Square root of 2

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The square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1.
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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, also known as Pythagoras' constant, often denoted by

\sqrt{2},

is the positive real number that, when multiplied by itself, gives the number 2. Its numerical value approximated to 65 decimal places (sequence A002193 in OEIS) is:

1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799.

The square root of 2 was probably the first known irrational number. 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 inexpensive calculators with no SQRT function, the quick approximation 99/ 70 for the square root of two is better than the quick approximation 22/ 7 for pi, probably the most widely known irrational number.

The silver ratio is

1+\sqrt{2}.\,

Contents

[edit] History

The Babylonian clay tablet YBC 7289 (c. 2000–1650 BC) gives an approximation of \sqrt{2} in four sexagesimal figures, which is about six decimal figures:[1]

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 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.[2] That is,

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

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. [1] Other legends report that Hippasus was drowned by fanatical Pythagoreans [2], or merely expelled from their circle. [3]

[edit] Computation algorithm

There are a number of algorithms used in approximating the square root of 2, which - as an infinite nonrepeating decimal - can only ever be approximated. The most common algorithm for this, one used as a basis in many computers and calculators, is the Babylonian method[3] of computing square roots, which is one of many methods of computing square roots. It goes as follows:

First, pick an arbitrary guess, F0; 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_{n+1} = \frac{F_n + \frac{2}{F_n}}{2}

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. Among mathematical constants with nonrepeating decimal expansions, only π has been calculated more accurately. [4]

[edit] Proof of irrationality

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.

  1. Assume that √2 is a rational number, meaning that there exists an integer a and an integer b such that a / b = √2.
  2. Then √2 can be written as an irreducible fraction (the fraction is reduced as much as possible) 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.
  4. Therefore a2 is even because it is equal to 2 b2 which is obviously even.
  5. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
  6. Because a is even, there exists an integer k that fulfills: a = 2k.
  7. We insert the last equation of (3) in (6): 2b2 = (2k)2 is equivalent to 2b2 = 4k2 is equivalent to b2 = 2k2.
  8. Because 2k2 is even it follows that b2 is also even which means that b is even because only even numbers have even squares.
  9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

Since we have found 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.

[edit] A different proof

Another reductio ad absurdum showing that √2 is irrational is less well-known. 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) = 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.

[edit] Properties of the square root of two

One-half of √2, approximately 0.70710 67811 86548, is a common quantity in geometry and trigonometry, due to the fact that the unit vector that makes a 45° angle with the axes in a plane has the coordinates

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

This number satisfies

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

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

\!\ {1 \over {\sqrt{2} - 1}} = \sqrt{2} + 1.

This is a result of a property of silver means.

The square root of two also has the continued fraction

\!\ 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \cdots}}}.

[edit] Series 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{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

\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,

\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

\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 gives

\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

\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. [5]

[edit] See also

[edit] Notes

  1. ^ Fowler and Robson, p. 368.
    Photograph, illustration, and description of the root(2) tablet from the Yale Babylonian Collection
    High resolution photographs, descriptions, and analysis of the root(2) tablet (YBC 7289) from the Yale Babylonian Collection
  2. ^ Henderson.
  3. ^ Although the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of \sqrt{2} seen on tablet YBC 7289. Fowler and Robson offer informed and detailed conjectures.
    Fowler and Robson, p. 376. Flannery, p. 32, 158.

[edit] References

[edit] External links

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