Square root of 2

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.[11]

There are a number of algorithms for approximating √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[12] for computing square roots, which is one of many methods of computing square roots. It goes as follows:

First, pick a guess, a₀ > 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:

${\displaystyle a_{n+1}={\frac {a_{n}+{\frac {2}{a_{n}}}}{2}}={\frac {a_{n}}{2}}+{\frac {1}{a_{n}}}.}$

The more iterations through the algorithm (that is, the more computations performed and the greater "n"), the better the approximation. Each iteration roughly doubles the number of correct digits. Starting with a₀ = 1, the results of the algorithm are as follows:

• 1 (a₀)
• 3/2 = 1.5 (a₁)
• 17/12 = 1.416... (a₂)
• 577/408 = 1.414215... (a₃)
• 665857/470832 = 1.4142135623746... (a₄)

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). Since it is a convergent of the continued fraction representation of the square root of two, any better rational approximation has a denominator not less than 169, since 239/169 (≈ 1.4142012) is the next convergent 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 a₀ = 1 (665,857/470,832) is too large by about 1.6×10⁻¹²; its square is ≈ 2.0000000000045.

In 1997 the value of √2 was calculated to 137,438,953,444 decimal places by Yasumasa Kanada's team. In February 2006 the record for the calculation of √2 was eclipsed with the use of a home computer. Shigeru Kondo calculated 1 trillion decimal places in 2010.[13] Among mathematical constants with computationally challenging decimal expansions, only π has been calculated more precisely.[14] Such computations aim to check empirically whether such numbers are normal.

This is a table of recent records in calculating the digits of √2.[15]

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.

1. Assume that √2 is a rational number, meaning that there exists a pair of integers whose ratio is exactly √2.
2. If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.
3. Then √2 can be written as an irreducible fraction 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 .
4. It follows that a²/b² = 2 and a² = 2b².   ( (a/b)ⁿ = aⁿ/bⁿ  )   ( a² and b² are integers)
5. Therefore, a² is even because it is equal to 2b². (2b² is necessarily even because it is 2 times another whole number and multiples of 2 are even.)
6. It follows that a must be even (as squares of odd integers are never even).
7. Because a is even, there exists an integer k that fulfills: a = 2k.
8. Substituting 2k from step 7 for a in the second equation of step 4: 2b² = (2k)² is equivalent to 2b² = 4k², which is equivalent to b² = 2k².
9. Because 2k² is divisible by two and therefore even, and because 2k² = b², it follows that b² is also even which means that b is even.
10. By steps 5 and 8 a and b are both even, which contradicts that a/b is irreducible as stated in step 3.

Q.E.D.

Because there is a contradiction, the assumption (1) that √2 is a rational number must be false. This means that √2 is not a rational number. That is, √2 is irrational.

This proof was hinted at by Aristotle, in his Analytica Priora, §I.23.[16] 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.[17]

As with the proof by infinite descent, we obtain ${\displaystyle a^{2}=2b^{2}}$. 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.

Figure 1. Stanley Tennenbaum's geometric proof of the irrationality of √2

A simple proof is attributed by John Horton Conway to Stanley Tennenbaum when the latter was a student in the early 1950s[18] and whose most recent appearance is in an article by Noson Yanofsky in the May–June 2016 issue of American Scientist.[19] 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 ((2b − a)²) must equal the sum of the two uncovered squares (2(a − b)²). However, these squares on the diagonal have positive integer sides that are smaller than the original squares. Repeating this process, there are 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.

Figure 2. Tom Apostol's geometric proof of the irrationality of √2

Another geometric reductio ad absurdum argument showing that √2 is irrational appeared in 2000 in the American Mathematical Monthly.[20] 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 △ABC be a right isosceles triangle with hypotenuse length m and legs n as shown in Figure 2. By the Pythagorean theorem, m/n = √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 the ∠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 √2 is irrational.

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 a and b, because the valuation (i.e., highest power of 2 dividing a number) of 2b² is odd, while the valuation of a² is even, they must be distinct integers; thus |2b² − a²| ≥ 1. Then[21]

${\displaystyle \left|{\sqrt {2}}-{\frac {a}{b}}\right|={\frac {|2b^{2}-a^{2}|}{b^{2}\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{b^{2}\left({\sqrt {2}}+{\frac {a}{b}}\right)}}\geq {\frac {1}{3b^{2}}},}$

the latter inequality being true because it is assumed that a/b ≤ 3 − √2 (otherwise the quantitative apartness can be trivially established). This gives a lower bound of 1/3b² for the difference |√2 − a/b|, yielding a direct proof of irrationality not relying on the law of excluded middle; see Errett Bishop (1985, p. 18). This proof constructively exhibits a discrepancy between √2 and any rational.

• Lemma: For the Diophantine equation ${\displaystyle x^{2}+y^{2}=z^{2}}$ in its primitive (simplest) form, integer solutions exist if and only if either ${\displaystyle x}$ or ${\displaystyle y}$ is odd, but never when both ${\displaystyle x}$ and ${\displaystyle y}$ are odd.[22]

Proof: For the given equation, there are only six possible combinations of oddness and evenness for whole-number values of ${\displaystyle x}$ and ${\displaystyle y}$ that produce a whole-number value for ${\displaystyle z}$. 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.

The fifth possibility (both ${\displaystyle x}$ and ${\displaystyle y}$ odd and ${\displaystyle z}$ even) can be shown to contain no solutions as follows.

Since ${\displaystyle z}$ is even, ${\displaystyle z^{2}}$ must be divisible by ${\displaystyle 4}$, hence

${\displaystyle x^{2}+y^{2}\equiv 0\mod 4}$

The square of any odd number is always ${\displaystyle \equiv 1{\bmod {4}}}$. The square of any even number is always ${\displaystyle \equiv 0{\bmod {4}}}$. Since both ${\displaystyle x}$ and ${\displaystyle y}$ are odd and ${\displaystyle z}$ is even:

${\displaystyle 1+1\equiv 0\mod 4}$

${\displaystyle 2\equiv 0\mod 4}$

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: ${\displaystyle {\sqrt {2}}}$ is irrational.

Proof: Assume ${\displaystyle {\sqrt {2}}}$ is rational. Therefore,

${\displaystyle {\sqrt {2}}={a \over b}}$

where ${\displaystyle a,b\in \mathrm {Z} }$

Squaring both sides,

${\displaystyle 2={a^{2} \over b^{2}}}$

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

${\displaystyle b^{2}+b^{2}=a^{2}}$

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

Therefore, the assumption that ${\displaystyle {\sqrt {2}}}$ is rational is contradicted.

${\displaystyle {\sqrt {2}}}$ is irrational. Q. E. D.

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

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

and

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

or equivalently,

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

The number can also be expressed by taking the Taylor series of a trigonometric function. For example, the series for cos π/4 gives

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

The Taylor series of √1 + x with x = 1 and using the double factorial n!! gives

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

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

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

It is not known whether √2 can be represented with a BBP-type formula. BBP-type formulas are known for π√2 and √2ln(1+√2), however.[26]

The number can be represented by an infinite series of Egyptian fractions, with denominators defined by 2ⁿth terms of a Fibonacci-like recurrence relation a(n)=34a(n-1)-a(n-2), a(0)=0, a(1)=6.[27]

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

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

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

${\displaystyle \!\ {\sqrt {2}}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+\ddots }}}}}}}}.}$

The 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 numbers (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: 1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408. The convergent p/q differs from √2 by almost exactly 1/2q²√2 and then the next convergent is p + 2q/p + q.

The following nested square expressions converge to √2:

${\displaystyle {\begin{aligned}{\sqrt {2}}&={\tfrac {3}{2}}-2\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-\cdots \right)^{2}\right)^{2}\right)^{2}\right)^{2}\\&={\tfrac {3}{2}}-4\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+\cdots \right)^{2}\right)^{2}\right)^{2}\right)^{2}.\end{aligned}}}$

In 1786, German physics professor Georg Lichtenberg[28] found that any sheet of paper whose long edge is √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.[28] Today, the (approximate) aspect ratio of paper sizes under ISO 216 (A4, A0, etc.) is 1:√2.

Proof:
Let ${\displaystyle S=}$ shorter length and ${\displaystyle L=}$ longer length of the sides of a sheet of paper, with

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

Let ${\displaystyle R'={\frac {L'}{S'}}}$ be the analogue ratio of the halved sheet, then

${\displaystyle R'={\frac {S}{L/2}}={\frac {2S}{L}}={\frac {2}{(L/S)}}={\frac {2}{\sqrt {2}}}={\sqrt {2}}=R}$.

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 √2.

The number has applications in the area of video games. In particular, the popularity of MOBAs with three lanes on a square map means that the geometry of the map is such that the middle lane is shorter than the top and bottom lanes by ~70% as given by the ratio √2/2, the reciprocal. This means a player can traverse the map diagonally from base to base in less than three quarters of the time it would take to use the top or bottom lanes.