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, often known as root 2 or radical 2 and written as √2, is the positive algebraic number that, when multiplied by itself, gives the number 2. It is more precisely 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 square with sides of one unit of length; this follows from the Pythagorean theorem. It was probably the first number known to be irrational. Its numerical value truncated to 65 decimal places is:

1.41421356237309504880168872420969807856967187537694807317667973799... (sequence A002193 in OEIS)

The square root of 2.

The quick approximation 99/70 (≈ 1.41429) for the square root of two is frequently used. Despite having a denominator of only 70, it differs from the correct value by less than 1/10,000.

History[link]

Babylonian clay tablet YBC 7289 with annotations.

The Babylonian clay tablet YBC 7289 (c. 1800–1600 BCE) gives an approximation of √2 in four sexagesimal figures, which is about six decimal figures:^[1]

Failed to parse (Missing texvc executable; please see math/README to configure.): 1 + \frac{24}{60} + \frac{51}{60^2} + \frac{10}{60^3} = 1.41421\overline{296}.

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.^[2] That is,

Failed to parse (Missing texvc executable; please see math/README to configure.): 1 + \frac{1}{3} + \frac{1}{3 \cdot 4} - \frac{1}{3 \cdot4 \cdot 34} = \frac{577}{408} \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 √2.

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, ancient Greeks 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.^[3][4][5] The square root of two is occasionally called "Pythagoras's number" or "Pythagoras's Constant", for example Conway & Guy (1996).^[6]

Computation algorithms[link]

There are a number of algorithms for approximating √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^[7] of 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:

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 approximation of the square root of 2 is achieved. Each iteration approximately doubles the number of correct digits. Starting with a₀ = 1 the next approximations are

• 3/2 = 1.5
• 17/12 = 1.416...
• 577/408 = 1.414215...
• 665857/470832 = 1.4142135623746...

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 3.6 GHz PC with 16 GiB of memory.^[8]

Among mathematical constants with computationally challenging decimal expansions, only π has been calculated more precisely.^[9]

Proofs of irrationality[link]

A short proof of the irrationality of √2 can be obtained from the rational root theorem, that is, 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) = x² − 2, it follows that √2 is either an integer or irrational. Because √2 is not an integer (2 is not a perfect square), √2 must therefore be irrational.

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

Proof by infinite descent[link]

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 an integer a and an integer b in general such that a / b = √2.
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.
3. It follows that a² / b² = 2 and a² = 2 b².   ( (a / b)ⁿ = aⁿ / bⁿ  )
4. Therefore a² is even because it is equal to 2 b². (2 b² is necessarily even because it is 2 times another whole number and even numbers are multiples of 2.)
5. It follows that a must be even (as squares of odd integers are never even).
6. Because a is even, there exists an integer k that fulfills: a = 2k.
7. Substituting 2k from step 6 for a in the second equation of step 3: 2b² = (2k)² is equivalent to 2b² = 4k², which is equivalent to b² = 2k².
8. 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.
9. By steps 5 and 8 a and b are both even, which contradicts that a / b is irreducible as stated in step 2.

Q.E.D.

Because there is a contradiction, the assumption (1) that √2 is a rational number must be false. By the law of excluded middle, the opposite is proven: √2 is irrational.

This proof was hinted by Aristotle, in his Analytica Priora, §I.23.^[10] It appeared first as a full proof in Euclid's Elements, as proposition 117 of Book X. However, since the early 19th century historians agree that this proof is an interpolation and not attributable to Euclid.^[11]

Proof by unique factorization[link]

An alternative proof uses the same approach with the fundamental theorem of arithmetic which says every integer greater than 1 has a unique factorization into powers of primes.

1. Assume that √2 is a rational number. Then there are integers a and b such that a is coprime to b and √2 = a / b. In other words, √2 can be written as an irreducible fraction.
2. The value of b cannot be 1 as there is no integer a the square of which is 2.
3. There must be a prime p which divides b and which does not divide a otherwise the fraction would not be irreducible.
4. The square of a can be factored as the product of the primes into which a is factored but with each power doubled.
5. Therefore by unique factorization the prime p which divides b, and also its square, cannot divide the square of a.
6. Therefore the square of an irreducible fraction cannot be reduced to an integer.
7. Therefore the square root of 2 cannot be a rational number.

This proof can be generalized to show that any root of any natural number which is not the square of a natural number is irrational. The article quadratic irrational gives a proof of the same result but not using the fundamental theorem of arithmetic.

Proof by infinite descent, not involving factoring[link]

The following reductio ad absurdum argument showing the irrationality of √2 is less well-known. It uses the additional information 2 > √2 > 1 so that 1 > √2 − 1 > 0.^[12]

1. Assume that √2 is a rational number. This would mean that there exist positive integers m and n with n ≠ 0 such that m/n = √2. Then m = n√2 and m√2 = 2n.
2. We may assume that n is the smallest integer so that n√2 is an integer. That is, that the fraction m/n is in lowest terms.
3. Then Failed to parse (Missing texvc executable; please see math/README to configure.): \sqrt{2} = \frac{m}{n}=\frac{m(\sqrt{2}-1)}{n(\sqrt{2}-1)}=\frac{2n-m}{m-n}

1. Because 1 > √2 − 1 > 0, it follows that n > n(√2 − 1) = m − n > 0.
2. 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[link]

Another reductio ad absurdum showing that √2 is irrational is less well-known.^[13] 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 previous proof viewed geometrically.

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.

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 we have 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.

Analytic proof[link]

• Lemma: let Failed to parse (Missing texvc executable; please see math/README to configure.): \alpha \in \mathbb{R}^+

and Failed to parse (Missing texvc executable; please see math/README to configure.): p_1,p_2,\dots, q_1, q_2, \ldots \in \mathbb{N}
such that Failed to parse (Missing texvc executable; please see math/README to configure.): \left|\alpha q_n - p_n \right| \neq 0
for all Failed to parse (Missing texvc executable; please see math/README to configure.): n \in \mathbb{N}
and

Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{n \rightarrow \infty} p_n = \lim_{n \rightarrow \infty} q_n = \infty\,

Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{n \rightarrow \infty} \left| \alpha q_n - p_n \right| = 0.\,

Then α is irrational.

Proof: suppose α = a/b with a, b ∈ N⁺.

For sufficiently big n,

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \alpha q_n - p_n \right| < \frac{1}{b}

then

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| a q_n /\ \! b - p_n \right| < \frac{1}{b}

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| a q_n - b p_n \right| < 1 \,

but aq_n − bp_n is an integer, absurd, then α is irrational.

• √2 is irrational.

Proof: let p₁ = q₁ = 1 and

Failed to parse (Missing texvc executable; please see math/README to configure.): p_{n+1} = p_n^2 + 2 q_n^2 \,

Failed to parse (Missing texvc executable; please see math/README to configure.): q_{n+1} = 2 p_n q_n \,\!

for all n ∈ N.

By induction,

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \sqrt{2} q_n - p_n \right| < \frac{1}{2^{2^{n-1}}}

for all n ∈ N. For n = 1,

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \sqrt{2} q_1 - p_1 \right| < \frac{1}{2}

and if is true for n then is true for n + 1. In fact

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \sqrt{2} q_n - p_n \right|^2 < \frac{1}{2^{2^n}}

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \sqrt{2} (2 p_n q_n) - (p_n^2 + 2 q_n^2) \right| < \frac{1}{2^{2^n}}

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < \left| \sqrt{2} q_{n + 1} - p_{n + 1} \right| < \frac{1}{2^{2^n}}.

By application of the lemma, √2 is irrational.

Constructive proof[link]

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 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, so that, applying the law of trichotomy in the context of an effectively computable predicate over Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbb{N} , we obtain Failed to parse (Missing texvc executable; please see math/README to configure.): |2 b^2 - a^2| \geq 1 . Then^[14]

Failed to parse (Missing texvc executable; please see math/README to configure.): |\sqrt2 - a / b| = \frac{|2b^2-a^2|}{b^2(\sqrt{2}+a/b)} \ge \frac{1}{b^2(\sqrt2 + a / b)} \ge \frac{1}{3b^2},

the latter inequality being true because we assume Failed to parse (Missing texvc executable; please see math/README to configure.): \tfrac{a}{b} \le 3- \sqrt{2}

(otherwise the quantitative apartness can be trivially established). This gives a lower bound of Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{3b^2}
for the difference Failed to parse (Missing texvc executable; please see math/README to configure.): |\sqrt2 - 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 Failed to parse (Missing texvc executable; please see math/README to configure.): \sqrt{2}

and any rational.

Properties of the square root of two[link]

One-half of √2, also 1 divided by the square root 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

Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).

This number satisfies

Failed to parse (Missing texvc executable; please see math/README to configure.): \tfrac{1}{2}\sqrt{2} = \sqrt{\tfrac{1}{2}} = \frac{1}{\sqrt{2}} = \cos(45^{\circ}) = \sin(45^{\circ}).

One interesting property of the square root of 2 is as follows:

Failed to parse (Missing texvc executable; please see math/README to configure.): \!\ {1 \over {\sqrt{2} - 1}} = \sqrt{2} + 1

since Failed to parse (Missing texvc executable; please see math/README to configure.): (\sqrt{2}+1)(\sqrt{2}-1)=2-1=1.

This is related to the property of silver means.

The square root of 2 can also be expressed in terms of the copies of the imaginary unit i using only the square root and arithmetic operations:

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\sqrt{i}+i \sqrt{i}}{i}\text{ and }\frac{\sqrt{-i}-i \sqrt{-i}}{-i}.

The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square.

Failed to parse (Missing texvc executable; please see math/README to configure.): \sqrt{2}^ {\sqrt{2}^ {\sqrt{2}^ {\ \cdot^ {\cdot^ \cdot}}}} = 2

The square root of 2 appears in Viète's formula for π:

Failed to parse (Missing texvc executable; please see math/README to configure.): 2^m\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}} \to \pi\text{ as }m \to \infty\,

for m square roots and only one minus sign.^[15]

Similarly in appearance but with a finite number of terms, the square root of 2 appears in various trigonometric constants:^[16]

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(5\tfrac58 ^\circ) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(11\tfrac14 ^\circ) = \frac12\sqrt{2-\sqrt{2+\sqrt{2}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(16\tfrac78 ^\circ) = \frac12\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(22\tfrac12 ^\circ) = \frac12\sqrt{2-\sqrt{2}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(28\tfrac18 ^\circ) = \frac12\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(33\tfrac34 ^\circ) = \frac12\sqrt{2-\sqrt{2-\sqrt{2}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(39\tfrac38 ^\circ) = \frac12\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(45 ^\circ) = \frac12\sqrt{2};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(50\tfrac58 ^\circ) = \frac12\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(56\tfrac14 ^\circ) = \frac12\sqrt{2+\sqrt{2-\sqrt{2}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(61\tfrac78 ^\circ) = \frac12\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(67\tfrac12 ^\circ) = \frac12\sqrt{2+\sqrt{2}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(73\tfrac18 ^\circ) = \frac12\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2}}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(78\tfrac34 ^\circ) = \frac12\sqrt{2+\sqrt{2+\sqrt{2}}};

Failed to parse (Missing texvc executable; please see math/README to configure.): \sin(84\tfrac38 ^\circ) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}.

It is not known whether √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.^[17]

Series and product representations[link]

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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,

Failed to parse (Missing texvc executable; please see math/README to configure.): \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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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.

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \sqrt{2} = \sum_{k=0}^\infty \frac{(2k+1)!}{(k!)^2 2^{3k+1}} = \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 √2 ln(1 + √2), however. [1]

Continued fraction representation[link]

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \!\ \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 the square root of 2 by almost exactly Failed to parse (Missing texvc executable; please see math/README to configure.): \scriptstyle{\frac{1}{q^2\sqrt{2}}} ^{[citation needed]} and then the next convergent is (p + 2q)/(p + q).

Paper size[link]

The square root of two is the aspect ratio of paper sizes under ISO 216 (A4, A0, etc.). This ratio guarantees that cutting a sheet in half along a line parallel to its short side results in two sheets having the same ratio.

See also[link]

• Square root of 3
• Square root of 5
• Silver ratio, 1 + √2
• 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, which in turn means that the ratio of areas between two successive apertures is 2.
• Viète's formula

Notes[link]

References[link]

• Apostol, Tom M. (2000), "Irrationality of the square root of two – A geometric proof", American Mathematical Monthly 107 (9): 841–842, DOI:10.2307/2695741, JSTOR 2695741 .
• Aristotle (2007), Analytica priora, 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.
• Flannery, David (2005), The Square Root of Two, Springer-Verlag, ISBN 0-387-20220-X .
• Fowler, David; Robson, Eleanor (1998), "Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context", Historia Mathematica 25 (4): 366–378, DOI:10.1006/hmat.1998.2209, http://www.hps.cam.ac.uk/dept/robson-fowler-square.pdf .
• Good, I. J.; Gover, T. N. (1967), "The generalized serial test and the binary expansion of √2", Journal of the Royal Statistical Society, Series A 130 (1): 102–107, DOI:10.2307/2344040, JSTOR 2344040 .
• Henderson, David W. (2000), "Square roots in the Śulba Sūtras", in Gorini, Catherine A., Geometry At Work: Papers in Applied Geometry, Cambridge University Press, pp. 39–45, ISBN 978-0-88385-164-7, http://www.math.cornell.edu/~dwh/papers/sulba/sulba.html .

External links[link]

• Gourdon, X.; Sebah, P. (2001), "Pythagoras' Constant: √2", Numbers, Constants and Computation, http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html .
• Weisstein, Eric W., "Pythagoras's Constant" from MathWorld.
• The Square Root of Two to 5 million digits by Jerry Bonnell and Robert Nemiroff. May, 1994.
• Square root of 2 is irrational, a collection of proofs
• √2.net, enthusiast site with realtime computation
• Square Root of 2 Search Engine 2 billion searchable digits of √2, π, and e

David Rudisha

Rudisha in Daegu, 2011.
Personal information
Nationality	Kenya
Born	(1988-12-17) 17 December 1988 (age 23) Kilgoris
Height	1.90 m (6 ft 3 in)[1]
Sport
Sport	Track and field
Event(s)	800 metres
Achievements and titles
Personal best(s)	400 m: 45.50 (2010) 800 m: 1:41.01 WR (2010)
Medal record Men's athletics Competitor for  Kenya World Championships Gold 2011 Daegu 800 m African Championships Gold 2010 Nairobi 800 m Gold 2008 Addis Ababa 800 m World Junior Championships Gold 2006 Beijing 800 m Competitor for Africa Continental Cup Gold 2010 Split 800 m

David Lekuta Rudisha (born 17 December 1988) is a Kenyan middle distance runner and world record holder in the 800 metres with a time of 1:41.01. Rudisha holds four of the ten fastest times for this event.

Early life[link]

Born in Kilgoris, Trans Mara District, Rudisha went to St. Francis, Kimuron Secondary School in Iten, Keiyo District, which is known for nurturing several top runners. Initially he was a 400 metres runner, but his coach, Colm O’Connell, prompted him to try 800 metres. In 2006 he became the world junior champion over that distance.^[2]

Career[link]

Rudisha competed at the 2009 World Athletics Championships, reaching the 800 metres semifinals. In September 2009, Rudisha won the IAAF Grand Prix meeting in Rieti, Italy, posting a new African record of 1:42.01, beating the 25-year old record of 1:42.28 set by compatriot Sammy Koskei. That effort put him in fourth place on the all-time list.^[3] In the 2010 IAAF Diamond League, he took on Abubaker Kaki at the Bislett Games in June. He defeated Sebastian Coe's 31-year-old meet record with a run of 1:42.04, giving him another place in the top-ten fastest ever 800 m and leaving Kaki the consolation of the fastest ever non-winning time.^[4] On 10 July 2010, Rudisha ran the 800 m in 1:41.51 at the KBC Night of Athletics in Heusden, Belgium; this new personal record placed him #2 all-time in the world for the 800 m.^[5]

Rudisha currently holds the world record for the 800 m. He broke the record at the ISTAF IAAF World Challenge meet in Berlin on 22 August 2010, with a time of 1:41.09.^[6] The previous record, held by Kenyan-born Danish Citizen runner Wilson Kipketer, had stood since 1997; Rudisha broke the record by 0.02 seconds.^[7] Rudisha claimed that it had been his "first real attempt" at breaking the record, and that he was capable of improving the time. This was proven one week later when he set a new WR of 1:41.01 in Rieti, Italy.^[8] In November 2010, at the age of 21, he became the youngest ever athlete to win the IAAF World Athlete of the Year award. He also won the Kenyan Sportsman of the Year award ^[9]

Personal life[link]

Rudisha is a member of the Maasai ethnic group in Kenya.^[2] Rudisha's father, Daniel Rudisha, is a former runner who won silver at the 1968 Olympics as part of the Kenyan 4 × 400 m relay team, while his mother Naomi is a former 400 m hurdler.^[10]

He is married to Lizzy Naanyu with a daughter (as of 2010).^[10]

Achievements[link]

Rudisha at the 2010 Memorial van Damme.

Year	Competition	Venue	Position	Notes
Representing  Kenya
2006	World Junior Championships	Beijing, China	1st	800 m (1:47.40)
2007	African Junior Championships	Ouagadougou, Burkina Faso	1st	800 m (1:46.41)[11]
2008	African Championships	Addis Ababa, Ethiopia	1st	800 m (1:44.20)
2009	World Athletics Final	Thessaloniki, Greece	1st	800 m (1:44.85)
2010	African Championships	Nairobi, Kenya	1st	800 m (1:42.84)
2011	World Championships	Daegu, South Korea	1st	800 m (1:43.91)

Personal bests[link]

References[link]

External links[link]

Wikimedia Commons has media related to: David Rudisha

	Biography portal
	Kenya portal

• IAAF profile for David Rudisha

Records
Preceded by Wilson Kipketer	Men's 800 metres world record holder 29 August 2010 – present	Succeeded by Incumbent
Awards
Preceded by Usain Bolt	IAAF World Athlete of the Year 2010, 2011	Succeeded by Usain Bolt
Preceded by Usain Bolt	Men's Track & Field Athlete of the Year 2010	Succeeded by Incumbent
Sporting positions
Preceded by Abubaker Kaki	Men's 800 metres best year performance 2009–2011	Succeeded by Incumbent