DOSEN PROFIL LENGKAP

~A^{c}\cup B~} is empty - elements can be only in   A ∖ B   {\displaystyle ~A\setminus B~} . In propositional logic it tells, that   ¬ A ∨ B   {\displaystyle...

B^{c}~} is empty - elements can be only in   B ∖ A   {\displaystyle ~B\setminus A~} . In propositional logic it tells, that   A ∨ ¬ B   {\displaystyle...

DescriptionSetminus included.svg English: Cutting a set B out of A, where B is included in A Date 25 March 2021 Source Own work Author Sascha Lill 95...

DescriptionSetminus infinite.svg English: Cutting two exhaustions of a set Date 25 March 2021 Source Own work Author Sascha Lill 95...

DescriptionSetminus disjoint.svg English: removing a disjoint set B doesn't change A Date 25 March 2021 Source Own work Author Sascha Lill 95...

{\displaystyle \Psi _{c}:\mathbb {\hat {C}} \setminus {\overline {\mathbb {D} }}\to \mathbb {\hat {C}} \setminus Kc} For c = − 2 {\displaystyle c=-2\,} :...

DescriptionSetminus.png English: setminus Date 4 March 2017 Source Own work Author Svjo...

{\displaystyle \Psi _{c}:\mathbb {\hat {C}} \setminus {\overline {\mathbb {D} }}\to \mathbb {\hat {C}} \setminus Kc} For c = − 2 {\displaystyle c=-2\,} :...

DescriptionVenn A setminus B.svg Русский: Изображение содержит диаграмму Эйлера-Венна демонстрирующую разность множеств A и B. Date 27 September 2009 Source...

{\displaystyle \Psi _{c}:\mathbb {\hat {C}} \setminus {\overline {\mathbb {D} }}\to \mathbb {\hat {C}} \setminus Kc} Para c = − 2 {\displaystyle c=-2} : Ψ...

constitui como uma “quebra”de linha para o editor. Neste caso utiliza-se $\setminus$ ou $\backslash$ que produzirão \. 2.5.6 Curiosidades Para produzir...

A_{f}(\infty )} Fatou set F ( f ) = C ∖ J ( f ) {\displaystyle F(f)=\mathbb {C} \setminus J(f)} Fatou set consist of components ( subsets) : attracting components...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

    ⇔ {\displaystyle \Leftrightarrow }     A ∖ B = ∅ {\displaystyle A\setminus B=\emptyset }     ⇔ {\displaystyle \Leftrightarrow }     =        A ⊆ B...

C ∖ { − 1 } {\displaystyle h(z)={\frac {z-1}{z+1}},\;z\in \mathbb {C} \setminus \{-1\}} maps right half plane : { z ∈ C : R e ( z ) ≥ 0 } {\displaystyle...

\Leftrightarrow }   A c ∪ B {\displaystyle ~A^{c}\cup B} =   A c {\displaystyle ~A^{c}} =   B {\displaystyle ~B} =   B ∖ A {\displaystyle ~B\setminus A}...

\Leftrightarrow }   A Δ B {\displaystyle ~A\Delta B} =   A ∖ B {\displaystyle ~A\setminus B} =   B ∖ A {\displaystyle ~B\setminus A} = ∅ {\displaystyle \emptyset }...

    ⇔ {\displaystyle \Leftrightarrow }     B = B ∖ A {\displaystyle B=B\setminus A}     ⇔ {\displaystyle \Leftrightarrow }     =        A ⊆ B c {\displaystyle...

{\displaystyle \Leftrightarrow }     A c = B ∖ A {\displaystyle A^{c}=B\setminus A}     ⇔ {\displaystyle \Leftrightarrow }     =        A c ⊆ B {\displaystyle...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

{\displaystyle \setminus ~} D   {\displaystyle D~}      =   {\displaystyle =~}      D C   {\displaystyle D^{C}~} ∖   {\displaystyle \setminus ~}      =  ...

R ∖ { 0 } → R : x ↦ { 1 x > 0 − 1 x < 0 {\displaystyle f:\mathbb {R} \setminus \{0\}\to \mathbb {R} :x\mapsto {\begin{cases}1&x>0\\-1&x<0\end{cases}}}...

1 3 − x } {\displaystyle \left\{\left(x,y\right)\in \left(\mathbb {R} \setminus \{3\}\times \mathbb {R} \right)~|~y={\frac {(4\,x^{3}-6\,x^{2}+1)\,{\sqrt...

A_{f}(\infty )} Fatou set F ( f ) = C ∖ J ( f ) {\displaystyle F(f)=\mathbb {C} \setminus J(f)} Fatou set consist of components ( subsets) : attracting components...

y(x):={\frac {(4\,x^{3}-6\,x^{2}+1)\,{\sqrt {x+1}}}{3-x}}\right|x\in \mathbb {R} \setminus 3\right\}} for the range x ∈ [ − 1 ; 1 , 5 ] {\displaystyle x\in \left[-1;1...

Essendo il numero di componenti c ( G ∖ < X , Y > ) = 2 {\displaystyle c(G\setminus <X,Y>)=2} allora questo VCP è un ECS, inoltre è un EDS minimo, infatti...

2 3 4 5 {\displaystyle \scriptscriptstyle {\begin{array}{c|ccccc}_{x}\setminus ^{y}&1&2&3&4&5\\\hline 1\\2\\3\\4\\5\\\end{array}}} Each of the four relations...