Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://en.wikipedia.org/wiki/Square_root_of_two below:

Square root of 2 - Wikipedia

Unique positive real number which when multiplied by itself gives 2

"Pythagoras's constant" redirects here and is not to be confused with

.

The square root of 2 is equal to the length of the

of an

with legs of length 1.

The square root of 2 (approximately 1.4142) is the positive real number that, when multiplied by itself or squared, equals the number 2. It may be written as 2 {\displaystyle {\sqrt {2}}} or 2 1 / 2 {\displaystyle 2^{1/2}} . It is an algebraic number, and therefore not a transcendental number. Technically, it should be 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.^[1] The fraction ⁠99/70⁠ (≈ 1.4142857) is sometimes used as a good rational approximation with a reasonably small denominator.

Sequence A002193 in the On-Line Encyclopedia of Integer Sequences consists of the digits in the decimal expansion of the square root of 2, here truncated to 60 decimal places:^[2]

1.414213562373095048801688724209698078569671875376948073176679

The Babylonian clay tablet YBC 7289 (c. 1800–1600 BC) gives an approximation of 2 {\displaystyle {\sqrt {2}}} in four sexagesimal figures, 1 24 51 10, which is accurate to about six decimal digits,^[3] and is the closest possible three-place sexagesimal representation of 2 {\displaystyle {\sqrt {2}}} , representing a margin of error of only –0.000042%:

1 + 24 60 + 51 60 2 + 10 60 3 = 305470 216000 = 1.41421 296 ¯ . {\displaystyle 1+{\frac {24}{60}}+{\frac {51}{60^{2}}}+{\frac {10}{60^{3}}}={\frac {305470}{216000}}=1.41421{\overline {296}}.}

Another early approximation 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.^[4] That is,

1 + 1 3 + 1 3 × 4 − 1 3 × 4 × 34 = 577 408 = 1.41421 56862745098039 ¯ . {\displaystyle 1+{\frac {1}{3}}+{\frac {1}{3\times 4}}-{\frac {1}{3\times 4\times 34}}={\frac {577}{408}}=1.41421{\overline {56862745098039}}.}

This approximation, diverging from the actual value of 2 {\displaystyle {\sqrt {2}}} by approximately +0.07%, is the seventh in a sequence of increasingly accurate approximations based on the sequence of Pell numbers, which can be derived from the continued fraction expansion of 2 {\displaystyle {\sqrt {2}}} . Despite having a smaller denominator, it is only slightly less accurate than the Babylonian approximation.

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, the Pythagoreans 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, though this has little to any substantial evidence in traditional historian practice.^[5][6] The square root of two is occasionally called Pythagoras's number^[7] or Pythagoras's constant.

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.^[8]

There are many algorithms for approximating 2 {\displaystyle {\sqrt {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^[9] for computing square roots, an example of Newton's method for computing roots of arbitrary functions. It goes as follows:

First, pick a guess, a 0 > 0 {\displaystyle a_{0}>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:

a n + 1 = 1 2 ( a n + 2 a n ) = a n 2 + 1 a n . {\displaystyle a_{n+1}={\frac {1}{2}}\left(a_{n}+{\dfrac {2}{a_{n}}}\right)={\frac {a_{n}}{2}}+{\frac {1}{a_{n}}}.}

Each iteration improves the approximation, roughly doubling the number of correct digits. Starting with a 0 = 1 {\displaystyle a_{0}=1} , the subsequent iterations yield:

a 1 = 3 2 = 1 .5 , a 2 = 17 12 = 1.41 6 … , a 3 = 577 408 = 1.41421 5 … , a 4 = 665857 470832 = 1.41421356237 46 … , ⋮ {\displaystyle {\begin{alignedat}{3}a_{1}&={\tfrac {3}{2}}&&=\mathbf {1} .5,\\a_{2}&={\tfrac {17}{12}}&&=\mathbf {1.41} 6\ldots ,\\a_{3}&={\tfrac {577}{408}}&&=\mathbf {1.41421} 5\ldots ,\\a_{4}&={\tfrac {665857}{470832}}&&=\mathbf {1.41421356237} 46\ldots ,\\&\qquad \vdots \end{alignedat}}}

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).

The next two better rational approximations are ⁠140/99⁠ (≈ 1.4141414...) with a marginally smaller error (approx. −0.72×10⁻⁴), and ⁠239/169⁠ (≈ 1.4142012) 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 {\displaystyle {\sqrt {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 {\displaystyle {\sqrt {2}}} was eclipsed with the use of a home computer. Shigeru Kondo calculated one trillion decimal places in 2010.^[10] Other mathematical constants whose decimal expansions have been calculated to similarly high precision include π, e, and the golden ratio.^[11] Such computations provide empirical evidence of whether these numbers are normal.

This is a table of recent records in calculating the digits of 2 {\displaystyle {\sqrt {2}}} .^[11]

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof of a negation by refutation: it proves the statement " 2 {\displaystyle {\sqrt {2}}} is not rational" by assuming that it is rational and then deriving a falsehood.

1. Assume that 2 {\displaystyle {\sqrt {2}}} is a rational number, meaning that there exists a pair of integers whose ratio is exactly 2 {\displaystyle {\sqrt {2}}} .
2. If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.
3. Then 2 {\displaystyle {\sqrt {2}}} can be written as an irreducible fraction a b {\displaystyle {\frac {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 a 2 b 2 = 2 {\displaystyle {\frac {a^{2}}{b^{2}}}=2} and a 2 = 2 b 2 {\displaystyle a^{2}=2b^{2}} .   ( (⁠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.)
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 = 2 k {\displaystyle a=2k} .
8. Substituting 2k from step 7 for a in the second equation of step 4: 2 b 2 = a 2 = ( 2 k ) 2 = 4 k 2 {\displaystyle 2b^{2}=a^{2}=(2k)^{2}=4k^{2}} , which is equivalent to b 2 = 2 k 2 {\displaystyle b^{2}=2k^{2}} .
9. Because 2k² is divisible by two and therefore even, and because 2 k 2 = b 2 {\displaystyle 2k^{2}=b^{2}} , 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 step 3 (that a b {\displaystyle {\frac {a}{b}}} is irreducible).

Since we have derived a falsehood, the assumption (1) that 2 {\displaystyle {\sqrt {2}}} is a rational number must be false. This means that 2 {\displaystyle {\sqrt {2}}} is not a rational number; that is to say, 2 {\displaystyle {\sqrt {2}}} is irrational.

This proof was hinted at by Aristotle, in his Analytica Priora, §I.23.^[12] 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.^[13]

Assume by way of contradiction that 2 {\displaystyle {\sqrt {2}}} were rational. Then we may write 2 + 1 = q p {\displaystyle {\sqrt {2}}+1={\frac {q}{p}}} as an irreducible fraction in lowest terms, with coprime positive integers q > p {\displaystyle q>p} . Since ( 2 − 1 ) ( 2 + 1 ) = 2 − 1 2 = 1 {\displaystyle ({\sqrt {2}}-1)({\sqrt {2}}+1)=2-1^{2}=1} , it follows that 2 − 1 {\displaystyle {\sqrt {2}}-1} can be expressed as the irreducible fraction p q {\displaystyle {\frac {p}{q}}} . However, since 2 − 1 {\displaystyle {\sqrt {2}}-1} and 2 + 1 {\displaystyle {\sqrt {2}}+1} differ by an integer, it follows that the denominators of their irreducible fraction representations must be the same, i.e. q = p {\displaystyle q=p} . This gives the desired contradiction.

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

The irrationality of 2 {\displaystyle {\sqrt {2}}} also follows from the rational root theorem, which states that a rational root of a polynomial, if it exists, must be the quotient of a factor of the constant term and a factor of the leading coefficient. In the case of p ( x ) = x 2 − 2 {\displaystyle p(x)=x^{2}-2} , the only possible rational roots are ± 1 {\displaystyle \pm 1} and ± 2 {\displaystyle \pm 2} . As 2 {\displaystyle {\sqrt {2}}} is not equal to ± 1 {\displaystyle \pm 1} or ± 2 {\displaystyle \pm 2} , it follows that 2 {\displaystyle {\sqrt {2}}} is irrational. This application also invokes the integer root theorem, a stronger version of the rational root theorem for the case when p ( x ) {\displaystyle p(x)} is a monic polynomial with integer coefficients; for such a polynomial, all roots are necessarily integers (which 2 {\displaystyle {\sqrt {2}}} is not, as 2 is not a perfect square) or irrational.

The rational root theorem (or integer root theorem) may be used to show that any square root of any natural number that is not a perfect square is irrational. For other proofs that the square root of any non-square natural number is irrational, see Quadratic irrational number or Infinite descent.

A simple proof is attributed to Stanley Tennenbaum when he was a student in the early 1950s.^[14][15] Assume that 2 = a / b {\displaystyle {\sqrt {2}}=a/b} , where a {\displaystyle a} and b {\displaystyle b} are coprime positive integers. Then a {\displaystyle a} and b {\displaystyle b} are the smallest positive integers for which a 2 = 2 b 2 {\displaystyle a^{2}=2b^{2}} . Geometrically, this implies that a square with side length a {\displaystyle a} will have an area equal to two squares of (lesser) side length b {\displaystyle b} . Call these squares A and B. We can draw these squares and compare their areas - the simplest way to do so is to fit the two B squares into the A squares. When we try to do so, we end up with the arrangement in Figure 1., in which the two B squares overlap in the middle and two uncovered areas are present in the top left and bottom right. In order to assert a 2 = 2 b 2 {\displaystyle a^{2}=2b^{2}} , we would need to show that the area of the overlap is equal to the area of the two missing areas, i.e. ( 2 b − a ) 2 {\displaystyle (2b-a)^{2}} = 2 ( a − b ) 2 {\displaystyle 2(a-b)^{2}} . In other terms, we may refer to the side lengths of the overlap and missing areas as p = 2 b − a {\displaystyle p=2b-a} and q = a − b {\displaystyle q=a-b} , respectively, and thus we have p 2 = 2 q 2 {\displaystyle p^{2}=2q^{2}} . But since we can see from the diagram that p < a {\displaystyle p<a} and q < b {\displaystyle q<b} , and we know that p {\displaystyle p} and q {\displaystyle q} are integers from their definitions in terms of a {\displaystyle a} and b {\displaystyle b} , this means that we are in violation of the original assumption that a {\displaystyle a} and b {\displaystyle b} are the smallest positive integers for which a 2 = 2 b 2 {\displaystyle a^{2}=2b^{2}} .

Hence, even in assuming that a {\displaystyle a} and b {\displaystyle b} are the smallest positive integers for which a 2 = 2 b 2 {\displaystyle a^{2}=2b^{2}} , we may prove that there exists a smaller pair of integers p {\displaystyle p} and q {\displaystyle q} which satisfy the relation. This contradiction within the definition of a {\displaystyle a} and b {\displaystyle b} implies that they cannot exist, and thus 2 {\displaystyle {\sqrt {2}}} must be irrational.

Tom M. Apostol made another geometric reductio ad absurdum argument showing that 2 {\displaystyle {\sqrt {2}}} is irrational.^[16] 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 same algebraic proof as Tennebaum's proof, viewed geometrically in 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 {\displaystyle {\frac {m}{n}}={\sqrt {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 ∠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 {\displaystyle {\sqrt {2}}} is irrational.

While the proofs by infinite descent are constructively valid when "irrational" is defined to mean "not rational", we can obtain a constructively stronger statement by using a positive definition of "irrational" as "quantifiably apart from every rational". Let a and b be positive integers such that 1<⁠a/b⁠< 3/2 (as 1<2< 9/4 satisfies these bounds). Now 2b² and a² cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus |2b² − a²| ≥ 1. Multiplying the absolute difference |√2 − ⁠a/b⁠| by b²(√2 + ⁠a/b⁠) in the numerator and denominator, we get^[17]

| 2 − a b | = | 2 b 2 − a 2 | b 2 ( 2 + a b ) ≥ 1 b 2 ( 2 + a b ) ≥ 1 3 b 2 , {\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 1<⁠a/b⁠< 3/2, giving ⁠a/b⁠ + √2 ≤ 3 (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 in its constructively stronger form, not relying on the law of excluded middle.^[18] This proof constructively exhibits an explicit discrepancy between 2 {\displaystyle {\sqrt {2}}} and any rational.

This proof uses the following property of primitive Pythagorean triples:

If a, b, and c are coprime positive integers such that a² + b² = c², then c is never even.^[19]

This lemma can be used to show that two identical perfect squares can never be added to produce another perfect square.

Suppose the contrary that 2 {\displaystyle {\sqrt {2}}} is rational. Therefore,

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

where a , b ∈ Z {\displaystyle a,b\in \mathbb {Z} } and gcd ( a , b ) = 1 {\displaystyle \gcd(a,b)=1}

Squaring both sides,

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

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

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

Here, (b, b, a) is a primitive Pythagorean triple, and from the lemma a is never even. However, this contradicts the equation 2b² = a² which implies that a must be even.

The multiplicative inverse (reciprocal) of the square root of two is a widely used constant, with the decimal value:^[20]

0.70710678118654752440084436210484903928483593768847...

It is often encountered in geometry and trigonometry because the unit vector, which makes a 45° angle with the axes in a plane, has the coordinates

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

Each coordinate satisfies

2 2 = 1 2 = 1 2 = sin ⁡ 45 ∘ = cos ⁡ 45 ∘ . {\displaystyle {\frac {\sqrt {2}}{2}}={\sqrt {\tfrac {1}{2}}}={\frac {1}{\sqrt {2}}}=\sin 45^{\circ }=\cos 45^{\circ }.}

One interesting property of 2 {\displaystyle {\sqrt {2}}} is

  1 2 − 1 = 2 + 1 {\displaystyle \!\ {1 \over {{\sqrt {2}}-1}}={\sqrt {2}}+1}

since

( 2 + 1 ) ( 2 − 1 ) = 2 − 1 = 1. {\displaystyle \left({\sqrt {2}}+1\right)\!\left({\sqrt {2}}-1\right)=2-1=1.}

This is related to the property of silver ratios.

2 {\displaystyle {\sqrt {2}}} can also be expressed in terms of copies of the imaginary unit i using only the square root and arithmetic operations, if the square root symbol is interpreted suitably for the complex numbers i and −i:

i + i i i  and  − i − i − i − i {\displaystyle {\frac {{\sqrt {i}}+i{\sqrt {i}}}{i}}{\text{ and }}{\frac {{\sqrt {-i}}-i{\sqrt {-i}}}{-i}}}

2 {\displaystyle {\sqrt {2}}} is also the only real number other than 1 whose infinite tetrate (i.e., infinite exponential tower) is equal to its square. In other words: if for c > 1, x₁ = c and x_n+1 = c^x_n for n > 1, the limit of x_n as n → ∞ will be called (if this limit exists) f(c). Then 2 {\displaystyle {\sqrt {2}}} is the only number c > 1 for which f(c) = c². Or symbolically:

2 2 2   ⋅   ⋅   ⋅ = 2. {\displaystyle {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{~\cdot ^{~\cdot ^{~\cdot }}}}}=2.}

2 {\displaystyle {\sqrt {2}}} appears in Viète's formula for π,

2 π = 1 2 ⋅ 1 2 + 1 2 1 2 ⋅ 1 2 + 1 2 1 2 + 1 2 1 2 ⋯ , {\displaystyle {\frac {2}{\pi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}}}\cdots ,}

which is related to the formula^[21]

π = lim m → ∞ 2 m 2 − 2 + 2 + 2 + ⋯ + 2 ⏟ m  square roots . {\displaystyle \pi =\lim _{m\to \infty }2^{m}\underbrace {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+\cdots +{\sqrt {2}}}}}}}}}} _{m{\text{ square roots}}}\,.}

Similar in appearance but with a finite number of terms, 2 {\displaystyle {\sqrt {2}}} appears in various trigonometric constants:^[22]

sin ⁡ π 32 = 1 2 2 − 2 + 2 + 2 sin ⁡ 3 π 16 = 1 2 2 − 2 − 2 sin ⁡ 11 π 32 = 1 2 2 + 2 − 2 − 2 sin ⁡ π 16 = 1 2 2 − 2 + 2 sin ⁡ 7 π 32 = 1 2 2 − 2 − 2 + 2 sin ⁡ 3 π 8 = 1 2 2 + 2 sin ⁡ 3 π 32 = 1 2 2 − 2 + 2 − 2 sin ⁡ π 4 = 1 2 2 sin ⁡ 13 π 32 = 1 2 2 + 2 + 2 − 2 sin ⁡ π 8 = 1 2 2 − 2 sin ⁡ 9 π 32 = 1 2 2 + 2 − 2 + 2 sin ⁡ 7 π 16 = 1 2 2 + 2 + 2 sin ⁡ 5 π 32 = 1 2 2 − 2 − 2 − 2 sin ⁡ 5 π 16 = 1 2 2 + 2 − 2 sin ⁡ 15 π 32 = 1 2 2 + 2 + 2 + 2 {\displaystyle {\begin{aligned}\sin {\frac {\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {11\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}&\quad \sin {\frac {7\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {3\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\\[6pt]\sin {\frac {3\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {\pi }{4}}&={\tfrac {1}{2}}{\sqrt {2}}&\quad \sin {\frac {13\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}}}\\[6pt]\sin {\frac {\pi }{8}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}&\quad \sin {\frac {9\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}}&\quad \sin {\frac {7\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}\\[6pt]\sin {\frac {5\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2-{\sqrt {2-{\sqrt {2}}}}}}}}&\quad \sin {\frac {5\pi }{16}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2-{\sqrt {2}}}}}}&\quad \sin {\frac {15\pi }{32}}&={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}\end{aligned}}}

It is not known whether 2 {\displaystyle {\sqrt {2}}} is a normal number, which is a stronger property than irrationality, but statistical analyses of its binary expansion are consistent with the hypothesis that it is normal to base two.^[23]

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

1 2 = ∏ k = 0 ∞ ( 1 − 1 ( 4 k + 2 ) 2 ) = ( 1 − 1 4 ) ( 1 − 1 36 ) ( 1 − 1 100 ) ⋯ {\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

2 = ∏ k = 0 ∞ ( 4 k + 2 ) 2 ( 4 k + 1 ) ( 4 k + 3 ) = ( 2 ⋅ 2 1 ⋅ 3 ) ( 6 ⋅ 6 5 ⋅ 7 ) ( 10 ⋅ 10 9 ⋅ 11 ) ( 14 ⋅ 14 13 ⋅ 15 ) ⋯ {\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,

2 = ∏ k = 0 ∞ ( 1 + 1 4 k + 1 ) ( 1 − 1 4 k + 3 ) = ( 1 + 1 1 ) ( 1 − 1 3 ) ( 1 + 1 5 ) ( 1 − 1 7 ) ⋯ . {\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

1 2 = ∑ k = 0 ∞ ( − 1 ) k ( π 4 ) 2 k ( 2 k ) ! . {\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 {\displaystyle {\sqrt {1+x}}} with x = 1 and using the double factorial n!! gives

2 = ∑ k = 0 ∞ ( − 1 ) k + 1 ( 2 k − 3 ) ! ! ( 2 k ) ! ! = 1 + 1 2 − 1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + ⋯ = 1 + 1 2 − 1 8 + 1 16 − 5 128 + 7 256 + ⋯ . {\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 =1+{\frac {1}{2}}-{\frac {1}{8}}+{\frac {1}{16}}-{\frac {5}{128}}+{\frac {7}{256}}+\cdots .}

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

2 = ∑ k = 0 ∞ ( 2 k + 1 ) ! 2 3 k + 1 ( k ! ) 2 = 1 2 + 3 8 + 15 64 + 35 256 + 315 4096 + 693 16384 + ⋯ . {\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 {\displaystyle {\sqrt {2}}} can be represented with a BBP-type formula. BBP-type formulas are known for π√2 and 2 ln ⁡ ( 1 + 2 ) {\displaystyle {\sqrt {2}}\ln(1+{\sqrt {2}})} , however.^[24]

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:^[25]

2 = 3 2 − 1 2 ∑ n = 0 ∞ 1 a ( 2 n ) = 3 2 − 1 2 ( 1 6 + 1 204 + 1 235416 + … ) . {\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 two has the following continued fraction representation:

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

The convergents ⁠p/q⁠ 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 (i.e., p² − 2q² = ±1). The first convergents are: ⁠1/1⁠, ⁠3/2⁠, ⁠7/5⁠, ⁠17/12⁠, ⁠41/29⁠, ⁠99/70⁠, ⁠239/169⁠, ⁠577/408⁠ and the convergent following ⁠p/q⁠ is ⁠p + 2q/p + q⁠. The convergent ⁠p/q⁠ differs from 2 {\displaystyle {\sqrt {2}}} by almost exactly 1 2 2 q 2 {\displaystyle {\frac {1}{2{\sqrt {2}}q^{2}}}} , which follows from:

| 2 − p q | = | 2 q 2 − p 2 | q 2 ( 2 + p q ) = 1 q 2 ( 2 + p q ) ≈ 1 2 2 q 2 {\displaystyle \left|{\sqrt {2}}-{\frac {p}{q}}\right|={\frac {|2q^{2}-p^{2}|}{q^{2}\!\left({\sqrt {2}}+{\frac {p}{q}}\right)}}={\frac {1}{q^{2}\!\left({\sqrt {2}}+{\frac {p}{q}}\right)}}\thickapprox {\frac {1}{2{\sqrt {2}}q^{2}}}}

The following nested square expressions converge to 2 {\textstyle {\sqrt {2}}} :

2 = 3 2 − 2 ( 1 4 − ( 1 4 − ( 1 4 − ⋯ ) 2 ) 2 ) 2 = 3 2 − 4 ( 1 8 + ( 1 8 + ( 1 8 + ⋯ ) 2 ) 2 ) 2 . {\displaystyle {\begin{aligned}{\sqrt {2}}&={\tfrac {3}{2}}-2\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-{\bigl (}{\tfrac {1}{4}}-\cdots {\bigr )}^{2}\right)^{2}\right)^{2}\\[10mu]&={\tfrac {3}{2}}-4\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+{\bigl (}{\tfrac {1}{8}}+\cdots {\bigr )}^{2}\right)^{2}\right)^{2}.\end{aligned}}}

In 1786, German physics professor Georg Christoph Lichtenberg^[26] found that any sheet of paper whose long edge is 2 {\displaystyle {\sqrt {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.^[26] Today, the (approximate) aspect ratio of paper sizes under ISO 216 (A4, A0, etc.) is 1: 2 {\displaystyle {\sqrt {2}}} .

Proof:

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

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

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

R ′ = S L / 2 = 2 S L = 2 ( L / S ) = 2 2 = 2 = R . {\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 {\displaystyle {\sqrt {2}}} .

• In the brain there are lattice cells, discovered in 2005 by a group led by May-Britt and Edvard Moser. "The grid cells were found in the cortical area located right next to the hippocampus [...] At one end of this cortical area the mesh size is small and at the other it is very large. However, the increase in mesh size is not left to chance, but increases by the squareroot of two from one area to the next."^[27]

• List of mathematical constants
• Square root of 3
• Square root of 5
• Gelfond–Schneider constant, 2^√2
• Silver ratio, 1 + √2

1. ^ Fowler, David H. (1994). "The Story of the Discovery of Incommensurability, Revisited". In Gavroglu, Kostas; Christianidis, Jean; Nicolaidis, Efthymios (eds.). Trends in the Historiography of Science. Boston Studies in the Philosophy of Science. Vol. 151. Dortrecht: Springer. pp. 221–236. doi:10.1007/978-94-017-3596-4. ISBN 978-9048142644.
2. ^ Sloane, N. J. A. (ed.). "Sequence A002193 (Decimal expansion of square root of 2)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2020-08-10.
3. ^ 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. MR 1662496. See p. 368.
   Photograph, illustration, and description of the root(2) tablet from the Yale Babylonian Collection Archived 2012-08-13 at the Wayback Machine
   High resolution photographs, descriptions, and analysis of the root(2) tablet (YBC 7289) from the Yale Babylonian Collection
4. ^ Henderson, David W. (2000). "Square roots in the Śulba Sūtras". In Gorini, Catherine A. (ed.). Geometry At Work: Papers in Applied Geometry. Mathematical Association of America Notes. Vol. 53. Washington, D.C.: The Mathematical Association of America. pp. 39–45. ISBN 978-0883851647.
5. ^ "The Dangerous Ratio". nrich.maths.org. Retrieved 2023-09-18.
6. ^ Von Fritz, Kurt (1945). "The Discovery of Incommensurability by Hippasus of Metapontum". Annals of Mathematics. 46 (2): 242–264. doi:10.2307/1969021. ISSN 0003-486X. JSTOR 1969021.
7. ^ Conway, John H.; Guy, Richard K. (1996). The Book of Numbers. New York: Copernicus. p. 25. ISBN 978-1461240723.
8. ^ Williams, Kim; Ostwald, Michael (2015). Architecture and Mathematics from Antiquity to the Future: Volume I: Antiquity to the 1500s. Birkhäuser. p. 204. ISBN 9783319001371.
9. ^ Although the term "Babylonian method" is common in modern usage, there is no direct evidence showing how the Babylonians computed the approximation of 2 {\displaystyle {\sqrt {2}}} seen on tablet YBC 7289. Fowler and Robson offer informed and detailed conjectures.
   Fowler and Robson, p. 376. Flannery, p. 32, 158.
10. ^ Constants and Records of Computation. Numbers.computation.free.fr. 12 Aug 2010. Archived from the original on 2012-03-01. Retrieved 2012-09-07.
11. ^ ^{a b} Records set by y-cruncher. Archived from the original on 2022-04-07. Retrieved 2022-04-07.
12. ^ All that Aristotle says, while writing about proofs by contradiction, is that "the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate".
13. ^ The edition of the Greek text of the Elements published by E. F. August in Berlin in 1826–1829 already relegates this proof to an Appendix. The same thing occurs with J. L. Heiberg's edition (1883–1888).
14. ^ Miller, Steven J.; Montague, David (Apr 2012). "Picturing Irrationality". Mathematics Magazine. Vol. 85, no. 2. pp. 110–114. doi:10.4169/math.mag.85.2.110. JSTOR 10.4169/math.mag.85.2.110.
15. ^ Yanofsky, Noson S. (May–Jun 2016). "Paradoxes, Contradictions, and the Limits of Science". American Scientist. Vol. 103, no. 3. pp. 166–173. JSTOR 44808923.
16. ^ Apostol, Tom M. (2000). "Irrationality of The Square Root of Two – A Geometric Proof". The American Mathematical Monthly. 107 (9): 841–842. doi:10.2307/2695741. JSTOR 2695741.
17. ^ See Katz, Karin Usadi; Katz, Mikhail G. (2011). "Meaning in Classical Mathematics: Is it at Odds with Intuitionism?". Intellectica. 56 (2): 223–302 (see esp. Section 2.3, footnote 15). arXiv:1110.5456. Bibcode:2011arXiv1110.5456U.
18. ^ Bishop, Errett (1985). "Schizophrenia in Contemporary Mathematics.". In Rosenblatt, Murray (ed.). Errett Bishop: Reflections on Him and His Research. Contemporary Mathematics. Vol. 39. Providence, RI: American Mathematical Society. pp. 1–32. doi:10.1090/conm/039/788163. ISBN 0821850407. ISSN 0271-4132.
19. ^ Sierpiński, Wacław (2003). Pythagorean Triangles. Translated by Sharma, Ambikeshwa. Mineola, NY: Dover. pp. 4–6. ISBN 978-0486432786.
20. ^ Sloane, N. J. A. (ed.). "Sequence A010503 (Decimal expansion of 1/sqrt(2))". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2024-11-03.
21. ^ Courant, Richard; Robbins, Herbert (1941). What is mathematics? An Elementary Approach to Ideas and Methods. London: Oxford University Press. p. 124.
22. ^ Julian D. A. Wiseman Sin and cos in surds Archived 2009-05-06 at the Wayback Machine
23. ^ Good, I. J.; Gover, T. N. (1967). "The generalized serial test and the binary expansion of 2 {\displaystyle {\sqrt {2}}} ". Journal of the Royal Statistical Society, Series A. 130 (1): 102–107. doi:10.2307/2344040. JSTOR 2344040.
24. ^ Bailey, David H. (13 Feb 2011). A Compendium of BBP-Type Formulas for Mathematical Constants (PDF). Archived (PDF) from the original on 2011-06-10. Retrieved 2010-04-30.
25. ^ Sloane, N. J. A. (ed.). "Sequence A082405 (a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=6)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-09-05.
26. ^ ^{a b} Houston, Keith (2016). The Book: A Cover-to-Cover Exploration of the Most Powerful Object of Our Time. W. W. Norton & Company. p. 324. ISBN 978-0393244809.
27. ^ Nordengen, Kaja (2016). The Book: Hjernen er sternen. 2016 Kagge Forlag AS. p. 81. ISBN 978-82-489-2018-2.

• Aristotle (1938) [c. 350 BC]. Categories; On Interpretation; Prior Analytics. Greek text with translation. Loeb Classical Library. Vol. 325. Translated by H. P. Cooke; Hugh Tredennick. Cambridge, MA: Harvard University Press. Prior Analytics § I.23. ISBN 9780674993594.
• Flannery, David (2006). The Square Root of 2: A Dialogue Concerning a Number and a Sequence. New York: Copernicus Books. ISBN 978-0387202204.
• 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.

• Gourdon, X.; Sebah, P. (2001). "Pythagoras' Constant: 2 {\displaystyle {\sqrt {2}}} ". Numbers, Constants and Computation..
• The Square Root of Two to 5 million digits by Jerry Bonnell and Robert J. Nemiroff. May, 1994.
• Square root of 2 is irrational, a collection of proofs
• Haran, Brady (27 Jan 2012). Root 2 (video). Numberphile. featuring Grime, James; Bowley, Roger.
• ⁠ 2 {\displaystyle {\sqrt {2}}} ⁠ Search Engine 2 billion searchable digits of √2, π, and e

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4