Well-definition

In mathematics and computer science, well-definedness denotes the property of an object to be clearly defined. The term is primarily used when there is a possibility that the object is otherwise ambiguous .

By definition , a well-defined expression provides exactly one value or one possible interpretation.

In a broader sense, this term is sometimes used to say that an object is consistent, i.e. H. is formally correctly defined.

The question of whether an object is well defined, there is often characterized in mathematics that an object not only by a definition equation (explicit), but also by a characteristic property (implicitly) defined can be. In the case of functions or links in particular, it happens that they can only be "implicitly defined". This is done by first defining a relation (as a subset of a Cartesian product ) with the same number of places (explicitly). This relation is expressly asserted that it is of a specific type, for example function or link. However, the entire "definition" is not complete and valid until proof of the claim has been established. One then says: the object or the concept is (as this specific type) well-defined . Otherwise one speaks of ambiguity u. Ä., And the math object remains undefined.

Simple examples

analogy

1. The definition of a species of goat A is:

"Horned mammal, with property A".

This type of goat A is not well defined as a type of goat because there are other mammals with horns that may have property A.

However, if we prove that property A occurs exclusively in goats, then goat species A is well-defined because there can then be exactly one species of mammal that fulfills property A and the definition is therefore unambiguous.

mathematics

1. "For everyone is " defined "as the number for which applies ." ${\displaystyle x\in \mathbb {R} _{\geq 0}}$ ${\displaystyle f_{1}(x)}$ ${\displaystyle y\in \mathbb {R} _{\geq 0}}$ ${\displaystyle y^{2}=x}$
2. "For everyone is " defined "as the number for which applies ." ${\displaystyle x\in \mathbb {R} }$ ${\displaystyle f_{2}(x)}$ ${\displaystyle y\in \mathbb {R} }$ ${\displaystyle y^{2}=x}$
3. "For everyone is " defined "as the number for which applies ." ${\displaystyle x\in \mathbb {R} _{\geq 0}}$ ${\displaystyle f_{3}(x)}$ ${\displaystyle y\in \mathbb {R} }$ ${\displaystyle y^{2}=x}$

This is supposed to be the "definition" of functions with a specified set of definitions and values . ${\displaystyle f_{1},f_{2},f_{3}}$

To 1:	For every number in the definition set ${\displaystyle x}$ ${\displaystyle \mathbb {R} _{\geq 0}}$ exists one( Left totality ) andjust one( Right uniqueness ) Number in the set of values with the property . (The square function of to is bijective .) The function is thus well-defined. is the square root function. ${\displaystyle y}$ ${\displaystyle \mathbb {R} _{\geq 0}}$ ${\displaystyle y^{2}=x}$ ${\displaystyle \mathbb {R} _{\geq 0}}$ ${\displaystyle \mathbb {R} _{\geq 0}}$ ${\displaystyle f_{1}}$ ${\displaystyle f_{1}}$
To 2:	The two-digit relation is not left total. Because it is an element of the left set that is supposed to represent the definition set. But there is none , the right amount, with . The ${\displaystyle f_{2}:=\{(x,y)\mid x\in \mathbb {R} ,y\in \mathbb {R} ,y^{2}=x\}\subseteq \mathbb {R} \times \mathbb {R} }$ ${\displaystyle x=-1}$ ${\displaystyle \in \mathbb {R} }$ ${\displaystyle y\in \mathbb {R} }$ ${\displaystyle y^{2}=x=-1}$existenceis hurt. So (as a function) is not well defined and not a function. ${\displaystyle f_{2}}$
To 3:	The two-digit relation is not legally clear. Because it applies to two different elements from the right set that should represent the set of values. The ${\displaystyle f_{3}:=\{(x,y)\mid x\in \mathbb {R} _{\geq 0},y\in \mathbb {R} ,y^{2}=x\}\subseteq \mathbb {R} _{\geq 0}\times \mathbb {R} }$ ${\displaystyle y_{1}^{\;\!2}=4=y_{2}^{\;\!2}}$ ${\displaystyle y_{1}:=2,y_{2}:=-2}$ ${\displaystyle \mathbb {R} }$Uniquenessis hurt. So (as a function) is not well defined. ${\displaystyle f_{3}}$

Definition without anticipation

The quotation marks for "defined" and "definition" can be avoided if you do not immediately define a function. Instead, the first step is to define only a two-digit relationship - whatever works. (This is what happened in the comments on the simple examples 2 and 3.)

In a second step, it is shown that the two-digit relation defined in this way has the properties left totality and right uniqueness, i.e. is a function. This second step corresponds exactly to the usual checking of the well-definedness.

The same mathematical objects can therefore also be formed without the term "well-defined," which means that this term turns out to be dispensable in mathematics.

At the same time, the anticipation of the functional property in the "definition" is common practice, above all because the object of the definition is immediately made known as a function . And since the purpose of a "definition" is not its failure, there is no lack of well-defined definition in mathematical texts.

Representative independence

In the literature, there is often the definition of well-definedness as independence from representatives. Occasionally it is expressly pointed out that there is no further meaning.

Typically, the question of the well-defined character of a function must be asked if the equation defining the function refers not (only) to the arguments themselves, but (also) to elements of the arguments. This is sometimes unavoidable when the arguments are equivalence classes . A member of an equivalence class is representative called, and to such a reference.

This will be explained using an example. Every rational number can be written as a fraction of two whole numbers, the numerator and the denominator. So we "define" it as a "function" that assigns each rational number to its numerator. ${\displaystyle f\colon \mathbb {Q} \to \mathbb {Z} \colon a/b\mapsto a}$

Now applies , so should apply , a contradiction! So the "definition" of cannot be correct. The "definition" of is not well defined. Let us take a closer look at the "definition" : The fraction stands for the equivalence class of all pairs for which applies. The definition of should therefore be more precise: For all rational numbers is "defined" as the value for which there is a with . The equivalence class is the argument of the reference being made to the representative. Now it turns out that there are several such - for example, this is or is not well-defined and the "definition" is not one. ${\displaystyle 1/2=2/4}$ ${\displaystyle 1=f(1/2)=f(2/4)=2}$ ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle a/b}$ ${\displaystyle [a/b]}$ ${\displaystyle (x,y)}$ ${\displaystyle ay=xb}$ ${\displaystyle f}$ ${\displaystyle q}$ ${\displaystyle f(q)}$ ${\displaystyle x\in \mathbb {Z} }$ ${\displaystyle y\in \mathbb {N} }$ ${\displaystyle [x/y]=q}$ ${\displaystyle q}$ ${\displaystyle f;}$ ${\displaystyle x/y\in q.}$ ${\displaystyle x/y}$ ${\displaystyle q=[1/2]}$ ${\displaystyle x=1,y=2}$ ${\displaystyle x=2,y=4.}$ ${\displaystyle f}$

If an element so multiple representations (in this example , , , ...), then has a function of this element has a value assigned that of the representation of independent. The "definition", for example, fulfills this condition. ${\displaystyle a\in A}$ ${\displaystyle 1/2}$ ${\displaystyle 2/4}$ ${\displaystyle 3/6}$ ${\displaystyle f\colon A\to B}$ ${\displaystyle f(a)}$ ${\displaystyle a}$ ${\displaystyle f\colon \mathbb {Q} \setminus \{0\}\to \mathbb {Q} \setminus \{0\}\colon a/b\mapsto b/a}$

Representative independence must be proven for the following two mathematical concepts:

Induced images

Definition of the induced mapping

Two sets and as well as equivalence relations on and on are given . With which was equivalence class of the element with respect to and are adjusted with the equivalence class of the element with respect to . The set of equivalence classes is called the factor set of (after the equivalence relation ). ${\displaystyle A_{1}}$ ${\displaystyle A_{2}}$ ${\displaystyle \sim _{1}}$ ${\displaystyle A_{1}}$ ${\displaystyle \sim _{2}}$ ${\displaystyle A_{2}}$ ${\displaystyle [a_{1}]_{1}}$ ${\displaystyle \{a\mid a\sim _{1}a_{1}\}}$ ${\displaystyle a_{1}\in A_{1}}$ ${\displaystyle \sim _{1}}$ ${\displaystyle [a_{2}]_{2}}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle \sim _{2}}$ ${\displaystyle {A_{1}}_{/\sim _{1}}:=\{[a_{1}]_{1}\mid a_{1}\in A_{1}\}}$ ${\displaystyle A_{1}}$ ${\displaystyle \sim _{1}}$

If one has now given a function (or mapping) , a (two-digit) relation can always be found on the pair ${\displaystyle f\colon A_{1}\to A_{2}}$ ${\displaystyle {\tilde {f}}}$

	${\displaystyle {\color {OliveGreen}{A_{1}}_{/\sim _{1}}}}$	${\displaystyle \times }$	${\displaystyle {\color {red}{A_{2}}_{/\sim _{2}}}}$
the factor sets according to the regulation
${\displaystyle {\tilde {f}}:=\{\;(}$	${\displaystyle {\color {OliveGreen}[a_{1}]_{1}}}$	${\displaystyle ,}$	${\displaystyle {\color {red}[f(a_{1})]_{2}}}$	${\displaystyle )\,\mid \,a_{1}\in A_{1}\}}$

define. This definition is valid and fully-fledged as a definition of a relation . But their purpose is (mostly) to define an image . This is also the name of the mapping induced by , although the use of the term mapping actually represents an anticipation of the as yet unproven well-definition. ${\displaystyle {\tilde {f}}}$ ${\displaystyle f}$

Well-definition of an induced mapping

First of all, there is only a two-digit relation that fulfills the (remaining) requirements for the (also two-digit relation of the) function or mapping if there is only one (single) function value for each argument value . The following must apply here: ${\displaystyle {\tilde {f}}}$ ${\displaystyle \subseteq {A_{1}}_{/\sim _{1}}\times {A_{2}}_{/\sim _{2}}}$ ${\displaystyle [a_{1}]_{1}}$ ${\displaystyle {\tilde {f}}([a_{1}]_{1})}$

${\displaystyle \forall x,y\in A_{1}\quad ([x]_{1}=[y]_{1}\Longrightarrow [f(x)]_{2}=[f(y)]_{2})}$.

Just when they ( representatives independence called) requirement is met, the induced "image" is well defined , so it remains only a relation, but really a picture . ${\displaystyle {\tilde {f}}}$

Examples of induced mappings

• Be and . As the equivalence relation we choose the "equivalence modulo 3", i. i.e. it applies ${\displaystyle A_{1}=\mathbb {Z} }$ ${\displaystyle A_{2}=\{0,1\}}$ ${\displaystyle \sim _{1}}$

${\displaystyle x\sim _{1}y,\quad \mathrm {falls} \quad {\frac {x-y}{3}}\in \mathbb {Z} \ .}$

Let the equivalence relation be the usual equality, so if . (An equivalence class therefore consists of exactly one element.) ${\displaystyle \sim _{2}}$ ${\displaystyle x\sim _{2}y\;}$ ${\displaystyle x=y}$

As a function we choose

${\displaystyle f\colon \mathbb {Z} \to \{0,1\},\ x\mapsto {\begin{cases}0,&{\text{wenn }}x{\text{ gerade,}}\\1,&{\text{wenn }}x{\text{ ungerade.}}\end{cases}}}$

The induced "mapping" is then

${\displaystyle {\tilde {f}}\colon \mathbb {Z} _{/\sim _{1}}\to \{0,1\}_{/\sim _{2}},\ [x]_{1}\mapsto [f(x)]_{2}={\begin{cases}[0]_{2},&{\text{wenn }}x{\text{ gerade,}}\\{[}1{]}_{2},&{\text{wenn }}x{\text{ ungerade.}}\end{cases}}}$

It applies now , though . In this case, the "induced mapping" is  not well-defined and not a mapping. ${\displaystyle {\tilde {f}}([5]_{1})=[1]_{2}=\{1\}\neq \{0\}=[0]_{2}={\tilde {f}}([8]_{1})}$ ${\displaystyle [5]_{1}=[8]_{1}}$ ${\displaystyle {\tilde {f}}}$

• Be . The equivalence relation is explained by ${\displaystyle A_{1}=A_{2}=\mathbb {R} }$ ${\displaystyle \sim _{1}}$

${\displaystyle x\sim _{1}y,\quad \mathrm {falls} \quad {\frac {x-y}{2\pi }}\in \mathbb {Z} \ ,}$

and be the ordinary likeness again. The real cosine now induces the mapping ${\displaystyle \sim _{2}}$

${\displaystyle {\tilde {f}}\colon \mathbb {R} _{/\sim _{1}}\to \mathbb {R} _{/\sim _{2}},[x]_{1}\mapsto [\cos(x)]_{2}}$.

This mapping is well-defined, as shown by: Be with the property . According to the definition of, there now exists an with , and therefore it follows , using the fact that the cosine has a period of . ${\displaystyle x,y\in \mathbb {R} }$ ${\displaystyle [x]_{1}=[y]_{1}}$ ${\displaystyle \sim _{1}}$ ${\displaystyle k\in \mathbb {Z} }$ ${\displaystyle x=y+k\cdot 2\pi }$ ${\displaystyle {\tilde {f}}([x]_{1})=\cos(x)=\cos(y+k\cdot 2\pi )=\cos(y)={\tilde {f}}([y]_{1})}$ ${\displaystyle 2\pi }$

Induced linkage

Definition of the induced link

Let be a nonempty set with an equivalence relation and an inner connection  . Using one can visit its factor structure, the ternary relation ${\displaystyle A}$ ${\displaystyle \sim }$ ${\displaystyle *\colon A\times A\to A}$ ${\displaystyle *}$

${\displaystyle {\begin{array}{rcccccl}&A_{/\sim }&\times &A_{/\sim }&\times &A_{/\sim }\\{\tilde {*}}:=\{(&[a]&,&[b]&,&[a*b]&)\mid a,b\in A\}\end{array}}}$

define. In anticipation of the well-definedness still to be proven, the link induced by the factor structure is mentioned. ${\displaystyle {\tilde {*}}}$ ${\displaystyle *}$

Well-definition for induced connections

So that this relation is really a connection, the result must not depend on the choice of the representative in a class. That means it must apply to everyone with the property : ${\displaystyle a_{1},a_{2},b_{1},b_{2}\in A}$ ${\displaystyle a_{1}\sim a_{2},\;b_{1}\sim b_{2}}$

${\displaystyle a_{1}*b_{1}\sim a_{2}*b_{2}\ .}$

If this is the case, the induced link is a (real) link (which is said to be well-defined). ${\displaystyle {\tilde {*}}}$

Examples of induced links

• The connection , given by , is not well-defined: [5] = [2] and [3] = [6], but ${\displaystyle p\colon \mathbb {Z} /3\mathbb {Z} \times \mathbb {Z} /3\mathbb {Z} \to \mathbb {Z} /3\mathbb {Z} }$ ${\displaystyle p([a],[b]):=[{|a|}^{|b|}]}$

${\displaystyle p([5],[3])=[5^{3}]=[125]=[3\cdot 41+2]=[2]\neq [1]=[3\cdot 21+1]=[64]=[2^{6}]=p([2],[6])\ }$.

• Look at the symmetrical group and within it the subgroup . The linkage induced on the factor set is not well defined. It is and of course however ${\displaystyle S_{3}=\{id,(1,2),(1,3),(2,3),(1,2,3),(1,3,2)\}}$ ${\displaystyle U:=\{id,(1,2)\}}$ ${\displaystyle S_{3}/U}$ ${\displaystyle [id]=[(1,2)]}$ ${\displaystyle [(1,3,2)]=[(1,3,2)],}$

${\displaystyle [(1,2)*(1,3,2)]=[(1,3)]\neq [(1,3,2)]=[id*(1,3,2)].}$

• The addition and the multiplication in a remainder class ring are well defined. The remainder class addition is precisely the link induced by the addition in and the normal divisor . ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ ${\displaystyle (n\in \mathbb {N} )}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle n\mathbb {Z} }$
• If the group is a normal divisor , then the link induced on is well-defined and is called the factor group from to . The property of being a normal divisor is even equivalent to the fact that the induced link is well-defined on the factor set . Because be and arbitrary. For the well-definition of the induced group linkage on the left secondary classes, the following must apply: ${\displaystyle N}$ ${\displaystyle G}$ ${\displaystyle G/N}$ ${\displaystyle G/N}$ ${\displaystyle G}$ ${\displaystyle N}$ ${\displaystyle G/N}$ ${\displaystyle g_{1},g_{2}\in G}$ ${\displaystyle n_{1},n_{2}\in N}$

${\displaystyle [g_{1}*n_{1}]\;{\tilde {*}}\;[g_{2}*n_{2}]=[g_{1}*g_{2}]\ ,}$

so . However, this corresponds to definition 2 of the normal divisor. The same result is obtained with the right secondary classes. ${\displaystyle g_{2}*n_{2}*g_{2}^{-1}\in N}$

Well-definition in mathematical notation

For real numbers, the notation for the product is considered to be well-defined, since the multiplication fulfills the associative law. In accordance with the rest of the mathematical notation, it is unique because the product for three real numbers always provides a unique value. ${\displaystyle abc}$ ${\displaystyle (ab)c=a(bc)}$ ${\displaystyle abc}$ ${\displaystyle a,b,c}$

This also applies to the non-commutative quaternions in the multiplication .

The subtraction is not associative. Nevertheless, the representation is considered to be well-defined. ${\displaystyle a-b-c}$ ${\displaystyle -b:=+(-b)}$

For real numbers and the notation for the quotient is well defined. For the quaternions that are not commutative in the multiplication , this notation is not considered to be well-defined. ${\displaystyle x}$ ${\displaystyle y\neq 0}$ ${\displaystyle {\frac {x}{y}}}$ ${\displaystyle xy^{-1}=y^{-1}x}$

Programming languages

In the case of notations with operators in mathematics and computer science, however, additional rules for operator precedence and associativity can usually be used to achieve uniqueness even without brackets.

In the C programming language , for example, the subtraction - operator is left-associative; H. it is evaluated from left to right: a-b-c= (a-b)-c. However, the assignment operator = is right-associative; H. a=b=c= a=(b=c).

There is only one rule of priority in the APL programming language : First the brackets are processed, then the rest from right to left.

Completeness and consistency

In a broader sense, well-definedness is also extended to other areas. It then designates a meaningful and consistent definition. Synonym for “not well defined” in this sense are also used as “not defined” or “not fully defined”.

Domain of definition of a function

In the definition range of the figure , the zero must not be included in the definition range, since it provides the value . However, dividing by zero is not explained in real numbers; H. there is no real number " ". (In a broader sense, one could set. But that does nothing to the example, since it is not a real number! In addition, one would have to identify and with one another, as for versus diverges.) ${\displaystyle f\colon \mathbb {R} \to \mathbb {R} ,x\mapsto {\frac {1}{x}}}$ ${\displaystyle x=0}$ ${\displaystyle f(0)={\frac {1}{0}}}$ ${\displaystyle 1/0}$ ${\displaystyle 1/0:=\infty }$ ${\displaystyle \infty }$ ${\displaystyle \infty }$ ${\displaystyle -\infty }$ ${\displaystyle f(x)}$ ${\displaystyle x\nearrow 0}$ ${\displaystyle -\infty }$

Likewise, it is not explained in the real numbers to take the square root of negative numbers. In other words, the “function” is not well-defined, but the function is. ${\displaystyle f\colon \mathbb {R} \to \mathbb {R} ,x\mapsto {\sqrt {x}}}$ ${\displaystyle f\colon \mathbb {R} _{\geq 0}\to \mathbb {R} ,x\mapsto {\sqrt {x}}}$

Range of values ​​of a function

If you write the formula as a "function" , the value is assigned to the value . In this case, however, this is not permitted, since it is not a natural number and is therefore not in the value range. ${\displaystyle f(x)=2x}$ ${\displaystyle f\colon \mathbb {Z} \to \mathbb {N} ,x\mapsto 2\cdot x}$ ${\displaystyle x=-2}$ ${\displaystyle f(-2)=-4}$ ${\displaystyle -4}$

Links in groups

Internal links of an algebraic structure (e.g. a group ) are also functions (usually with two arguments). The same conditions apply to them: The combination of elements of the structure must result in a clearly defined element of . The term isolation is often used incorrectly here, but it refers to the definition of substructures. ${\displaystyle G}$ ${\displaystyle G}$ ${\displaystyle G}$

Well-definition of sets

A set is well-defined if the defining clearly states for any object that it is either a member of the set or is not a member of the set. In particular, certain forms of impredicative definitions are excluded.

See also

• Well-defined problems in cognitive psychology

Web links

• Eric W. Weisstein : well-defined . In: MathWorld (English).
• well defined . In: PlanetMath . (English)
• Definition and well-defined on Maths-Online

Individual evidence

1. ↑ Similarly, a function with arguments is initially defined as a -digit relation and left totality and right uniqueness are related to the pair . ${\displaystyle D_{1}\times \dotsb \times D_{n}\to W}$ ${\displaystyle n}$ ${\displaystyle (n+1)}$ ${\displaystyle R\subseteq D_{1}\times \dotsb \times D_{n}\times W}$ ${\displaystyle (D_{1}\times \dotsb \times D_{n})\times W}$
2. ^ Serge Lang : Algebra . 3. Edition. 1993, p. X (Prerequisites).
3. ↑ Albrecht Beutelspacher : That is trivial oBdA! Braunschweig 1997, p. 9.