closest possible three-place sexagesimal representation of ${\sqrt {2}}$ , representing a margin of error of only –0.000042%:^[3] $1+{\frac {24}{60}}+{\frac {51}{60^{2}}}+{\frac {10}{60^{3}}}={\frac {305470}{216000}}=1.41421{\overline {296}}.$

Another early approximation is given inancient Indian mathematical texts, theSulbasutras (c. 800–200 BC), as follows: "Increase the length [of the side] by its third and this third by its own fourth less the thirty-fourth part of that fourth." That is,[4] $1+{\frac {1}{3}}+{\frac {1}{3\times 4}}-{\frac {1}{3\times 4\times 34}}={\frac {577}{408}}=1.41421{\overline {56862745098039}}.$

This approximation is the seventh in a sequence of increasingly accurate approximations based on the sequence ofPell numbers, which can be derived from thecontinued fraction expansion of ${\sqrt {2}}$ . Despite having a smaller denominator, it is only slightly less accurate than the Babylonian approximation.

Pythagoreans discovered that the diagonal of asquare is incommensurable with its side, or in modern language, that the square root of two isirrational. Little is known with certainty about the time or circumstances of this discovery, but the name ofHippasus of Metapontum is often mentioned. For a while, the Pythagoreans treated as an official secret the discovery that the square root of two is irrational, and, according to legend, Hippasus was murdered for divulging it, though this has little if any substantial evidence in traditional historical practice.^[5]^[6] The square root of two is occasionally calledPythagoras's number^[7] orPythagoras's constant.

Ancient Roman architecture

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Inancient Roman architecture,Vitruvius describes the use of the square root of 2 progression orad quadratum technique. It consists basically in a geometric, rather than arithmetic, method to double a square, in which the diagonal of the original square is equal to the side of the resulting square. Vitruvius attributes the idea toPlato. The system was employed to build pavements by creating a squaretangent to the corners of the original square at 45 degrees of it. The proportion was also used to designatria by giving them a length equal to a diagonal taken from a square, whose sides are equivalent to the intended atrium's width.^[8]

Decimal value

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Computation algorithms

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Further information:Methods of computing square roots

There are manyalgorithms for approximating ${\sqrt {2}}$ as a ratio ofintegers or as a decimal. The most common algorithm for this, which is used as a basis in many computers and calculators, is theBabylonian method^[9] for computing square roots, an example ofNewton's method for computing roots of arbitrary functions. It goes as follows:

First, pick a guess, $a_{0}>0$ ; the value of the guess affects only how many iterations are required to reach an approximation of a certain accuracy. Then, using that guess, iterate through the followingrecursive computation:

a_{n+1}={\frac {1}{2}}\left(a_{n}+{\dfrac {2}{a_{n}}}\right)={\frac {a_{n}}{2}}+{\frac {1}{a_{n}}}.

Each iteration improves the approximation, roughly doubling the number of correct digits. Starting with $a_{0}=1$ , the subsequent iterations yield:

{\begin{alignedat}{3}a_{1}&={\tfrac {3}{2}}&&=\mathbf {1} .5,\\a_{2}&={\tfrac {17}{12}}&&=\mathbf {1.41} 6\ldots ,\\a_{3}&={\tfrac {577}{408}}&&=\mathbf {1.41421} 5\ldots ,\\a_{4}&={\tfrac {665857}{470832}}&&=\mathbf {1.41421356237} 46\ldots ,\\&\qquad \vdots \end{alignedat}}

Rational approximations

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The Babylonians had approximated the number as $1+{\frac {24}{60}}+{\frac {51}{60^{2}}}+{\frac {10}{60^{3}}}=1.41421{\overline {296}}$ .^[3]^[10]

The rational approximation⁠99/70⁠ (≈1.4142857) differs from the correct value by less than⁠1/10,000⁠ (approx.+0.72×10⁻⁴). Likewise,⁠140/99⁠ (≈1.4141414...) has a marginally smaller error (approx.−0.72×10⁻⁴), and⁠239/169⁠ (≈1.4142012) has an error of approximately−0.12×10⁻⁴.

The rational approximation of the square root of two derived from four iterations of the Babylonian method after starting witha₀ = 1 (⁠665,857/470,832⁠) is too large by about1.6×10⁻¹²; its square is ≈ 2.0000000000045.^[10]

Records in computation

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In 1997, the value of ${\sqrt {2}}$ was calculated to 137,438,953,444 decimal places byYasumasa Kanada's team. In February 2006, the record for the calculation of ${\sqrt {2}}$ was eclipsed with the use of a home computer. Shigeru Kondo calculated onetrillion decimal places in 2010.^[11] Othermathematical constants whose decimal expansions have been calculated to similarly high precision includeπ,e, and thegolden ratio.^[12] Such computations provide empirical evidence of whether these numbers arenormal.

This is a table of recent records in calculating the digits of ${\sqrt {2}}$ .^[12]

Date	Name	Number of digits
4 April 2025	Teck Por Lim	24000000000000
26 December 2023	Jordan Ranous	20000000000000
5 January 2022	Tizian Hanselmann	10000000001000
28 June 2016	Ron Watkins	10000000000000
3 April 2016	Ron Watkins	5000000000000
20 January 2016	Ron Watkins	2000000000100
9 February 2012	Alexander Yee	2000000000050
22 March 2010	Shigeru Kondo	1000000000000

Proofs of irrationality

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Proof by infinite descent

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One proof of the number's irrationality is the followingproof by infinite descent. It is also aproof of a negation by refutation: it proves the statement " ${\sqrt {2}}$ is not rational" by assuming that it is rational and then deriving a falsehood.

Assume that ${\sqrt {2}}$ is a rational number, meaning that there exists a pair of integers whose ratio is exactly ${\sqrt {2}}$ .
If the two integers have a commonfactor, it can be eliminated using theEuclidean algorithm.
Then ${\sqrt {2}}$ can be written as anirreducible fraction ${\frac {a}{b}}$ such thata andb arecoprime integers (having no common factor) which additionally means that at least one ofa orb must beodd.
It follows that ${\frac {a^{2}}{b^{2}}}=2$ and $a^{2}=2b^{2}$ . ( (⁠a/b⁠)ⁿ =⁠aⁿ/bⁿ⁠ ) (a² andb² are integers)
Therefore,a² iseven because it is equal to2b². (2b² is necessarily even because it is 2 times another whole number.)
It follows thata must be even (as squares of odd integers are never even).
Becausea is even, there exists an integerk that fulfills $a=2k$ .
Substituting2k from step 7 fora in the second equation of step 4: $2b^{2}=a^{2}=(2k)^{2}=4k^{2}$ , which is equivalent to $b^{2}=2k^{2}$ .
Because2k² is divisible by two and therefore even, and because $2k^{2}=b^{2}$ , it follows thatb² is also even which means thatb is even.
By steps 5 and 8,a andb are both even, which contradicts step 3 (that ${\frac {a}{b}}$ is irreducible).

Since a falsehood has been derived, the assumption (1) that ${\sqrt {2}}$ is a rational number must be false. This means that ${\sqrt {2}}$ is not a rational number; that is to say, ${\sqrt {2}}$ is irrational.

This proof was hinted at byAristotle, in hisAnalytica Priora, §I.23.^[13] It appeared first as a full proof inEuclid'sElements, as proposition 117 of Book X. However, since the early 19th century, historians have agreed that this proof is aninterpolation and not attributable to Euclid.^[14]

Proof using reciprocals

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Assume by way of contradiction that ${\sqrt {2}}$ were rational. Then we may write ${\sqrt {2}}+1={\frac {q}{p}}$ as an irreducible fraction in lowest terms, with coprime positive integers $q>p$ . Since $({\sqrt {2}}-1)({\sqrt {2}}+1)=2-1^{2}=1$ , it follows that ${\sqrt {2}}-1$ can be expressed as the irreducible fraction ${\frac {p}{q}}$ . However, since ${\sqrt {2}}-1$ and ${\sqrt {2}}+1$ differ by an integer, it follows that the denominators of their irreducible fraction representations must be the same, i.e. $q=p$ . This gives the desired contradiction.

Proof by unique factorization

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As with the proof by infinite descent, we obtain $a^{2}=2b^{2}$ . Being the same quantity, each side has the sameprime factorization by thefundamental theorem of arithmetic, and in particular, would have to have the factor 2 occur the same number of times. However, the factor 2 appears an odd number of times on the right, but an even number of times on the left—a contradiction.

Application of the rational root theorem

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The irrationality of ${\sqrt {2}}$ also follows from therational root theorem, which states that a rationalroot of apolynomial, if it exists, must be thequotient of a factor of the constant term and a factor of theleading coefficient. In the case of $p(x)=x^{2}-2$ , the only possible rational roots are $\pm 1$ and $\pm 2$ . As ${\sqrt {2}}$ is not equal to $\pm 1$ or $\pm 2$ , it follows that ${\sqrt {2}}$ is irrational. This application also invokes the integer root theorem, a stronger version of the rational root theorem for the case when $p(x)$ is amonic polynomial with integercoefficients; for such a polynomial, all roots are necessarily integers (which ${\sqrt {2}}$ is not, as 2 is not a perfect square) or irrational.

The rational root theorem (or integer root theorem) may be used to show that any square root of anynatural number that is not a perfect square is irrational. For other proofs that the square root of any non-square natural number is irrational, seeQuadratic irrational number orInfinite descent.

Geometric proofs

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Tennenbaum's proof

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Figure 1. Stanley Tennenbaum's geometric proof of theirrationality of√2

A simple proof is attributed toStanley Tennenbaum when he was a student in the early 1950s.^[15]^[16] Assume that ${\sqrt {2}}=a/b$ , where $a {\displaystyle a}$ and $b {\displaystyle b}$ are coprime positive integers. Then $a {\displaystyle a}$ and $b {\displaystyle b}$ are the smallest positive integers for which $a^{2}=2b^{2}$ . Geometrically, this implies that a square with side length $a {\displaystyle a}$ will have an area equal to two squares of (lesser) side length $b {\displaystyle b}$ . Call these squares A and B. We can draw these squares and compare their areas - the simplest way to do so is to fit the two B squares into the A squares. When we try to do so, we end up with the arrangement in Figure 1., in which the two B squares overlap in the middle and two uncovered areas are present in the top left and bottom right. In order to assert $a^{2}=2b^{2}$ , we would need to show that the area of the overlap is equal to the area of the two missing areas, i.e. $(2b-a)^{2}$ = $2(a-b)^{2}$ . In other terms, we may refer to the side lengths of the overlap and missing areas as $p=2b-a$ and $q=a-b$ , respectively, and thus we have $p^{2}=2q^{2}$ . But since we can see from the diagram that $p<a$ and $q<b$ , and we know that $p {\displaystyle p}$ and $q {\displaystyle q}$ are integers from their definitions in terms of $a {\displaystyle a}$ and $b {\displaystyle b}$ , this means that we are in violation of the original assumption that $a {\displaystyle a}$ and $b {\displaystyle b}$ are the smallest positive integers for which $a^{2}=2b^{2}$ .

Hence, even in assuming that $a {\displaystyle a}$ and $b {\displaystyle b}$ are the smallest positive integers for which $a^{2}=2b^{2}$ , we may prove that there exists a smaller pair of integers $p {\displaystyle p}$ and $q {\displaystyle q}$ which satisfy the relation. This contradiction within the definition of $a {\displaystyle a}$ and $b {\displaystyle b}$ implies that they cannot exist, and thus ${\sqrt {2}}$ must be irrational.

Apostol's proof

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Figure 2. Tom Apostol's geometric proof of the irrationality of√2

Tom M. Apostol made another geometricreductio ad absurdum argument showing that ${\sqrt {2}}$ is irrational.^[17] It is also an example of proof by infinite descent. It makes use ofcompass and straightedge construction, proving the theorem by a method similar to that employed by ancient Greek geometers. It is essentially the same algebraic proof as Tennebaum's proof, viewed geometrically in another way.

Let△ ABC be a right isosceles triangle with hypotenuse lengthm and legsn as shown in Figure 2. By thePythagorean theorem, ${\frac {m}{n}}={\sqrt {2}}$ . Supposem andn are integers. Letm:n be aratio given in itslowest terms.

Draw the arcsBD andCE with centreA. JoinDE. It follows thatAB =AD,AC =AE and∠BAC and∠DAE coincide. Therefore, thetrianglesABC andADE arecongruent bySAS.

Because∠EBF is a right angle and∠BEF is half a right angle,△ BEF is also a right isosceles triangle. HenceBE =m −n impliesBF =m −n. By symmetry,DF =m −n, and△ FDC is also a right isosceles triangle. It also follows thatFC =n − (m −n) = 2n −m.

Hence, there is an even smaller right isosceles triangle, with hypotenuse length2n −m and legsm −n. These values are integers even smaller thanm andn and in the same ratio, contradicting the hypothesis thatm:n is in lowest terms. Therefore,m andn cannot be both integers; hence, ${\sqrt {2}}$ is irrational.

Constructive proof

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While the proofs by infinite descent are constructively valid when "irrational" is defined to mean "not rational", we can obtain a constructively stronger statement by using a positive definition of "irrational" as "quantifiably apart from every rational". Leta andb be positive integers such that1<⁠a/b⁠< 3/2 (as1<2< 9/4 satisfies these bounds). Now2b² anda² cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus|2b² −a²| ≥ 1. Multiplying the absolute difference|√2 −⁠a/b⁠| byb²(√2 +⁠a/b⁠) in the numerator and denominator, we get^[18]

\left|{\sqrt {2}}-{\frac {a}{b}}\right|={\frac {|2b^{2}-a^{2}|}{b^{2}\!\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{b^{2}\!\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{3b^{2}}},

the latterinequality being true because it is assumed that1<⁠a/b⁠< 3/2, giving⁠a/b⁠ + √2 ≤ 3 (otherwise the quantitative apartness can be trivially established). This gives a lower bound of⁠1/3b²⁠ for the difference|√2 −⁠a/b⁠|, yielding a direct proof of irrationality in its constructively stronger form, not relying on thelaw of excluded middle.^[19] This proof constructively exhibits an explicit discrepancy between ${\sqrt {2}}$ and any rational.

Proof by Pythagorean triples

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This proof uses the following property of primitivePythagorean triples:

Ifa,b, andc are coprime positive integers such thata² +b² =c², thenc is never even.^[20]

This lemma can be used to show that two identical perfect squares can never be added to produce another perfect square.

Suppose the contrary that ${\sqrt {2}}$ is rational. Therefore,

{\sqrt {2}}={a \over b}

where

a,b\in \mathbb {Z}

and

\gcd(a,b)=1

Squaring both sides,

2={a^{2} \over b^{2}}

2b^{2}=a^{2}

b^{2}+b^{2}=a^{2}

Here,(b,b,a) is a primitive Pythagorean triple, and from the lemmaa is never even. However, this contradicts the equation2b² =a² which implies thata must be even.

Multiplicative inverse

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Themultiplicative inverse (reciprocal) of the square root of two is a widely usedconstant, with the decimal value:^[21]

0.70710678118654752440084436210484903928483593768847...

It is often encountered ingeometry andtrigonometry because theunit vector, which makes a 45°angle with the axes in aplane, has the coordinates

\left({\frac {\sqrt {2}}{2}},{\frac {\sqrt {2}}{2}}\right)\!.

Each coordinate satisfies

{\frac {\sqrt {2}}{2}}={\sqrt {\tfrac {1}{2}}}={\frac {1}{\sqrt {2}}}=\sin 45^{\circ }=\cos 45^{\circ }.

Properties

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Angle size and sectorarea are the same when the conic radius is√2. This diagram illustrates the circular and hyperbolic functions based on sector areasu.

One interesting property of ${\sqrt {2}}$ is

\!\ {1 \over {{\sqrt {2}}-1}}={\sqrt {2}}+1

since

\left({\sqrt {2}}+1\right)\!\left({\sqrt {2}}-1\right)=2-1=1.

This is related to the property ofsilver ratios.

${\sqrt {2}}$ can also be expressed in terms of copies of theimaginary uniti using only thesquare root andarithmetic operations, if the square root symbol is interpreted suitably for thecomplex numbersi and−i:

{\frac {{\sqrt {i}}+i{\sqrt {i}}}{i}}{\text{ and }}{\frac {{\sqrt {-i}}-i{\sqrt {-i}}}{-i}}

${\sqrt {2}}$ is also the only real number other than 1 whose infinitetetration (i.e., infinite exponential tower) is equal to its square. In other words: if forc > 1,x₁ =c andx_n+1 =c^x_n forn > 1, thelimit ofx_n asn → ∞ will be called (if this limit exists)f(c). Then ${\sqrt {2}}$ is the only numberc > 1 for whichf(c) =c². Or symbolically:

{\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{~\cdot ^{~\cdot ^{~\cdot }}}}}=2.

${\sqrt {2}}$ appears inViète's formula forπ,

{\frac {2}{\pi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}}}\cdots ,

which is related to the formula^[22]

\pi =\lim _{m\to \infty }2^{m}\underbrace {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+\cdots +{\sqrt {2}}}}}}}}}} _{m{\text{ square roots}}}\,.

Similar in appearance but with a finite number of terms, ${\sqrt {2}}$ appears in varioustrigonometric constants:^[23]

{\begin{aligned}\sin {\frac {\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {11\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}&\quad \sin {\frac {7\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\\[6pt]\sin {\frac {3\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {\pi }{4}}&={\tfrac {1}{2}}{\sqrt {2}}&\quad \sin {\frac {13\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}&\quad \sin {\frac {9\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {7\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}\\[6pt]\sin {\frac {5\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {5\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {15\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}\end{aligned}}

It is not known whether ${\sqrt {2}}$ is anormal number, which is a stronger property than irrationality, but statistical analyses of itsbinary expansion are consistent with the hypothesis that it is normal tobase two.^[24]

Representations

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Series and product

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The identitycos ⁠π/4⁠ = sin ⁠π/4⁠ =⁠1/√2⁠, along with the infinite product representations for thesine and cosine, leads to products such as

{\frac {1}{\sqrt {2}}}=\prod _{k=0}^{\infty }\left(1-{\frac {1}{(4k+2)^{2}}}\right)=\left(1-{\frac {1}{4}}\right)\!\left(1-{\frac {1}{36}}\right)\!\left(1-{\frac {1}{100}}\right)\cdots

and

{\sqrt {2}}=\prod _{k=0}^{\infty }{\frac {(4k+2)^{2}}{(4k+1)(4k+3)}}=\left({\frac {2\cdot 2}{1\cdot 3}}\right)\!\left({\frac {6\cdot 6}{5\cdot 7}}\right)\!\left({\frac {10\cdot 10}{9\cdot 11}}\right)\!\left({\frac {14\cdot 14}{13\cdot 15}}\right)\cdots

or equivalently,

{\sqrt {2}}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\!\left(1-{\frac {1}{3}}\right)\!\left(1+{\frac {1}{5}}\right)\!\left(1-{\frac {1}{7}}\right)\cdots .

The number can also be expressed by taking theTaylor series of atrigonometric function. For example, the series forcos ⁠π/4⁠ gives

{\frac {1}{\sqrt {2}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}{\bigl (}{\frac {\pi }{4}}{\bigr )}^{2k}}{(2k)!}}.

The Taylor series of ${\sqrt {1+x}}$ withx = 1 and using thedouble factorialn!! gives

{\sqrt {2}}=\sum _{k=0}^{\infty }(-1)^{k+1}{\frac {(2k-3)!!}{(2k)!!}}=1+{\frac {1}{2}}-{\frac {1}{2\cdot 4}}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}}+\cdots =1+{\frac {1}{2}}-{\frac {1}{8}}+{\frac {1}{16}}-{\frac {5}{128}}+{\frac {7}{256}}+\cdots .

Theconvergence of this series can be accelerated with anEuler transform, producing

{\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{2^{3k+1}{(k!)}^{2}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .

It is not known whether ${\sqrt {2}}$ can be represented with aBBP-type formula. BBP-type formulas are known for⁠ $\pi {\sqrt {2}}$ ⁠ and ${\sqrt {2}}\ln \left(1+{\sqrt {2}}~\!\right)$ , however.^[25]

The number can be represented by an infinite series ofEgyptian fractions, with denominators defined by 2ⁿth terms of aFibonacci-likerecurrence relationa(n) = 34a(n−1) −a(n−2),a(0) = 0,a(1) = 6:^[26]

{\sqrt {2}}={\frac {3}{2}}-{\frac {1}{2}}\sum _{n=0}^{\infty }{\frac {1}{a(2^{n})}}={\frac {3}{2}}-{\frac {1}{2}}\left({\frac {1}{6}}+{\frac {1}{204}}+{\frac {1}{235416}}+\dots \right).

Continued fraction

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The square root of two has the followingcontinued fraction representation:

{\sqrt {2}}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{\ddots }}}}}}}}.

Theconvergents⁠p/q⁠ formed by truncating this representation form a sequence of fractions that approximate the square root of two to increasing accuracy, and that are described by thePell numbers (i.e.,p² − 2q² = ±1). The first convergents are:⁠1/1⁠,⁠3/2⁠,⁠7/5⁠,⁠17/12⁠,⁠41/29⁠,⁠99/70⁠,⁠239/169⁠,⁠577/408⁠ and the convergent following⁠p/q⁠ is⁠p + 2q/p +q⁠. The convergent⁠p/q⁠ differs from ${\sqrt {2}}$ by almost exactly ${\frac {1}{2{\sqrt {2}}q^{2}}}$ , which follows from:

\left|{\sqrt {2}}-{\frac {p}{q}}\right|={\frac {|2q^{2}-p^{2}|}{q^{2}\!\left({\sqrt {2}}+{\frac {p}{q}}\right)}}={\frac {1}{q^{2}\!\left({\sqrt {2}}+{\frac {p}{q}}\right)}}\thickapprox {\frac {1}{2{\sqrt {2}}q^{2}}}

Nested square

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The following nested square expressions converge to ${\textstyle {\sqrt {2}}}$ :

{\begin{aligned}{\sqrt {2}}&={\tfrac {3}{2}}-2\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-{\bigl (}{\tfrac {1}{4}}-\cdots {\bigr )}^{2}\right)^{2}\right)^{2}\\[10mu]&={\tfrac {3}{2}}-4\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+{\bigl (}{\tfrac {1}{8}}+\cdots {\bigr )}^{2}\right)^{2}\right)^{2}.\end{aligned}}

^{[citation needed]}

Applications

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Paper size

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In 1786, German physics professorGeorg Christoph Lichtenberg^[27] found that any sheet of paper whose long edge is ${\sqrt {2}}$ times longer than its short edge could be folded in half and aligned with its shorter side to produce a sheet with exactly the same proportions as the original. This ratio of lengths of the longer over the shorter side guarantees that cutting a sheet in half along a line results in the smaller sheets having the same (approximate) ratio as the original sheet. When Germany standardisedpaper sizes at the beginning of the 20th century, they used Lichtenberg's ratio to create the"A" series of paper sizes.^[27] Today, the (approximate)aspect ratio of paper sizes underISO 216 (A4, A0, etc.) is 1: ${\sqrt {2}}$ . A0 is 841 mm × 1189 mm, giving a ratio of 0.707317..., around 0.0297% larger than the exact value.^[28]

Proof:

Let $S=$ shorter length and $L=$ longer length of the sides of a sheet of paper, with

R={\frac {L}{S}}={\sqrt {2}}

as required by ISO 216.

Let $R'={\frac {L'}{S'}}$ be the analogous ratio of the halved sheet, then

R'={\frac {S}{L/2}}={\frac {2S}{L}}={\frac {2}{(L/S)}}={\frac {2}{\sqrt {2}}}={\sqrt {2}}=R.

Physical sciences

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Distances between vertices of a doubleunit cube are square roots of the first sixnatural numbers. (√7 is not possible due toLegendre's three-square theorem.)

There are some interesting properties involving the square root of 2 in thephysical sciences:

The square root of two is thefrequency ratio of atritone interval in twelve-toneequal temperament music.
The square root of two forms the relationship off-stops in photographic lenses, which in turn means that the ratio ofareas between two successiveapertures is 2.
The celestial latitude (declination) of the Sun during a planet's astronomicalcross-quarter day points equals the tilt of the planet's axis divided by ${\sqrt {2}}$ .
In the brain there are lattice cells, discovered in 2005 by a group led by May-Britt and Edvard Moser. "The grid cells were found in the cortical area located right next to the hippocampus [...] At one end of this cortical area the mesh size is small and at the other it is very large. However, the increase in mesh size is not left to chance, but increases by the squareroot of two from one area to the next."^[29]

Notes

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^Fowler, David H. (1994). "The Story of the Discovery of Incommensurability, Revisited". In Gavroglu, Kostas; Christianidis, Jean; Nicolaidis, Efthymios (eds.).Trends in the Historiography of Science. Boston Studies in the Philosophy of Science. Vol. 151. Dortrecht: Springer. pp. 221–236.doi:10.1007/978-94-017-3596-4.ISBN 978-9048142644.
^Sloane, N. J. A. (ed.)."Sequence A002193 (Decimal expansion of square root of 2)".TheOn-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved2020-08-10.
^^a ^bFowler, David;Robson, Eleanor (1998)."Square root approximations in old Babylonian mathematics: YBC 7289 in context".Historia Mathematica.25 (4):366–378.doi:10.1006/hmat.1998.2209.MR 1662496. See p. 368.
Photograph, illustration, and description of theroot(2) tablet from the Yale Babylonian Collection Archived 2012-08-13 at theWayback Machine
High resolution photographs, descriptions, and analysis of theroot(2) tablet (YBC 7289) from the Yale Babylonian Collection
^Henderson, David W. (2000). "Square roots in the Śulba Sūtras". In Gorini, Catherine A. (ed.).Geometry At Work: Papers in Applied Geometry. Mathematical Association of America Notes. Vol. 53. Washington, D.C.:The Mathematical Association of America. pp. 39–45.ISBN 978-0883851647.
^"The Dangerous Ratio".nrich.maths.org. Retrieved2023-09-18.
^Von Fritz, Kurt (1945). "The Discovery of Incommensurability by Hippasus of Metapontum".Annals of Mathematics.46 (2):242–264.doi:10.2307/1969021.ISSN 0003-486X.JSTOR 1969021.
^Conway, John H.;Guy, Richard K. (1996).The Book of Numbers. New York: Copernicus. p. 25.ISBN 978-1461240723.
^Williams, Kim; Ostwald, Michael (2015).Architecture and Mathematics from Antiquity to the Future: Volume I: Antiquity to the 1500s. Birkhäuser. p. 204.ISBN 9783319001371.
^Although the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of ${\sqrt {2}}$ seen on tablet YBC 7289. Fowler and Robson offer informed and detailed conjectures.
Fowler and Robson, p. 376. Flannery, p. 32, 158.
^^a ^bBaez, John C. (2 Dec 2011)."Babylon and the Square Root of 2".Azimuth. Retrieved2025-12-25.
^Constants and Records of Computation. Numbers.computation.free.fr. 12 Aug 2010.Archived from the original on 2012-03-01. Retrieved2012-09-07.
^^a ^bRecords set by y-cruncher.Archived from the original on 2022-04-07. Retrieved2022-04-07.
^All that Aristotle says, while writing aboutproofs by contradiction, is that "the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate".
^The edition of the Greek text of theElements published by E. F. August inBerlin in 1826–1829 already relegates this proof to an Appendix. The same thing occurs withJ. L. Heiberg's edition (1883–1888).
^Miller, Steven J.; Montague, David (Apr 2012). "Picturing Irrationality".Mathematics Magazine. Vol. 85, no. 2. pp. 110–114.doi:10.4169/math.mag.85.2.110.JSTOR 10.4169/math.mag.85.2.110.
^Yanofsky, Noson S. (May–Jun 2016). "Paradoxes, Contradictions, and the Limits of Science".American Scientist. Vol. 103, no. 3. pp. 166–173.JSTOR 44808923.
^Apostol, Tom M. (2000). "Irrationality of The Square Root of Two – A Geometric Proof".The American Mathematical Monthly.107 (9):841–842.doi:10.2307/2695741.JSTOR 2695741.
^SeeKatz, Karin Usadi;Katz, Mikhail G. (2011). "Meaning in Classical Mathematics: Is it at Odds with Intuitionism?".Intellectica.56 (2): 223–302 (see esp. Section 2.3, footnote 15).arXiv:1110.5456.Bibcode:2011arXiv1110.5456U.
^Bishop, Errett (1985). "Schizophrenia in Contemporary Mathematics.". InRosenblatt, Murray (ed.).Errett Bishop: Reflections on Him and His Research. Contemporary Mathematics. Vol. 39. Providence, RI:American Mathematical Society. pp. 1–32.doi:10.1090/conm/039/788163.ISBN 0821850407.ISSN 0271-4132.
^Sierpiński, Wacław (2003).Pythagorean Triangles. Translated by Sharma, Ambikeshwa. Mineola, NY: Dover. pp. 4–6.ISBN 978-0486432786.
^Sloane, N. J. A. (ed.)."Sequence A010503 (Decimal expansion of 1/sqrt(2))".TheOn-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved2024-11-03.
^Courant, Richard; Robbins, Herbert (1941).What is mathematics? An Elementary Approach to Ideas and Methods. London: Oxford University Press. p. 124.
^Julian D. A. WisemanSin and cos in surds Archived 2009-05-06 at theWayback Machine
^Good, I. J.; Gover, T. N. (1967). "The generalized serial test and the binary expansion of ${\sqrt {2}}$ ".Journal of the Royal Statistical Society, Series A.130 (1):102–107.doi:10.2307/2344040.JSTOR 2344040.
^Bailey, David H. (13 Feb 2011).A Compendium of BBP-Type Formulas for Mathematical Constants(PDF).Archived(PDF) from the original on 2011-06-10. Retrieved2010-04-30.
^Sloane, N. J. A. (ed.)."Sequence A082405 (a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=6)".TheOn-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved2016-09-05.
^^a ^bHouston, Keith (2016).The Book: A Cover-to-Cover Exploration of the Most Powerful Object of Our Time. W. W. Norton & Company. p. 324.ISBN 978-0393244809.
^"International Paper Sizes & Formats".Paper Sizes. Retrieved2020-06-29.
^Nordengen, Kaja (2016).The Book: Hjernen er sternen. 2016 Kagge Forlag AS. p. 81.ISBN 978-82-489-2018-2.

References

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Aristotle (1938) [c. 350 BC].Categories; On Interpretation; Prior Analytics. Greek text with translation. Loeb Classical Library. Vol. 325. Translated by H. P. Cooke; Hugh Tredennick. Cambridge, MA: Harvard University Press.Prior Analytics § I.23.ISBN 9780674993594.{{citation}}:ISBN / Date incompatibility (help)
Flannery, David (2006).The Square Root of 2: A Dialogue Concerning a Number and a Sequence. New York: Copernicus Books.ISBN 978-0387202204.
Fowler, David;Robson, Eleanor (1998). "Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context".Historia Mathematica.25 (4):366–378.doi:10.1006/hmat.1998.2209.

External links

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Gourdon, X.; Sebah, P. (2001). "Pythagoras' Constant: ${\sqrt {2}}$ ".Numbers, Constants and Computation..
The Square Root of Two to 5 million digits by Jerry Bonnell andRobert J. Nemiroff. May, 1994.
Square root of 2 is irrational, a collection of proofs
Haran, Brady (27 Jan 2012).Root 2 (video). Numberphile. featuring Grime, James; Bowley, Roger.
⁠ ${\sqrt {2}}$ ⁠ Search Engine 2 billion searchable digits of √2, π, ande

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