Square root of 2
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Image:Square root of 2 triangle.png The square root of 2, <math>\sqrt{2}</math>, also known as Pythagoras' constant, is the positive real number which, when multiplied by itself, gives the product 2. Its numerical value approximated to 65 decimal places Template:OEIS is:
- 1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799.
<math>\sqrt{2}</math> was the first known irrational number. Geometrically, <math>\sqrt{2}</math> is the length of a diagonal across a square with sides of one unit of length; this follows from Pythagoras' theorem.
The silver ratio is <math>1+\sqrt{2}</math>.
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History
The first approximation of this number was given in ancient Indian mathematical texts, the Sulbasutras (800 B.C. to 200 B.C.) as follows: Increase a unit length by its third and this third by its own fourth less the thirty-fourth part of that fourth. That is,
- <math>1 + \frac{1}{3} + \frac{1}{3 \cdot 4} - \frac{1}{3 \cdot4 \cdot 34} = \frac{577}{408} \approx 1.414215686.</math>
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. The story goes that Hippasus discovered irrational numbers when trying to represent the square root of 2 as a fraction . However 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.
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 of computing square roots, which is one of many methods of computing square roots. It goes as follows:
First, pick an arbitrary guess, <math>F_0</math>; 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:
- <math>F_{n+1} = \frac{F_n + \frac{2}{F_n}}{2} </math>
The more iterations through the algorithm (that is, the more computations performed and the greater "n"), the better approximation of <math>\sqrt{2}</math> achieved.
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.
- Assume that <math>\sqrt{2}</math> is a rational number, meaning that there exist an integer a and an integer b such that a / b = <math>\sqrt{2}</math>.
- Then <math>\sqrt{2}</math> can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)2 = 2.
- It follows that a2 / b2 = 2 and a2 = 2 b2.
- Therefore a2 is even because it is equal to 2 b2 which is obviously even.
- It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
- Because a is even, there exists a k that fulfills: a = 2k.
- We insert the last equation of (3) in (6): 2b2 = (2k)2 is equivalent to 2b2 = 4k2 is equivalent to b2 = 2k2.
- Because 2k2 is even it follows that b2 is also even which means that b is even because only even numbers have even squares.
- 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 <math>\sqrt{2}</math> is a rational number must be false. The opposite is proven: <math>\sqrt{2}</math> is irrational.
This proof can be generalized to show that any root of any natural number is either a natural number or irrational.
A different proof
Image:Irrationality of sqrt2.png Another reductio ad absurdum showing that <math>\sqrt{2}</math> 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 = <math>\sqrt{2}</math>. 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.
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 <math>\sqrt{2}</math> is irrational.
See also
- The square root of two is the aspect ratio of paper sizes under ISO 216.
Further reading
- Template:MathWorld
- Square Roots in the Sulbasutra
- Gourdon, X. & Sebah, P. Pythagoras' Constant: √2. Includes information on how to compute digits of <math>\sqrt{2}</math>.
- 5 million digits of √2fr:Racine carrée de deux