Square root of 2

The square root of 2 (approximately 1.4142) is the positive real number that, when multiplied by itself or squared, equals the number 2. It may be written in mathematics as $\sqrt{2}$ or $2^{1/2}$. It is an algebraic number, and therefore not a transcendental number. Technically, it should be called the principal square 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 a square with sides of one unit of length; this follows from the Pythagorean theorem. It was probably the first number known to be irrational. The fraction Template:Sfrac (≈ 1.4142857) is sometimes used as a good rational approximation with a reasonably small denominator.

Sequence Template:OEIS link in the On-Line Encyclopedia of Integer Sequences consists of the digits in the decimal expansion of the square root of 2, here truncated to 65 decimal places:

Template:Gaps

History

File:Ybc7289-bw.jpg

Babylonian clay tablet YBC 7289 with annotations. Besides showing the square root of 2 in sexagesimal (Template:Nowrap), the tablet also gives an example where one side of the square is 30 and the diagonal then is Template:Nowrap. The sexagesimal digit 30 can also stand for Template:Nowrap = Template:Sfrac, in which case Template:Nowrap is approximately 0.7071065.

The Babylonian clay tablet YBC 7289 (Template:Circa–1600 BC) gives an approximation of $\sqrt{2}$ in four sexagesimal figures, Template:Nowrap, which is accurate to about six decimal digits, and is the closest possible three-place sexagesimal representation of $\sqrt{2}$, representing a margin of error of only –0.000042%:

$1+\frac{24}{60}+\frac{51}{60^2}+\frac{10}{60^3}=\frac{305470}{216000}=1.41421\overline{296}.$

Another early approximation is given in ancient Indian mathematical texts, the Sulbasutras (Template:Circa–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,

$1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}=\frac{577}{408}=1.41421\overline{56862745098039}.$

This approximation, diverging from the actual value of $\sqrt{2}$ by approximately +0.07%, is the seventh in a sequence of increasingly accurate approximations based on the sequence of Pell numbers, which can be derived from the continued 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 a square is incommensurable with its side, or in modern language, that the square root of two is irrational. Little is known with certainty about the time or circumstances of this discovery, but the name of Hippasus 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 to any substantial evidence in traditional historian practice. The square root of two is occasionally called Pythagoras's number or Pythagoras's constant.

Ancient Roman architecture

In ancient Roman architecture, Vitruvius describes the use of the square root of 2 progression or ad 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 to Plato. The system was employed to build pavements by creating a square tangent to the corners of the original square at 45 degrees of it. The proportion was also used to design atria by giving them a length equal to a diagonal taken from a square, whose sides are equivalent to the intended atrium's width.

Decimal value

Computation algorithms

Template:Further There are many algorithms for approximating $\sqrt{2}$ as a ratio of integers or as a decimal. The most common algorithm for this, which is used as a basis in many computers and calculators, is the Babylonian method for computing square roots, an example of Newton'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 following recursive computation:

$a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})=\frac{a_n}{2}+\frac{1}{a_n}.$

Each iteration improves the approximation, roughly doubling the number of correct digits. Starting with $a_0=1$, the subsequent iterations yield:

$\begin{array}{cccc}{a}_1& =\frac{3}{2}& & =\mathrm{𝟏}.5,\\ a_2& =\frac{17}{12}& & =\mathrm{𝟏}\mathbf{.}\mathrm{𝟒}\mathrm{𝟏}6\dots ,\\ a_3& =\frac{577}{408}& & =\mathrm{𝟏}\mathbf{.}\mathrm{𝟒}\mathrm{𝟏}\mathrm{𝟒}\mathrm{𝟐}\mathrm{𝟏}5\dots ,\\ a_4& =\frac{665857}{470832}& & =\mathrm{𝟏}\mathbf{.}\mathrm{𝟒}\mathrm{𝟏}\mathrm{𝟒}\mathrm{𝟐}\mathrm{𝟏}\mathrm{𝟑}\mathrm{𝟓}\mathrm{𝟔}\mathrm{𝟐}\mathrm{𝟑}\mathrm{𝟕}46\dots ,\\ & ⋮\end{array}$

Rational approximations

A simple rational approximation Template:Sfrac (≈ 1.4142857) is sometimes used. Despite having a denominator of only 70, it differs from the correct value by less than Template:Sfrac (approx. Template:Val).

The next two better rational approximations are Template:Sfrac (≈ 1.4141414...) with a marginally smaller error (approx. Template:Val), and Template:Sfrac (≈ 1.4142012) with an error of approx Template:Val.

The rational approximation of the square root of two derived from four iterations of the Babylonian method after starting with Template:Math (Template:Sfrac) is too large by about Template:Val; its square is ≈ Template:Val.

Records in computation

In 1997, the value of $\sqrt{2}$ was calculated to 137,438,953,444 decimal places by Yasumasa 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 one trillion decimal places in 2010. Other mathematical constants whose decimal expansions have been calculated to similarly high precision include [[pi|Template:Pi]], [[e (mathematical constant)|Template:Mvar]], and the golden ratio. Such computations provide empirical evidence of whether these numbers are normal.

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

Date	Name	Number of digits
26 December 2023	Jordan Ranous	Template:Val
5 January 2022	Tizian Hanselmann	Template:Val
28 June 2016	Ron Watkins	Template:Val
3 April 2016	Ron Watkins	Template:Val
20 January 2016	Ron Watkins	Template:Val
9 February 2012	Alexander Yee	Template:Val
22 March 2010	Shigeru Kondo	Template:Val

Proofs of irrationality

Proof by infinite descent

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof 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.

1. Assume that $\sqrt{2}$ is a rational number, meaning that there exists a pair of integers whose ratio is exactly $\sqrt{2}$.
2. If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.
3. Then $\sqrt{2}$ can be written as an irreducible fraction $\frac{a}{b}$ such that Template:Math and Template:Math are coprime integers (having no common factor) which additionally means that at least one of Template:Math or Template:Math must be odd.
4. It follows that $\frac{a^2}{b^2}=2$ and $a^2=2b^2$.   ( Template:Math )   ( Template:Math are integers)
5. Therefore, Template:Math is even because it is equal to Template:Math. (Template:Math is necessarily even because it is 2 times another whole number.)
6. It follows that Template:Math must be even (as squares of odd integers are never even).
7. Because Template:Math is even, there exists an integer Template:Math that fulfills $a=2k$.
8. Substituting Template:Math from step 7 for Template:Math in the second equation of step 4: $2b^2=a^2=(2k{)}^2=4k^2$, which is equivalent to $b^2=2k^2$.
9. Because Template:Math is divisible by two and therefore even, and because $2k^2=b^2$, it follows that Template:Math is also even which means that Template:Math is even.
10. By steps 5 and 8, Template:Math and Template:Math are both even, which contradicts step 3 (that $\frac{a}{b}$ is irreducible).

Since we have derived a falsehood, 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 by Aristotle, in his Analytica Priora, §I.23. It appeared first as a full proof in Euclid's Elements, as proposition 117 of Book X. However, since the early 19th century, historians have agreed that this proof is an interpolation and not attributable to Euclid.

Proof using reciprocals

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

As with the proof by infinite descent, we obtain $a^2=2b^2$. Being the same quantity, each side has the same prime factorization by the fundamental 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

The irrationality of $\sqrt{2}$ also follows from the rational root theorem, which states that a rational root of a polynomial, if it exists, must be the quotient of a factor of the constant term and a factor of the leading 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 a monic polynomial with integer coefficients; 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 any natural number that is not a perfect square is irrational. For other proofs that the square root of any non-square natural number is irrational, see Quadratic irrational number or Infinite descent.

Geometric proofs

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Figure 1. Stanley Tennenbaum's geometric proof of the irrationality of Template:Sqrt

A simple proof is attributed to Stanley Tennenbaum when he was a student in the early 1950s. Assume that $\sqrt{2}=a/b$, where $a$ and $b$ are coprime positive integers. Then $a$ and $b$ are the smallest positive integers for which $a^2=2b^2$. Now consider two squares with sides $a$ and $b$, and place two copies of the smaller square inside the larger one as shown in Figure 1. The area of the square overlap region in the centre must equal the sum of the areas of the two uncovered squares. Hence there exist positive integers $p=2b-a$ and $q=a-b$ such that $p^2=2q^2$. Since it can be seen geometrically that $p<a$ and $q<b$, this contradicts the original assumption.

Figure 2. Tom Apostol's geometric proof of the irrationality of Template:Math

Tom M. Apostol made another geometric reductio ad absurdum argument showing that $\sqrt{2}$ is irrational. It is also an example of proof by infinite descent. It makes use of classic compass 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 in the previous paragraph, viewed geometrically in another way.

Let Template:Math be a right isosceles triangle with hypotenuse length Template:Math and legs Template:Math as shown in Figure 2. By the Pythagorean theorem, $\frac{m}{n}=\sqrt{2}$. Suppose Template:Math and Template:Math are integers. Let Template:Math be a ratio given in its lowest terms.

Draw the arcs Template:Math and Template:Math with centre Template:Math. Join Template:Math. It follows that Template:Math, Template:Math and Template:Math and Template:Math coincide. Therefore, the triangles Template:Math and Template:Math are congruent by SAS.

Because Template:Math is a right angle and Template:Math is half a right angle, Template:Math is also a right isosceles triangle. Hence Template:Math implies Template:Math. By symmetry, Template:Math, and Template:Math is also a right isosceles triangle. It also follows that Template:Math.

Hence, there is an even smaller right isosceles triangle, with hypotenuse length Template:Math and legs Template:Math. These values are integers even smaller than Template:Math and Template:Math and in the same ratio, contradicting the hypothesis that Template:Math is in lowest terms. Therefore, Template:Math and Template:Math cannot be both integers; hence, $\sqrt{2}$ is irrational.

Constructive proof

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". Let Template:Math and Template:Math be positive integers such that Template:Math (as Template:Math satisfies these bounds). Now Template:Math and Template:Math cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus Template:Math. Multiplying the absolute difference Template:Math by Template:Math in the numerator and denominator, we get

$|\sqrt{2}-\frac{a}{b}|=\frac{|2b^2-a^2|}{b^2(\sqrt{2}+\frac{a}{b})}\ge \frac{1}{b^2(\sqrt{2}+\frac{a}{b})}\ge \frac{1}{3b^2},$

the latter inequality being true because it is assumed that Template:Math, giving Template:Math (otherwise the quantitative apartness can be trivially established). This gives a lower bound of Template:Math for the difference Template:Math, yielding a direct proof of irrationality in its constructively stronger form, not relying on the law of excluded middle. This proof constructively exhibits an explicit discrepancy between $\sqrt{2}$ and any rational.

Proof by Pythagorean triples

This proof uses the following property of primitive Pythagorean triples:

If Template:Math, Template:Math, and Template:Math are coprime positive integers such that Template:Math, then Template:Math is never even.

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,

$\sqrt{2}=\frac{a}{b}$

where $a,b\in \mathbb{Z}$ and $\gcd (a,b)=1$

Squaring both sides,

$2=\frac{a^2}{b^2}$

$2b^2=a^2$

$b^2+b^2=a^2$

Here, Template:Math is a primitive Pythagorean triple, and from the lemma Template:Math is never even. However, this contradicts the equation Template:Math which implies that Template:Math must be even.

Multiplicative inverse

The multiplicative inverse (reciprocal) of the square root of two is a widely used constant, with the decimal value:

Template:Gaps

It is often encountered in geometry and trigonometry because the unit vector, which makes a 45° angle with the axes in a plane, has the coordinates

$(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}).$

Each coordinate satisfies

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

Properties

File:Circular and hyperbolic angle.svg

Angle size and sector area are the same when the conic radius is Template:Math. This diagram illustrates the circular and hyperbolic functions based on sector areas Template:Math.

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

$\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$

since

$(\sqrt{2}+1)(\sqrt{2}-1)=2-1=1.$

This is related to the property of silver ratios.

$\sqrt{2}$ can also be expressed in terms of copies of the imaginary unit Template:Math using only the square root and arithmetic operations, if the square root symbol is interpreted suitably for the complex numbers Template:Math and Template:Math:

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

$\sqrt{2}$ is also the only real number other than 1 whose infinite tetrate (i.e., infinite exponential tower) is equal to its square. In other words: if for Template:Math, Template:Math and Template:Math for Template:Math, the limit of Template:Math as Template:Math will be called (if this limit exists) Template:Math. Then $\sqrt{2}$ is the only number Template:Math for which Template:Math. Or symbolically:

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

$\sqrt{2}$ appears in Viète's formula for Template:Pi,

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

which is related to the formula

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

Similar in appearance but with a finite number of terms, $\sqrt{2}$ appears in various trigonometric constants:

$\begin{array}{cccccc}\sin \frac{\pi }{32}& =\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}& \sin \frac{3\pi }{16}& =\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2}}}& \sin \frac{11\pi }{32}& =\frac{1}{2}\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\\ \sin \frac{\pi }{16}& =\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}}& \sin \frac{7\pi }{32}& =\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2}}}}& \sin \frac{3\pi }{8}& =\frac{1}{2}\sqrt{2+\sqrt{2}}\\ \sin \frac{3\pi }{32}& =\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}}& \sin \frac{\pi }{4}& =\frac{1}{2}\sqrt{2}& \sin \frac{13\pi }{32}& =\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2}}}}\\ \sin \frac{\pi }{8}& =\frac{1}{2}\sqrt{2-\sqrt{2}}& \sin \frac{9\pi }{32}& =\frac{1}{2}\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2}}}}& \sin \frac{7\pi }{16}& =\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}\\ \sin \frac{5\pi }{32}& =\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}& \sin \frac{5\pi }{16}& =\frac{1}{2}\sqrt{2+\sqrt{2-\sqrt{2}}}& \sin \frac{15\pi }{32}& =\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}\end{array}$

It is not known whether $\sqrt{2}$ is a normal number, which is a stronger property than irrationality, but statistical analyses of its binary expansion are consistent with the hypothesis that it is normal to base two.

Representations

Series and product

The identity Template:Math, along with the infinite product representations for the sine and cosine, leads to products such as

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

and

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

or equivalently,

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

The number can also be expressed by taking the Taylor series of a trigonometric function. For example, the series for Template:Math gives

$\frac{1}{\sqrt{2}}=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{\left(\frac{\pi }{4}\right)}^{2k}}{(2k)!}.$

The Taylor series of Template:Math with Template:Math and using the double factorial Template:Math gives

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

The convergence of this series can be accelerated with an Euler transform, producing

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

It is not known whether $\sqrt{2}$ can be represented with a BBP-type formula. BBP-type formulas are known for Template:Math and Template:Math, however.

The number can be represented by an infinite series of Egyptian fractions, with denominators defined by 2ⁿ th terms of a Fibonacci-like recurrence relation a(n) = 34a(n−1) − a(n−2), a(0) = 0, a(1) = 6.

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

Continued fraction

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The square root of 2 and approximations by convergents of continued fractions

The square root of two has the following continued fraction representation:

$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{\ddots }}}}.$

The convergents Template:Math 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 the Pell numbers (i.e., Template:Math). The first convergents are: Template:Math and the convergent following Template:Math is Template:Math. The convergent Template:Math differs from $\sqrt{2}$ by almost exactly Template:Math, which follows from:

$|\sqrt{2}-\frac{p}{q}|=\frac{|2q^2-p^2|}{q^2(\sqrt{2}+\frac{p}{q})}=\frac{1}{q^2(\sqrt{2}+\frac{p}{q})}\approx \frac{1}{2\sqrt{2}{q}^2}$

Nested square

The following nested square expressions converge to Template:Nobr

$\begin{array}{cc}\sqrt{2}& =\frac{3}{2}-2{(\frac{1}{4}-{(\frac{1}{4}-(\frac{1}{4}-\cdots {)}^2)}^2)}^2\\ & =\frac{3}{2}-4{(\frac{1}{8}+{(\frac{1}{8}+(\frac{1}{8}+\cdots {)}^2)}^2)}^2.\end{array}$

Applications

Paper size

File:A size illustration2.svg

The A series of paper sizes

In 1786, German physics professor Georg Christoph Lichtenberg 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 standardised paper sizes at the beginning of the 20th century, they used Lichtenberg's ratio to create the "A" series of paper sizes. Today, the (approximate) aspect ratio of paper sizes under ISO 216 (A4, A0, etc.) is 1:$\sqrt{2}$.

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

$R^{\prime }=\frac{S}{L/2}=\frac{2S}{L}=\frac{2}{(L/S)}=\frac{2}{\sqrt{2}}=\sqrt{2}=R.$

Physical sciences

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

• The square root of two is the frequency ratio of a tritone interval in twelve-tone equal temperament music.
• The square root of two forms the relationship of f-stops in photographic lenses, which in turn means that the ratio of areas between two successive apertures is 2.
• The celestial latitude (declination) of the Sun during a planet's astronomical cross-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."

See also

• List of mathematical constants
• Square root of 3, Template:Math
• Square root of 5, Template:Math
• Gelfond–Schneider constant, Template:Math
• Silver ratio, Template:Math

References

• Template:Citation
• Template:Citation
• Template:Citation

External links

• Template:Citation.
• The Square Root of Two to 5 million digits by Jerry Bonnell and Robert J. Nemiroff. May, 1994.
• Square root of 2 is irrational, a collection of proofs
• Template:Citation
• Template:Tmath Search Engine 2 billion searchable digits of Template:Radic, Template:Pi and Template:Mvar