Square root of 2#Constructive proof

File:Ybc7289-bw.jpg with annotations. Besides showing the square root of 2 in sexagesimal ({{nowrap|1 24 51 10}}), the tablet also gives an example where one side of the square is 30 and the diagonal then is {{nowrap|42 25 35}}. The sexagesimal digit 30 can also stand for {{nowrap|0 30}} = {{sfrac|1|2}}, in which case {{nowrap|0 42 25 35}} is approximately 0.7071065.]]

The Babylonian clay tablet YBC 7289 ({{Circa|1800}}–1600 BC) gives an approximation of $\backslash sqrt\{2\}$ in four sexagesimal figures, {{nowrap|1 24 51 10}}, which is accurate to about six decimal digits,Fowler and Robson, p. 368.
[http://it.stlawu.edu/%7Edmelvill/mesomath/tablets/YBC7289.html Photograph, illustration, and description of the root(2) tablet from the Yale Babylonian Collection] {{webarchive|url=https://web.archive.org/web/20120813054036/http://it.stlawu.edu/%7Edmelvill/mesomath/tablets/YBC7289.html |date=2012-08-13 }}
[http://www.math.ubc.ca/%7Ecass/Euclid/ybc/ybc.html High resolution photographs, descriptions, and analysis of the root(2) tablet (YBC 7289) from the Yale Babylonian Collection] and is the closest possible three-place sexagesimal representation of $\backslash sqrt\{2\}$, representing a margin of error of only –0.000042%:

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

Another early approximation is given in ancient Indian mathematical texts, the Sulbasutras ({{Circa|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.{{citation |last=Henderson |first=David W. |author-link=David W. Henderson |editor-last=Gorini |editor-first=Catherine A. |date=2000 |chapter=Square roots in the Śulba Sūtras |title=Geometry At Work: Papers in Applied Geometry |series=Mathematical Association of America Notes |volume=53 |location=Washington, D.C. |publisher=The Mathematical Association of America |pages=39–45 |isbn=978-0883851647 |url=http://www.math.cornell.edu/~dwh/papers/sulba/sulba.html}} That is,

:$1\; +\; \backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{3\; \backslash times\; 4\}\; -\; \backslash frac\{1\}\{3\; \backslash times4\; \backslash times\; 34\}\; =\; \backslash frac\{577\}\{408\}\; =\; 1.41421\backslash overline\{56862745098039\}.$

This approximation, diverging from the actual value of $\backslash sqrt\{2\}$ 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 $\backslash sqrt\{2\}$. 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.{{citation |title=The Dangerous Ratio |url=https://nrich.maths.org/2671 |access-date=2023-09-18 |website=nrich.maths.org}}{{citation |last=Von Fritz |first=Kurt |date=1945 |title=The Discovery of Incommensurability by Hippasus of Metapontum |journal=Annals of Mathematics |volume=46 |issue=2 |pages=242–264 |doi=10.2307/1969021 |jstor=1969021 |issn=0003-486X}} The square root of two is occasionally called Pythagoras's number{{citation |last1=Conway |first1=John H. |author1-link=John H. Conway |last2=Guy |first2=Richard K. |author2-link = Richard K. Guy |date=1996 |title=The Book of Numbers |title-link=The Book of Numbers (math book) |location=New York |publisher=Copernicus |page=25 |isbn=978-1461240723}} or Pythagoras's constant.

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.{{citation |title=Architecture and Mathematics from Antiquity to the Future: Volume I: Antiquity to the 1500s|last1=Williams|first1=Kim|author1-link=Kim Williams (architect)|last2=Ostwald|first2=Michael|publisher=Birkhäuser|year=2015|isbn=9783319001371|pages=204}}

{{Further|Methods of computing square roots}}

There are many algorithms for approximating $\backslash sqrt\{2\}$ 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 methodAlthough the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of $\backslash sqrt\{2\}$ seen on tablet YBC 7289. Fowler and Robson offer informed and detailed conjectures.
Fowler and Robson, p. 376. Flannery, p. 32, 158. 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:

:$a\_\{n+1\}\; =\; \backslash frac12\backslash left(a\_n\; +\; \backslash dfrac\{2\}\{a\_n\}\backslash right)=\backslash frac\{a\_n\}\{2\}+\backslash frac\{1\}\{a\_n\}.$

Each iteration improves the approximation, roughly doubling the number of correct digits. Starting with $a\_0=1$, the subsequent iterations yield:

:$\backslash begin\{alignat\}\{3\}$

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{alignat}

A simple rational approximation {{sfrac|99|70}} (≈ 1.4142857) is sometimes used. Despite having a denominator of only 70, it differs from the correct value by less than {{sfrac|1|10,000}} (approx. {{val|+0.72e-4}}).

The next two better rational approximations are {{sfrac|140|99}} (≈ 1.4141414...) with a marginally smaller error (approx. {{val|-0.72e-4}}), and {{sfrac|239|169}} (≈ 1.4142012) with an error of approx {{val|-0.12e-4}}.

The rational approximation of the square root of two derived from four iterations of the Babylonian method after starting with {{math|a₀ {{=}} 1}} ({{sfrac|665,857|470,832}}) is too large by about {{val|1.6e-12}}; its square is ≈ {{val|2.0000000000045}}.

In 1997, the value of $\backslash sqrt\{2\}$ was calculated to 137,438,953,444 decimal places by Yasumasa Kanada's team. In February 2006, the record for the calculation of $\backslash sqrt\{2\}$ was eclipsed with the use of a home computer. Shigeru Kondo calculated one trillion decimal places in 2010.{{citation |url=http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html |title=Constants and Records of Computation |publisher=Numbers.computation.free.fr |date=2010-08-12 |access-date=2012-09-07 |url-status=live |archive-url=https://web.archive.org/web/20120301190937/http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html |archive-date=2012-03-01 }} Other mathematical constants whose decimal expansions have been calculated to similarly high precision include {{pi}}, e (mathematical constant), and the golden ratio.{{citation |url=http://www.numberworld.org/y-cruncher/records.html |title=Records set by y-cruncher |access-date=2022-04-07 |url-status=live |archive-url=https://web.archive.org/web/20220407052022/http://www.numberworld.org/y-cruncher/records.html |archive-date=2022-04-07 }} Such computations provide empirical evidence of whether these numbers are normal.

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

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 "$\backslash sqrt\{2\}$ is not rational" by assuming that it is rational and then deriving a falsehood.

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

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

This proof was hinted at by Aristotle, in his Analytica Priora, §I.23.All that Aristotle says, while writing about proofs 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". 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.The edition of the Greek text of the Elements published by E. F. August in Berlin in 1826–1829 already relegates this proof to an Appendix. The same thing occurs with J. L. Heiberg's edition (1883–1888).

Assume by way of contradiction that $\backslash sqrt\; 2$ were rational. Then we may write $\backslash sqrt\; 2\; +\; 1\; =\; \backslash frac\{q\}\{p\}$ as an irreducible fraction in lowest terms, with coprime positive integers $q>p$. Since $(\backslash sqrt\; 2-1)(\backslash sqrt\; 2+1)=2-1^2=1$, it follows that $\backslash sqrt\; 2-1$ can be expressed as the irreducible fraction $\backslash frac\{p\}\{q\}$. However, since $\backslash sqrt\; 2-1$ and $\backslash sqrt\; 2+1$ 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.

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.

The irrationality of $\backslash sqrt\{2\}$ 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 $\backslash pm\; 1$ and $\backslash pm\; 2$. As $\backslash sqrt\{2\}$ is not equal to $\backslash pm\; 1$ or $\backslash pm\; 2$, it follows that $\backslash sqrt\{2\}$ 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 $\backslash sqrt\{2\}$ 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.

File:NYSqrt2.svg of {{sqrt|2}}]]

A simple proof is attributed to Stanley Tennenbaum when he was a student in the early 1950s.{{citation |last1=Miller |first1=Steven J. |last2=Montague |first2=David |date=April 2012 |title=Picturing Irrationality |jstor=10.4169/math.mag.85.2.110 |magazine=Mathematics Magazine |volume=85 |issue=2 |pages=110–114 |doi=10.4169/math.mag.85.2.110 }}{{citation |last=Yanofsky |first=Noson S. |date=May–June 2016 |title=Paradoxes, Contradictions, and the Limits of Science |jstor=44808923 |magazine=American Scientist |volume=103 |issue=3 |pages=166–173 }} Assume that $\backslash sqrt\{2\}\; =\; a/b$, where $a$ and $b$ are coprime positive integers. Then $a$ and $b$ are the smallest positive integers for which $a^2\; =\; 2b^2$. Geometrically, this implies that a square with side length $a$ will have an area equal to two squares of (lesser) side length $b$. 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 and , and we know that $p$ and $q$ are integers from their definitions in terms of $a$ and $b$, this means that we are in violation of the original assumption that $a$ and $b$ are the smallest positive integers for which $a^2\; =\; 2b^2$.

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

File:Irrationality of sqrt2.svg

Tom M. Apostol made another geometric reductio ad absurdum argument showing that $\backslash sqrt\{2\}$ is irrational.{{citation |last=Apostol |first=Tom M. |author-link=Tom M. Apostol |date=2000 |title=Irrationality of The Square Root of Two – A Geometric Proof |jstor=2695741 |journal=The American Mathematical Monthly |volume=107 |number=9 |pages=841–842 |doi=10.2307/2695741 }} 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 {{math|△ ABC}} be a right isosceles triangle with hypotenuse length {{math|m}} and legs {{math|n}} as shown in Figure 2. By the Pythagorean theorem, $\backslash frac\{m\}\{n\}=\backslash sqrt\{2\}$. Suppose {{math|m}} and {{math|n}} are integers. Let {{math|m:n}} be a ratio given in its lowest terms.

Draw the arcs {{math|BD}} and {{math|CE}} with centre {{math|A}}. Join {{math|DE}}. It follows that {{math|AB {{=}} AD}}, {{math|AC {{=}} AE}} and {{math|∠BAC}} and {{math|∠DAE}} coincide. Therefore, the triangles {{math|ABC}} and {{math|ADE}} are congruent by SAS.

Because {{math|∠EBF}} is a right angle and {{math|∠BEF}} is half a right angle, {{math|△ BEF}} is also a right isosceles triangle. Hence {{math|BE {{=}} m − n}} implies {{math|BF {{=}} m − n}}. By symmetry, {{math|DF {{=}} m − n}}, and {{math|△ FDC}} is also a right isosceles triangle. It also follows that {{math|FC {{=}} n − (m − n) {{=}} 2n − m}}.

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

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 {{math|a}} and {{math|b}} be positive integers such that {{math|1<{{sfrac|a|b}}< 3/2}} (as {{math|1<2< 9/4}} satisfies these bounds). Now {{math|2b{{sup|2}} }} and {{math|a{{sup|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 {{math|{{abs|2b{{sup|2}} − a{{sup|2}}}} ≥ 1}}. Multiplying the absolute difference {{math|{{abs|{{sqrt|2}} − {{sfrac|a|b}}}}}} by {{math| b{{sup|2}}({{sqrt|2}} + {{sfrac|a|b}})}} in the numerator and denominator, we getSee {{citation

| last1 = Katz | first1 = Karin Usadi

| last2 = Katz | first2 = Mikhail G.

| author2-link = Mikhail Katz

| arxiv = 1110.5456

| issue = 2

| journal = Intellectica

| pages = 223–302 (see esp. Section 2.3, footnote 15)

| title = Meaning in Classical Mathematics: Is it at Odds with Intuitionism?

| volume = 56

| year = 2011| bibcode = 2011arXiv1110.5456U}}

:$\backslash left|\backslash sqrt2\; -\; \backslash frac\{a\}\{b\}\backslash right|\; =\; \backslash frac$

the latter inequality being true because it is assumed that {{math|1<{{sfrac|a|b}}< 3/2}}, giving {{math|{{sfrac|a|b}} + {{sqrt|2}} ≤ 3 }} (otherwise the quantitative apartness can be trivially established). This gives a lower bound of {{math|{{sfrac|1|3b{{sup|2}}}}}} for the difference {{math|{{abs|{{sqrt|2}} − {{sfrac|a|b}}}}}}, yielding a direct proof of irrationality in its constructively stronger form, not relying on the law of excluded middle.{{citation |last=Bishop |first=Errett |author-link=Errett Bishop |editor-last=Rosenblatt |editor-first=Murray |editor-link=Murray Rosenblatt |date=1985 |chapter=Schizophrenia in Contemporary Mathematics. |title=Errett Bishop: Reflections on Him and His Research |series=Contemporary Mathematics |volume=39 |location=Providence, RI |publisher=American Mathematical Society |pages=1–32 |doi=10.1090/conm/039/788163 |isbn=0821850407 |issn=0271-4132}} This proof constructively exhibits an explicit discrepancy between $\backslash sqrt\{2\}$ and any rational.

This proof uses the following property of primitive Pythagorean triples:

: If {{math|a}}, {{math|b}}, and {{math|c}} are coprime positive integers such that {{math|1=a² + b² = c²}}, then {{math|c}} is never even.{{citation |last=Sierpiński |first=Wacław |author-link=Waclaw Sierpinski |translator-last=Sharma |translator-first=Ambikeshwa |date=2003 |title=Pythagorean Triangles |location=Mineola, NY |publisher=Dover |pages=4–6 |isbn=978-0486432786}}

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 $\backslash sqrt2$ is rational. Therefore,

:$\backslash sqrt2\; =\; \{a\; \backslash over\; b\}$

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

:Squaring both sides,

:$2\; =\; \{a^2\; \backslash over\; b^2\}$

:$2b^2\; =\; a^2$

:$b^2+b^2\; =\; a^2$

Here, {{math|(b, b, a)}} is a primitive Pythagorean triple, and from the lemma {{math|a}} is never even. However, this contradicts the equation {{math|1=2b² = a²}} which implies that {{math|a}} must be even.

The multiplicative inverse (reciprocal) of the square root of two is a widely used constant, with the decimal value:{{cite OEIS |1=A010503 |2=Decimal expansion of 1/sqrt(2) |access-date=3 November 2024}}

:{{gaps|0.70710|67811|86547|52440|08443|62104|84903|92848|35937|68847|...}}

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

:$\backslash left(\backslash frac\{\backslash sqrt\{2\}\}\{2\},\; \backslash frac\{\backslash sqrt\{2\}\}\{2\}\backslash right)\backslash !.$

Each coordinate satisfies

:$\backslash frac\{\backslash sqrt\{2\}\}\{2\}\; =\; \backslash sqrt\{\backslash tfrac\{1\}\{2\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\; =\; \backslash sin\; 45^\backslash circ\; =\; \backslash cos\; 45^\backslash circ.$

File:Circular and hyperbolic angle.svg size and sector area are the same when the conic radius is {{Math|{{sqrt|2}}}}. This diagram illustrates the circular and hyperbolic functions based on sector areas {{math|u}}.]]

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

:$\backslash !\backslash \; \{1\; \backslash over\; \{\backslash sqrt\{2\}\; -\; 1\}\}\; =\; \backslash sqrt\{2\}\; +\; 1$

since

:$\backslash left(\backslash sqrt\{2\}+1\backslash right)\backslash !\backslash left(\backslash sqrt\{2\}-1\backslash right)\; =\; 2-1\; =\; 1.$

This is related to the property of silver ratios.

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

:$\backslash frac\{\backslash sqrt\{i\}+i\; \backslash sqrt\{i\}\}\{i\}\backslash text\{\; and\; \}\backslash frac\{\backslash sqrt\{-i\}-i\; \backslash sqrt\{-i\}\}\{-i\}$

$\backslash sqrt\{2\}$ 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 {{math|c > 1}}, {{math|x₁ {{=}} c}} and {{math|x_n+1 {{=}} c^x_n}} for {{math|n > 1}}, the limit of {{math|x_n}} as {{math|n → ∞}} will be called (if this limit exists) {{math|f(c)}}. Then $\backslash sqrt\{2\}$ is the only number {{math|c > 1}} for which {{math|f(c) {{=}} c²}}. Or symbolically:

:$\backslash sqrt\{2\}^\{\backslash sqrt\{2\}^\{\backslash sqrt\{2\}^\{~\backslash cdot^\{~\backslash cdot^\{~\backslash cdot\}\}\}\}\}\; =\; 2.$

$\backslash sqrt\{2\}$ appears in Viète's formula for {{pi}},

:$$

\frac2\pi = \sqrt\frac12 \cdot \sqrt{\frac12 + \frac12\sqrt\frac12} \cdot \sqrt{\frac12 + \frac12\sqrt{\frac12 + \frac12\sqrt\frac12}} \cdots,

which is related to the formula{{Citation |title=What is mathematics? An Elementary Approach to Ideas and Methods |first1=Richard |last1=Courant |first2=Herbert |last2=Robbins |location=London |publisher=Oxford University Press |year=1941 |page=124 }}

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

Similar in appearance but with a finite number of terms, $\backslash sqrt\{2\}$ appears in various trigonometric constants:Julian D. A. Wiseman [http://www.jdawiseman.com/papers/easymath/surds_sin_cos.html Sin and cos in surds] {{webarchive|url=https://web.archive.org/web/20090506080636/http://www.jdawiseman.com/papers/easymath/surds_sin_cos.html |date=2009-05-06 }}

:$\backslash begin\{align\}$

\sin\frac{\pi}{32} &= \tfrac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}} &\quad

\sin\frac{3\pi}{16} &= \tfrac12\sqrt{2-\sqrt{2-\sqrt{2}}} &\quad

\sin\frac{11\pi}{32} &= \tfrac12\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2}}}} \\[6pt]

\sin\frac{\pi}{16} &= \tfrac12\sqrt{2-\sqrt{2+\sqrt{2}}} &\quad

\sin\frac{7\pi}{32} &= \tfrac12\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2}}}} &\quad

\sin\frac{3\pi}{8} &= \tfrac12\sqrt{2+\sqrt{2}} \\[6pt]

\sin\frac{3\pi}{32} &= \tfrac12\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}} &\quad

\sin\frac{\pi}{4} &= \tfrac12\sqrt{2} &\quad

\sin\frac{13\pi}{32} &= \tfrac12\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2}}}} \\[6pt]

\sin\frac{\pi}{8} &= \tfrac12\sqrt{2-\sqrt{2}} &\quad

\sin\frac{9\pi}{32} &= \tfrac12\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2}}}} &\quad

\sin\frac{7\pi}{16} &= \tfrac12\sqrt{2+\sqrt{2+\sqrt{2}}} \\[6pt]

\sin\frac{5\pi}{32} &= \tfrac12\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}} &\quad

\sin\frac{5\pi}{16} &= \tfrac12\sqrt{2+\sqrt{2-\sqrt{2}}} &\quad

\sin\frac{15\pi}{32} &= \tfrac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}

\end{align}

It is not known whether $\backslash sqrt\{2\}$ 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.{{citation |last1=Good |first1=I. J. |author1-link=I. J. Good |last2=Gover |first2=T. N. |date=1967 |title=The generalized serial test and the binary expansion of $\backslash sqrt\{2\}$ |journal=Journal of the Royal Statistical Society, Series A |jstor=2344040 |volume=130 |issue=1 |pages=102–107 |doi=10.2307/2344040 }}

The identity {{math|cos {{sfrac|π|4}} {{=}} sin {{sfrac|π|4}} {{=}} {{sfrac|1|{{sqrt|2}}}}}}, along with the infinite product representations for the sine and cosine, leads to products such as

:$\backslash frac\{1\}\{\backslash sqrt\; 2\}\; =\; \backslash prod\_\{k=0\}^\backslash infty\; \backslash left(1-\backslash frac\{1\}\{(4k+2)^2\}\backslash right)\; =$

\left(1-\frac{1}{4}\right)\!\left(1-\frac{1}{36}\right)\!\left(1-\frac{1}{100}\right) \cdots

and

:$\backslash sqrt\{2\}\; =\; \backslash prod\_\{k=0\}^\backslash infty\backslash frac\{(4k+2)^2\}\{(4k+1)(4k+3)\}\; =$

\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,

:$\backslash sqrt\{2\}\; =\; \backslash prod\_\{k=0\}^\backslash infty\backslash left(1+\backslash frac\{1\}\{4k+1\}\backslash right)\backslash left(1-\backslash frac\{1\}\{4k+3\}\backslash right)\; =$

\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 {{math|cos {{sfrac|π|4}}}} gives

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

The Taylor series of {{math|{{sqrt|1 + x}}}} with {{math|x {{=}} 1}} and using the double factorial {{math|n!!}} gives

:$\backslash sqrt\{2\}\; =\; \backslash sum\_\{k=0\}^\backslash infty\; (-1)^\{k+1\}\; \backslash frac\{(2k-3)!!\}\{(2k)!!\}\; =$

1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \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

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

It is not known whether $\backslash sqrt\{2\}$ can be represented with a BBP-type formula. BBP-type formulas are known for {{math|π{{sqrt|2}}}} and {{math|{{sqrt|2}} ln(1+{{sqrt|2}})}}, however.{{citation |url=http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf |title=A Compendium of BBP-Type Formulas for Mathematical Constants |last1=Bailey |first1=David H. |date=13 February 2011 |access-date=2010-04-30 |url-status=live |archive-url=https://web.archive.org/web/20110610050911/http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf |archive-date=2011-06-10 }}

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.{{Cite OEIS |1=A082405|2=a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=6 |access-date=2016-09-05}}

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

File:Dedekind cut- square root of two.png]]

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

:$\backslash sqrt2\; =\; 1\; +\; \backslash cfrac\{1\}\{2\; +\; \backslash cfrac\{1\}\{2\; +\; \backslash cfrac\{1\}\{2\; +\; \backslash cfrac1\backslash ddots\}\}\}.$

The convergents {{math|{{sfrac|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., {{Math|1=p² − 2q² = ±1}}). The first convergents are: {{math|{{sfrac|1|1}}, {{sfrac|3|2}}, {{sfrac|7|5}}, {{sfrac|17|12}}, {{sfrac|41|29}}, {{sfrac|99|70}}, {{sfrac|239|169}}, {{sfrac|577|408}}}} and the convergent following {{math|{{sfrac|p|q}}}} is {{math|{{sfrac|p + 2q|p + q}}}}. The convergent {{math|{{sfrac|p|q}}}} differs from $\backslash sqrt\{2\}$ by almost exactly {{math|{{sfrac|1|2{{sqrt|2}}q{{sup|2}}}}}}, which follows from:

:$\backslash left|\backslash sqrt2\; -\; \backslash frac\{p\}\{q\}\backslash right|\; =\; \backslash frac$

The following nested square expressions converge to {{nobr|$\backslash sqrt2$:}}

:$\backslash begin\{align\}$

\sqrt{2}

&= \tfrac32 - 2 \left( \tfrac14 - \left( \tfrac14 - \bigl( \tfrac14 - \cdots \bigr)^2 \right)^2 \right)^2 \\[10mu]

&= \tfrac32 - 4 \left( \tfrac18 + \left( \tfrac18 + \bigl( \tfrac18 + \cdots \bigr)^2 \right)^2 \right)^2.

\end{align}

File:A size illustration2.svg

In 1786, German physics professor Georg Christoph Lichtenberg found that any sheet of paper whose long edge is $\backslash sqrt\{2\}$ 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.{{citation |title=The Book: A Cover-to-Cover Exploration of the Most Powerful Object of Our Time|last=Houston|first=Keith|publisher=W. W. Norton & Company|year=2016|isbn=978-0393244809|pages=324}} Today, the (approximate) aspect ratio of paper sizes under ISO 216 (A4, A0, etc.) is 1:$\backslash sqrt\{2\}$.

Proof:

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

:$R\; =\; \backslash frac\{L\}\{S\}\; =\; \backslash sqrt\{2\}$ as required by ISO 216.

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

:$R\text{'}\; =\; \backslash frac\{S\}\{L/2\}\; =\; \backslash frac\{2S\}\{L\}\; =\; \backslash frac\{2\}\{(L/S)\}\; =\; \backslash frac\{2\}\{\backslash sqrt\{2\}\}\; =\; \backslash sqrt\{2\}\; =\; R.$

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 $\backslash sqrt\{2\}$.

{{distances_between_double_cube_corners.svg}}

• 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."{{citation |title=The Book: Hjernen er sternen|last=Nordengen|first=Kaja|publisher=2016 Kagge Forlag AS|year=2016|isbn=978-82-489-2018-2|page=81}}

• List of mathematical constants
• Square root of 3, {{math|{{sqrt|3}}}}
• Square root of 5, {{math|{{sqrt|5}}}}
• Gelfond–Schneider constant, {{math|2^{{sqrt|2}}}}
• Silver ratio, {{math|1 + {{sqrt|2}}}}

{{clear}}

{{Reflist|25em}}

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}}

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| last2 = Robson | first2 = Eleanor | author2-link = Eleanor Robson

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| title = Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context

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}}

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| last1 = Gourdon | first1 = X.

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• [https://www.gutenberg.org/ebooks/129 The Square Root of Two to 5 million digits] by Jerry Bonnell and Robert J. Nemiroff. May, 1994.
• [http://www.cut-the-knot.org/proofs/sq_root.shtml Square root of 2 is irrational], a collection of proofs
• {{citation |last=Haran |first=Brady |author-link=Brady Haran |others=featuring Grime, James; Bowley, Roger |title=Root 2 |url=https://www.numberphile.com/videos/root-2 |series=Numberphile |type=video |date=27 January 2012 }}
• [http://pisearch.org/sqrt2 {{tmath|\sqrt2}} Search Engine] 2 billion searchable digits of {{radic|2}}, {{pi}} and {{mvar|e}}

{{Algebraic numbers}}

{{Irrational number}}

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Category:Quadratic irrational numbers

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