# 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 ⁠99/70⁠ (≈ 1.4142857) is sometimes used as a good rational approximation with a reasonably small denominator.
Sequence A002193 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:
    1.41421356237309504880168872420969807856967187537694807317667973799

## History

The Babylonian clay tablet YBC 7289 (c. 1800–1600 BC) gives an approximation of ${\sqrt {2}}$ in four sexagesimal figures, 1 24 51 10, 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 (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,
    $1+{\frac {1}{3}}+{\frac {1}{3\times 4}}-{\frac {1}{3\times 4\times 34}}={\frac {577}{408}}=1.41421{\overline {56862745098039}}.$
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

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

A simple rational approximation ⁠99/70⁠ (≈ 1.4142857) is sometimes used. Despite having a denominator of only 70, it differs from the correct value by less than ⁠1/10,000⁠ (approx. +0.72×10$^{−4}$).
The next two better rational approximations are ⁠140/99⁠ (≈ 1.4141414...) with a marginally smaller error (approx. −0.72×10$^{−4}$), and ⁠239/169⁠ (≈ 1.4142012) with an error of approx −0.12×10$^{−4}$.
The rational approximation of the square root of two derived from four iterations of the Babylonian method after starting with a$_0$ = 1 (⁠665,857/470,832⁠) is too large by about 1.6×10$^{−12}$; its square is ≈ 2.0000000000045.

### 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 π, e, 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 | 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
### 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.
* 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 common factor, it can be eliminated using the Euclidean algorithm.
* Then ${\sqrt {2}}$ can be written as an irreducible fraction ${\frac {a}{b}}$ such that a and b are coprime integers (having no common factor) which additionally means that at least one of a or b must be odd.
* It follows that ${\frac {a^{2}}{b^{2}}}=2$ and $a^{2}=2b^{2}$.   ( (⁠a/b⁠)$^n$ = ⁠a$^n$/b$^n$⁠ )   ( a$^2$ and b$^2$ are integers)
* Therefore, a$^2$ is even because it is equal to 2b$^2$. (2b$^2$ is necessarily even because it is 2 times another whole number.)
* It follows that a must be even (as squares of odd integers are never even).
* Because a is even, there exists an integer k that fulfills $a=2k$.
* Substituting 2k from step 7 for a in the second equation of step 4: $2b^{2}=a^{2}=(2k)^{2}=4k^{2}$, which is equivalent to $b^{2}=2k^{2}$.
* Because 2k$^2$ is divisible by two and therefore even, and because $2k^{2}=b^{2}$, it follows that b$^2$ is also even which means that b is even.
* By steps 5 and 8, a and b 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

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.
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 △ ABC be a right isosceles triangle with hypotenuse length m and legs n as shown in Figure 2. By the Pythagorean theorem, ${\frac {m}{n}}={\sqrt {2}}$. Suppose m and n are integers. Let m:n be a ratio given in its lowest terms.
Draw the arcs BD and CE with centre A. Join DE. It follows that AB = AD, AC = AE and ∠BAC and ∠DAE coincide. Therefore, the triangles ABC and ADE are congruent by SAS.
Because ∠EBF is a right angle and ∠BEF is half a right angle, △ BEF is also a right isosceles triangle. Hence BE = m − n implies BF = m − n. By symmetry, DF = m − n, and △ FDC is also a right isosceles triangle. It also follows that FC = n − (m − n) = 2n − m.
Hence, there is an even smaller right isosceles triangle, with hypotenuse length 2n − m and legs m − n. These values are integers even smaller than m and n and in the same ratio, contradicting the hypothesis that m:n is in lowest terms. Therefore, m and n 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 a and b be positive integers such that 1<⁠a/b⁠< 3/2 (as 1<2< 9/4 satisfies these bounds). Now 2b$^2$ and a$^2$ cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus |2b$^2$ − a$^2$| ≥ 1. Multiplying the absolute difference |√2 − ⁠a/b⁠| by b$^2$(√2 + ⁠a/b⁠) in the numerator and denominator, we get
    $\left|{\sqrt {2}}-{\frac {a}{b}}\right|={\frac {|2b^{2}-a^{2}|}{b^{2}\!\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{b^{2}\!\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{3b^{2}}},$
the latter inequality being true because it is assumed that 1<⁠a/b⁠< 3/2, giving ⁠a/b⁠ + √2 ≤ 3 (otherwise the quantitative apartness can be trivially established). This gives a lower bound of ⁠1/3b$^2$⁠ for the difference |√2 − ⁠a/b⁠|, 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 a, b, and c are coprime positive integers such that a$^2$ + b$^2$ = c$^2$, then c 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,
    ${\sqrt {2}}={a \over b}$
    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 lemma a is never even. However, this contradicts the equation 2b$^2$ = a$^2$ which implies that a must be even.

## Multiplicative inverse

The multiplicative inverse (reciprocal) of the square root of two is a widely used constant, with the decimal value:
    0.70710678118654752440084436210484903928483593768847...
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
    $\left({\frac {\sqrt {2}}{2}},{\frac {\sqrt {2}}{2}}\right)\!.$
Each coordinate satisfies
    ${\frac {\sqrt {2}}{2}}={\sqrt {\tfrac {1}{2}}}={\frac {1}{\sqrt {2}}}=\sin 45^{\circ }=\cos 45^{\circ }.$

## Properties

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 of silver ratios.
${\sqrt {2}}$ can also be expressed in terms of copies of the imaginary unit i using only the square root and arithmetic operations, if the square root symbol is interpreted suitably for the complex numbers i 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 infinite tetrate (i.e., infinite exponential tower) is equal to its square. In other words: if for c > 1, x$_1$ = c and x$_{n+1}$ = c$^{xn}$ for n > 1, the limit of x$_n$ as n → ∞ will be called (if this limit exists) f(c). Then ${\sqrt {2}}$ is the only number c > 1 for which f(c) = c$^2$. Or symbolically:
    ${\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{~\cdot ^{~\cdot ^{~\cdot }}}}}=2.$
${\sqrt {2}}$ appears in Viète's formula for π,
    ${\frac {2}{\pi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}}}\cdots ,$
which is related to the formula
    $\pi =\lim _{m\to \infty }2^{m}\underbrace {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+\cdots +{\sqrt {2}}}}}}}}}} _{m{\text{ square roots}}}\,.$
Similar in appearance but with a finite number of terms, ${\sqrt {2}}$ appears in various trigonometric constants:
    ${\begin{aligned}\sin {\frac {\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {11\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}&\quad \sin {\frac {7\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\\[6pt]\sin {\frac {3\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {\pi }{4}}&={\tfrac {1}{2}}{\sqrt {2}}&\quad \sin {\frac {13\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}&\quad \sin {\frac {9\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {7\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}\\[6pt]\sin {\frac {5\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {5\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {15\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}\end{aligned}}$
It is not known whether ${\sqrt {2}}$ is 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 cos ⁠π/4⁠ = sin ⁠π/4⁠ = ⁠1/√2⁠, along with the infinite product representations for the sine 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 the Taylor series of a trigonometric function. For example, the series for cos ⁠π/4⁠ gives
    ${\frac {1}{\sqrt {2}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}\left({\frac {\pi }{4}}\right)^{2k}}{(2k)!}}.$
The Taylor series of √1 + x with x = 1 and using the double factorial n!! gives
    ${\sqrt {2}}=\sum _{k=0}^{\infty }(-1)^{k+1}{\frac {(2k-3)!!}{(2k)!!}}=1+{\frac {1}{2}}-{\frac {1}{2\cdot 4}}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}}+\cdots =1+{\frac {1}{2}}-{\frac {1}{8}}+{\frac {1}{16}}-{\frac {5}{128}}+{\frac {7}{256}}+\cdots .$
The convergence of this series can be accelerated with an Euler 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 a BBP-type formula. BBP-type formulas are known for π√2 and √2 ln(1+√2), however.
The number can be represented by an infinite series of Egyptian fractions, with denominators defined by 2$^n$ th terms of a Fibonacci-like recurrence relation a(n) = 34a(n−1) − a(n−2), a(0) = 0, a(1) = 6.
    ${\sqrt {2}}={\frac {3}{2}}-{\frac {1}{2}}\sum _{n=0}^{\infty }{\frac {1}{a(2^{n})}}={\frac {3}{2}}-{\frac {1}{2}}\left({\frac {1}{6}}+{\frac {1}{204}}+{\frac {1}{235416}}+\dots \right)$

### Continued fraction

The square root of two has the following continued fraction representation:
    ${\sqrt {2}}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{\ddots }}}}}}}}.$
The convergents ⁠p/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 the Pell numbers (i.e., p$^2$ − 2q$^2$ = ±1). The first convergents are: ⁠1/1⁠, ⁠3/2⁠, ⁠7/5⁠, ⁠17/12⁠, ⁠41/29⁠, ⁠99/70⁠, ⁠239/169⁠, ⁠577/408⁠ and the convergent following ⁠p/q⁠ is ⁠p + 2q/p + q⁠. The convergent ⁠p/q⁠ differs from ${\sqrt {2}}$ by almost exactly ⁠1/2√2q$^2$⁠, 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

The following nested square expressions converge to √2{\textstyle {\sqrt {2}}}:
    ${\begin{aligned}{\sqrt {2}}&={\tfrac {3}{2}}-2\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-{\bigl{\tfrac {1}{4}}-\cdots {\bigr )}^{2}\right)^{2}\right)^{2}\\[10mu]&={\tfrac {3}{2}}-4\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+{\bigl{\tfrac {1}{8}}+\cdots {\bigr )}^{2}\right)^{2}\right)^{2}.\end{aligned}}$

## Applications
### Paper size

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 $R'={\frac {L'}{S'}}$ 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

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."