Editing Square root of 2 (section)

==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 by contradiction#Refutation_by_contradiction|proof of a negation by refutation]]: it proves the statement "<math>\sqrt{2}</math> is not rational" by assuming that it is rational and then deriving a falsehood.

# Assume that <math>\sqrt{2}</math> is a rational number, meaning that there exists a pair of integers whose ratio is exactly <math>\sqrt{2}</math>.
# If the two integers have a common [[divisor|factor]], it can be eliminated using the [[Euclidean algorithm]].
# Then <math>\sqrt{2}</math> can be written as an [[irreducible fraction]] <math>\frac{a}{b}</math> 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 [[parity (mathematics)|odd]].
# It follows that <math>\frac{a^2}{b^2}=2</math> and <math>a^2=2b^2</math>. &emsp; (&thinsp;{{math|[[Exponent#Identities and properties|({{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 [[parity (mathematics)|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.)
# 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 = 2k</math>.
# Substituting {{math|2''k''}} from step 7 for {{math|''a''}} in the second equation of step 4: <math>2b^2 = a^2 = (2k)^2 = 4k^2</math>, which is equivalent to <math>b^2=2k^2</math>.
# Because {{math|2''k''{{sup|2}}}} is divisible by two and therefore even, and because <math>2k^2=b^2</math>, 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 step 3 (that <math>\frac{a}{b}</math> is irreducible).

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

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

===Proof by unique factorization===
As with the proof by infinite descent, we obtain <math>a^2 = 2b^2</math>. 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 <math>\sqrt{2}</math> also follows from the [[rational root theorem]], which states that a rational [[root of a function|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 <math>p(x) = x^2 - 2</math>, the only possible rational roots are <math>\pm 1</math> and <math>\pm 2</math>. As <math>\sqrt{2}</math> is not equal to <math>\pm 1</math> or <math>\pm 2</math>, it follows that <math>\sqrt{2}</math> is irrational. This application also invokes the integer root theorem, a stronger version of the rational root theorem for the case when <math>p(x)</math> is a [[monic polynomial]] with integer [[coefficient]]s; for such a polynomial, all roots are necessarily integers (which <math>\sqrt{2}</math> 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#Square_root_of_non-square_is_irrational|Quadratic irrational number]] or [[proof by infinite descent#Irrationality of √k if it is not an integer|Infinite descent]].

===Geometric proofs===
==== Tennenbaum's proof ====
[[File:NYSqrt2.svg|thumb|Figure 1. Stanley Tennenbaum's geometric proof of the [[Irrational number|irrationality]] of {{math|√2}}]]
A simple proof is attributed to [[Stanley Tennenbaum]] when he was a student in the early 1950s.<ref>{{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 }}</ref><ref>{{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 }}</ref> Assume that <math>\sqrt{2} = a/b</math>, where <math>a</math> and <math>b</math> are coprime positive integers. Then <math>a</math> and <math>b</math> are the smallest positive integers for which <math>a^2 = 2b^2</math>. Geometrically, this implies that a square with side length <math>a</math> will have an area equal to two squares of (lesser) side length <math>b</math>. 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 <math>a^2 = 2b^2</math>, we would need to show that the area of the overlap is equal to the area of the two missing areas, i.e. <math>(2b-a)^2</math> = <math>2(a-b)^2</math>. In other terms, we may refer to the side lengths of the overlap and missing areas as <math>p = 2b-a</math> and <math>q = a-b</math>, respectively, and thus we have <math>p^2 = 2q^2</math>. But since we can see from the diagram that <math>p < a</math> and <math>q < b</math>, and we know that <math>p</math> and <math>q</math> are integers from their definitions in terms of <math>a</math> and <math>b</math>, this means that we are in violation of the original assumption that <math>a</math> and <math>b</math> are the smallest positive integers for which <math>a^2 = 2b^2</math>. 

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

==== Apostol's proof ====
[[File:Irrationality of sqrt2.svg|left|thumb|Figure 2. Tom Apostol's geometric proof of the irrationality of {{math|√2}}]]
[[Tom M. Apostol]] made another geometric ''[[reductio ad absurdum]]'' argument showing that <math>\sqrt{2}</math> is irrational.<ref>{{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 }}</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 same algebraic proof as Tennebaum's proof, viewed geometrically in another way.

Let {{math|△&hairsp;''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>\frac{m}{n}=\sqrt{2}</math>. 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 [[triangle]]s {{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|△&hairsp;''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|△&hairsp;''FDC''}} is also a right isosceles triangle. It also follows that {{math|''FC'' {{=}} ''n'' − (''m'' − ''n'') {{=}} 2''n'' − ''m''}}.

Hence, there is 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}</math> is irrational.
{{clear}}

===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  {{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|2''b''{{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|2''b''{{sup|2}} − ''a''{{sup|2}}}} ≥ 1}}. Multiplying the absolute difference {{math|{{abs|√2 − {{sfrac|''a''|''b''}}}}}} by {{math| ''b''{{sup|2}}(√2 + {{sfrac|''a''|''b''}})}} in the numerator and denominator, we get<ref>See {{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}}</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 (mathematics)|inequality]] being true because it is assumed that {{math|1<{{sfrac|''a''|''b''}}< 3/2}}, giving {{math|{{sfrac|''a''|''b''}} + √2 ≤ 3 }} (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|√2 − {{sfrac|''a''|''b''}}}}}}, yielding a direct proof of irrationality in its constructively stronger form, not relying on the [[law of excluded middle]].<ref>{{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}}</ref> This proof constructively exhibits an explicit discrepancy between <math>\sqrt{2}</math> and any rational.

===Proof by Pythagorean triples===
This proof uses the following property of primitive [[Pythagorean triple]]s:

: If {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} are coprime positive integers such that {{math|1=''a''<sup>2</sup> + ''b''<sup>2</sup> = ''c''<sup>2</sup>}}, then {{math|''c''}} is never even.<ref name=Sierpinski>{{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}}</ref>

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 <math>\sqrt2</math> is rational. Therefore,

:<math>\sqrt2 = {a \over b}</math>
:where <math>a,b \in \mathbb{Z}</math> and <math>\gcd(a,b) = 1</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>

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=2''b''<sup>2</sup> = ''a''<sup>2</sup>}} which implies that {{math|''a''}} must be even.