Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Algebraic number

From Wikipedia, the free encyclopedia
Type of complex number
Not to be confused withAlgebraic solution.

The square root of 2 is an algebraic number equal to the length of thehypotenuse of aright triangle with legs of length 1.

Inmathematics, analgebraic number is a number that is aroot of a non-zeropolynomial in one variable withinteger (or, equivalently,rational) coefficients. For example, thegolden ratio(1+5)/2{\displaystyle (1+{\sqrt {5}})/2} is an algebraic number, because it is a root of the polynomialx2x1{\displaystyle x^{2}-x-1}, i.e., a solution to the equationx2x1=0{\displaystyle x^{2}-x-1=0}, and thecomplex number1+i{\displaystyle 1+i} is algebraic because it is a root of the polynomialx4+4{\displaystyle x^{4}+4}. Algebraic numbers include allintegers,rational numbers, andn-th roots of integers.

Algebraiccomplex numbers are closed under addition, subtraction, multiplication and division, and hence form afield, denotedQ¯{\displaystyle {\overline {\mathbb {Q} }}}. The set of algebraicreal numbersQ¯R{\displaystyle {\overline {\mathbb {Q} }}\cap \mathbb {R} } is also a field.

Numbers which are not algebraic are calledtranscendental and includeπ ande. There arecountably infinite algebraic numbers, hencealmost all real (or complex) numbers (in the sense ofLebesgue measure) are transcendental.

Examples

[edit]
  • Allrational numbers are algebraic. Any rational number, expressed as the quotient of anintegera and a (non-zero)natural numberb, satisfies the above definition, becausex =a/b is the root of a non-zero polynomial, namelybxa.[1]
  • Quadratic irrational numbers, irrational solutions of a quadratic polynomialax2 +bx +c with integer coefficientsa,b, andc, are algebraic numbers. If the quadratic polynomial is monic (a = 1), the roots are further qualified asquadratic integers.
    • Gaussian integers, complex numbersa +bi for which botha andb are integers, are also quadratic integers. This is becausea +bi andabi are the two roots of the quadraticx2 − 2ax +a2 +b2.
  • Aconstructible number can be constructed from a given unit length using a straightedge and compass. It includes all quadratic irrational roots, all rational numbers, and all numbers that can be formed from these using thebasic arithmetic operations and the extraction of square roots. (By designating cardinal directions for +1, −1, +i, and −i, complex numbers such as3+i2{\displaystyle 3+i{\sqrt {2}}} are considered constructible.)
  • Any expression formed from algebraic numbers using any finite combination of the basic arithmetic operations and extraction ofnth roots gives another algebraic number.
  • Polynomial roots that cannot be expressed in terms of the basic arithmetic operations and extraction ofnth roots (such as the roots ofx5x + 1).That happens with many but not all polynomials of degree 5 or higher.
  • Values oftrigonometric functions of rational multiples ofπ (except when undefined): for example,cosπ/7,cos3π/7, andcos5π/7 satisfy8x3 − 4x2 − 4x + 1 = 0. This polynomial isirreducible over the rationals and so the three cosines areconjugate algebraic numbers. Likewise,tan3π/16,tan7π/16,tan11π/16, andtan15π/16 satisfy the irreducible polynomialx4 − 4x3 − 6x2 + 4x + 1 = 0, and so are conjugatealgebraic integers. This is the equivalent of angles which, when measured in degrees, have rational numbers.[2]
  • Some but not all irrational numbers are algebraic:

Properties

[edit]
Algebraic numbers on thecomplex plane colored by degree (bright orange/red = 1, green = 2, blue = 3, yellow = 4). The larger points come from polynomials with smaller integer coefficients.
  • If a polynomial with rational coefficients is multiplied through by theleast common denominator, the resulting polynomial with integer coefficients has the same roots. This shows that an algebraic number can be equivalently defined as a root of a polynomial with either integer or rational coefficients.
  • Given an algebraic number, there is a uniquemonic polynomial with rational coefficients of leastdegree that has the number as a root. This polynomial is called itsminimal polynomial. If its minimal polynomial has degreen, then the algebraic number is said to be ofdegreen. For example, allrational numbers have degree 1, and an algebraic number of degree 2 is aquadratic irrational.
  • The algebraic numbers aredensein the reals. This follows from the fact they contain the rational numbers, which are dense in the reals themselves.
  • The set of algebraic numbers is countable,[4][5] and therefore itsLebesgue measure as a subset of the complex numbers is 0 (essentially, the algebraic numbers take up no space in the complex numbers). That is to say,"almost all" real and complex numbers are transcendental.
  • All algebraic numbers arecomputable and thereforedefinable andarithmetical.
  • For real numbersa andb, the complex numbera +bi is algebraic if and only if botha andb are algebraic.[6]

Degree of simple extensions of the rationals as a criterion to algebraicity

[edit]

For anyα{\displaystyle \alpha }, thesimple extension of the rationals byα{\displaystyle \alpha }, denoted byQ(α){\displaystyle \mathbb {Q} (\alpha )} (whose elements are thef(α){\displaystyle f(\alpha )} forf{\displaystyle f} arational function with rational coefficients which is defined atα{\displaystyle \alpha }), is of finitedegree if and only ifα{\displaystyle \alpha } is an algebraic number.

The condition of finite degree means that there is a fixed set of numbers{ai}{\displaystyle \{a_{i}\}} of finitecardinalityk{\displaystyle k} with elements inQ(α){\displaystyle \mathbb {Q} (\alpha )} such thatQ(α)=i=1kaiQ{\displaystyle \textstyle \mathbb {Q} (\alpha )=\sum _{i=1}^{k}a_{i}\mathbb {Q} }; that is, each element ofQ(α){\displaystyle \mathbb {Q} (\alpha )} can be written as a sumi=1kaiqi{\displaystyle \textstyle \sum _{i=1}^{k}a_{i}q_{i}} for some rational coefficients{qi}{\displaystyle \{q_{i}\}}.

Since theai{\displaystyle a_{i}} are themselves members ofQ(α){\displaystyle \mathbb {Q} (\alpha )}, each can be expressed as sums of products of rational numbers and powers ofα{\displaystyle \alpha }, and therefore this condition is equivalent to the requirement that for some finiten{\displaystyle n},Q(α)={i=nnαiqi|qiQ}.{\displaystyle \mathbb {Q} (\alpha )={\biggl \lbrace }\sum _{i=-n}^{n}\alpha ^{i}q_{i}\mathbin {\bigg |} q_{i}\in \mathbb {Q} {\biggr \rbrace }.}

The latter condition is equivalent toαn+1{\displaystyle \alpha ^{n+1}}, itself a member ofQ(α){\displaystyle \mathbb {Q} (\alpha )}, being expressible asi=nnαiqi{\displaystyle \textstyle \sum _{i=-n}^{n}\alpha ^{i}q_{i}} for some rationals{qi}{\displaystyle \{q_{i}\}}, soα2n+1=i=02nαiqin{\displaystyle \textstyle \alpha ^{2n+1}=\sum _{i=0}^{2n}\alpha ^{i}q_{i-n}} or, equivalently,α{\displaystyle \alpha } is a root ofx2n+1i=02nxiqin{\displaystyle \textstyle x^{2n+1}-\sum _{i=0}^{2n}x^{i}q_{i-n}}; that is, an algebraic number with a minimal polynomial of degree not larger than2n+1{\displaystyle 2n+1}.

It can similarly be proven that for any finite set of algebraic numbersα1{\displaystyle \alpha _{1}},α2{\displaystyle \alpha _{2}}...αn{\displaystyle \alpha _{n}}, the field extensionQ(α1,α2,...αn){\displaystyle \mathbb {Q} (\alpha _{1},\alpha _{2},...\alpha _{n})} has a finite degree.

Field

[edit]
Algebraic numbers colored by degree (blue = 4, cyan = 3, red = 2, green = 1). The unit circle is black.[further explanation needed]

The sum, difference, product, and quotient (if the denominator is nonzero) of two algebraic numbers is again algebraic:

For any two algebraic numbersα{\displaystyle \alpha },β{\displaystyle \beta }, this follows directly from the fact that thesimple extensionQ(γ){\displaystyle \mathbb {Q} (\gamma )}, forγ{\displaystyle \gamma } being eitherα+β{\displaystyle \alpha +\beta },αβ{\displaystyle \alpha -\beta },αβ{\displaystyle \alpha \beta } or (forβ0{\displaystyle \beta \neq 0})α/β{\displaystyle \alpha /\beta }, is alinear subspace of the finite-degree field extensionQ(α,β){\displaystyle \mathbb {Q} (\alpha ,\beta )}, and therefore has a finite degree itself, from which it follows (as shownabove) thatγ{\displaystyle \gamma } is algebraic.

An alternative way of showing this is constructively, by using theresultant.

Algebraic numbers thus form afield[7]Q¯{\displaystyle {\overline {\mathbb {Q} }}} (sometimes denoted byA{\displaystyle \mathbb {A} }, but that usually denotes theadele ring).

Algebraic closure

[edit]

Every root of a polynomial equation whose coefficients arealgebraic numbers is again algebraic. That can be rephrased by saying that the field of algebraic numbers isalgebraically closed. In fact, it is the smallest algebraically closed field containing the rationals and so it is called thealgebraic closure of the rationals.

That the field of algebraic numbers is algebraically closed can be proven as follows: Letβ{\displaystyle \beta } be a root of a polynomialα0+α1x+α2x2...+αnxn{\displaystyle \alpha _{0}+\alpha _{1}x+\alpha _{2}x^{2}...+\alpha _{n}x^{n}} with coefficients that are algebraic numbersα0{\displaystyle \alpha _{0}},α1{\displaystyle \alpha _{1}},α2{\displaystyle \alpha _{2}}...αn{\displaystyle \alpha _{n}}. The field extensionQQ(α1,α2,...αn){\displaystyle \mathbb {Q} ^{\prime }\equiv \mathbb {Q} (\alpha _{1},\alpha _{2},...\alpha _{n})} then has a finite degree with respect toQ{\displaystyle \mathbb {Q} }. The simple extensionQ(β){\displaystyle \mathbb {Q} ^{\prime }(\beta )} then has a finite degree with respect toQ{\displaystyle \mathbb {Q} ^{\prime }} (since all powers ofβ{\displaystyle \beta } can be expressed by powers of up toβn1{\displaystyle \beta ^{n-1}}). Therefore,Q(β)=Q(β,α1,α2,...αn){\displaystyle \mathbb {Q} ^{\prime }(\beta )=\mathbb {Q} (\beta ,\alpha _{1},\alpha _{2},...\alpha _{n})} also has a finite degree with respect toQ{\displaystyle \mathbb {Q} }. SinceQ(β){\displaystyle \mathbb {Q} (\beta )} is a linear subspace ofQ(β){\displaystyle \mathbb {Q} ^{\prime }(\beta )}, it must also have a finite degree with respect toQ{\displaystyle \mathbb {Q} }, soβ{\displaystyle \beta } must be an algebraic number.

Related fields

[edit]

Numbers defined by radicals

[edit]

Any number that can be obtained from the integers using afinite number ofadditions,subtractions,multiplications,divisions, and taking (possibly complex)nth roots wheren is a positive integer are algebraic. The converse, however, is not true: there are algebraic numbers that cannot be obtained in this manner. These numbers are roots of polynomials of degree 5 or higher, a result ofGalois theory (seeQuintic equations and theAbel–Ruffini theorem). For example, the equation:

x5x1=0{\displaystyle x^{5}-x-1=0}

has a unique real root, ≈ 1.1673, that cannot be expressed in terms of only radicals and arithmetic operations.

Closed-form number

[edit]
Main article:Closed-form number

Algebraic numbers are all numbers that can be defined explicitly or implicitly in terms of polynomials, starting from the rational numbers. One may generalize this to "closed-form numbers", which may be defined in various ways. Most broadly, all numbers that can be defined explicitly or implicitly in terms of polynomials, exponentials, and logarithms are called "elementary numbers", and these include the algebraic numbers, plus some transcendental numbers. Most narrowly, one may consider numbersexplicitly defined in terms of polynomials, exponentials, and logarithms – this does not include all algebraic numbers, but does include some simple transcendental numbers such ase orln 2.

Algebraic integers

[edit]
Visualisation of the (countable) field of algebraic numbers in the complex plane. Colours indicate the leading integer coefficient of the polynomial the number is a root of (red = 1 i.e. the algebraic integers, green = 2, blue = 3, yellow = 4...). Points becomes smaller as the other coefficients and number of terms in the polynomial become larger. View shows integers 0,1 and 2 at bottom right, +i near top.
Main article:Algebraic integer

Analgebraic integer is an algebraic number that is a root of a polynomial with integer coefficients with leading coefficient 1 (amonic polynomial). Examples of algebraic integers are5+132,{\displaystyle 5+13{\sqrt {2}},}26i,{\displaystyle 2-6i,} and12(1+i3).{\textstyle {\frac {1}{2}}(1+i{\sqrt {3}}).} Therefore, the algebraic integers constitute a propersuperset of theintegers, as the latter are the roots of monic polynomialsxk for allkZ{\displaystyle k\in \mathbb {Z} }. In this sense, algebraic integers are to algebraic numbers whatintegers are torational numbers.

The sum, difference and product of algebraic integers are again algebraic integers, which means that the algebraic integers form aring. The namealgebraic integer comes from the fact that the only rational numbers that are algebraic integers are the integers, and because the algebraic integers in anynumber field are in many ways analogous to the integers. IfK is a number field, itsring of integers is the subring of algebraic integers inK, and is frequently denoted asOK. These are the prototypical examples ofDedekind domains.

Special classes

[edit]

Notes

[edit]
  1. ^Some of the following examples come fromHardy & Wright (1972, pp. 159–160, 178–179)
  2. ^Garibaldi 2008.
  3. ^Also,Liouville's theorem can be used to "produce as many examples of transcendental numbers as we please," cf.Hardy & Wright (1972, p. 161ff)
  4. ^Hardy & Wright 1972, p. 160, 2008:205.
  5. ^Niven 1956, Theorem 7.5..
  6. ^Niven 1956, Corollary 7.3..
  7. ^Niven 1956, p. 92.

References

[edit]
Number systems
Sets ofdefinable numbers
Composition algebras
Split
types
Otherhypercomplex
Infinities andinfinitesimals
Other types
International
National
Other
Retrieved from "https://en.wikipedia.org/w/index.php?title=Algebraic_number&oldid=1320017125"
Category:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp