Continued Fraction: | 1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}} |
The square root of 2 (approximately 1.4142) is a real number that, when multiplied by itself or squared, equals the number 2. It may be written in mathematics as
\sqrt{2}
21/2
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. The fraction (≈ 1.4142857) is sometimes used as a good rational approximation with a reasonably small denominator.
Sequence 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 65 decimal places:[1]
The Babylonian clay tablet YBC 7289 (–1600 BC) gives an approximation of
\sqrt{2}
\sqrt{2}
1+
24 | |
60 |
+
51 | |
602 |
+
10 | |
603 |
=
305470 | |
216000 |
=1.41421\overline{296}.
Another early approximation is given in ancient Indian mathematical texts, the Sulbasutras (–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.[3] That is,
1+
1 | |
3 |
+
1 | |
3 x 4 |
-
1 | |
3 x 4 x 34 |
=
577 | |
408 |
=1.41421\overline{56862745098039}.
This approximation, diverging from the actual value of
\sqrt{2}
\sqrt{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, 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.[4] [5] The square root of two is occasionally called Pythagoras's number 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.[6]
There are many algorithms for approximating
\sqrt{2}
First, pick a guess,
a0>0
an+1=
12\left(a | |
n |
+
\dfrac{2}{a | + | ||||
|
1 | |
an |
.
Each iteration improves the approximation, roughly doubling the number of correct digits. Starting with
a0=1
\begin{alignat}{3} a1&=\tfrac{3}{2}&&=1.5,\\ a2&=\tfrac{17}{12}&&=1.416\ldots,\\ a3&=\tfrac{577}{408}&&=1.414215\ldots,\\ a4&=\tfrac{665857}{470832}&&=1.4142135623746\ldots,\\ & \vdots \end{alignat}
A simple rational approximation (≈ 1.4142857) is sometimes used. Despite having a denominator of only 70, it differs from the correct value by less than (approx.).
The next two better rational approximations are (≈ 1.4141414...) with a marginally smaller error (approx.), and (≈ 1.4142012) with an error of approx .
The rational approximation of the square root of two derived from four iterations of the Babylonian method after starting with is too large by about ; its square is ≈ .
In 1997, the value of
\sqrt{2}
\sqrt{2}
This is a table of recent records in calculating the digits of
\sqrt{2}
data-sort-type="usLongDate" | Date ! | Name | data-sort-type="number" | Number of digits |
---|---|---|---|---|
January 5, 2022 | data-sort-value="H" | Tizian Hanselmann | ||
June 28, 2016 | data-sort-value="W" | Ron Watkins | ||
April 3, 2016 | data-sort-value="W" | Ron Watkins | ||
January 20, 2016 | data-sort-value="W" | Ron Watkins | ||
February 9, 2012 | data-sort-value="Y" | Alexander Yee | ||
March 22, 2010 | data-sort-value="K" | Shigeru Kondo |
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 "
\sqrt{2}
\sqrt{2}
\sqrt{2}
\sqrt{2}
a | |
b |
a2 | |
b2 |
=2
a2=2b2
a=2k
2b2=a2=(2k)2=4k2
b2=2k2
2k2=b2
a | |
b |
Since we have derived a falsehood, the assumption (1) that
\sqrt{2}
\sqrt{2}
\sqrt{2}
This proof was hinted at 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 have agreed that this proof is an interpolation and not attributable to Euclid.[11]
As with the proof by infinite descent, we obtain
a2=2b2
The irrationality of
\sqrt{2}
p(x)=x2-2
\pm1
\pm2
\sqrt{2}
\pm1
\pm2
\sqrt{2}
p(x)
\sqrt{2}
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.[12] [13] Given two squares with integer sides respectively a and b, one of which has twice the area of the other, place two copies of the smaller square in the larger as shown in Figure 1. The square overlap region in the middle (
(2b-a)2
2(a-b)2
Tom M. Apostol made another geometric reductio ad absurdum argument showing that
\sqrt{2}
Let be a right isosceles triangle with hypotenuse length and legs as shown in Figure 2. By the Pythagorean theorem,
m | |
n |
=\sqrt{2}
Draw the arcs and with centre . Join . It follows that, and and coincide. Therefore, the triangles and are congruent by SAS.
Because is a right angle and is half a right angle, is also a right isosceles triangle. Hence implies . By symmetry,, and is also a right isosceles triangle. It also follows that .
Hence, there is an even smaller right isosceles triangle, with hypotenuse length and legs . These values are integers even smaller than and and in the same ratio, contradicting the hypothesis that is in lowest terms. Therefore, and cannot be both integers; hence,
\sqrt{2}
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 and be positive integers such that (as satisfies these bounds). Now and cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus . Multiplying the absolute difference by in the numerator and denominator, we get[14]
\left|\sqrt2-
a | |
b |
\right|=
|2b2-a2| | + | |
b2\left(\sqrt{2 |
a | |
b |
\right)}\ge
1 | |||||||||
|
\ge
1 | |
3b2 |
,
the latter inequality being true because it is assumed that, giving (otherwise the quantitative apartness can be trivially established). This gives a lower bound of for the difference, yielding a direct proof of irrationality in its constructively stronger form, not relying on the law of excluded middle; see Errett Bishop (1985, p. 18). This proof constructively exhibits an explicit discrepancy between
\sqrt{2}
This proof uses the following property of primitive Pythagorean triples:
If,, and are coprime positive integers such that, then is never even.
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
\sqrt2
\sqrt2={a\overb}
where
a,b\inZ
\gcd(a,b)=1
Squaring both sides,
2={a2\overb2}
2b2=a2
b2+b2=a2
Here, is a primitive Pythagorean triple, and from the lemma is never even. However, this contradicts the equation which implies that must be even.
The multiplicative inverse (reciprocal) of the square root of two (i.e., the square root of) is a widely used constant.
1{\sqrt{2}} | |
= |
\sqrt{2 | |
One-half of
\sqrt{2}
\sqrt{2}
\left( | \sqrt{2 |
\tfrac{1}{2}\sqrt{2}=\sqrt{\tfrac{1}{2}}=
1 | |
\sqrt{2 |
One interesting property of
\sqrt{2}
{1\over{\sqrt{2}-1}}=\sqrt{2}+1
\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=2-1=1.
\sqrt{2}
\sqrt{i | \sqrt{i}}{i}and | |
+i |
\sqrt{-i | |
-i |
\sqrt{-i}}{-i}
\sqrt{2}
\sqrt{2}
\sqrt{2}\sqrt{2\sqrt{2
| |||||
\sqrt{2}
2\pi | \sqrt | |
= |
12 | \sqrt{ | |
⋅ |
12 | |
+ |
| ||||
|
| ||||||
|
which is related to the formula
\pi=\limm\toinfty2m\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+ … +\sqrt{2}}}}}}msquareroots.
Similar in appearance but with a finite number of terms,
\sqrt{2}
\begin{align} \sin | \pi |
32 |
&=\tfrac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}& \sin
3\pi | |
16 |
&=\tfrac12\sqrt{2-\sqrt{2-\sqrt{2}}}& \sin
11\pi | |
32 |
&=\tfrac12\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\\[6pt] \sin
\pi | |
16 |
&=\tfrac12\sqrt{2-\sqrt{2+\sqrt{2}}}& \sin
7\pi | |
32 |
&=\tfrac12\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2}}}}& \sin
3\pi | |
8 |
&=\tfrac12\sqrt{2+\sqrt{2}}\\[6pt] \sin
3\pi | |
32 |
&=\tfrac12\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}}& \sin
\pi | |
4 |
&=\tfrac12\sqrt{2}& \sin
13\pi | |
32 |
&=\tfrac12\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2}}}}\\[6pt] \sin
\pi | |
8 |
&=\tfrac12\sqrt{2-\sqrt{2}}& \sin
9\pi | |
32 |
&=\tfrac12\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2}}}}& \sin
7\pi | |
16 |
&=\tfrac12\sqrt{2+\sqrt{2+\sqrt{2}}}\\[6pt] \sin
5\pi | |
32 |
&=\tfrac12\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}& \sin
5\pi | |
16 |
&=\tfrac12\sqrt{2+\sqrt{2-\sqrt{2}}}& \sin
15\pi | |
32 |
&=\tfrac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \end{align}
It is not known whether
\sqrt{2}
The identity, along with the infinite product representations for the sine and cosine, leads to products such as
1 | |
\sqrt2 |
=
infty | ||
\prod | \left(1- | |
k=0 |
1 | |
(4k+2)2 |
\right)= \left(1-
1 | \right)\left(1- | |
4 |
1 | \right)\left(1- | |
36 |
1 | |
100 |
\right) …
\sqrt{2}=
| |||||
\prod | = \left( | ||||
k=0 |
2 ⋅ 2 | \right)\left( | |
1 ⋅ 3 |
6 ⋅ 6 | \right)\left( | |
5 ⋅ 7 |
10 ⋅ 10 | \right)\left( | |
9 ⋅ 11 |
14 ⋅ 14 | |
13 ⋅ 15 |
\right) …
\sqrt{2}=
| |||||
\prod | \right)\left(1- | ||||
k=0 |
1 | |
4k+3 |
\right)= \left(1+
1 | \right)\left(1- | |
1 |
1 | \right)\left(1+ | |
3 |
1 | \right)\left(1- | |
5 |
1 | |
7 |
\right) … .
The number can also be expressed by taking the Taylor series of a trigonometric function. For example, the series for gives
1 | |
\sqrt{2 |
\sqrt{2}=
infty | |
\sum | |
k=0 |
(-1)k+1
(2k-3)!! | |
(2k)!! |
= 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 |
+ … .
The convergence of this series can be accelerated with an Euler transform, producing
\sqrt{2}=
infty | |
\sum | |
k=0 |
(2k+1)! | |
23k+1(k!)2 |
=
1 | + | |
2 |
3 | |
8 |
+
15 | |
64 |
+
35 | |
256 |
+
315 | |
4096 |
+
693 | |
16384 |
+ … .
\sqrt{2}
The number can be represented by an infinite series of Egyptian fractions, with denominators defined by 2n th terms of a Fibonacci-like recurrence relation a(n) = 34a(n−1) − a(n−2), a(0) = 0, a(1) = 6.[18]
\sqrt{2}= | 3 | - |
2 |
1 | |
2 |
infty | |
\sum | |
n=0 |
1 | = | |
a(2n) |
3 | - | |
2 |
1 | \left( | |
2 |
1 | + | |
6 |
1 | + | |
204 |
1 | |
235416 |
+...\right)
The square root of two has the following continued fraction representation:
\sqrt2=1+\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 (i.e.,). The first convergents are: and the convergent following is . The convergent differs from
\sqrt{2}
\left|\sqrt2-
p | |
q |
\right|=
|2q2-p2| | + | |
q2\left(\sqrt{2 |
p | |
q |
\right)}=
1 | |||||||||
|
\thickapprox
1 | |
2\sqrt{2 |
q2}
The following nested square expressions converge to
\begin{align} \sqrt{2} &=\tfrac32-2\left(\tfrac14-\left(\tfrac14-l(\tfrac14- … r)2\right)2\right)2\\[10mu] &=\tfrac32-4\left(\tfrac18+\left(\tfrac18+l(\tfrac18+ … r)2\right)2\right)2. \end{align}
In 1786, German physics professor Georg Christoph Lichtenberg found that any sheet of paper whose long edge is
\sqrt{2}
\sqrt{2}
Proof:
Let
S=
L=
R=
L | |
S |
=\sqrt{2}
R'=
L' | |
S' |
R'=
S | |
L/2 |
=
2S | |
L |
=
2 | |
(L/S) |
=
2 | |
\sqrt{2 |
There are some interesting properties involving the square root of 2 in the physical sciences:
\sqrt{2}
\sqrt{2}
\sqrt{2}