Square root of 3

 Template:Irrational numbers Binary 1.1011101101100111101... Decimal 1.7320508075688772935... Hexadecimal 1.BB67AE8584CAA73B... Continued fraction ${\displaystyle 1+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+\ddots }}}}}}}}}}}$

The square root of 3 is the positive real number that, when multiplied by itself, gives the number 3. It is more precisely called the principal square root of 3, to distinguish it from the negative number with the same property. It is denoted by

${\displaystyle {\sqrt {3}}.}$

The first sixty digits of its decimal expansion are:

1.73205 08075 68877 29352 74463 41505 87236 69428 05253 81038 06280 5580... (sequence A002194 in OEIS)

As of December 2013, its numerical value in decimal has been computed to at least ten billion digits.[1] The rounded value of 1.732 is correct to within 0.01% of the actual value. A close fraction is ${\displaystyle {\tfrac {97}{56}}}$ (1.732142857...).

Archimedes reported (1351/780)2 > 3 > (265/153)2,{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} accurate to 1/608400 (6-places) and 2/23409 (4-places), respectively.

The square root of 3 is an irrational number. It is also known as Theodorus' constant, named after Theodorus of Cyrene.

It can be expressed as the continued fraction [1; 1, 2, 1, 2, 1, 2, 1, ...] (sequence A040001 in OEIS), expanded on the right.

It can also be expressed by generalized continued fractions such as

${\displaystyle [2;-4,-4,-4,...]=2-{\cfrac {1}{4-{\cfrac {1}{4-{\cfrac {1}{4-\ddots }}}}}}}$

which is [1;1, 2,1, 2,1, 2,1, ...] evaluated at every second term.

Proof of irrationality

This irrationality proof for the square root of 3 uses Fermat's method of infinite descent:

Suppose that Template:Sqrt is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as ${\displaystyle {\frac {m}{n}}}$ for natural numbers Template:Mvar and Template:Mvar.

Therefore, multiplying by 1 will give an equal expression:

${\displaystyle {\frac {m({\sqrt {3}}-q)}{n({\sqrt {3}}-q)}}}$

where Template:Mvar is the largest integer smaller than Template:Sqrt. Note that both the numerator and the denominator have been multiplied by a number smaller than 1.

Through this, and by multiplying out both the numerator and the denominator, we get:

${\displaystyle {\frac {m{\sqrt {3}}-mq}{n{\sqrt {3}}-nq}}}$

It follows that Template:Mvar can be replaced with ${\displaystyle {\sqrt {3}}n}$:

${\displaystyle {\frac {n{\sqrt {3}}^{2}-mq}{n{\sqrt {3}}-nq}}}$

Then, Template:Sqrt can also be replaced with ${\displaystyle {\frac {m}{n}}}$ in the denominator:

${\displaystyle {\frac {n{\sqrt {3}}^{2}-mq}{n{\frac {m}{n}}-nq}}}$

The square of Template:Sqrt can be replaced by 3. As ${\displaystyle {\frac {m}{n}}}$ is multiplied by Template:Mvar, their product equals Template:Mvar:

${\displaystyle {\frac {3n-mq}{m-nq}}}$

Then Template:Sqrt can be expressed in lower terms than ${\displaystyle m/n}$ (since the first step reduced the sizes of both the numerator and the denominator, and subsequent steps did not change them) as ${\displaystyle {\frac {3n-mq}{m-nq}}}$, which is a contradiction to the hypothesis that ${\displaystyle m/n}$ was in lowest terms.[2]

An alternate proof of this is, assuming ${\displaystyle {\sqrt {3}}={\frac {m}{n}}}$ with ${\displaystyle {\frac {m}{n}}}$ being a fully reduced fraction:

Multiplying by Template:Mvar both terms, and then squaring both gives

${\displaystyle 3n^{2}=m^{2}.}$

Since the left side is divisible by 3, so is the right side, requiring that Template:Mvar is divisible by 3. Then, Template:Mvar can be expressed as ${\displaystyle 3k}$:

${\displaystyle 3n^{2}=(3k)^{2}}$
${\displaystyle 3n^{2}=9k^{2}}$

Therefore, dividing both terms by 3 gives:

${\displaystyle n^{2}=3k^{2}}$

Since the right side is divisible by 3, so is the left side and hence so is Template:Mvar. Thus, as both Template:Mvar and Template:Mvar are divisible by 3, they have a common factor and ${\displaystyle m/n}$ is not a fully reduced fraction, contradicting the original premise.

Geometry and trigonometry

An equilateral triangle with a side length of 2 has a height of Template:Sqrt.
The square root of 3 is equal to the length between parallel sides of a regular hexagon with sides of length 1.

The square root of 3 can be found as the leg length of an equilateral triangle that encompasses a circle with a diameter of 1.

If an equilateral triangle with sides of length 1 is cut into two equal halves, by bisecting an internal angle across to make a right angle with one side, the right angle triangle's hypotenuse is length one and the sides are of length 1/2 and Template:Sqrt/2. From this the trigonometric function tangent of 60 degrees equals Template:Sqrt, and the sine of 60° and the cosine of 30° both equal half of Template:Sqrt.

The square root of 3 also appears in algebraic expressions for various other trigonometric constants, including[3] the sines of 3°, 12°, 15°, 21°, 24°, 33°, 39°, 48°, 51°, 57°, 66°, 69°, 75°, 78°, 84°, and 87°.

It is the distance between parallel sides of a regular hexagon with sides of length 1. On the complex plane, this distance is expressed as iTemplate:Sqrt mentioned below.

It is the length of the space diagonal of a unit cube.

The vesica piscis has a major axis: minor axis ratio equal to the square root of three, this can be shown by constructing two equilateral triangles within it.

Square root of −3

Multiplication of Template:Sqrt by the imaginary unit gives a square root of −3, an imaginary number. More exactly,

${\displaystyle {\sqrt {-3}}=\pm {\sqrt {3}}i}$ (see square root of negative numbers).

It is an Eisenstein integer. Namely, it is expressed as the difference between two non-real cubic roots of 1 (which are Eisenstein integers).

Other uses

Power engineering

In power engineering, the voltage between two phases in a three-phase system equals Template:Sqrt times the line to neutral voltage. This is because any two phases are 120 degrees apart, and two points on a circle 120 degrees apart are separated by Template:Sqrt times the radius (see geometry examples above).

Notes

1. Lukasz Komsta: Computations page
2. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
3. Julian D. A. Wiseman [Sin and Cos in Surds http://www.jdawiseman.com/papers/easymath/surds_sin_cos.html]

References

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