# Square root of 2

Square root of 2

The square root of 2, also known as Pythagoras' constant, often denoted by

:$sqrt\left\{2\right\}$ or √2

but can also be written as

:$2^\left\{1/2\right\},,$

is the positive real number that, when multiplied by itself, gives the number 2. Its numerical value approximated to 65 decimal places OEIS|id=A002193 is::1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799...

The square root of 2 was probably the first number known to be irrational. Geometrically, it is the length of a diagonal across a square with sides of one unit of length; this follows from the Pythagorean theorem. On basic calculators with no square-root function, the quick approximation $frac\left\{99\right\}\left\{70\right\}$ for the square root may be used. Despite having a denominator of only 70, it differs from the correct value by less than 1/10 000.

The silver ratio is

:$1+sqrt\left\{2\right\}.,$

History

The Babylonian clay tablet YBC 7289 (c. 1800–1600 BCE) gives an approximation of $sqrt\left\{2\right\}$ in four sexagesimal figures, which is about six decimal figures: [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]
[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]
]

:$1 + frac\left\{24\right\}\left\{60\right\} + frac\left\{51\right\}\left\{60^2\right\} + frac\left\{10\right\}\left\{60^3\right\} = 1.41421overline\left\{296\right\}.$

Another early close approximation of this number is given in ancient Indian mathematical texts, the Sulbasutras (c. 800–200 BCE) 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." [ Henderson.] That is,

:$1 + frac\left\{1\right\}\left\{3\right\} + frac\left\{1\right\}\left\{3 cdot 4\right\} - frac\left\{1\right\}\left\{3 cdot4 cdot 34\right\} = frac\left\{577\right\}\left\{408\right\} approx 1.414215686.$

This ancient Indian approximation is the seventh in a sequence of increasingly accurate approximations based on the sequence of Pell numbers, that can be derived from the continued fraction expansion of $sqrt\left\{2\right\}.$

The discovery of the irrational numbers is usually attributed to the Pythagorean Hippasus of Metapontum, who produced a (most likely geometrical) proof of the irrationality of the square root of 2. According to one legend, Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so he sentenced Hippasus to death by drowning. [http://www.washingtonpost.com/wp-srv/style/longterm/books/chap1/mysteryaleph.htm Washingtonpost.com: The Mystery Of The Aleph: Mathematics, the Kabbalah, and the Search for Infinity] ] Other legends report that Hippasus was drowned by some Pythagoreans, [ [http://scienceworld.wolfram.com/biography/Hippasus.html Hippasus of Metapontum (ca. 500 BC) - from Eric Weisstein's World of Scientific Biography] ] or merely expelled from their circle.

Computation algorithm

There are a number of algorithms for approximating the square root of 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 [Although the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of √2 seen on tablet YBC 7289. Fowler and Robson offer informed and detailed conjectures.
Fowler and Robson, p. 376. Flannery, p. 32, 158.
] of computing square roots, which is one of many methods of computing square roots. It goes as follows:

First, pick an arbitrary guess, $F_0$; the guess doesn't matter, as it only affects how many iterations are required to reach an approximation of a certain accuracy. Then, using that guess, iterate through the following recursive computation:

:$F_\left\{n+1\right\} = frac\left\{F_n + frac\left\{2\right\}\left\{F_n\left\{2\right\}.$

The more iterations through the algorithm (that is, the more computations performed and the greater "n"), the better approximation of the square root of 2 is achieved.

The value of √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 √2 was eclipsed with the use of a home computer. Shigeru Kondo calculated 200,000,000,000 decimal places in slightly over 13 days and 14 hours using a 3.6GHz PC with 16GB of memory. [http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html Constants and Records of Computation]

Among mathematical constants with nonrepeating decimal expansions, only &pi; has been calculated more accurately. [ [http://numbers.computation.free.fr/Constants/Miscellaneous/Records.html Number of known digits] ]

Proofs of irrationality

A short proof of this result is to obtain it from Gauss's lemma, that if "p"("x") is a monic polynomial with integer coefficients, then any rational root of "p"("x") is necessarily an integer. Applying this to the polynomial "p"("x") = "x"2 − 2, it follows that √2 is either an integer or irrational. Since √2 is not an integer (2 is not a perfect square), √2 must therefore be irrational.

See quadratic irrational for a proof that the square root of any non-square natural number is irrational.

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, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, which means that the proposition must be true.

# Assume that √2 is a rational number, meaning that there exists an integer "a" and an integer "b" such that "a" / "b" = √2.
# Then √2 can be written as an irreducible fraction "a" / "b" such that "a" and "b" are coprime integers and ("a" / "b")2 = 2.
# It follows that "a"2 / "b"2 = 2 and "a"2 = 2 "b"2. ("("a" / "b")n = "a"n / "b"n")
# Therefore "a"2 is even because it is equal to 2 "b"2. (2 "b"2 is necessarily even because it's divisible by 2—that is, (2 "b"2)/2 = "b"2 — and numbers divisible by two are even by definition.")
# It follows that "a" must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
# Because "a" is even, there exists an integer "k" that fulfills: "a" = 2"k".
# Substituting 2"k" from (6) for "a" in the second equation of (3): 2"b"2 = (2"k")2 is equivalent to 2"b"2 = 4"k"2 is equivalent to "b"2 = 2"k"2.
# Because 2"k"2 is divisible by two and therefore even, and because 2"k"2 = "b"2, it follows that "b"2 is also even which means that "b" is even.
# By (5) and (8) "a" and "b" are both even, which contradicts that "a" / "b" is irreducible as stated in (2).::"Q.E.D."

Since there is a contradiction, the assumption (1) that √2 is a rational number must be false. The opposite is proven: √2 is irrational.

This proof can be generalized to show that any root of any natural number is either a natural number or irrational.
# Assume that $sqrt\left\{2\right\}$ is a rational number. Then $exists a,b in mathbb\left\{Z\right\}$ such that a is coprime to "b" and $sqrt\left\{2\right\}=\left\{a over b\right\}$. In other words, $sqrt\left\{2\right\}$ can be written as an irreducible fraction.
# It follows that $2 = \left\{a^2 over b^2\right\}$, and that $a^2=2b^2$.
# So, by the properties of exponents along with the unique factorization theorem, 2 divides both "a""n" and "a".
# Factoring out 2 from (2), we have $a^2=2b\text{'}$ for some $b\text{'}in mathbb\left\{Z\right\}$.
# Therefore 2 divides "a".
# But this contradicts the assumption that "a" and "b" are coprime.
# Therefore $sqrt\left\{2\right\}$ is not a rational number.::"Q.E.D." -->

Another proof

The following reductio ad absurdum argument showing the irrationality of √2 is less well-known. It uses the additional information √2 > 1.
# Assume that √2 is a rational number. This would mean that there exist integers "m" and "n" with "n" ≠ 0 such that "m"/"n" = √2.
# Then √2 can also be written as an irreducible fraction "m"/"n" with "positive" integers, because √2 > 0.
# Then $sqrt\left\{2\right\} = frac\left\{sqrt\left\{2\right\}cdot n\left(sqrt\left\{2\right\}-1\right)\right\}\left\{n\left(sqrt\left\{2\right\}-1\right)\right\} = frac\left\{2n-sqrt\left\{2\right\}n\right\}\left\{sqrt\left\{2\right\}n-n\right\} = frac\left\{2n-m\right\}\left\{m-n\right\}, ext\left\{ because \right\}sqrt\left\{2\right\},n,=,m.$
# Since √2 > 1, it follows that "m" > "n", which in turn implies that "m" > 2"n" – "m".
# So the fraction "m"/"n" for √2, which according to (2) is already in lowest terms, is represented by (3) in strictly lower terms. This is a contradiction, so the assumption that √2 is rational must be false.

Geometric proof

Another reductio ad absurdum showing that √2 is irrational is less well-known. [Apostol (2000), p. 841] 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.

Let "ABC" be a right isosceles triangle with hypotenuse length "m" and legs "n". 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.

Since ∠"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") = 2"n" − "m".

Hence we have an even smaller right isosceles triangle, with hypotenuse length 2"n" − "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.

Properties of the square root of two

One-half of √2, approximately 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

:$left\left(frac\left\{sqrt\left\{2\left\{2\right\}, frac\left\{sqrt\left\{2\left\{2\right\} ight\right).$

This number satisfies

:$frac\left\{sqrt\left\{2\left\{2\right\} = sqrt\left\{frac\left\{1\right\}\left\{2 = frac\left\{1\right\}\left\{sqrt\left\{2 = cos\left(45^circ\right) = sin\left(45^circ\right).$

One interesting property of the square root of two is as follows:

:$! \left\{1 over \left\{sqrt\left\{2\right\} - 1 = sqrt\left\{2\right\} + 1.$

This is a result of a property of silver means.

Another interesting property of the square root of two:

:$sqrt\left\{2+sqrt\left\{2+sqrt\left\{2 cdots\right\} = 2.$

The square root of two can also be expressed in terms of the copies of the imaginary unit "i" using only the square root and arithmetic operations:

:$frac\left\{sqrt\left\{i\right\}+i sqrt\left\{i\left\{i\right\}$ and $frac\left\{sqrt\left\{-i\right\}-i sqrt\left\{-i\left\{-i\right\}.$

The square root of two is also the only real number whose infinite tetrate is equal to its square.

:$sqrt\left\{2\right\}^ \left\{sqrt\left\{2\right\}^ \left\{sqrt\left\{2\right\}^ \left\{ cdot^ \left\{cdot^ cdot = 2$

eries and product representations

The identity cos(π/4) = sin(π/4) = 1/√2, along with the infinite product representations for the sine and cosine, leads to products such as

:$frac\left\{1\right\}\left\{sqrt 2\right\} = prod_\left\{k=0\right\}^inftyleft\left(1-frac\left\{1\right\}\left\{\left(4k+2\right)^2\right\} ight\right) =left\left(1-frac\left\{1\right\}\left\{4\right\} ight\right)left\left(1-frac\left\{1\right\}\left\{36\right\} ight\right)left\left(1-frac\left\{1\right\}\left\{100\right\} ight\right) cdots$

and

:$sqrt\left\{2\right\} =prod_\left\{k=0\right\}^inftyfrac\left\{\left(4k+2\right)^2\right\}\left\{\left(4k+1\right)\left(4k+3\right)\right\} =left\left(frac\left\{2 cdot 2\right\}\left\{1 cdot 3\right\} ight\right)left\left(frac\left\{6 cdot 6\right\}\left\{5 cdot 7\right\} ight\right)left\left(frac\left\{10 cdot 10\right\}\left\{9 cdot 11\right\} ight\right)left\left(frac\left\{14 cdot 14\right\}\left\{13 cdot 15\right\} ight\right) cdots$

or equivalently,

:$sqrt\left\{2\right\} =prod_\left\{k=0\right\}^inftyleft\left(1+frac\left\{1\right\}\left\{4k+1\right\} ight\right)left\left(1-frac\left\{1\right\}\left\{4k+3\right\} ight\right)=left\left(1+frac\left\{1\right\}\left\{1\right\} ight\right)left\left(1-frac\left\{1\right\}\left\{3\right\} ight\right)left\left(1+frac\left\{1\right\}\left\{5\right\} ight\right)left\left(1-frac\left\{1\right\}\left\{7\right\} ight\right) cdots.$

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

:$frac\left\{1\right\}\left\{sqrt\left\{2 = sum_\left\{k=0\right\}^infty frac\left\{\left(-1\right)^k left\left(frac\left\{pi\right\}\left\{4\right\} ight\right)^\left\{2k\left\{\left(2k\right)!\right\}.$

The Taylor series of √(1+"x") with "x" = 1 gives

:$sqrt\left\{2\right\} = sum_\left\{k=0\right\}^infty \left(-1\right)^\left\{k+1\right\} frac\left\{\left(2k-3\right)!!\right\}\left\{\left(2k\right)!!\right\} =1 + frac\left\{1\right\}\left\{2\right\} - frac\left\{1\right\}\left\{2cdot4\right\} + frac\left\{1cdot3\right\}\left\{2cdot4cdot6\right\} -frac\left\{1cdot3cdot5\right\}\left\{2cdot4cdot6cdot8\right\} + cdots.$

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

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

It is not known whether √2 can be represented with a BBP-type formula. BBP-type formulas are known for π√2 and √2 ln(1+√2), however. [http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf]

Continued fraction representation

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

:$! sqrt\left\{2\right\} = 1 + cfrac\left\{1\right\}\left\{2 + cfrac\left\{1\right\}\left\{2 + cfrac\left\{1\right\}\left\{2 + cfrac\left\{1\right\}\left\{ddots.$

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 due to their use in approximating the ratio between the sides and diagonal of a square).

Paper size

The square root of two is the aspect ratio of paper sizes under ISO 216. This ratio guarantees that cutting in half a sheet by a line parallel to its short side results in two sheets having the same ratio.

Indeed, if a rectangle has sides $x$ and $x sqrt\left\{2\right\}$, its half has sides $x$ and $x sqrt\left\{2\right\}/2$, the latter being the same as $x/sqrt\left\{2\right\}$. Therefore, the proportion between the long side ($x$) and the short side ($x/sqrt\left\{2\right\}$) is again $sqrt\left\{2\right\}$.

ee also

* Square root of 3
* Square root of 5
* The square root of two is the frequency ratio of a tritone interval in twelve-tone equal temperament music.
* The square root of two also forms the relationship of f stops in photographic lenses.

Notes

References

*
*
*
* Gourdon, X. & Sebah, P. [http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html Pythagoras' Constant: √2] . Includes information on how to compute digits of $sqrt\left\{2\right\}$.
* Henderson, David W., " [http://www.math.cornell.edu/~dwh/papers/sulba/sulba.html Square Roots in the Sulbasutra] "
*

External links

* [http://www.gutenberg.org/etext/129 The Square Root of Two to 5 million digits by Jerry Bonnell and Robert Nemiroff. May, 1994.]
* [http://www.cut-the-knot.org/proofs/sq_root.shtml Square root of 2 is irrational] , a collection of proofs
* [http://xn--2-tbo.net √2.net] , enthusiast site with realtime computation

Wikimedia Foundation. 2010.

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