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# √2

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 Title: √2 Author: World Heritage Encyclopedia Language: English Subject: Standard deviation Collection: Publisher: World Heritage Encyclopedia Publication Date:

### √2

The square root of 2, often known as root 2 or radical 2 and written as $\sqrt\left\{2\right\}$, 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 OEIS).

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 (approx. 7.2 × 10 −5).

 Binary 1.0110101000001001111... Decimal 1.4142135623730950488... Hexadecimal 1.6A09E667F3BCC908B2F... Continued fraction $1 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \ddots\right\}\right\}\right\}\right\}$

## History

The Babylonian clay tablet YBC 7289 (c. 1800–1600 BC) gives an approximation of $\sqrt\left\{2\right\}$ in four sexagesimal figures, which is about six decimal figures:

$1 + \frac\left\{24\right\}\left\{60\right\} + \frac\left\{51\right\}\left\{60^2\right\} + \frac\left\{10\right\}\left\{60^3\right\} = \frac\left\{30547\right\}\left\{21600\right\} = 1.41421\overline\left\{296\right\}.$

Another early close 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. That is,

$1 + \frac\left\{1\right\}\left\{3\right\} + \frac\left\{1\right\}\left\{3 \cdot 4\right\} - \frac\left\{1\right\}\left\{3 \cdot4 \cdot 34\right\} = \frac\left\{577\right\}\left\{408\right\} = 1.414\overline\left\{2156862745098039\right\}.$

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 $\sqrt\left\{2\right\}$. 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. The square root of two is occasionally called "Pythagoras' number" or "Pythagoras' Constant", for example Conway & Guy (1996).

## Computation algorithms

There are a number of algorithms for approximating $\sqrt\left\{2\right\}$, 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 of computing square roots, which is one of many methods of computing square roots. It goes as follows:

First, pick a guess, $a_0 > 0$; 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:

$a_\left\{n+1\right\} = \frac\left\{a_n + \frac\left\{2\right\}\left\{a_n\right\}\right\}\left\{2\right\}=\frac\left\{a_n\right\}\left\{2\right\}+\frac\left\{1\right\}\left\{a_n\right\}.$

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 $\sqrt\left\{2\right\}$ 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 $\sqrt\left\{2\right\}$ 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. Among mathematical constants with computationally challenging decimal expansions, only has been calculated more precisely. Such computations aim to check empirically whether such numbers are normal.

## Proofs of irrationality

A short proof of the irrationality of $\sqrt\left\{2\right\}$ can be obtained from the rational root theorem, that is, if $p\left(x\right)$ is a monic polynomial with integer coefficients, then any rational root of $p\left(x\right)$ is necessarily an integer. Applying this to the polynomial $p\left(x\right) = x^2 - 2$, it follows that $\sqrt\left\{2\right\}$ is either an integer or irrational. Because $\sqrt\left\{2\right\}$ is not an integer (2 is not a perfect square), $\sqrt\left\{2\right\}$ must therefore be irrational. 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.

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

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 $\sqrt\left\{2\right\}$ is a rational number, meaning that there exists an integer $a$ and an integer $b$ in general such that $a/b = \sqrt\left\{2\right\}$.
2. Then $\sqrt\left\{2\right\}$ can be written as an irreducible fraction $a/b$ such that $a$ and $b$ are coprime integers.
3. It follows that $a^2 / b^2 = 2$ and $a^2 = 2 b^2$.   ( $\left(a/b\right)^n = a^n/b^n$  )
4. Therefore $a^2$ is even because it is equal to $2 b^2$. ($2 b^2$ is necessarily even because it is 2 times another whole number and multiples of 2 are even.)
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: $2 b^2 = \left(2k\right)^2$ is equivalent to $2b^2 = 4k^2$, which is equivalent to $b^2 = 2k^2$.
8. Because $2k^2$ is divisible by two and therefore even, and because $2k^2 = b^2$, it follows that $b^2$ 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 $\sqrt\left\{2\right\}$ is a rational number must be false. This means that $\sqrt\left\{2\right\}$ is not a rational number; i.e., $\sqrt\left\{2\right\}$ is irrational.

This proof was hinted at by Aristotle, in his Analytica Priora, §I.23. 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.

### Proof by unique factorization

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 $\sqrt\left\{2\right\}$ is a rational number. Then there are integers a and b such that a is coprime to b and $\sqrt\left\{2\right\} =a/b$. In other words, $\sqrt\left\{2\right\}$ 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 if an integer is not an exact kth power of another integer then its kth root 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

The following reductio ad absurdum argument showing the irrationality of $\sqrt\left\{2\right\}$ is less well-known. It uses the additional information $2 > \sqrt\left\{2\right\} > 1$ so that $1 > \sqrt\left\{2\right\} - 1 > 0$.

1. Assume that $\sqrt\left\{2\right\}$ is a rational number. This would mean that there exist positive integers m and n with $n \ne 0$ such that $m/n = \sqrt\left\{2\right\}$. Then $m = n \sqrt\left\{2\right\}$ and $m\sqrt\left\{2\right\} = 2n$.
2. We may assume that n is the smallest integer so that $n\sqrt\left\{2\right\}$ is an integer. That is, that the fraction m/n is in lowest terms.
3. Then $\sqrt\left\{2\right\} = \frac\left\{m\right\}\left\{n\right\}=\frac\left\{m\left(\sqrt\left\{2\right\}-1\right)\right\}\left\{n\left(\sqrt\left\{2\right\}-1\right)\right\}=\frac\left\{2n-m\right\}\left\{m-n\right\}$
4. Because $1 > \sqrt\left\{2\right\} - 1 > 0$, it follows that $n > n\left(\sqrt\left\{2\right\} - 1\right) = m - n > 0$.
5. So the fraction m/n for $\sqrt\left\{2\right\}$, 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 $\sqrt\left\{2\right\}$ is rational must be false.

### Geometric proof

Another reductio ad absurdum showing that $\sqrt\left\{2\right\}$ 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. 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, $m/n = \sqrt\left\{2\right\}$. 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 $\sqrt\left\{2\right\}$ is irrational.

### Analytic proof

• Lemma: let $\alpha \in \mathbb\left\{R\right\}^+$ and $p_1,p_2,\dots, q_1, q_2, \ldots \in \mathbb\left\{N\right\}$ such that $\left|\alpha q_n - p_n \right| \neq 0$ for all $n \in \mathbb\left\{N\right\}$ and
$\lim_\left\{n \rightarrow \infty\right\} p_n = \lim_\left\{n \rightarrow \infty\right\} q_n = \infty\,$
$\lim_\left\{n \rightarrow \infty\right\} \left| \alpha q_n - p_n \right| = 0.\,$
Then α is irrational.

Proof: suppose α = a/b with ab ∈ N+.

For sufficiently big n,

$0 < \left| \alpha q_n - p_n \right| < \frac\left\{1\right\}\left\{b\right\}$

then

$0 < \left| a q_n /\ \! b - p_n \right| < \frac\left\{1\right\}\left\{b\right\}$
$0 < \left| a q_n - b p_n \right| < 1 \,$

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

• $\sqrt\left\{2\right\}$ is irrational.

Proof: let $p_1 = q_1 = 1$ and

$p_\left\{n+1\right\} = p_n^2 + 2 q_n^2 \,$
$q_\left\{n+1\right\} = 2 p_n q_n \,\!$

for all $n \in \mathbf\left\{N\right\}$.

By induction,

$0 < \left| \sqrt\left\{2\right\} q_n - p_n \right| < \frac\left\{1\right\}\left\{2^\left\{2^\left\{n-1\right\}\right\}\right\}$

for all $n \in \mathbf\left\{N\right\}$. For $n = 1$,

$0 < \left| \sqrt\left\{2\right\} q_1 - p_1 \right| < \frac\left\{1\right\}\left\{2\right\}$

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

$0 < \left| \sqrt\left\{2\right\} q_n - p_n \right|^2 < \frac\left\{1\right\}\left\{2^\left\{2^n\right\}\right\}$
$0 < \left| \sqrt\left\{2\right\} \left(2 p_n q_n\right) - \left(p_n^2 + 2 q_n^2\right) \right| < \frac\left\{1\right\}\left\{2^\left\{2^n\right\}\right\}$
$0 < \left| \sqrt\left\{2\right\} q_\left\{n + 1\right\} - p_\left\{n + 1\right\} \right| < \frac\left\{1\right\}\left\{2^\left\{2^n\right\}\right\}.$

By application of the lemma, $\sqrt\left\{2\right\}$ is irrational.

### Constructive proof

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 positive 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; thus $|2 b^2 - a^2| \geq 1$. Then

$\left|\sqrt2 - \frac\left\{a\right\}\left\{b\right\}\right| = \frac\left\{|2b^2-a^2|\right\}\left\{b^2\left(\sqrt\left\{2\right\}+a/b\right)\right\} \ge \frac\left\{1\right\}\left\{b^2\left(\sqrt2 + a / b\right)\right\} \ge \frac\left\{1\right\}\left\{3b^2\right\},$

the latter inequality being true because we assume $\tfrac\left\{a\right\}\left\{b\right\} \le 3- \sqrt\left\{2\right\}$ (otherwise the quantitative apartness can be trivially established). This gives a lower bound of $\frac\left\{1\right\}\left\{3b^2\right\}$ for the difference $|\sqrt2 - a / b|$, 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 $\sqrt\left\{2\right\}$ and any rational.

## Properties of the square root of two

One-half of $\sqrt\left\{2\right\}$, 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

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

This number satisfies

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

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

$\!\ \left\{1 \over \left\{\sqrt\left\{2\right\} - 1\right\}\right\} = \sqrt\left\{2\right\} + 1$

since $\left(\sqrt\left\{2\right\}+1\right)\left(\sqrt\left\{2\right\}-1\right)=2-1=1.$ This is related to the property of silver ratios.

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:

$\frac\left\{\sqrt\left\{i\right\}+i \sqrt\left\{i\right\}\right\}\left\{i\right\}\text\left\{ and \right\}\frac\left\{\sqrt\left\{-i\right\}-i \sqrt\left\{-i\right\}\right\}\left\{-i\right\}.$

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

$\sqrt\left\{2\right\}^ \left\{\sqrt\left\{2\right\}^ \left\{\sqrt\left\{2\right\}^ \left\{\ \cdot^ \left\{\cdot^ \cdot\right\}\right\}\right\}\right\} = 2$

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

$2^m\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2+\cdots+\sqrt\left\{2\right\}\right\}\right\}\right\} \to \pi\text\left\{ as \right\}m \to \infty\,$

for m square roots and only one minus sign.

Similar in appearance but with a finite number of terms, the square root of 2 appears in various trigonometric constants:

$\sin\left(5\tfrac58 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(11\tfrac14 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\};$
$\sin\left(16\tfrac78 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(22\tfrac12 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2\right\}\right\};$
$\sin\left(28\tfrac18 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2-\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(33\tfrac34 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\};$
$\sin\left(39\tfrac38 ^\circ\right) = \frac12\sqrt\left\{2-\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(45 ^\circ\right) = \frac12\sqrt\left\{2\right\};$
$\sin\left(50\tfrac58 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2-\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(56\tfrac14 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\};$
$\sin\left(61\tfrac78 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2-\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(67\tfrac12 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2\right\}\right\};$
$\sin\left(73\tfrac18 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2+\sqrt\left\{2-\sqrt\left\{2\right\}\right\}\right\}\right\};$
$\sin\left(78\tfrac34 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\};$
$\sin\left(84\tfrac38 ^\circ\right) = \frac12\sqrt\left\{2+\sqrt\left\{2+\sqrt\left\{2+\sqrt\left\{2\right\}\right\}\right\}\right\}.$

It is not known whether $\sqrt\left\{2\right\}$ 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.

## Series and product representations

The identity $\cos\left(\pi/4\right) = \sin\left(\pi/4\right) = 1/\sqrt\left\{2\right\}$, along with the infinite product representations for the sine and cosine, leads to products such as

$\frac\left\{1\right\}\left\{\sqrt 2\right\} = \prod_\left\{k=0\right\}^\infty$

\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

$\sqrt\left\{2\right\} =$

\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,

$\sqrt\left\{2\right\} =$

\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 $\cos\left(\pi/4\right)$ gives

$\frac\left\{1\right\}\left\{\sqrt\left\{2\right\}\right\} = \sum_\left\{k=0\right\}^\infty \frac\left\{\left(-1\right)^k \left\left(\frac\left\{\pi\right\}\left\{4\right\}\right\right)^\left\{2k\right\}\right\}\left\{\left(2k\right)!\right\}.$

The Taylor series of $\sqrt\left\{\left(1 + x\right)\right\}$ with $x = 1$ and using the double factorial $n!!$ gives

$\sqrt\left\{2\right\} = \sum_\left\{k=0\right\}^\infty \left(-1\right)^\left\{k+1\right\} \frac\left\{\left(2k-3\right)!!\right\}\left\{\left(2k\right)!!\right\} =$

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

$\sqrt\left\{2\right\} = \sum_\left\{k=0\right\}^\infty \frac\left\{\left(2k+1\right)!\right\}\left\{\left(k!\right)^2 2^\left\{3k+1\right\}\right\} = \frac\left\{1\right\}\left\{2\right\} +\frac\left\{3\right\}\left\{8\right\} +$

\frac{15}{64} + \frac{35}{256} + \frac{315}{4096} + \frac{693}{16384} + \cdots.

It is not known whether $\sqrt\left\{2\right\}$ can be represented with a BBP-type formula. BBP-type formulas are known for π$\sqrt\left\{2\right\}$ and $\sqrt\left\{2\right\} \ln\left(1 + \sqrt\left\{2\right\}\right)$, however.

## Continued fraction representation

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

$\!\ \sqrt\left\{2\right\} = 1 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \ddots\right\}\right\}\right\}\right\}.$

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 $\scriptstyle\left\{\frac\left\{1\right\}\left\{2 q^2\sqrt\left\{2\right\}\right\}\right\}$ and then the next convergent is (p + 2q)/(p + q).

## Paper size

The square root of two is the approximate 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 the smaller sheets having the same ratio as the original sheet.