View source for Square root of 2

{{redirect-distinguish|Pythagoras's constant|Pythagoras number}}
[[File:Square root of 2 triangle.svg|thumb|200px|The square root of 2 is equal to the length of the [[hypotenuse]] of a [[right triangle]] with legs of length 1]]

The '''square root of 2''', or the (1/2)th [[power of 2]], written in mathematics as {{math|{{sqrt|2}}}} or {{math|2{{sup|{{frac|1|2}}}}}}, is the positive [[algebraic number]] that, when multiplied by itself, gives the number [[2 (number)|2]]. Technically, it is 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 [[Unit square|square with sides of one unit of length]]; this follows from the [[Pythagorean theorem]]. It was probably the first number known to be [[irrational number|irrational]].

As a good rational approximation for the square root of two, with a reasonably small denominator, the fraction {{sfrac|99|70}} (≈ 1.4142857) is sometimes used.

The sequence {{OEIS link|A002193}} in the [[OEIS]] consists of the digits in the decimal expansion of the square root of 2, here truncated to 65 [[decimal|decimal places]]:

:{{gaps|1.41421|35623|73095|04880|16887|24209|69807|85696|71875|37694|80731|76679|73799|}}... 

{| class="infobox bordered" cellpadding=5
| colspan="2" align="center" | {{Irrational numbers}}
|-
|[[Binary numeral system|Binary]]
| {{gaps|1.0110|1010|0000|1001|1110|…}}
|-
| [[Decimal]]
| {{gaps|1.41421|35623|73095|0488…}}
|-
| [[Hexadecimal]]
| {{gaps|1.6A09|E667|F3BC|C908|B2F…}}
|-
| [[Square root of 2#Continued fraction representation|Continued fraction]]
| <math>1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}</math>
|}

==History==
[[File:Ybc7289-bw.jpg|right|thumb|200px|Babylonian clay tablet [[YBC 7289]] 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 [[Babylonia]]n clay tablet [[YBC 7289]] (c. 1800–1600 BC) gives an approximation of {{math|{{sqrt|2}}}} in four [[sexagesimal]] figures, {{nowrap|1 24 51 10}}, which is accurate to about six [[decimal]] digits,<ref>Fowler and Robson, p. 368.<br />[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 }}<br />[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]</ref> and is the closest possible three-place sexagesimal representation of {{math|{{sqrt|2}}}}:
:<math>1 + \frac{24}{60} + \frac{51}{60^2} + \frac{10}{60^3} = \frac{305470}{216000} = 1.41421\overline{296}.</math>

Another early close approximation is given in [[History of India|ancient Indian]] mathematical texts, the [[Sulba Sutras|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.''<ref>Henderson.</ref> That is,
:<math>1 + \frac{1}{3} + \frac{1}{3 \times 4} - \frac{1}{3 \times4 \times 34} = \frac{577}{408} = 1.41421\overline{56862745098039}.</math>

This approximation is the seventh in a sequence of increasingly accurate approximations based on the sequence of [[Pell number]]s, which can be derived from the [[continued fraction]] expansion of {{math|{{sqrt|2}}}}. Despite having a smaller denominator, it is only slightly less accurate than the Babylonian approximation.

[[Pythagoreanism|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 number|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.<ref>Stephanie J. Morris, [http://jwilson.coe.uga.edu/emt669/student.folders/morris.stephanie/emt.669/essay.1/pythagorean.html "The Pythagorean Theorem"] {{webarchive|url=https://web.archive.org/web/20130530001616/http://jwilson.coe.uga.edu/emt669/Student.Folders/Morris.Stephanie/EMT.669/Essay.1/Pythagorean.html |date=2013-05-30 }}, Dept. of Math. Ed., [[University of Georgia]].</ref><ref>Brian Clegg, [http://nrich.maths.org/2671 "The Dangerous Ratio ..."] {{webarchive|url=https://web.archive.org/web/20130627142742/http://nrich.maths.org/2671 |date=2013-06-27 }}, Nrich.org, November 2004.</ref><ref>Kurt von Fritz, [https://www.jstor.org/pss/1969021 "The discovery of incommensurability by Hippasus of Metapontum"], Annals of Mathematics, 1945.</ref> The square root of two is occasionally called '''Pythagoras' number''' or '''Pythagoras' constant''', for example by {{harvtxt|Conway|Guy|1996}}.<ref>{{citation
 | last1 = Conway | first1 = John H. | author1-link = John H. Conway
 | last2 = Guy | first2 = Richard K. | author2-link = Richard K. Guy
 | page = 25
 | publisher = Copernicus
 | title = The Book of Numbers
 | year = 1996}}</ref>

=== Ancient Roman architecture ===
In [[Ancient Roman architecture|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 [[Atrium (architecture)|atria]] by giving them a length equal to a diagonal taken from a square which sides are equivalent to the intended atrium's width.<ref>{{Cite book|title=Architecture and Mathematics from Antiquity to the Future: Volume I: Antiquity to the 1500s|last=Williams|first=Kim|last2=Ostwald|first2=Michael|publisher=Birkhäuser|year=2015|isbn=9783319001371|location=|pages=204}}</ref>

==Computation algorithms==
{{further|Methods of computing square roots}}
There are a number of algorithms for approximating {{math|{{sqrt|2}}}}, which in expressions as a ratio of integers or as a decimal can only be approximated. The most common algorithm for this, one used as a basis in many computers and calculators, is the [[Babylonian method]]<ref>Although the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of {{math|{{sqrt|2}}}} seen on tablet YBC&nbsp;7289. Fowler and Robson offer informed and detailed conjectures.<br />Fowler and Robson, p. 376. Flannery, p. 32, 158.</ref> of computing square roots, which is one of many [[methods of computing square roots]]. It goes as follows:

First, pick a guess, {{math|''a''{{sub|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 [[Recursion|recursive]] computation:
:<math>a_{n+1} = \frac{a_n + \frac{2}{a_n}}{2}=\frac{a_n}{2}+\frac{1}{a_n}. </math>

The more iterations through the algorithm (that is, the more computations performed and the greater "{{math|''n''}}"), the better approximation of the square root of 2 is achieved. Each iteration approximately doubles the number of correct digits. Starting with {{math|''a''<sub>0</sub> {{=}} 1}} the next approximations are
* {{sfrac|3|2}}     = '''1'''.5
* {{sfrac|17|12}}   = '''1.41'''6...
* {{sfrac|577|408}} = '''1.41421'''5...
* {{sfrac|665857|470832}} = '''1.41421356237'''46...

The value of {{math|{{sqrt|2}}}} was calculated to 137,438,953,444 decimal places by [[Yasumasa Kanada]]'s team in 1997. In February 2006 the record for the calculation of {{math|{{sqrt|2}}}} was eclipsed with the use of a home computer. Shigeru Kondo calculated 1 [[Trillion (short scale)|trillion]] decimal places in 2010.<ref>{{cite web |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 |accessdate=2012-09-07 |url-status=live |archiveurl=https://web.archive.org/web/20120301190937/http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html |archivedate=2012-03-01 }}</ref> For a development of this record, see the table below. Among mathematical constants with computationally challenging decimal expansions, only [[pi|{{pi}}]] has been calculated more precisely.<ref>{{cite web |url=http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html |title=Number of known digits |publisher=Numbers.computation.free.fr |date=2010-08-12 |accessdate=2012-09-07 |url-status=live |archiveurl=https://web.archive.org/web/20120301190937/http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html |archivedate=2012-03-01 }}</ref>
Such computations aim to check empirically whether such numbers are [[normal number|normal]].

===Rational approximations===
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}}).  Since it is a convergent of the [[Square root of 2#Continued fraction representation|continued fraction representation]] of the square root of two, any better rational approximation has a denominator not less than 169, since {{sfrac|239|169}} (≈ 1.4142012) is the next convergent with an error of approx. {{val|-0.12e-4}}. 

The rational approximation of the square root of two, {{sfrac|665,857|470,832}}, derived from the fourth step in the Babylonian method starting with {{math|''a''<sub>0</sub> {{=}} 1}}, is too large by approx. {{val|1.6e-12}}: its square is {{val|2.0000000000045}}…

===Record progression===
This is a table of recent records in calculating digits of {{math|{{sqrt|2}}}}.
{| class="wikitable" style="text-align:center; caption-side:bottom"
|-
! Date !! Name !! Number of digits
|-
| June 28, 2016 || Ron Watkins || 10 trillion
|-
| April 3, 2016 || Ron Watkins || 5 trillion
|-
| February 9, 2012 || Alexander Yee || 2 trillion
|-
| March 22, 2010 || Shigeru Kondo || 1 trillion (10<sup>12</sup>)
|-
|+ Reference:<ref name="y-cruncher">{{cite web |url=http://www.numberworld.org/y-cruncher/records.html |title=Archived copy |accessdate=2015-12-03 |url-status=live |archiveurl=https://web.archive.org/web/20151020062050/http://www.numberworld.org/y-cruncher/records.html |archivedate=2015-10-20 }}</ref>
|}

==Proofs of irrationality==
A short proof of the irrationality of {{math|{{sqrt|2}}}} can be obtained from the [[rational root theorem]], that is, if {{math|''p''(''x'')}} is a [[monic polynomial]] with integer coefficients, then any [[rational number|rational]] [[root of a function|root]] of {{math|''p''(''x'')}} is necessarily an integer. Applying this to the polynomial {{math|''p''(''x'') {{=}} ''x''{{sup|2}} − 2}}, it follows that {{math|{{sqrt|2}}}} is either an integer or irrational. Because {{math|{{sqrt|2}}}} is not an integer (2 is not a perfect square), {{math|{{sqrt|2}}}} must therefore be irrational. This proof can be generalized to show that any square root of any natural number that is not the square of a natural number is irrational.

See [[quadratic irrational]] or [[proof by infinite descent#Irrationality of √k if it is not an integer|infinite descent]] for a proof that the square root of any non-square natural number is irrational.

===Proof by induction===
{{Copy edit|for=compliance with [[MOS:MATH]] mathematical formatting style guidelines|date=November 2019}}
For proving {{math|{{sqrt|2}}}} ∉ ℚ one has to show that ∀ x, y : ℤ y ≠ 0 → {{math|{{sqrt|2}}}} ≠ x / y which is equivalent to 

     (1) ∀ x, y : ℕ y ≠ 0 → 2⋅y² ≠ x². 

Statement (1) is be proved indirectly as follows:<ref>{{Cite book|title=Elementary Number Theory and Its Applications|last=Rosen|first=Kenneth H.|publisher=Pearson Addison Wesley|year=2005|isbn=978-0-07-305188-8|location=Boston|pages=721}}</ref> Assume that (1) is false. Then some numbers a, b ∈ ℕ exist such that
 
     (2) b ≠ 0 and 2⋅b² = a². 

We may assume that a and b are chosen such that a + b is minimal which means that 

     (3) ∀ x, y : ℕ x + y < a + b ∧ y ≠ 0 → 2⋅y² ≠ x² 

holds. It follows from (2) that a is even and different from 0 (as a² would be either odd or b = 0 otherwise), and therefore a = 2⋅n + 2 = 2⋅(n + 1) for some n ∈ ℕ. Now replacing a by 2⋅(n + 1) in (2) yields 2⋅b² = (2⋅(n + 1))², hence 

     (4) b² = 2⋅(a/2)²

(as n + 1 = a / 2). Finally, we replace x by b and y by a / 2 in (3), so that '''2⋅(a / 2)² ≠ b²''' is obtained (because b + a / 2 < a + b and a / 2 ≠ 0). But this is a contradiction to (4), hence the assumption was false and consequently (1) is true. 

This proof is an indirect ''induction'' proof in fact, which can be seen by the demand that "a and b are chosen such that (2) is satisfied and a + b is ''minimal''". A direct induction proof uses the induction schema
  
  	∀ x, y : ℕ x = 0 → φ(x, y)
   	∀ x, y : ℕ x ≠ 0 ∧ φ(y, x/2) → φ(x, y)
        -----------------------------------------
         	∀ x, y : ℕ φ(x, y)
         
where φ(x, y) := y ≠ 0 → 2⋅y² ≠ x² so that ∀ x, y : ℕ x ≠ 0 ∧ x / 2 ≠ 0 → 2⋅(x / 2)² ≠ y² becomes the induction hypothesis. This induction schema is sound, because x ≠ 0 entails y + x / 2 < x + y. The induction conclusion 2⋅y² ≠ x² is rewritten in the direct proof to y² ≠ 2⋅(x / 2)² by the same argumentation as in the indirect proof, and then follows immediately from the induction hypothesis.

===Proof by infinite descent===
One proof of the number's irrationality is the following proof by [[infinite descent]]. It is also a [[proof by contradiction]], also known as an indirect proof, in that the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.

# Assume that {{math|{{sqrt|2}}}} is a rational number, meaning that there exists a pair of integers whose ratio is {{math|{{sqrt|2}}}}.
# If the two integers have a common factor, it can be eliminated using the [[Euclidean algorithm]].
# Then {{math|{{sqrt|2}}}} can be written as an [[irreducible fraction]] {{math|{{sfrac|''a''|''b''}}}} such that {{math|''a''}} and {{math|''b''}} are [[coprime]] integers (having no common factor).
# It follows that {{math|{{sfrac|''a''{{sup|2}}|''b''{{sup|2}}}} {{=}} 2}} and {{math|''a''{{sup|2}} {{=}} 2''b''{{sup|2}}}}. &emsp; (&thinsp;[[Exponent#Identities and properties|{{math|({{sfrac|''a''|''b''}}){{sup|''n''}} {{=}} {{sfrac|''a''{{sup|''n''}}|''b''{{sup|''n''}}}}}}]] &thinsp;) &emsp; ( {{math|''a''{{sup|2}} and b''{{sup|2}}}} are integers)
# Therefore, {{math|''a''{{sup|2}}}} is even because it is equal to {{math|2''b''{{sup|2}}}}. ({{math|2''b''{{sup|2}}}} is necessarily even because it is 2 times another whole number and multiples of 2 are even.)
# It follows that {{math|''a''}} must be even (as squares of odd integers are never even).
# Because {{math|''a''}} is even, there exists an integer {{math|''k''}} that fulfills: {{math|''a'' {{=}} 2''k''}}.
# Substituting {{math|2''k''}} from step 7 for {{math|''a''}} in the second equation of step 4: {{math|2''b''{{sup|2}} {{=}} (2''k''){{sup|2}}}} is equivalent to {{math|2''b''{{sup|2}} {{=}} 4''k''{{sup|2}}}}, which is equivalent to {{math|''b''{{sup|2}} {{=}} 2''k''{{sup|2}}}}.
# Because {{math|2''k''{{sup|2}}}} is divisible by two and therefore even, and because {{math|2''k''{{sup|2}} {{=}} ''b''{{sup|2}}}}, it follows that {{math|''b''{{sup|2}}}} is also even which means that {{math|''b''}} is even.
# By steps 5 and 8 {{math|''a''}} and {{math|''b''}} are both even, which contradicts that {{math|{{sfrac|''a''|''b''}}}} is irreducible as stated in step 3.
::'''''[[Q.E.D.]]'''''

Because there is a contradiction, the assumption (1) that {{math|{{sqrt|2}}}} is a rational number must be false. This means that {{math|{{sqrt|2}}}} is not a rational number; i.e., {{math|{{sqrt|2}}}} is irrational.

This proof was hinted at by [[Aristotle]], in his ''[[Prior Analytics|Analytica Priora]]'', §I.23.<ref>All that Aristotle says, while writing about [[Proof by contradiction|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”.</ref> It appeared first as a full proof in [[Euclid]]'s ''[[Euclid's Elements|Elements]]'', as proposition 117 of Book X. However, since the early 19th century historians have agreed that this proof is an [[Interpolation (manuscripts)|interpolation]] and not attributable to Euclid.<ref>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 [[Johan Ludvig Heiberg (historian)|J. L. Heiberg's]] edition (1883–1888).</ref>

===Geometric proof===
[[File:NYSqrt2.svg|thumb|Figure 1. Stanley Tennenbaum's geometric proof of the irrationality of {{sqrt|2}}.]]
A simple proof is attributed by [[John Horton Conway]] to [[Stanley Tennenbaum]] when the latter was a student in the early 1950s<ref>[http://www.cut-the-knot.org/proofs/sq_root.shtml Proof 8‴] {{webarchive|url=https://web.archive.org/web/20160422220756/http://www.cut-the-knot.org/proofs/sq_root.shtml |date=2016-04-22 }}</ref> and whose most recent appearance is in an article by Noson Yanofsky in the May–June 2016 issue of ''[[American Scientist]]''.<ref>{{cite web|author=Yanofsky, N.|year=2016|title=Paradoxes, Contradictions, and the Limits of Science|url=http://www.americanscientist.org/issues/feature/paradoxes-contradictions-and-the-limits-of-science|url-status=live|archiveurl=https://web.archive.org/web/20160630035856/http://www.americanscientist.org/issues/feature/paradoxes-contradictions-and-the-limits-of-science|archivedate=2016-06-30}}</ref>  Given two squares with integer sides respectively ''a'' and ''b'', one of which has twice the area of the other, place two copies of the smaller square in the larger as shown in Figure 1.  The square overlap region in the middle ({{math|(2''b'' − ''a'')<sup>2</sup>}}) must equal the sum of the two uncovered squares ({{math|2(''a'' − ''b'')<sup>2</sup>}}).  But these squares on the diagonal have positive integer sides that are smaller than the original squares.  Repeating this process we can find arbitrarily small squares one twice the area of the other, yet both having positive integer sides, which is impossible since positive integers cannot be less than 1.

[[File:Irrationality of sqrt2.svg|left|thumb|Figure 2.  Tom Apostol's geometric proof of the irrationality of sqrt(2).]]
Another geometric [[reductio ad absurdum]] argument showing that {{math|{{sqrt|2}}}} is irrational appeared in 2000 in the [[American Mathematical Monthly]].<ref>{{citation|title=Irrationality of The Square Root of Two -- A Geometric Proof|author=Tom M. Apostol|journal=The American Mathematical Monthly|volume=107|number=9|date=Nov 2000|pages=841–842|doi=10.2307/2695741}}</ref> 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 algebraic proof of the previous section viewed geometrically in yet 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]], {{math|{{sfrac|''m''|''n''}} {{=}} {{sqrt|2}}}}. Suppose {{math|''m''}} and {{math|''n''}} are [[integer]]s. 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 the {{math|∠''BAC''}} and {{math|∠''DAE''}} coincide. Therefore, the triangles {{math|''ABC''}} and {{math|''ADE''}} are [[Congruence (geometry)|congruent]] by [[Side-angle-side|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'') {{=}} 2''n'' − ''m''}}.

Hence we have an even smaller right isosceles triangle, with hypotenuse length {{math|2''n'' − ''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 {{math|{{sqrt|2}}}} is irrational.

===Constructive proof===
In a constructive approach, one distinguishes between on the one hand not being rational, and on the other hand being irrational (i.e., being quantifiably apart from every rational), the latter being a stronger property. Given positive integers {{math|''a''}} and {{math|''b''}}, because the [[Singly and doubly even#Definitions|valuation]] (i.e., highest power of 2 dividing a number) of {{math|2''b''<sup>2</sup>}} is odd, while the valuation of {{math|''a''<sup>2</sup>}} is even, they must be distinct integers; thus {{math|{{abs|2''b''{{sup|2}} − ''a''{{sup|2}}}} ≥ 1}}. Then<ref>See {{citation
 | last1 = Katz | first1 = Karin Usadi
 | last2 = Katz | first2 = Mikhail G.
 | author2-link = Mikhail Katz
 | arxiv = 1110.5456
 | doi =
 | 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}}</ref>
:<math>\left|\sqrt2 - \frac{a}{b}\right| = \frac{|2b^2-a^2|}{b^2\left(\sqrt{2}+\frac{a}{b}\right)} \ge \frac{1}{b^2\left(\sqrt2 + \frac{a}{b}\right)} \ge \frac{1}{3b^2},</math>

the latter inequality being true because we assume {{math|{{sfrac|''a''|''b''}} ≤ 3 − {{sqrt|2}}}} (otherwise the quantitative apartness can be trivially established). This gives a lower bound of {{math|{{sfrac|1|3''b''{{sup|2}}}}}} for the difference {{math|{{abs|{{sqrt|2}} − {{sfrac|''a''|''b''}}}}}}, yielding a direct proof of irrationality not relying on the [[law of excluded middle]]; see [[Errett Bishop]] (1985, p.&nbsp;18). This proof constructively exhibits a discrepancy between {{math|{{sqrt|2}}}} and any rational.

===Proof by Diophantine equations===

* ''Lemma'':   For the [[Diophantine equation]] <math>x^2+y^2=z^2</math> in its primitive (simplest) form, integer solutions exist if and only if either <math>x</math> or  <math>y</math> is odd, but never when both  <math>x</math> and  <math>y</math> are odd.<ref>{{Cite book|title=Elementary Number Theory|last=Burton|first=David|publisher=McGraw-Hill Education|year=2007|isbn=978-0-07-305188-8|location=|pages=247}}</ref>

''Proof'':  For the given equation, there are only six possible combinations of oddness and evenness for whole-number values of  <math>x</math> and  <math>y</math> that produce a whole-number value for <math>z</math>. A simple enumeration of all six possibilities shows why four of these six are impossible. Of the two remaining possibilities, one can be proven to not contain any solutions using modular arithmetic, leaving the sole remaining possibility as the only one to contain solutions, if any.
{| class="wikitable"
!x, y
!z
!
|-
|Both even
|Even
|Impossible. The given Diophantine equation is primitive and therefore contains no common factors throughout 
|-
|Both odd
|Odd
|Impossible. The sum of two odd numbers does not produce an odd number.
|-
|Both even
|Odd
|Impossible. The sum of two even numbers does not produce an odd number.
|-
|One even, another odd
|Even
|Impossible. The sum of an even number and an odd number does not produce an even number.
|-
|Both odd
|Even
|Possible.
|-
|One even, another odd
|Odd
|Possible.
|}
The fifth possibility (both <math>x</math> and <math>y</math> odd and <math>z</math> even) can be shown to contain no solutions as follows.

Since <math>z</math> is even, <math>z^2</math> must be divisible by <math>4</math>, hence

:<math>x^2+y^2 \equiv 0 \mod4</math>

The square of any odd number is always  <math>\equiv1\bmod4</math>. The square of any even number is always <math>\equiv 0\bmod4</math>. Since both <math>x</math> and <math>y</math> are odd and <math>z</math> is even, we have:

:<math>1+1\equiv0\mod4</math>
:<math>2\equiv 0\mod4</math>

which is impossible. Therefore, the fifth possibility is also ruled out, leaving the sixth to be the only possible combination to contain solutions, if any.

'''An extension of this lemma is the result that two identical whole-number squares can never be added to produce another whole-number square, even when the equation is not in its simplest form.'''

* ''Theorem:'' <math>\sqrt2</math> is irrational.

''Proof'':  Assume <math>\sqrt2</math> is rational. Therefore,

:<math>\sqrt2= {a \over b}</math>
:where <math>a,b\in\Zeta</math>
:Squaring both sides,
:<math>2={a^2 \over b^2}</math>
:<math>2b^2=a^2</math>
:<math>b^2+b^2=a^2</math>

But the lemma proves that the sum of two identical whole-number squares cannot produce another whole-number square.

Therefore, the assumption that <math>\sqrt2</math> is rational is contradicted.

<math>\sqrt2</math> is irrational.   '''''[[Q. E. D.]]'''''

==Properties of the square root of two==
[[File:Circular and hyperbolic angle.svg|350px|thumb|[[Angle]] size and sector [[area]] are the same when the conic radius is {{sqrt|2}}. This diagram illustrates the circular and hyperbolic functions based on sector areas {{math|''u''}}.]]
One-half of {{math|{{sqrt|2}}}}, also the [[reciprocal (mathematics)|reciprocal]] of {{math|{{sqrt|2}}}}, approximately {{gaps|0.70710|67811|86548}}, is a common quantity in geometry and [[trigonometry]] because the [[unit vector]] that makes a 45° angle with the axes in a plane has the coordinates
:<math>\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).</math>
This number satisfies
:<math>\tfrac{1}{2}\sqrt{2} = \sqrt{\tfrac{1}{2}} = \frac{1}{\sqrt{2}} = \cos 45^{\circ} = \sin 45^{\circ}.</math>

One interesting property of {{math|{{sqrt|2}}}} is as follows:
:<math> \!\ {1 \over {\sqrt{2} - 1}} = \sqrt{2} + 1 </math>
since
:<math>\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=2-1=1.</math>
This is related to the property of [[silver ratio]]s.

{{math|{{sqrt|2}}}} can also be expressed in terms of the copies of the [[imaginary unit]] {{math|''i''}} using only the [[square root]] and [[arithmetic operations]]:
:<math>\frac{\sqrt{i}+i \sqrt{i}}{i}\text{ and }\frac{\sqrt{-i}-i \sqrt{-i}}{-i}</math>
'''if''' the square root symbol is interpreted suitably for the complex numbers {{math|''i''}} and {{math|−''i''}}.

{{math|{{sqrt|2}}}} is also the only real number other than 1 whose infinite [[Tetration|tetrate]] (i.e., infinite exponential tower) is equal to its square. In other words: if for {{math|c > 1}} we define {{math|''x''<sub>1</sub> {{=}} ''c''}} and {{math|''x''<sub>''n''+1</sub> {{=}} ''c''<sup>''x''<sub>''n''</sub></sup>}} for {{math|''n'' > 1}}, we will call the limit of {{math|''x''<sub>''n''</sub>}} as {{math|''n'' → ∞}} (if this limit exists) {{math|''f''(''c'')}}. Then {{math|{{sqrt|2}}}} is the only number {{math|''c'' > 1}} for which {{math|''f''(''c'') {{=}} ''c''<sup>2</sup>}}. Or symbolically:
:<math>\sqrt{2}^ {(\sqrt{2}^ {(\sqrt{2}^ {(\ \cdot^ {\cdot^ \cdot)))}}}} = 2.</math>

{{math|{{sqrt|2}}}} appears in [[Viète's formula]] for {{pi}}:
: <math>2^m\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}} \to \pi\text{ as }m \to \infty</math>
for {{math|''m''}} square roots and only one minus sign.<ref>{{Citation |title=What is mathematics? An Elementary Approach to Ideas and Methods |first=Richard |last=Courant |first2=Herbert |last2=Robbins |location=London |publisher=Oxford University Press |year=1941 |page=124 }}</ref>

Similar in appearance but with a finite number of terms, {{math|{{sqrt|2}}}} appears in various trigonometric constants:<ref>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 }}</ref>
:<math>\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}</math>

It is not known whether {{math|{{sqrt|2}}}} is a [[normal number]], a stronger property than irrationality, but statistical analyses of its [[binary expansion]] are consistent with the hypothesis that it is normal to base two.<ref>{{harvtxt|Good|Gover|1967}}.</ref>

==Series and product representations==
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
:<math>\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</math>
and
:<math>\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</math>
or equivalently,
:<math>\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.</math>

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
:<math>\frac{1}{\sqrt{2}} = \sum_{k=0}^\infty \frac{(-1)^k \left(\frac{\pi}{4}\right)^{2k}}{(2k)!}.</math>
The Taylor series of {{math|{{sqrt|1 + ''x''}}}} with {{math|''x'' {{=}} 1}} and using the [[double factorial]] {{math|''n''!!}} gives

:<math>\sqrt{2} = \sum_{k=0}^\infty (-1)^{k+1} \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.</math>

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

:<math>\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.</math>

It is not known whether {{math|{{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.<ref>{{cite web |url=http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf |title=Archived copy |accessdate=2010-04-30 |url-status=live |archiveurl=https://web.archive.org/web/20110610050911/http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf |archivedate=2011-06-10 }}</ref>

The number can be represented by an infinite series of [[Egyptian fractions]], with denominators defined by 2<sup>n</sup>th terms of a [[Fibonacci]]-like recurrence relation a(n)=34a(n-1)-a(n-2), a(0)=0, a(1)=6.<ref>{{Cite OEIS|1=A082405|2=a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=6|accessdate=2016-09-05}}</ref>

:<math>\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) </math>

==Continued fraction representation==
[[File:Dedekind cut- square root of two.png|thumb|335px|The square root of 2 and approximations by convergents of continued fractions]]
The square root of two has the following [[continued fraction]] representation:
:<math> \!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}. </math>

The [[Convergent (continued fraction)|convergents]] 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 number]]s (known as side and diameter numbers to the ancient Greeks because of their use in approximating the ratio between the sides and diagonal of a square). 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}}}}. The convergent {{math|{{sfrac|''p''|''q''}}}} differs from {{math|{{sqrt|2}}}} by almost exactly {{math|{{sfrac|1|2''q''{{sup|2}}{{sqrt|2}}}}}}{{Citation needed|date=September 2011}} and then the next convergent is {{math|{{sfrac|''p'' + 2''q''|''p'' + ''q''}}}}.

==Nested square representations==
The following nested square expressions converge to {{sqrt|2}}:
:<math>\begin{align} \sqrt{2}
&=\tfrac{3}{2} - 2 \left( \tfrac{1}{4}- \left( \tfrac{1}{4}-\left( \tfrac{1}{4}- \left( \tfrac{1}{4}- \cdots \right)^2 \right)^2 \right)^2 \right)^2\\
&=\tfrac{3}{2} - 4 \left( \tfrac{1}{8}+ \left( \tfrac{1}{8}+\left( \tfrac{1}{8}+ \left( \tfrac{1}{8}+ \cdots \right)^2 \right)^2 \right)^2 \right)^2.
\end{align} </math>

==Derived constants==
The reciprocal of the square root of two (the square root of {{sfrac|1|2}}) is a widely used constant.

:<math>\frac1{\sqrt{2}} = \frac{\sqrt{2}}{2} = \sin 45^\circ = \cos 45^\circ = 0.70710\,67811\,86547\,52440\,08443\,62104\,84903\,928...</math> {{OEIS|id=A010503}}

==Paper size==
In 1786, German physics professor Georg Lichtenberg<ref name=":0" /> 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 standarised paper sizes at the beginning of the 20 century, they used Lichtenberg's ratio to create the [[ISO_216#A_series|"A" series]] of paper sizes.<ref name=":0">{{Cite book|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=0393244806|location=|pages=324}}</ref> Today, the (approximate) [[aspect ratio]] of [[paper size]]s under [[ISO 216]] (A4, A0, etc.) is 1:{{math|{{sqrt|2}}}}. 

Proof:<br />
Let <math>S = </math> shorter length and <math>L = </math> longer length of the sides of a sheet of paper, with<br />
:<math>R = \frac{L}{S} = \sqrt{2}</math> as required by ISO 216.
Let <math>R' = \frac{L'}{S'}</math> be the analogue ratio of the halved sheet, then<br />
:<math>R' = \frac{S}{L/2} = \frac{2S}{L} = \frac{2}{(L/S)} = \frac{2}{\sqrt{2}} = \sqrt{2} = R</math>.

==See also==
* [[Square root of 3]]
* [[Square root of 5]]
* [[Silver ratio]], {{math|1 + {{sqrt|2}}}}
* The square root of two is the [[interval (music)#Frequency ratios|frequency ratio]] of a [[tritone]] interval in twelve-tone [[equal temperament]] music.
* The square root of two forms the relationship of [[F-number|f-stops]] in photographic lenses, which in turn means that the ratio of ''areas'' between two successive [[aperture]]s 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 {{math|{{sqrt|2}}}}.
* [[Gelfond–Schneider constant]], {{math|2<sup>{{sqrt|2}}</sup>}}.
* [[Viète's formula]]

==Notes==
{{Reflist|2}}

==References==
* {{Citation
 | last = Apostol | first = Tom M. | author-link = Tom M. Apostol
 | doi = 10.2307/2695741
 | issue = 9
 | journal = [[American Mathematical Monthly]]
 | pages = 841–842
 | title = Irrationality of the square root of two – A geometric proof
 | volume = 107
 | year = 2000
 | jstor = 2695741}}.
* {{Citation | author = [[Aristotle]] | title = [[Prior Analytics|Analytica priora]] | year = 2007 | publisher = eBooks@Adelaide}}
* Bishop, Errett (1985), Schizophrenia in contemporary mathematics. Errett Bishop: reflections on him and his research (San Diego, Calif., 1983), 1–32, Contemp. Math. 39, Amer. Math. Soc., Providence, RI.
* {{Citation
 | last = Flannery | first = David
 | isbn = 0-387-20220-X
 | publisher = Springer-Verlag
 | title = The Square Root of Two
 | year = 2005}}.
* {{Citation
 |last1        = Fowler
 |first1       = David
 |author1-link = David Fowler (mathematician)
 |last2        = Robson
 |first2       = Eleanor
 |author2-link = Eleanor Robson
 |doi          = 10.1006/hmat.1998.2209
 |issue        = 4
 |journal      = Historia Mathematica
 |pages        = 366–378
 |title        = Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context
 |url          = http://www.hps.cam.ac.uk/dept/robson-fowler-square.pdf
 |archive-url  = https://web.archive.org/web/20060903161813/http://www.hps.cam.ac.uk/dept/robson-fowler-square.pdf
 |url-status     = dead
 |archive-date = 2006-09-03
 |volume       = 25
 |year         = 1998
}}.
* {{Citation
 | doi = 10.2307/2344040
 | last1 = Good | first1 = I. J. | author1-link = I. J. Good
 | last2 = Gover | first2 = T. N.
 | issue = 1
 | journal = Journal of the Royal Statistical Society, Series A
 | pages = 102–107
 | title = The generalized serial test and the binary expansion of {{math|{{sqrt|2}}}}
 | volume = 130
 | year = 1967
 | jstor = 2344040}}.
* {{Citation
 | last = Henderson | first = David W.
 | contribution = Square roots in the Śulba Sūtras
 | editor-last = Gorini | editor-first = Catherine A.
 | isbn = 978-0-88385-164-7
 | pages = 39–45
 | publisher = Cambridge University Press
 | title = Geometry At Work: Papers in Applied Geometry
 | url = http://www.math.cornell.edu/~dwh/papers/sulba/sulba.html
 | year = 2000}}.

==External links==
* {{Citation
 | last1 = Gourdon | first1 = X.
 | last2 = Sebah | first2 = P.
 | contribution = Pythagoras' Constant: {{math|{{sqrt|2}}}}
 | title = Numbers, Constants and Computation
 | url = http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html
 | year = 2001}}.
* {{MathWorld|urlname=PythagorassConstant|title=Pythagoras's Constant}}
* [http://www.gutenberg.org/etext/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
* {{cite web|last=Grime|first=James|title=The Square Root {{math|{{sqrt|2}}}} of Two|url=http://www.numberphile.com/videos/root2.html|work=Numberphile|publisher=[[Brady Haran]]|author2=Bowley, Roger }}

{{Algebraic numbers}}
{{Irrational number}}
{{authority control}}

{{DEFAULTSORT:Square Root Of Two}}
[[Category:Algebraic numbers]]
[[Category:Quadratic irrational numbers]]
[[Category:Mathematical constants]]
[[Category:Pythagorean theorem]]
[[Category:Articles containing proofs]]
[[Category:Irrational numbers]]