Math, asked by aditya4267, 1 year ago

what is the order of root 3​

Answers

Answered by rekhasushil78
0

Answer:

Step-by-step explanation:

The square root of 3 is an irrational number. It is also known as Theodorus' constant, named after Theodorus of Cyrene, who proved its irrationality. The first sixty digits of its decimal expansion are: 1.73205080756887729352744634150587236694280525381038062805580…

Answered by geetz1921
0

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 √3.

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

The first sixty digits of its decimal expansion are:

1.73205080756887729352744634150587236694280525381038062805580… (sequence A002194 in the OEIS)

As of December 2013, its numerical value in decimal has been computed to at least ten billion digits.[1]

The fraction

97

/

56

(1.732142857…) for the square root of three can be used as an approximation. Despite having a denominator of only 56, it differs from the correct value by less than

1

/

10,000

(approximately 9.2×10−5). The rounded value of 1.732 is correct to within 0.01% of the actual value.

Archimedes reported (

1351

/

780

)2

> 3 > (

265

/

153

)2

,[2] accurate to

1

/

608400

(six decimal places) and

2

/

23409

(four decimal places), respectively.

It can be expressed as the continued fraction [1; 1, 2, 1, 2, 1, 2, 1, …] (sequence A040001 in the 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 }}}}}}} [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.

The following nested square expressions converge to √3:

{\displaystyle \!\ {\sqrt {3}}=2-2\left({\frac {1}{2}}-\left({\frac {1}{2}}-\left({\frac {1}{2}}-\left({\frac {1}{2}}-\dots \right)^{2}\right)^{2}\right)^{2}\right)^{2}={\frac {7}{4}}-4\left({\frac {1}{16}}+\left({\frac {1}{16}}+\left({\frac {1}{16}}+\left({\frac {1}{16}}+\dots \right)^{2}\right)^{2}\right)^{2}\right)^{2}.} {\displaystyle \!\ {\sqrt {3}}=2-2\left({\frac {1}{2}}-\left({\frac {1}{2}}-\left({\frac {1}{2}}-\left({\frac {1}{2}}-\dots \right)^{2}\right)^{2}\right)^{2}\right)^{2}={\frac {7}{4}}-4\left({\frac {1}{16}}+\left({\frac {1}{16}}+\left({\frac {1}{16}}+\left({\frac {1}{16}}+\dots \right)^{2}\right)^{2}\right)^{2}\right)^{2}.}

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