id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-10300 | Empty Set is Open in Neighborhood Space | Let $\struct {S, \NN}$ be a neighborhood space.
Then the empty set $\O$ is an open set of $\struct {S, \NN}$. | Suppose $\O$ were not an open set of $\struct {S, \NN}$.
Then $\exists x \in \O$ such that $\O$ is not a neighborhood of $\O$.
By definition of empty set, such an $x$ does not exist.
Hence the result.
{{qed}} | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Then the [[Definition:Empty Set|empty set]] $\O$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$. | Suppose $\O$ were not an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$.
Then $\exists x \in \O$ such that $\O$ is not a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $\O$.
By definition of [[Definition:Empty Set|empty set]], such an $x$ does not exist.
Hence the resul... | Empty Set is Open in Neighborhood Space | https://proofwiki.org/wiki/Empty_Set_is_Open_in_Neighborhood_Space | https://proofwiki.org/wiki/Empty_Set_is_Open_in_Neighborhood_Space | [
"Neighborhood Spaces",
"Empty Set",
"Open Sets"
] | [
"Definition:Neighborhood Space",
"Definition:Empty Set",
"Definition:Open Set (Neighborhood Space)"
] | [
"Definition:Open Set (Neighborhood Space)",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Empty Set"
] |
proofwiki-10301 | Whole Space is Open in Neighborhood Space | Let $\struct {S, \NN}$ be a neighborhood space.
Then $S$ itself is an open set of $\struct {S, \NN}$. | Let $x \in S$.
Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$.
As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$.
As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$.
{{qed}} | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Then $S$ itself is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$. | Let $x \in S$.
Then by [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $\text N 1$]] there exists a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] $N$ of $x$.
As $N \subseteq S$ it follows from [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $\text N 3$]] that $S$ is a [[Definiti... | Whole Space is Open in Neighborhood Space | https://proofwiki.org/wiki/Whole_Space_is_Open_in_Neighborhood_Space | https://proofwiki.org/wiki/Whole_Space_is_Open_in_Neighborhood_Space | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Open Set (Neighborhood Space)"
] | [
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)"
] |
proofwiki-10302 | Intersection of two Open Sets of Neighborhood Space is Open | Let $\struct {S, \NN}$ be a neighborhood space.
Let $U$ and $V$ be open sets of $\struct {S, \NN}$.
Then $U \cap V$ is an open set of $\struct {S, \NN}$. | Let $U$ and $V$ be open sets of $\struct {S, \NN}$.
Let $x \in S$ such that $x \in U \cap V$.
Then $x \in U$ and $x \in V$, both of which are neighborhoods of $x$ by definition of open set.
By neighborhood space axiom $(\text N 4)$ it follows that $U \cap V$ is a neighborhood of $x$.
As $x$ is arbitrary, it follows tha... | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $U$ and $V$ be [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$.
Then $U \cap V$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$. | Let $U$ and $V$ be [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$.
Let $x \in S$ such that $x \in U \cap V$.
Then $x \in U$ and $x \in V$, both of which are [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $x$ by definition of [[Definition:Open Set (Neighborhood Space)|o... | Intersection of two Open Sets of Neighborhood Space is Open | https://proofwiki.org/wiki/Intersection_of_two_Open_Sets_of_Neighborhood_Space_is_Open | https://proofwiki.org/wiki/Intersection_of_two_Open_Sets_of_Neighborhood_Space_is_Open | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Open Set (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)"
] | [
"Definition:Open Set (Neighborhood Space)",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)",
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)"
] |
proofwiki-10303 | Union of Open Sets of Neighborhood Space is Open | Let $S$ be a neighborhood space.
Let $I$ be an indexing set.
Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a family of open sets of $\struct {S, \NN}$ indexed by $I$.
Then their union $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\struct {S, \NN}$. | Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
Then $x \in U_\beta$ for some $\beta \in I$.
By definition of open set, $U_\beta$ is a neighborhood of $x$.
But from Set is Subset of Union:
:$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$
By neighborhood space axiom $N 3$ it follows that $... | Let $S$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family]] of [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$ indexed by $I$.
... | Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
Then $x \in U_\beta$ for some $\beta \in I$.
By definition of [[Definition:Open Set (Neighborhood Space)|open set]], $U_\beta$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
But from [[Set is Subset of Union]]:
:$\ds U_\beta \subset... | Union of Open Sets of Neighborhood Space is Open | https://proofwiki.org/wiki/Union_of_Open_Sets_of_Neighborhood_Space_is_Open | https://proofwiki.org/wiki/Union_of_Open_Sets_of_Neighborhood_Space_is_Open | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Open Set (Neighborhood Space)",
"Definition:Set Union/Family of Sets",
"Definition:Open Set (Neighborhood Space)"
] | [
"Definition:Open Set (Neighborhood Space)",
"Definition:Neighborhood (Neighborhood Space)",
"Set is Subset of Union",
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)"
] |
proofwiki-10304 | Neighborhood Space is Topological Space | Let $\struct {S, \NN}$ be a neighborhood space.
Let $\tau = \set {N: N \in \NN}$ be the set of all open sets of $\struct {S, \NN}$.
Then $\struct {S, \tau}$ forms a topological space. | Each of the open set axioms is examined in turn: | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $\tau = \set {N: N \in \NN}$ be the [[Definition:Set|set]] of all [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$.
Then $\struct {S, \tau}$ forms a [[Definition:Topological Space|topological space]]. | Each of the [[Axiom:Open Set Axioms|open set axioms]] is examined in turn: | Neighborhood Space is Topological Space | https://proofwiki.org/wiki/Neighborhood_Space_is_Topological_Space | https://proofwiki.org/wiki/Neighborhood_Space_is_Topological_Space | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Set",
"Definition:Open Set (Neighborhood Space)",
"Definition:Topological Space"
] | [
"Axiom:Open Set Axioms",
"Axiom:Open Set Axioms"
] |
proofwiki-10305 | Subset in Neighborhood Space is Neighborhood iff it contains Open Set | Let $\struct {S, \NN}$ be a neighborhood space.
Let $x \in S$ be a point of $S$.
Let $N \subseteq S$ be a subset of $S$.
Then $N$ is a neighborhood of $x$ {{iff}} there exists an open set $U$ of $\struct {S, \NN}$ such that $x \in U \subseteq N$. | === Necessary Condition ===
Let $N$ be a neighborhood of $x$.
Then by neighborhood space axiom $\text N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.
By neighborhood space axiom $\text N 2$, $x \in U$.
{{qed|lemma}} | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $x \in S$ be a [[Definition:Element|point]] of $S$.
Let $N \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Then $N$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$ {{iff}} there exists an [[Definit... | === Necessary Condition ===
Let $N$ be a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
Then by [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $\text N 5$]], $N$ contains a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] $U$ of $x$ such that $U$ is a [[Definition:Neigh... | Subset in Neighborhood Space is Neighborhood iff it contains Open Set | https://proofwiki.org/wiki/Subset_in_Neighborhood_Space_is_Neighborhood_iff_it_contains_Open_Set | https://proofwiki.org/wiki/Subset_in_Neighborhood_Space_is_Neighborhood_iff_it_contains_Open_Set | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Element",
"Definition:Subset",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Open Set (Neighborhood Space)"
] | [
"Definition:Neighborhood (Neighborhood Space)",
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Element",
"Axiom:Neighborhood Space Axioms",
"Definition:Neighborhood (Neighborhood Space)",
"Axiom:Neighborhoo... |
proofwiki-10306 | Induced Neighborhood Space is Neighborhood Space | Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\struct {S, \tau}$.
Then $\struct {S, \NN}$ is a neighborhood space. | Let $x \in S$.
Let $\NN_x$ be the neighborhood filter of $x$.
From Basic Properties of Neighborhood in Topological Space, $\NN_x$ fulfils the neighborhood space axioms.
Hence the result.
{{qed}}
Category:Neighborhood Spaces
az6sirkh6uxvz9p958oypwr5nuh6on5 | Let $S$ be a [[Definition:Set|set]].
Let $\tau$ be a [[Definition:Topology|topology]] on $S$, thus forming the [[Definition:Topological Space|topological space]] $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\struct {S, \... | Let $x \in S$.
Let $\NN_x$ be the [[Definition:Neighborhood Filter of Point|neighborhood filter of $x$]].
From [[Basic Properties of Neighborhood in Topological Space]], $\NN_x$ fulfils the [[Axiom:Neighborhood Space Axioms|neighborhood space axioms]].
Hence the result.
{{qed}}
[[Category:Neighborhood Spaces]]
az6s... | Induced Neighborhood Space is Neighborhood Space | https://proofwiki.org/wiki/Induced_Neighborhood_Space_is_Neighborhood_Space | https://proofwiki.org/wiki/Induced_Neighborhood_Space_is_Neighborhood_Space | [
"Neighborhood Spaces"
] | [
"Definition:Set",
"Definition:Topology",
"Definition:Topological Space",
"Definition:Neighborhood Space Induced by Topological Space",
"Definition:Neighborhood Space"
] | [
"Definition:Neighborhood Filter/Point",
"Basic Properties of Neighborhood in Topological Space",
"Axiom:Neighborhood Space Axioms",
"Category:Neighborhood Spaces"
] |
proofwiki-10307 | Topological Space Induced by Neighborhood Space is Topological Space | Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\struct {S, \NN}$.
Then $\struct {S, \tau}$ is a topological space. | From Neighborhood Space is Topological Space, $\struct {S, \NN}$ is a topological space.
Consequently, the open sets of $\struct {S, \tau}$ are exactly the open sets of $\struct {S, \NN}$.
{{qed}}
Category:Neighborhood Spaces
lfo4lhob0eucnlcfixr1sexh9omba2s | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $\struct {S, \tau}$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\struct {S, \NN}$]].
Then $\struct {S, \tau}$ is a [[Definition:Topological Space|topological space]]. | From [[Neighborhood Space is Topological Space]], $\struct {S, \NN}$ is a [[Definition:Topological Space|topological space]].
Consequently, the [[Definition:Open Set (Topology)|open sets]] of $\struct {S, \tau}$ are exactly the [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$.
{{qed}}
[[Ca... | Topological Space Induced by Neighborhood Space is Topological Space | https://proofwiki.org/wiki/Topological_Space_Induced_by_Neighborhood_Space_is_Topological_Space | https://proofwiki.org/wiki/Topological_Space_Induced_by_Neighborhood_Space_is_Topological_Space | [
"Neighborhood Spaces"
] | [
"Definition:Neighborhood Space",
"Definition:Topological Space Induced by Neighborhood Space",
"Definition:Topological Space"
] | [
"Neighborhood Space is Topological Space",
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Open Set (Neighborhood Space)",
"Category:Neighborhood Spaces"
] |
proofwiki-10308 | Topological Space induced by Neighborhood Space induced by Topological Space | Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\struct {S, \tau'}$ be the topological space induced by $\NN$ on $S$.
Then $\tau = \tau'$. | Let $U \in \tau$ be an open set of $\struct {S, \tau}$.
By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.
By definition, $U$ is an open set of $\struct {S, \NN}$.
Thus by definition of neighborhood space induced by $\tau$ on $S$:
:$U \in \NN$
Then, by definition of the topo... | Let $S$ be a [[Definition:Set|set]].
Let $\tau$ be a [[Definition:Topology|topology]] on $S$, thus forming the [[Definition:Topological Space|topological space]] $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]... | Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau}$.
By [[Set is Open iff Neighborhood of all its Points]], $U$ is a [[Definition:Neighborhood of Point|neighborhood]] of each of its [[Definition:Element|points]].
By definition, $U$ is an [[Definition:Open Set (Neighborhood Space)... | Topological Space induced by Neighborhood Space induced by Topological Space | https://proofwiki.org/wiki/Topological_Space_induced_by_Neighborhood_Space_induced_by_Topological_Space | https://proofwiki.org/wiki/Topological_Space_induced_by_Neighborhood_Space_induced_by_Topological_Space | [
"Neighborhood Spaces",
"Topology"
] | [
"Definition:Set",
"Definition:Topology",
"Definition:Topological Space",
"Definition:Neighborhood Space Induced by Topological Space",
"Definition:Topological Space Induced by Neighborhood Space"
] | [
"Definition:Open Set/Topology",
"Set is Open iff Neighborhood of all its Points",
"Definition:Neighborhood (Topology)/Point",
"Definition:Element",
"Definition:Open Set (Neighborhood Space)",
"Definition:Neighborhood Space Induced by Topological Space",
"Definition:Topological Space Induced by Neighborh... |
proofwiki-10309 | Correspondence between Neighborhood Space and Topological Space | Let $S$ be a set.
Let $\struct {S, \tau}$ be a topological space.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\phi: \struct {S, \tau} \to \struct {S, \NN}$ be the mapping defined as:
:$\forall x \in S: \map \phi x = x$
:$\forall U \in \tau: \phi \sqbrk U = U \in \NN$
Let $\struct {T,... | From the construction of:
:the neighborhood space induced by $\tau$ on $S$
:the topological space induced by $\NN$ on $S$
the mappings $\phi$ and $\psi$ are well-defined mappings.
From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection.
From Neighborhood Space induced by... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\struct {S, \NN}$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]].
Let $\phi: \struct {S, \tau} \to \struct {S, \NN}$ be the [[D... | From the construction of:
:the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]]
:the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\NN$ on $S$]]
the mappings $\phi$ and $\psi$ are [[Definition:Well-Defined Mapping|we... | Correspondence between Neighborhood Space and Topological Space | https://proofwiki.org/wiki/Correspondence_between_Neighborhood_Space_and_Topological_Space | https://proofwiki.org/wiki/Correspondence_between_Neighborhood_Space_and_Topological_Space | [
"Neighborhood Spaces",
"Topology"
] | [
"Definition:Set",
"Definition:Topological Space",
"Definition:Neighborhood Space Induced by Topological Space",
"Definition:Mapping",
"Definition:Neighborhood Space",
"Definition:Topological Space Induced by Neighborhood Space",
"Definition:Mapping"
] | [
"Definition:Neighborhood Space Induced by Topological Space",
"Definition:Topological Space Induced by Neighborhood Space",
"Definition:Well-Defined/Mapping",
"Topological Space induced by Neighborhood Space induced by Topological Space",
"Definition:Bijection",
"Neighborhood Space induced by Topological ... |
proofwiki-10310 | Neighborhood Space induced by Topological Space induced by Neighborhood Space | Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\NN$ on $S$.
Let $\struct {S, \NN'}$ be the neighborhood space induced by $\tau$ on $S$.
Then $\NN = \NN'$. | Let $x \in S$.
Let $\NN_x$ be the set of all neighborhoods of $x$.
Let $N \in \NN_x$ be a neighborhood of $x$.
From Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is the superset of some open set $U$ in $\struct {S, \NN}$.
By Neighborhood Space is Topological Space we have that $U$ is an ope... | Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
Let $\struct {S, \tau}$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\NN$ on $S$]].
Let $\struct {S, \NN'}$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighbo... | Let $x \in S$.
Let $\NN_x$ be the [[Definition:Set|set]] of all [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $x$.
Let $N \in \NN_x$ be a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
From [[Subset in Neighborhood Space is Neighborhood iff it contains Open Set]], $N$ is t... | Neighborhood Space induced by Topological Space induced by Neighborhood Space | https://proofwiki.org/wiki/Neighborhood_Space_induced_by_Topological_Space_induced_by_Neighborhood_Space | https://proofwiki.org/wiki/Neighborhood_Space_induced_by_Topological_Space_induced_by_Neighborhood_Space | [
"Neighborhood Spaces",
"Topology"
] | [
"Definition:Neighborhood Space",
"Definition:Topological Space Induced by Neighborhood Space",
"Definition:Neighborhood Space Induced by Topological Space"
] | [
"Definition:Set",
"Definition:Neighborhood (Neighborhood Space)",
"Definition:Neighborhood (Neighborhood Space)",
"Subset in Neighborhood Space is Neighborhood iff it contains Open Set",
"Definition:Subset/Superset",
"Definition:Open Set (Neighborhood Space)",
"Neighborhood Space is Topological Space",
... |
proofwiki-10311 | Binomial Coefficient with Zero/Integer Coefficients | :$\forall n \in \N: \dbinom n 0 = 1$
where $\dbinom n 0$ denotes a binomial coefficient. | From the definition:
{{begin-eqn}}
{{eqn | l = \binom n 0
| r = \frac {n!} {0! \ n!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n!} {1 \cdot n!}
| c = {{Defof|Factorial}} of $0$
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
{{qed}} | :$\forall n \in \N: \dbinom n 0 = 1$
where $\dbinom n 0$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From the definition:
{{begin-eqn}}
{{eqn | l = \binom n 0
| r = \frac {n!} {0! \ n!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n!} {1 \cdot n!}
| c = {{Defof|Factorial}} of $0$
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
{{qed}} | Binomial Coefficient with Zero/Integer Coefficients | https://proofwiki.org/wiki/Binomial_Coefficient_with_Zero/Integer_Coefficients | https://proofwiki.org/wiki/Binomial_Coefficient_with_Zero/Integer_Coefficients | [
"Examples of Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [] |
proofwiki-10312 | Form of Geometric Sequence of Integers in Lowest Terms | Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of positive integers only.
Let $r$ be the common ratio of $G_n$.
Let the elements of $G_n$ be the smallest positive integers such that $G_n$ has common ratio $r$.
Then the $j$th term of $G_n$ is given by:
:$a_j ... | From Form of Geometric Sequence of Integers the $j$th term of $G_n$ is given by:
:$a_j = k p^{n - j} q^j$
where the common ratio is $\dfrac q p$.
Thus:
:$a_0 = k p^n$
:$a_n = k q^n$
From Geometric Sequence in Lowest Terms has Coprime Extremes it follows that $k = 1$.
Hence the result.
{{qed}}
Category:Geometric Sequenc... | Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] of [[Definition:Length of Sequence|length]] $n$ consisting of [[Definition:Positive Integer|positive integers]] only.
Let $r$ be the [[Definition:Common Ratio|common ratio]] of $G_n$.
Let the elements... | From [[Form of Geometric Sequence of Integers]] the $j$th [[Definition:Term of Geometric Sequence|term]] of $G_n$ is given by:
:$a_j = k p^{n - j} q^j$
where the [[Definition:Common Ratio|common ratio]] is $\dfrac q p$.
Thus:
:$a_0 = k p^n$
:$a_n = k q^n$
From [[Geometric Sequence in Lowest Terms has Coprime Extremes... | Form of Geometric Sequence of Integers in Lowest Terms | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_in_Lowest_Terms | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_in_Lowest_Terms | [
"Geometric Sequences of Integers"
] | [
"Definition:Geometric Sequence",
"Definition:Length of Sequence",
"Definition:Positive/Integer",
"Definition:Geometric Sequence/Common Ratio",
"Definition:Positive/Integer",
"Definition:Geometric Sequence/Common Ratio",
"Definition:Geometric Sequence/Term"
] | [
"Form of Geometric Sequence of Integers",
"Definition:Geometric Sequence/Term",
"Definition:Geometric Sequence/Common Ratio",
"Geometric Sequence in Lowest Terms has Coprime Extremes",
"Category:Geometric Sequences of Integers"
] |
proofwiki-10313 | Form of Geometric Sequence of Integers with Coprime Extremes | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of positive integers only.
Let $a_1$ and $a_n$ be coprime.
Then the $j$th term of $Q_n$ is given by:
:$a_j = q^j p^{n - j}$ | Let $r$ be the common ratio of $Q_n$.
Let the elements of $Q_n$ be the smallest positive integers such that $Q_n$ has common ratio $r$.
From Geometric Sequence with Coprime Extremes is in Lowest Terms, the elements of $Q_n$ are the smallest positive integers such that $Q_n$ has common ratio $r$.
From Form of Geometric ... | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] of [[Definition:Length of Sequence|length]] $n$ consisting of [[Definition:Positive Integer|positive integers]] only.
Let $a_1$ and $a_n$ be [[Definition:Coprime Integers|coprime]].
Then the $j$th [[... | Let $r$ be the [[Definition:Common Ratio|common ratio]] of $Q_n$.
Let the elements of $Q_n$ be the smallest [[Definition:Positive Integer|positive integers]] such that $Q_n$ has [[Definition:Common Ratio|common ratio]] $r$.
From [[Geometric Sequence with Coprime Extremes is in Lowest Terms]], the elements of $Q_n$ ar... | Form of Geometric Sequence of Integers with Coprime Extremes | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_with_Coprime_Extremes | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_with_Coprime_Extremes | [
"Geometric Sequences of Integers"
] | [
"Definition:Geometric Sequence",
"Definition:Length of Sequence",
"Definition:Positive/Integer",
"Definition:Coprime/Integers",
"Definition:Geometric Sequence/Term"
] | [
"Definition:Geometric Sequence/Common Ratio",
"Definition:Positive/Integer",
"Definition:Geometric Sequence/Common Ratio",
"Geometric Sequence with Coprime Extremes is in Lowest Terms",
"Definition:Positive/Integer",
"Definition:Geometric Sequence/Common Ratio",
"Form of Geometric Sequence of Integers i... |
proofwiki-10314 | Naturally Ordered Semigroup forms Peano Structure | Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $0 \in S$ be the zero of $S$.
Let $1 \in S$ be the one of $S$.
Let $s: S \to S$ be the mapping defined as:
:$\map s n := n \circ 1$
Then $\struct {S, 0, s}$ is a Peano structure. | We verify Peano's axioms in turn.
First, suppose that $\map s m = \map s n$ for some $m, n \in S$.
That is:
:$m \circ 1 = n \circ 1$
By Axiom $(\text {NO} 2)$, it follows that $m = n$.
Hence Axiom $(\text P 3)$ holds.
{{AimForCont}} that $\map s n = 0$ for some $n \in S$.
That is:
:$n \circ 1 = 0$
By Sum with One is Im... | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
Let $0 \in S$ be the [[Definition:Zero of Naturally Ordered Semigroup|zero]] of $S$.
Let $1 \in S$ be the [[Definition:One of Naturally Ordered Semigroup|one]] of $S$.
Let $s: S \to S$ be the [[Definition:M... | We verify [[Axiom:Peano's Axioms|Peano's axioms]] in turn.
First, suppose that $\map s m = \map s n$ for some $m, n \in S$.
That is:
:$m \circ 1 = n \circ 1$
By [[Definition:Naturally Ordered Semigroup|Axiom $(\text {NO} 2)$]], it follows that $m = n$.
Hence [[Axiom:Peano's Axioms|Axiom $(\text P 3)$]] holds.
{... | Naturally Ordered Semigroup forms Peano Structure | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_forms_Peano_Structure | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_forms_Peano_Structure | [
"Naturally Ordered Semigroup",
"Peano's Axioms"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Zero (Number)/Naturally Ordered Semigroup",
"Definition:Unit (One)/Naturally Ordered Semigroup",
"Definition:Mapping",
"Definition:Peano Structure"
] | [
"Axiom:Peano's Axioms",
"Definition:Naturally Ordered Semigroup",
"Axiom:Peano's Axioms",
"Sum with One is Immediate Successor in Naturally Ordered Semigroup",
"Definition:Contradiction",
"Definition:Zero (Number)/Naturally Ordered Semigroup",
"Axiom:Peano's Axioms",
"Definition:Naturally Ordered Semi... |
proofwiki-10315 | Divisibility of Elements in Geometric Sequence of Integers | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers.
Let $j \ne k$.
Then:
:$\paren {\exists j \in \set {0, 1, \ldots, n - 1}: a_j \divides a_{j + 1} } \iff \paren {\forall j, k \in \set {0, 1, \ldots, n}, j < k: a_j \divides a_k}$
where $\divides$ denotes integer divisibility... | Let $a_j \divides a_{j + 1}$ for some $j \in \set {0, 1, \ldots, n - 1}$.
Then by definition of integer divisibility:
:$\exists r \in \Z: r a_j = a_{j + 1}$
Thus the common ratio of $Q_n$ is $r$.
So by definition of geometric sequence:
:$\forall j, k \in \set {0, 1, \ldots, n}, j < k: r^{k - j} a_j = a_k$
and so $a_j \... | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence of Integers|geometric sequence of integers]].
Let $j \ne k$.
Then:
:$\paren {\exists j \in \set {0, 1, \ldots, n - 1}: a_j \divides a_{j + 1} } \iff \paren {\forall j, k \in \set {0, 1, \ldots, n}, j < k: a_j \divides a_k... | Let $a_j \divides a_{j + 1}$ for some $j \in \set {0, 1, \ldots, n - 1}$.
Then by definition of [[Definition:Divisor of Integer|integer divisibility]]:
:$\exists r \in \Z: r a_j = a_{j + 1}$
Thus the [[Definition:Common Ratio|common ratio]] of $Q_n$ is $r$.
So by definition of [[Definition:Geometric Sequence|geometr... | Divisibility of Elements in Geometric Sequence of Integers | https://proofwiki.org/wiki/Divisibility_of_Elements_in_Geometric_Sequence_of_Integers | https://proofwiki.org/wiki/Divisibility_of_Elements_in_Geometric_Sequence_of_Integers | [
"Geometric Sequences of Integers"
] | [
"Definition:Geometric Sequence/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Geometric Sequence/Term",
"Definition:Geometric Sequence/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Geometric Sequence/Term",
"Definition:Geometric Sequence/Term",
"Definition:Divisor (Alg... | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Geometric Sequence/Common Ratio",
"Definition:Geometric Sequence",
"Definition:Converse Statement",
"Category:Geometric Sequences of Integers"
] |
proofwiki-10316 | Form of Geometric Sequence of Integers from One | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence consisting of integers only.
Let $a_0 = 1$.
Then the $j$th term of $Q_n$ is given by:
:$a_j = a^j$
where:
:the common ratio of $Q_n$ is $a$
:$a = a_1$.
Thus:
:$Q_n = \tuple {1, a, a^2, \ldots, a^n}$ | From Form of Geometric Sequence of Integers, the $j$th term of $Q_n$ is given by:
:$a_j = k q^j p^{n - j}$
where:
:the common ratio of $Q_n$ expressed in canonical form is $\dfrac q p$
:$k$ is an integer.
As $a_0 = 1$ it follows that:
:$1 = k p^{n - j}$
from which it follows that:
:$k = 1$
:$p = 1$
and the common ratio... | Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] consisting of [[Definition:Integer|integers]] only.
Let $a_0 = 1$.
Then the $j$th [[Definition:Term of Geometric Sequence|term]] of $Q_n$ is given by:
:$a_j = a^j$
where:
:the [[Definition:Common Ra... | From [[Form of Geometric Sequence of Integers]], the $j$th [[Definition:Term of Geometric Sequence|term]] of $Q_n$ is given by:
:$a_j = k q^j p^{n - j}$
where:
:the [[Definition:Common Ratio|common ratio]] of $Q_n$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]] is $\dfrac q p$
:$k$ is an ... | Form of Geometric Sequence of Integers from One | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_from_One | https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers_from_One | [
"Geometric Sequences of Integers"
] | [
"Definition:Geometric Sequence",
"Definition:Integer",
"Definition:Geometric Sequence/Term",
"Definition:Geometric Sequence/Common Ratio"
] | [
"Form of Geometric Sequence of Integers",
"Definition:Geometric Sequence/Term",
"Definition:Geometric Sequence/Common Ratio",
"Definition:Rational Number/Canonical Form",
"Definition:Integer",
"Definition:Geometric Sequence/Common Ratio",
"Category:Geometric Sequences of Integers"
] |
proofwiki-10317 | Divisors of Power of Prime | Let $p$ be a prime number.
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Then the only divisors of $p^n$ are $1, p, p^2, \ldots, p^{n - 1}, p^n$. | First it is necessary to establish that every element of the set $\set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ is in fact a divisor of $p^n$.
For any $j \in \set {1, 2, \ldots, n}$:
:$p^n = p^j p^{n - j}$
and so each of $1, p, p^2, \ldots, p^{n - 1}, p^n$ is a divisor of $p^n$.
{{qed|lemma}}
Let:
:$a \in \Z_{>0}: a \notin... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Then the only [[Definition:Divisor of Integer|divisors]] of $p^n$ are $1, p, p^2, \ldots, p^{n - 1}, p^n$. | First it is necessary to establish that every [[Definition:Element|element]] of the [[Definition:Set|set]] $\set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ is in fact a [[Definition:Divisor of Integer|divisor]] of $p^n$.
For any $j \in \set {1, 2, \ldots, n}$:
:$p^n = p^j p^{n - j}$
and so each of $1, p, p^2, \ldots, p^{n -... | Divisors of Power of Prime | https://proofwiki.org/wiki/Divisors_of_Power_of_Prime | https://proofwiki.org/wiki/Divisors_of_Power_of_Prime | [
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Strictly Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Element",
"Definition:Set",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Set",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Divisor Relation is Transitive",
"Euclid's Lemma for... |
proofwiki-10318 | Chebyshev Distance is Limit of P-Product Metric | Let $M_{1'} = \struct {A_{1'}, d_{1'} }$ and $M_{2'} = \struct {A_{2'}, d_{2'} }$ be metric spaces.
Let $\AA = A_{1'} \times A_{2'}$ be the cartesian product of $A_{1'}$ and $A_{2'}$.
Let $p \in \R_{\ge 1}$.
Let $d_p: \AA \times \AA \to \R$ be the $p$-product metric on $\AA$:
:$\map {d_p} {x, y} := \paren {\paren {\map... | Let $x$ and $y$ be arbitrary.
Let $a = \map {d_{1'} } {x_1, y_1}, b = \map {d_{2'} } {x_2, y_2}$.
{{WLOG}}, suppose that $\max \set {a, b} = a$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{p \mathop \to \infty} \paren {a^p + b^p}^{1/p}
| o = \ge
| r = \lim_{p \mathop \to \infty} \paren {a^p}^{1/p}
| c =
}}... | Let $M_{1'} = \struct {A_{1'}, d_{1'} }$ and $M_{2'} = \struct {A_{2'}, d_{2'} }$ be [[Definition:Metric Space|metric spaces]].
Let $\AA = A_{1'} \times A_{2'}$ be the [[Definition:Cartesian Product|cartesian product]] of $A_{1'}$ and $A_{2'}$.
Let $p \in \R_{\ge 1}$.
Let $d_p: \AA \times \AA \to \R$ be the [[Defin... | Let $x$ and $y$ be arbitrary.
Let $a = \map {d_{1'} } {x_1, y_1}, b = \map {d_{2'} } {x_2, y_2}$.
{{WLOG}}, suppose that $\max \set {a, b} = a$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{p \mathop \to \infty} \paren {a^p + b^p}^{1/p}
| o = \ge
| r = \lim_{p \mathop \to \infty} \paren {a^p}^{1/p}
| c = ... | Chebyshev Distance is Limit of P-Product Metric | https://proofwiki.org/wiki/Chebyshev_Distance_is_Limit_of_P-Product_Metric | https://proofwiki.org/wiki/Chebyshev_Distance_is_Limit_of_P-Product_Metric | [
"Chebyshev Distance",
"P-Product Metrics"
] | [
"Definition:Metric Space",
"Definition:Cartesian Product",
"Definition:P-Product Metric",
"Definition:Chebyshev Distance"
] | [
"Squeeze Theorem/Functions",
"Category:Chebyshev Distance",
"Category:P-Product Metrics"
] |
proofwiki-10319 | Bertrand's Theorem | Let $U: \R_{>0} \to \R$ be analytic for $r > 0$.
Let $M > 0$ be a nonvanishing angular momentum such that a stable circular orbit exists.
Suppose that every orbit sufficiently close to the circular orbit is closed.
Then $U$ is either $k r^2$ or $-\dfrac k r$ (for $k > 0$) up to an additive constant. | The substitution $x = M / r$ gives:
:$\ds \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac {\d x} {\sqrt {E - \map W x}}$
where $\map W x \equiv \map U {\dfrac M x} + \dfrac 1 2 x^2$.
In general, the frequency of oscillations around a stable equilibrium at $x = x_0$ for a particle of mass $m$ in a potential $V$ is given by:... | Let $U: \R_{>0} \to \R$ be [[Definition:Analytic Function|analytic]] for $r > 0$.
Let $M > 0$ be a [[Definition:Non-Vanishing|nonvanishing]] [[Definition:Angular Momentum|angular momentum]] such that a [[Definition:Stable Orbit|stable]] [[Definition:Circular Orbit|circular]] [[Definition:Orbit (Phase Space)|orbit]] ex... | The substitution $x = M / r$ gives:
:$\ds \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac {\d x} {\sqrt {E - \map W x}}$
where $\map W x \equiv \map U {\dfrac M x} + \dfrac 1 2 x^2$.
In general, the frequency of oscillations around a stable equilibrium at $x = x_0$ for a particle of mass $m$ in a potential $V$ is given by... | Bertrand's Theorem/Asymptotic Proof | https://proofwiki.org/wiki/Bertrand's_Theorem | https://proofwiki.org/wiki/Bertrand's_Theorem/Asymptotic_Proof | [
"Classical Mechanics",
"Bertrand's Theorem"
] | [
"Definition:Analytic Function",
"Definition:Non-Vanishing",
"Definition:Angular Momentum",
"Definition:Stable Orbit",
"Definition:Circular Orbit",
"Definition:Orbit (Phase Space)",
"Definition:Orbit (Phase Space)",
"Definition:Circular Orbit",
"Definition:Closed Orbit",
"Definition:Additive Consta... | [
"Definition:Constant"
] |
proofwiki-10320 | Bertrand's Theorem | Let $U: \R_{>0} \to \R$ be analytic for $r > 0$.
Let $M > 0$ be a nonvanishing angular momentum such that a stable circular orbit exists.
Suppose that every orbit sufficiently close to the circular orbit is closed.
Then $U$ is either $k r^2$ or $-\dfrac k r$ (for $k > 0$) up to an additive constant. | In general $U_M$ is not monotonic on $\openint {r_\min} {r_\max}$.
{{explain|Link to a proof of that statement.}}
Therefore a unique inverse $\map r {U_M}$ does not exist.
However, suppose it is possible to construct separate inverse functions $r_{1, 2}$ for the intervals $\openint {r_0} {r_\min}$ and $\openint {r_0} {... | Let $U: \R_{>0} \to \R$ be [[Definition:Analytic Function|analytic]] for $r > 0$.
Let $M > 0$ be a [[Definition:Non-Vanishing|nonvanishing]] [[Definition:Angular Momentum|angular momentum]] such that a [[Definition:Stable Orbit|stable]] [[Definition:Circular Orbit|circular]] [[Definition:Orbit (Phase Space)|orbit]] ex... | In general $U_M$ is not [[Definition:Monotonic Real Function|monotonic]] on $\openint {r_\min} {r_\max}$.
{{explain|Link to a proof of that statement.}}
Therefore a unique [[Definition:Inverse Function|inverse]] $\map r {U_M}$ does not exist.
However, suppose it is possible to construct separate [[Definition:Inverse ... | Bertrand's Theorem/Non-Perturbative Proof | https://proofwiki.org/wiki/Bertrand's_Theorem | https://proofwiki.org/wiki/Bertrand's_Theorem/Non-Perturbative_Proof | [
"Classical Mechanics",
"Bertrand's Theorem"
] | [
"Definition:Analytic Function",
"Definition:Non-Vanishing",
"Definition:Angular Momentum",
"Definition:Stable Orbit",
"Definition:Circular Orbit",
"Definition:Orbit (Phase Space)",
"Definition:Orbit (Phase Space)",
"Definition:Circular Orbit",
"Definition:Closed Orbit",
"Definition:Additive Consta... | [
"Definition:Monotone (Order Theory)/Real Function",
"Definition:Inverse Function",
"Definition:Inverse Function",
"Definition:Minimum Value of Real Function/Absolute",
"Definition:Orbit (Phase Space)",
"Definition:Stable Orbit",
"Definition:Circular Orbit",
"Definition:Orbit (Phase Space)",
"Definit... |
proofwiki-10321 | Cube Root of 2 is Irrational | :$\sqrt [3] 2$ is irrational. | {{AimForCont}} that $\sqrt [3] 2$ is rational.
Then:
{{begin-eqn}}
{{eqn | l = \sqrt [3] 2
| r = \frac p q
| c = for some integer $p$ and $q$
}}
{{eqn | ll= \leadsto
| l = p^3
| r = q^3 + q^3
| c =
}}
{{end-eqn}}
which contradicts Fermat's Last Theorem.
{{qed}}
Category:Number Theory
nr8x... | :$\sqrt [3] 2$ is [[Definition:Irrational Number|irrational]]. | {{AimForCont}} that $\sqrt [3] 2$ is [[Definition:Rational Number|rational]].
Then:
{{begin-eqn}}
{{eqn | l = \sqrt [3] 2
| r = \frac p q
| c = for some [[Definition:Integer|integer]] $p$ and $q$
}}
{{eqn | ll= \leadsto
| l = p^3
| r = q^3 + q^3
| c =
}}
{{end-eqn}}
which contradicts [... | Cube Root of 2 is Irrational | https://proofwiki.org/wiki/Cube_Root_of_2_is_Irrational | https://proofwiki.org/wiki/Cube_Root_of_2_is_Irrational | [
"Number Theory"
] | [
"Definition:Irrational Number"
] | [
"Definition:Rational Number",
"Definition:Integer",
"Fermat's Last Theorem",
"Category:Number Theory"
] |
proofwiki-10322 | Piecewise Continuously Differentiable Function/Definition 2 is Continuous | Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ satisfy the definition Piecewise Continuously Differentiable Function/Closed Intervals.
Then $f$ is continuous. | $f$ satisfies the conditions of Piecewise Continuously Differentiable Function/Closed Intervals.
Therefore, there exists a finite subdivision $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that $f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, ... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ satisfy the definition [[Definition:Piecewise Continuously Differentiable Function/Closed Intervals|Piecewise Continuously Differentiable Function/Closed Intervals]].
The... | $f$ satisfies the conditions of [[Definition:Piecewise Continuously Differentiable Function/Closed Intervals|Piecewise Continuously Differentiable Function/Closed Intervals]].
Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_... | Piecewise Continuously Differentiable Function/Definition 2 is Continuous | https://proofwiki.org/wiki/Piecewise_Continuously_Differentiable_Function/Definition_2_is_Continuous | https://proofwiki.org/wiki/Piecewise_Continuously_Differentiable_Function/Definition_2_is_Continuous | [
"Real Analysis"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuously Differentiable Function/Closed Intervals",
"Definition:Continuous Real Function"
] | [
"Definition:Piecewise Continuously Differentiable Function/Closed Intervals",
"Definition:Subdivision of Interval/Finite",
"Definition:Continuously Differentiable/Real Function",
"Definition:Derivative/Real Function",
"Definition:One-Sided Derivative",
"Definition:Differentiable Mapping/Real Function",
... |
proofwiki-10323 | Homomorphism of Chain Complexes induces Homomorphism of Homology | Let $A_\bullet$ and $B_\bullet$ be chain complexes of abelian groups.
Let $f: A_\bullet \to B_\bullet$ be a homomorphism.
Then for every $n$, $f$ induces a morphism $\map {H_n} {A_\bullet} \to \map {H_n} {B_\bullet}$ of homology groups.
{{explain|Domain of $n$}} | Let $\partial^A_\bullet$, $\partial^B_\bullet$ denote the differential on $A_\bullet$, respectively $B_\bullet$.
First it will be demonstrated that:
:$\forall a \in \map \ker {\partial^A_n} \subseteq A_n: \map {f_n} a \in \map \ker {\partial^B_n}$
Thus:
:$\partial^B_n \map {f_n} a = \map {f_{n - 1} } {\partial^A_n a} ... | Let $A_\bullet$ and $B_\bullet$ be [[Definition:Complex (Homological Algebra)|chain complexes]] of abelian groups.
Let $f: A_\bullet \to B_\bullet$ be a [[Definition:Homomorphism of Complexes|homomorphism]].
Then for every $n$, $f$ induces a morphism $\map {H_n} {A_\bullet} \to \map {H_n} {B_\bullet}$ of [[Definitio... | Let $\partial^A_\bullet$, $\partial^B_\bullet$ denote the differential on $A_\bullet$, respectively $B_\bullet$.
First it will be demonstrated that:
:$\forall a \in \map \ker {\partial^A_n} \subseteq A_n: \map {f_n} a \in \map \ker {\partial^B_n}$
Thus:
:$\partial^B_n \map {f_n} a = \map {f_{n - 1} } {\partial^A_n a... | Homomorphism of Chain Complexes induces Homomorphism of Homology | https://proofwiki.org/wiki/Homomorphism_of_Chain_Complexes_induces_Homomorphism_of_Homology | https://proofwiki.org/wiki/Homomorphism_of_Chain_Complexes_induces_Homomorphism_of_Homology | [
"Homological Algebra"
] | [
"Definition:Null Sequence (Homological Algebra)",
"Definition:Homomorphism of Complexes",
"Definition:Homology of Chain Complex"
] | [
"Definition:Quotient Mapping",
"Category:Homological Algebra"
] |
proofwiki-10324 | Homotopic Chain Maps Induce Equal Maps on Homology | Let $A_\bullet$, $B_\bullet$ be chain complexes of abelian groups.
Let $f, g: A_\bullet \to B_\bullet$ be chain maps which are homotopic.
Then $f$ and $g$ induce equal maps on homology. | Let $\partial^A_\bullet, \partial^B_\bullet$ be the differentials on $A_\bullet$ and $B_{\bullet}$ respectively.
Let $h$ be a homotopy between $f$ and $g$.
Let:
:$a \in \map {H_n} A \cong \map \ker {\partial^A_n} / \Img {\partial^A_{n + 1} }$
There exists $\tilde a \in \map \ker {\partial^A_n}$ representing $a$.
It is... | Let $A_\bullet$, $B_\bullet$ be [[Definition:Complex (Homological Algebra)|chain complexes]] of abelian groups.
Let $f, g: A_\bullet \to B_\bullet$ be [[Definition:Homomorphism of Complexes|chain maps]] which are [[Definition:Homotopic (Homological Algebra)|homotopic]].
Then $f$ and $g$ [[Homomorphism of Chain Compl... | Let $\partial^A_\bullet, \partial^B_\bullet$ be the differentials on $A_\bullet$ and $B_{\bullet}$ respectively.
Let $h$ be a homotopy between $f$ and $g$.
Let:
:$a \in \map {H_n} A \cong \map \ker {\partial^A_n} / \Img {\partial^A_{n + 1} }$
There exists $\tilde a \in \map \ker {\partial^A_n}$ representing $a$.
I... | Homotopic Chain Maps Induce Equal Maps on Homology | https://proofwiki.org/wiki/Homotopic_Chain_Maps_Induce_Equal_Maps_on_Homology | https://proofwiki.org/wiki/Homotopic_Chain_Maps_Induce_Equal_Maps_on_Homology | [
"Homological Algebra"
] | [
"Definition:Null Sequence (Homological Algebra)",
"Definition:Homomorphism of Complexes",
"Definition:Homotopic (Homological Algebra)",
"Homomorphism of Chain Complexes induces Homomorphism of Homology"
] | [
"Category:Homological Algebra"
] |
proofwiki-10325 | Equivalence of Definitions of Piecewise Continuously Differentiable Function | A function satisfying Piecewise Continuously Differentiable Function With Closed Intervals is equivalent to being continuous and satisfying Piecewise Continuously Differentiable Function With One-Sided Limits.
Let $f$ be a real function defined on a closed interval $\closedint a b$. | === Piecewise Continuously Differentiable Function With Closed Intervals implies continuity and Piecewise Continuously Differentiable Function With One-Sided Limits ===
Assume that $f$ satisfies the requirement of Piecewise Continuously Differentiable Function With Closed Intervals.
We need to prove that $f$ satisfies ... | A [[Definition:Real Function|function]] satisfying [[Definition:Piecewise Continuously Differentiable Function/Closed Intervals|Piecewise Continuously Differentiable Function With Closed Intervals]] is [[Definition:Logical Equivalence|equivalent]] to being [[Definition:Continuous Real Function|continuous]] and satisfyi... | === Piecewise Continuously Differentiable Function With Closed Intervals implies continuity and Piecewise Continuously Differentiable Function With One-Sided Limits ===
Assume that $f$ satisfies the requirement of [[Definition:Piecewise Continuously Differentiable Function/Closed Intervals|Piecewise Continuously Diffe... | Equivalence of Definitions of Piecewise Continuously Differentiable Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Piecewise_Continuously_Differentiable_Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Piecewise_Continuously_Differentiable_Function | [
"Piecewise Continuously Differentiable Functions"
] | [
"Definition:Real Function",
"Definition:Piecewise Continuously Differentiable Function/Closed Intervals",
"Definition:Logical Equivalence",
"Definition:Continuous Real Function",
"Definition:Piecewise Continuously Differentiable Function/One-Sided Limits",
"Definition:Real Function",
"Definition:Real In... | [
"Definition:Piecewise Continuously Differentiable Function/Closed Intervals",
"Definition:Piecewise Continuously Differentiable Function/One-Sided Limits",
"Definition:Continuous Real Function",
"Definition:Continuous Real Function",
"Piecewise Continuously Differentiable Function/Definition 2 is Continuous... |
proofwiki-10326 | Bounded Function Continuous on Open Interval is Darboux Integrable | Let $f$ be a real function defined on an interval $\closedint a b$ such that $a < b$.
Let $f$ be continuous on $\openint a b$.
Let $f$ be bounded on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$. | By Condition for Darboux Integrability, it suffices to show that, for a given strictly positive $\epsilon$, a subdivision $S$ of $\closedint a b$ exists such that:
:$\map U S - \map L S < \epsilon$
where $\map U S$ and $\map L S$ are respectively the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ {{... | Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $\closedint a b$ such that $a < b$.
Let $f$ be [[Definition:Continuous Real Function on Interval|continuous]] on $\openint a b$.
Let $f$ be [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b... | By [[Condition for Darboux Integrability]], it suffices to show that, for a given [[Definition:Strictly Positive Real Number|strictly positive]] $\epsilon$, a [[Definition:Subdivision of Interval|subdivision]] $S$ of $\closedint a b$ exists such that:
:$\map U S - \map L S < \epsilon$
where $\map U S$ and $\map L S$ ... | Bounded Function Continuous on Open Interval is Darboux Integrable | https://proofwiki.org/wiki/Bounded_Function_Continuous_on_Open_Interval_is_Darboux_Integrable | https://proofwiki.org/wiki/Bounded_Function_Continuous_on_Open_Interval_is_Darboux_Integrable | [
"Integral Calculus",
"Bounded Real-Valued Functions",
"Continuous Real Functions",
"Darboux Integrable Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval",
"Definition:Continuous Real Function/Interval",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Darboux Integrable Function"
] | [
"Condition for Darboux Integrability",
"Definition:Strictly Positive/Real Number",
"Definition:Subdivision of Interval",
"Definition:Upper Darboux Sum",
"Definition:Lower Darboux Sum",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Strictly Positive/Real Number",
"Definition:Bound of Real-Value... |
proofwiki-10327 | Square Inscribed in Circle is greater than Half Area of Circle | A square inscribed in a circle has an area greater than half that of the circle. | :300px
Let $ABCD$ be a square inscribed in a circle.
Let $EFGH$ be a square circumscribed around the same circle.
We have that:
:$ABCD$ is twice the area of the triangle $ADB$.
:$EFGH$ is twice the area of the rectangle $EFBD$.
From Area of Rectangle, the area of $EFBD$ is $ED \cdot DB$.
From Area of Triangle in Terms ... | A [[Definition:Square (Geometry)|square]] [[Definition:Polygon Inscribed in Circle|inscribed]] in a [[Definition:Circle|circle]] has an [[Definition:Area|area]] greater than half that of the [[Definition:Circle|circle]]. | :[[File:InscribedCircumscribedSquare.png|300px]]
Let $ABCD$ be a [[Definition:Square (Geometry)|square]] [[Definition:Polygon Inscribed in Circle|inscribed]] in a [[Definition:Circle|circle]].
Let $EFGH$ be a [[Definition:Square (Geometry)|square]] [[Definition:Polygon Circumscribed around Circle|circumscribed]] aro... | Square Inscribed in Circle is greater than Half Area of Circle | https://proofwiki.org/wiki/Square_Inscribed_in_Circle_is_greater_than_Half_Area_of_Circle | https://proofwiki.org/wiki/Square_Inscribed_in_Circle_is_greater_than_Half_Area_of_Circle | [
"Circles",
"Squares"
] | [
"Definition:Quadrilateral/Square",
"Definition:Inscribe/Polygon in Circle",
"Definition:Circle",
"Definition:Area",
"Definition:Circle"
] | [
"File:InscribedCircumscribedSquare.png",
"Definition:Quadrilateral/Square",
"Definition:Inscribe/Polygon in Circle",
"Definition:Circle",
"Definition:Quadrilateral/Square",
"Definition:Circumscribe/Polygon around Circle",
"Definition:Circle",
"Definition:Area",
"Definition:Triangle (Geometry)",
"D... |
proofwiki-10328 | Bounded Piecewise Continuous Function is Darboux Integrable | Let $f$ be a real function defined on the closed interval $\closedint a b$.
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$. | We are given that $f$ is piecewise continuous and bounded on $\closedint a b$.
Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:$f$ is bou... | Let $f$ be a [[Definition:Real Function|real function]] defined on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
Then $f$ is [[Definition:Darboux Integrable Function|Darbo... | We are given that $f$ is [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in ... | Bounded Piecewise Continuous Function is Darboux Integrable | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_is_Darboux_Integrable | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_is_Darboux_Integrable | [
"Piecewise Continuous Functions",
"Integral Calculus",
"Darboux Integrable Functions",
"Proofs by Induction"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/Bounded",
"Definition:Darboux Integrable Function"
] | [
"Definition:Piecewise Continuous Function/Bounded",
"Definition:Subdivision of Interval/Finite",
"Definition:Continuous Real Function/Open Interval",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Real Interval/Open",
"Definition:Subdivision of Interval/Finite",
"Principle of Mathematical Inducti... |
proofwiki-10329 | Bounded Piecewise Continuous Function may not have One-Sided Limits | Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.
Let $f$ be a bounded piecewise continuous function.
{{:Definition:Bounded Piecewise Continuous Function}}
Then it is not necessarily the case that $f$ is a piecewise continuous function with one-sided limits:
{{:Definition:Piecewise Con... | Consider the function:
:$\map f x = \begin{cases}
0 & : x = a \\
\map \sin {\dfrac 1 {x - a} } & : x \in \hointr a b
\end{cases}$
Consider the (finite) subdivision $\set {a, b}$ of $\closedint a b$.
We observe that $\map \sin {\dfrac 1 {x - a} }$ is continuous on $\openint a b$.
Since $\map f x = \map \sin {\dfrac 1 {x... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, $a < b$.
Let $f$ be a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]].
{{:Definition:Bounded Piecewise Continuous Function}}
Then it i... | Consider the [[Definition:Real Function|function]]:
:$\map f x = \begin{cases}
0 & : x = a \\
\map \sin {\dfrac 1 {x - a} } & : x \in \hointr a b
\end{cases}$
Consider the [[Definition:Finite Subdivision|(finite) subdivision]] $\set {a, b}$ of $\closedint a b$.
We observe that $\map \sin {\dfrac 1 {x - a} }$ is [[De... | Bounded Piecewise Continuous Function may not have One-Sided Limits | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_may_not_have_One-Sided_Limits | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_may_not_have_One-Sided_Limits | [
"Piecewise Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/Bounded",
"Definition:Piecewise Continuous Function/One-Sided Limits"
] | [
"Definition:Real Function",
"Definition:Subdivision of Interval/Finite",
"Definition:Continuous Real Function/Open Interval",
"Definition:Continuous Real Function/Open Interval",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Bound of Real-Valued Function",
"Definition:Piecewise Continuous Functi... |
proofwiki-10330 | Rational Numbers form Subset of Real Numbers | The set $\Q$ of rational numbers forms a subset of the real numbers $\R$. | Let $x \in \Q$, where $\Q$ denotes the set of rational numbers.
Consider the rational sequence:
:$x, x, x, \ldots$
This sequence is trivially Cauchy.
Thus there exists a Cauchy sequence $\eqclass {\sequence {x_n} } {}$ which is identified with a rational number $x \in \Q$ such that:
So by the definition of a real numbe... | The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] forms a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] $\R$. | Let $x \in \Q$, where $\Q$ denotes the [[Definition:Rational Number|set of rational numbers]].
Consider the [[Definition:Rational Sequence|rational sequence]]:
:$x, x, x, \ldots$
This [[Definition:Rational Sequence|sequence]] is trivially [[Definition:Cauchy Sequence|Cauchy]].
Thus there exists a [[Definition:Cauch... | Rational Numbers form Subset of Real Numbers | https://proofwiki.org/wiki/Rational_Numbers_form_Subset_of_Real_Numbers | https://proofwiki.org/wiki/Rational_Numbers_form_Subset_of_Real_Numbers | [
"Rational Numbers",
"Real Numbers"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Subset",
"Definition:Real Number"
] | [
"Definition:Rational Number",
"Definition:Rational Sequence",
"Definition:Rational Sequence",
"Definition:Cauchy Sequence",
"Definition:Cauchy Sequence",
"Definition:Rational Number",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Subset"
] |
proofwiki-10331 | Real Number is not necessarily Rational Number | Let $x$ be a real number.
Then it is not necessarily the case that $x$ is also a rational number. | By Proof by Counterexample:
Let $x = \sqrt 2$.
From Square Root of 2 is Irrational:
:$\sqrt 2$ is an irrational number.
By definition:
:$x \in \R \setminus \Q$
where:
:$\R$ is the set of real numbers
:$\Q$ is the set of rational numbers
:$\setminus$ denotes the set difference.
Thus $x$, while being a real number, is no... | Let $x$ be a [[Definition:Real Number|real number]].
Then it is not necessarily the case that $x$ is also a [[Definition:Rational Number|rational number]]. | By [[Proof by Counterexample]]:
Let $x = \sqrt 2$.
From [[Square Root of 2 is Irrational]]:
:$\sqrt 2$ is an [[Definition:Irrational Number|irrational number]].
By definition:
:$x \in \R \setminus \Q$
where:
:$\R$ is the [[Definition:Real Number|set of real numbers]]
:$\Q$ is the [[Definition:Rational Number|set of ... | Real Number is not necessarily Rational Number | https://proofwiki.org/wiki/Real_Number_is_not_necessarily_Rational_Number | https://proofwiki.org/wiki/Real_Number_is_not_necessarily_Rational_Number | [
"Real Numbers",
"Rational Numbers"
] | [
"Definition:Real Number",
"Definition:Rational Number"
] | [
"Proof by Counterexample",
"Square Root of 2 is Irrational",
"Definition:Irrational Number",
"Definition:Real Number",
"Definition:Rational Number",
"Definition:Set Difference",
"Definition:Real Number",
"Definition:Rational Number"
] |
proofwiki-10332 | Zero is both Positive and Negative | The number $0$ (zero) is the only (real) number which is both:
:a positive (real) number
and
:a negative (real) number. | Let $x$ be a real number which is both positive and negative.
Thus:
:$x \in \set {x \in \R: x \ge 0}$
and:
:$x \in \set {x \in \R: x \le 0}$
and so:
:$0 \le x \le 0$
from which:
:$x = 0$
{{qed}} | The [[Definition:Number|number]] $0$ ([[Definition:Zero (Number)|zero]]) is the only [[Definition:Real Number|(real) number]] which is both:
:a [[Definition:Positive Real Number|positive (real) number]]
and
:a [[Definition:Negative Real Number|negative (real) number]]. | Let $x$ be a [[Definition:Real Number|real number]] which is both [[Definition:Positive Real Number|positive]] and [[Definition:Negative Real Number|negative]].
Thus:
:$x \in \set {x \in \R: x \ge 0}$
and:
:$x \in \set {x \in \R: x \le 0}$
and so:
:$0 \le x \le 0$
from which:
:$x = 0$
{{qed}} | Zero is both Positive and Negative | https://proofwiki.org/wiki/Zero_is_both_Positive_and_Negative | https://proofwiki.org/wiki/Zero_is_both_Positive_and_Negative | [
"Numbers"
] | [
"Definition:Number",
"Definition:Zero (Number)",
"Definition:Real Number",
"Definition:Positive/Real Number",
"Definition:Negative/Real Number"
] | [
"Definition:Real Number",
"Definition:Positive/Real Number",
"Definition:Negative/Real Number"
] |
proofwiki-10333 | Binary Operation on Subset is Binary Operation | Let $S$ be a set.
Let $\circ$ be a binary operation on $S$.
Let $T \subseteq S$.
Let $\circ {\restriction}_T$ be the restriction of $\circ$ to $T$.
Then $\circ {\restriction}_T$ is a binary operation on $T$. | Let $\Bbb U$ be a universal set.
Let $\circ: S \times S \to \Bbb U$ be a binary operation on $S$.
Let $T \subseteq S$.
Let $\tuple {a, b} \in T \times T$.
By definition of ordered pair and cartesian product:
:$a \in T$ and $b \in T$
As $T \subseteq S$, it follows that:
:$a \in S$ and $b \in S$
Thus:
:$\tuple {a, b} \in... | Let $S$ be a [[Definition:Set|set]].
Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on $S$.
Let $T \subseteq S$.
Let $\circ {\restriction}_T$ be the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $T$.
Then $\circ {\restriction}_T$ is a [[Definition:Binary Operation|binary ope... | Let $\Bbb U$ be a [[Definition:Universal Set|universal set]].
Let $\circ: S \times S \to \Bbb U$ be a [[Definition:Binary Operation|binary operation]] on $S$.
Let $T \subseteq S$.
Let $\tuple {a, b} \in T \times T$.
By definition of [[Definition:Ordered Pair|ordered pair]] and [[Definition:Cartesian Product|cartesi... | Binary Operation on Subset is Binary Operation | https://proofwiki.org/wiki/Binary_Operation_on_Subset_is_Binary_Operation | https://proofwiki.org/wiki/Binary_Operation_on_Subset_is_Binary_Operation | [
"Binary Operations"
] | [
"Definition:Set",
"Definition:Operation/Binary Operation",
"Definition:Restriction/Operation",
"Definition:Operation/Binary Operation"
] | [
"Definition:Universal Set",
"Definition:Operation/Binary Operation",
"Definition:Ordered Pair",
"Definition:Cartesian Product",
"Definition:Operation/Binary Operation",
"Definition:Restriction/Operation",
"Definition:Operation/Binary Operation",
"Category:Binary Operations"
] |
proofwiki-10334 | Matrix Entrywise Addition over Ring is Associative | Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is associative on $\map {\MM_R} {m, n}$.
That... | Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be elements of the $m \times n$ matrix space over $R$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C
| r = \paren {\sqbrk a_{m n} + \sqbrk b_{m n} } + \sqbrk c_{m n}
| c = Definition ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition over Rin... | Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be [[Definition:Element|elements]] of the [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C
| r = \paren {\sqbrk a_{m n} + \sqb... | Matrix Entrywise Addition over Ring is Associative/Proof 1 | https://proofwiki.org/wiki/Matrix_Entrywise_Addition_over_Ring_is_Associative | https://proofwiki.org/wiki/Matrix_Entrywise_Addition_over_Ring_is_Associative/Proof_1 | [
"Matrix Entrywise Addition",
"Examples of Associative Operations",
"Matrix Entrywise Addition is Associative"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix Space",
"Definition:Matrix Entrywise Addition/Ring",
"Definition:Associative Operation"
] | [
"Definition:Element",
"Definition:Matrix Space"
] |
proofwiki-10335 | Matrix Entrywise Addition over Ring is Associative | Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is associative on $\map {\MM_R} {m, n}$.
That... | By definition, matrix entrywise addition is the '''Hadamard product''' of $\mathbf A$ and $\mathbf B$ with respect to ring addition.
We have from {{Ring-axiom|A1}} that ring addition is associative.
The result then follows directly from Associativity of Hadamard Product.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition over Rin... | By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' of $\mathbf A$ and $\mathbf B$ with respect to [[Definition:Ring Addition|ring addition]].
We have from {{Ring-axiom|A1}} that [[Definition:Ring Addition|ring addition]] is [[... | Matrix Entrywise Addition over Ring is Associative/Proof 2 | https://proofwiki.org/wiki/Matrix_Entrywise_Addition_over_Ring_is_Associative | https://proofwiki.org/wiki/Matrix_Entrywise_Addition_over_Ring_is_Associative/Proof_2 | [
"Matrix Entrywise Addition",
"Examples of Associative Operations",
"Matrix Entrywise Addition is Associative"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix Space",
"Definition:Matrix Entrywise Addition/Ring",
"Definition:Associative Operation"
] | [
"Definition:Matrix Entrywise Addition",
"Definition:Hadamard Product",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Associative Operation",
"Associativity of Hadamard Product"
] |
proofwiki-10336 | Sum of Möbius Function over Divisors | :$\ds \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$
where $\floor {\dfrac 1 n}$ is the floor of $\dfrac 1 n$. | The theorem is clearly true if $n = 1$.
Assume, then, that $n > 1$ and write, by the Fundamental Theorem of Arithmetic:
:$n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$
In the sum $\ds \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the divisors of $n$ which are products of distinct prim... | :$\ds \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$
where $\floor {\dfrac 1 n}$ is the [[Definition:Floor Function|floor]] of $\dfrac 1 n$. | The theorem is clearly true if $n = 1$.
Assume, then, that $n > 1$ and write, by the [[Fundamental Theorem of Arithmetic]]:
:$n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$
In the sum $\ds \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the [[Definition:Divisor of Integer|divisors]] of... | Sum of Möbius Function over Divisors | https://proofwiki.org/wiki/Sum_of_Möbius_Function_over_Divisors | https://proofwiki.org/wiki/Sum_of_Möbius_Function_over_Divisors | [
"Sum of Möbius Function over Divisors",
"Möbius Function"
] | [
"Definition:Floor Function"
] | [
"Fundamental Theorem of Arithmetic",
"Definition:Divisor (Algebra)/Integer",
"Definition:Multiplication/Integers",
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Möbius Function",
"Definition:Möbius Function",
"Definition:Multiplication/Integers",
"Definition:Distinct",
"Definition:... |
proofwiki-10337 | Final Topology with respect to Mapping | Let $\struct {X, \tau_X}$ be a topological space.
Let $Y$ be a set.
Let $f: X \to Y$ be a mapping.
Let $\tau_Y$ be the final topology on $Y$ {{WRT}} $f$.
Then:
:$\tau_Y = \set {U \subseteq Y : f^{-1} \sqbrk U \in \tau_X}$
Observe that the set on the {{RHS}} of the equality is sometimes denoted $f\tau$.
Further, the fol... | {{proof wanted}}
Category:Topology
qthidr0arb1dy6iz4bp08pivrosem5e | Let $\struct {X, \tau_X}$ be a [[Definition:Topological Space|topological space]].
Let $Y$ be a [[Definition:Set|set]].
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $\tau_Y$ be the [[Definition:Final Topology|final topology]] on $Y$ {{WRT}} $f$.
Then:
:$\tau_Y = \set {U \subseteq Y : f^{-1} \sqbrk U \i... | {{proof wanted}}
[[Category:Topology]]
qthidr0arb1dy6iz4bp08pivrosem5e | Final Topology with respect to Mapping | https://proofwiki.org/wiki/Final_Topology_with_respect_to_Mapping | https://proofwiki.org/wiki/Final_Topology_with_respect_to_Mapping | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Mapping",
"Definition:Final Topology",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Topology",
"Definitio... | [
"Category:Topology"
] |
proofwiki-10338 | Composition of Symmetries is Symmetry | Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
Let $\phi$ and $\psi$ be symmetries of $P$.
Then $\phi \circ \psi$ is also a symmetry of $P$. | By definition of composition of mappings:
:$\phi \circ \psi$ is a mapping.
We have by definition of symmetry that:
:$\map \phi P$ is congruent to $P$
and:
:$\map \psi {\map \phi P}$ is congruent to $\map \phi P$
Therefore:
:$\phi \circ \psi$ is congruent to $P$
Thus $\phi \circ \psi$ is a symmetry of $P$.
{{qed}} | Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry (Geometry)|symmetries]] of $P$.
Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
Let $\phi$ and $\psi$ be [[Definition:Symmetry (Geometry)|symmetries]]... | By definition of [[Definition:Composition of Mappings|composition of mappings]]:
:$\phi \circ \psi$ is a [[Definition:Mapping|mapping]].
We have by definition of [[Definition:Symmetry (Geometry)|symmetry]] that:
:$\map \phi P$ is [[Definition:Congruence (Geometry)|congruent]] to $P$
and:
:$\map \psi {\map \phi P}$ is... | Composition of Symmetries is Symmetry | https://proofwiki.org/wiki/Composition_of_Symmetries_is_Symmetry | https://proofwiki.org/wiki/Composition_of_Symmetries_is_Symmetry | [
"Symmetries (Geometry)"
] | [
"Definition:Geometric Figure",
"Definition:Set",
"Definition:Symmetry (Geometry)",
"Definition:Composition of Mappings",
"Definition:Symmetry (Geometry)",
"Definition:Symmetry (Geometry)"
] | [
"Definition:Composition of Mappings",
"Definition:Mapping",
"Definition:Symmetry (Geometry)",
"Definition:Congruence (Geometry)",
"Definition:Congruence (Geometry)",
"Definition:Congruence (Geometry)",
"Definition:Symmetry (Geometry)"
] |
proofwiki-10339 | Composition of Symmetries is Associative | Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
Let $\phi, \psi, \chi$ be symmetries of $P$.
Then:
:$\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
That is, composition of symmetries is associative. | From Composition of Symmetries is Symmetry:
:$\paren {\phi \circ \psi} \circ \chi$ is a symmetry
and:
:$\phi \circ \paren {\psi \circ \chi}$ is a symmetry.
It follows from Composition of Mappings is Associative that:
:$\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
{{qed}} | Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry (Geometry)|symmetries]] of $P$.
Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
Let $\phi, \psi, \chi$ be [[Definition:Symmetry (Geometry)|symmetries]... | From [[Composition of Symmetries is Symmetry]]:
:$\paren {\phi \circ \psi} \circ \chi$ is a [[Definition:Symmetry (Geometry)|symmetry]]
and:
:$\phi \circ \paren {\psi \circ \chi}$ is a [[Definition:Symmetry (Geometry)|symmetry]].
It follows from [[Composition of Mappings is Associative]] that:
:$\paren {\phi \circ \ps... | Composition of Symmetries is Associative | https://proofwiki.org/wiki/Composition_of_Symmetries_is_Associative | https://proofwiki.org/wiki/Composition_of_Symmetries_is_Associative | [
"Symmetries (Geometry)"
] | [
"Definition:Geometric Figure",
"Definition:Set",
"Definition:Symmetry (Geometry)",
"Definition:Composition of Mappings",
"Definition:Symmetry (Geometry)",
"Definition:Composition of Mappings",
"Definition:Symmetry (Geometry)",
"Definition:Associative Operation"
] | [
"Composition of Symmetries is Symmetry",
"Definition:Symmetry (Geometry)",
"Definition:Symmetry (Geometry)",
"Composition of Mappings is Associative"
] |
proofwiki-10340 | Square of Modulo less One equals One | Let $m \in \Z$ be an integer.
Let $\Z_m$ be the set of integers modulo $m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
Then:
:$\eqclass {m - 1} m \times_m \eqclass {m - 1} m = \eqclass 1 m$
where $\times_m$ denotes multiplication modulo $m$. | {{begin-eqn}}
{{eqn | l = \eqclass {m - 1} m \times_m \eqclass {m - 1} m
| r = \eqclass {\paren {m - 1}^2} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {m^2 - 2 m + 1} m
| c =
}}
{{eqn | r = \eqclass 1 m
| c =
}}
{{end-eqn}}
{{qed}} | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]]:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
Then:
:$\eqclass {m - 1} m \times_m \eqclass {m - 1} m = \eqclass 1 m$
where $\times_m$ denotes [[Definition:Modu... | {{begin-eqn}}
{{eqn | l = \eqclass {m - 1} m \times_m \eqclass {m - 1} m
| r = \eqclass {\paren {m - 1}^2} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {m^2 - 2 m + 1} m
| c =
}}
{{eqn | r = \eqclass 1 m
| c =
}}
{{end-eqn}}
{{qed}} | Square of Modulo less One equals One | https://proofwiki.org/wiki/Square_of_Modulo_less_One_equals_One | https://proofwiki.org/wiki/Square_of_Modulo_less_One_equals_One | [
"Modulo Arithmetic"
] | [
"Definition:Integer",
"Definition:Integers Modulo m",
"Definition:Modulo Multiplication"
] | [] |
proofwiki-10341 | Existence of Non-Square Residue | Let $m \in \Z$ be an integer such that $m > 2$.
Let $\Z_m$ be the set of integers modulo $m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
Then there exists at least one residue in $\Z_m$ which is not the product modulo $m$ of a residue with itself:
:$\exists p \in \Z_m: \forall x \in \Z_m: x... | We have that $1 \in \Z_m$ and that:
:$1 \cdot_m 1 = 1$
We have that $m - 1 \in \Z_m$ and that:
:$\paren {m - 1} \cdot_m \paren {m - 1} = 1$
Thus unless $m - 1 = 1$, that is, $m = 2$, there exist $2$ residues of $\Z_m$ whose product modulo $m$ with itself equals $1$.
There are $m - 2$ residues which, when multiplied mod... | Let $m \in \Z$ be an [[Definition:Integer|integer]] such that $m > 2$.
Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]]:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
Then there exists at least one [[Definition:Residue (Modulo Arithmetic)|residue]] in $\Z_m$ whi... | We have that $1 \in \Z_m$ and that:
:$1 \cdot_m 1 = 1$
We have that $m - 1 \in \Z_m$ and that:
:$\paren {m - 1} \cdot_m \paren {m - 1} = 1$
Thus unless $m - 1 = 1$, that is, $m = 2$, there exist $2$ [[Definition:Residue (Modulo Arithmetic)|residues]] of $\Z_m$ whose [[Definition:Modulo Multiplication|product modulo $... | Existence of Non-Square Residue | https://proofwiki.org/wiki/Existence_of_Non-Square_Residue | https://proofwiki.org/wiki/Existence_of_Non-Square_Residue | [
"Modulo Arithmetic"
] | [
"Definition:Integer",
"Definition:Integers Modulo m",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Modulo Multiplication",
"Definition:Congruence (Number Theory)/Residue"
] | [
"Definition:Congruence (Number Theory)/Residue",
"Definition:Modulo Multiplication",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Modulo Multiplication",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Modulo Multiplication",
... |
proofwiki-10342 | Symmetry Group of Equilateral Triangle is Symmetric Group | Let $D_3$ denote the symmetry group of the equilateral triangle.
Let $S_3$ denote the symmetric group on $3$ letters.
Then $D_3$ is isomorphic to $S_3$. | {{proofread}}
Follows from Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3.
{{qed}} | Let $D_3$ denote the [[Definition:Symmetry Group of Equilateral Triangle|symmetry group of the equilateral triangle]].
Let $S_3$ denote the [[Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
Then $D_3$ is [[Definition:Group Isomorphism|isomorphic]] to $S_3$. | {{proofread}}
Follows from [[Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3]].
{{qed}} | Symmetry Group of Equilateral Triangle is Symmetric Group | https://proofwiki.org/wiki/Symmetry_Group_of_Equilateral_Triangle_is_Symmetric_Group | https://proofwiki.org/wiki/Symmetry_Group_of_Equilateral_Triangle_is_Symmetric_Group | [
"Symmetry Group of Equilateral Triangle",
"Symmetric Group on 3 Letters"
] | [
"Definition:Symmetry Group of Equilateral Triangle",
"Symmetric Group on 3 Letters",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3"
] |
proofwiki-10343 | De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection | :$\ds \relcomp S {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \relcomp S H$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcap \mathbb T}
| r = S \setminus \paren {\bigcap \mathbb T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {S \setminus H}
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \bigcup_{H \mathop \in \... | :$\ds \relcomp S {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \relcomp S H$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcap \mathbb T}
| r = S \setminus \paren {\bigcap \mathbb T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {S \setminus H}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersecti... | De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/General_Case/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/General_Case/Complement_of_Intersection | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection"
] |
proofwiki-10344 | De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union | :$\ds \relcomp S {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \relcomp S H$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcup \mathbb T}
| r = S \setminus \paren {\bigcup \mathbb T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {S \setminus H}
| c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb ... | :$\ds \relcomp S {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \relcomp S H$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcup \mathbb T}
| r = S \setminus \paren {\bigcup \mathbb T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {S \setminus H}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De M... | De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/General_Case/Complement_of_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/General_Case/Complement_of_Union | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union"
] |
proofwiki-10345 | De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Union | :$\ds \relcomp S {\bigcup_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \relcomp S {S_i}$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcup_{i \mathop \in I} S_i}
| r = S \setminus \paren {\bigcup_{i \mathop \in I} S_i}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcap_{i \mathop \in I} \paren {S \setminus S_i}
| c = De Morgan's Laws for Set Difference: Difference with Union
}}
{{eqn ... | :$\ds \relcomp S {\bigcup_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \relcomp S {S_i}$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcup_{i \mathop \in I} S_i}
| r = S \setminus \paren {\bigcup_{i \mathop \in I} S_i}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcap_{i \mathop \in I} \paren {S \setminus S_i}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Differ... | De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Family_of_Sets/Complement_of_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Family_of_Sets/Complement_of_Union | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union"
] |
proofwiki-10346 | De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Intersection | :$\ds \relcomp S {\bigcap_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \relcomp S {S_i}$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcap_{i \mathop \in I} S_i}
| r = S \setminus \paren {\bigcap_{i \mathop \in I} S_i}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcup_{i \mathop \in I} \paren {S \setminus S_i}
| c = De Morgan's Laws for Set Difference: Difference with Intersection
}}... | :$\ds \relcomp S {\bigcap_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \relcomp S {S_i}$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\bigcap_{i \mathop \in I} S_i}
| r = S \setminus \paren {\bigcap_{i \mathop \in I} S_i}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \bigcup_{i \mathop \in I} \paren {S \setminus S_i}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Differ... | De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Family_of_Sets/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Family_of_Sets/Complement_of_Intersection | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection"
] |
proofwiki-10347 | De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union | :$\relcomp S {T_1 \cup T_2} = \relcomp S {T_1} \cap \relcomp S {T_2}$ | Let $x \in S$ througout.
{{begin-eqn}}
{{eqn | o =
| r = x \in \relcomp S {T_1 \cup T_2}
}}
{{eqn | o = \leadsto
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | o = \leadsto
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eqn... | :$\relcomp S {T_1 \cup T_2} = \relcomp S {T_1} \cap \relcomp S {T_2}$ | Let $x \in S$ througout.
{{begin-eqn}}
{{eqn | o =
| r = x \in \relcomp S {T_1 \cup T_2}
}}
{{eqn | o = \leadsto
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | o = \leadsto
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eq... | De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union/Proof 2 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Complement_of_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Complement_of_Union/Proof_2 | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations",
"De Morgan's Laws (Logic)/Conjunction of Negations",
"Definition:Set Equality/Definition 1"
] |
proofwiki-10348 | De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection | :$\relcomp S {T_1 \cap T_2} = \relcomp S {T_1} \cup \relcomp S {T_2}$ | Let $T_1, T_2 \subseteq S$.
Then from Intersection is Subset and Subset Relation is Transitive:
:$T_1 \cap T_2 \subseteq S$
Hence:
{{begin-eqn}}
{{eqn | l = \relcomp S {T_1 \cap T_2}
| r = S \setminus \paren {T_1 \cap T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \paren {S \setminus T_1} \cup \par... | :$\relcomp S {T_1 \cap T_2} = \relcomp S {T_1} \cup \relcomp S {T_2}$ | Let $T_1, T_2 \subseteq S$.
Then from [[Intersection is Subset]] and [[Subset Relation is Transitive]]:
:$T_1 \cap T_2 \subseteq S$
Hence:
{{begin-eqn}}
{{eqn | l = \relcomp S {T_1 \cap T_2}
| r = S \setminus \paren {T_1 \cap T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \paren {S \setminus T_1... | De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Complement_of_Intersection | [
"De Morgan's Laws"
] | [] | [
"Intersection is Subset",
"Subset Relation is Transitive",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection"
] |
proofwiki-10349 | De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection | :$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i := \set {x: \exists i \in I: x \in T_i}$
that is, the union of $\family {T_i}_{i \mathop \in I}$. | Suppose:
:$\ds x \in S \setminus \bigcap_{i \mathop \in I} T_i$
Note that by Set Difference is Subset we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o =... | :$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i := \set {x: \exists i \in I: x \in T_i}$
that is, the [[Definition:Union of Family|union]] of $\family {T_i}_{i \mathop \in I}$. | Suppose:
:$\ds x \in S \setminus \bigcap_{i \mathop \in I} T_i$
Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
... | De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Family_of_Sets/Difference_with_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Family_of_Sets/Difference_with_Intersection | [
"De Morgan's Laws",
"Indexed Families"
] | [
"Definition:Set Union/Family of Sets"
] | [
"Set Difference is Subset",
"De Morgan's Laws (Predicate Logic)/Denial of Universality"
] |
proofwiki-10350 | De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union | :$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
where:
:$\ds \bigcap_{i \mathop \in I} T_i := \set {x: \forall i \in I: x \in T_i}$
that is, the intersection of $\family {T_i}_{i \mathop \in I}$. | Suppose:
:$\ds x \in S \setminus \bigcup_{i \mathop \in I} T_i$
Note that by Set Difference is Subset we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcup_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o =... | :$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
where:
:$\ds \bigcap_{i \mathop \in I} T_i := \set {x: \forall i \in I: x \in T_i}$
that is, the [[Definition:Intersection of Family|intersection]] of $\family {T_i}_{i \mathop \in I}$. | Suppose:
:$\ds x \in S \setminus \bigcup_{i \mathop \in I} T_i$
Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcup_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
... | De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Family_of_Sets/Difference_with_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Family_of_Sets/Difference_with_Union | [
"De Morgan's Laws",
"Indexed Families"
] | [
"Definition:Set Intersection/Family of Sets"
] | [
"Set Difference is Subset",
"De Morgan's Laws (Predicate Logic)/Denial of Existence"
] |
proofwiki-10351 | De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Intersection | :$\ds \map \complement {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \map \complement H$ | {{begin-eqn}}
{{eqn | l = \map \complement {\bigcap \mathbb T}
| r = \mathbb U \setminus \paren {\bigcap \mathbb T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \bigcup... | :$\ds \map \complement {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \map \complement H$ | {{begin-eqn}}
{{eqn | l = \map \complement {\bigcap \mathbb T}
| r = \mathbb U \setminus \paren {\bigcap \mathbb T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Differenc... | De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/General_Case/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/General_Case/Complement_of_Intersection | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection"
] |
proofwiki-10352 | De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Union | :$\ds \map \complement {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \map \complement H$ | {{begin-eqn}}
{{eqn | l = \map \complement {\bigcup \mathbb T}
| r = \mathbb U \setminus \paren {\bigcup \mathbb T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
| c = De Morgan's Laws for Set Difference: Difference with Union
}}
{{eqn |... | :$\ds \map \complement {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \map \complement H$ | {{begin-eqn}}
{{eqn | l = \map \complement {\bigcup \mathbb T}
| r = \mathbb U \setminus \paren {\bigcup \mathbb T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Differenc... | De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/General_Case/Complement_of_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/General_Case/Complement_of_Union | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union"
] |
proofwiki-10353 | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | l = \overline {T_1 \cap T_2}
| r = \mathbb U \setminus \paren {T_1 \cap T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \paren {\mathbb U \setminus T_1} \cup \paren {\mathbb U \setminus T_2}
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \overline {T_1} \c... | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | l = \overline {T_1 \cap T_2}
| r = \mathbb U \setminus \paren {T_1 \cap T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \paren {\mathbb U \setminus T_1} \cup \paren {\mathbb U \setminus T_2}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De M... | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Proof 1 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection/Proof_1 | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection"
] |
proofwiki-10354 | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \overline {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = x \notin \paren {T_1 \cap T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \paren {x \in T_1 \land x \in T_2}
| c = {{Defof|Set Intersection}}
}}
{{eqn | o = \l... | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \overline {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = x \notin \paren {T_1 \cap T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \paren {x \in T_1 \land x \in T_2}
| c = {{Defof|Set Intersection}}
}}
{{eqn | o = \l... | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Proof 2 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection/Proof_2 | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Disjunction of Negations",
"Definition:Set Equality/Definition 1"
] |
proofwiki-10355 | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | l = \map \complement {\map \complement A \cup \map \complement B}
| r = \map \complement {\map \complement A} \cap \map \complement {\map \complement B}
| c = De Morgan's Laws: Complement of Union
}}
{{eqn | r = A \cap B
| c = Complement of Complement
}}
{{eqn | ll= \leadstoandfr... | :$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$ | {{begin-eqn}}
{{eqn | l = \map \complement {\map \complement A \cup \map \complement B}
| r = \map \complement {\map \complement A} \cap \map \complement {\map \complement B}
| c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan's Laws: Complement of Union]]
}}
{{eqn | r = A \ca... | De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Proof 3 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Intersection/Proof_3 | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Union",
"Complement of Complement",
"Definition:Set Complement",
"Complement of Complement"
] |
proofwiki-10356 | De Morgan's Laws (Set Theory)/Set Complement/Complement of Union | :$\overline {T_1 \cup T_2} = \overline T_1 \cap \overline T_2$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \overline {T_1 \cup T_2}
}}
{{eqn | o = \leadstoandfrom
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoan... | :$\overline {T_1 \cup T_2} = \overline T_1 \cap \overline T_2$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \overline {T_1 \cup T_2}
}}
{{eqn | o = \leadstoandfrom
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Set Complement}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoan... | De Morgan's Laws (Set Theory)/Set Complement/Complement of Union/Proof 2 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Complement/Complement_of_Union/Proof_2 | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations",
"Definition:Set Equality/Definition 1"
] |
proofwiki-10357 | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof | Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a set and $I$ is some finite indexing set.
Then:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the cardinality $\card I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcap_{i \mathop = 1}^n ... | Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Then:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the [[Definition:Cardinality|cardinality]] $\card I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection/Proof | [
"De Morgan's Laws"
] | [
"Definition:Set",
"Definition:Finite Set",
"Definition:Indexing Set"
] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"In... |
proofwiki-10358 | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection | :$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the cardinality $\card I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcap_{i \mathop = 1}^n ... | :$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the [[Definition:Cardinality|cardinality]] $\card I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection/Proof | [
"De Morgan's Laws"
] | [] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"In... |
proofwiki-10359 | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof | Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a set and $I$ is some finite indexing set.
Then:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the cardinality $\size I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcup_{i \mathop = 1}^n ... | Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Then:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the [[Definition:Cardinality|cardinality]] $\size I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union/Proof | [
"De Morgan's Laws"
] | [
"Definition:Set",
"Definition:Finite Set",
"Definition:Indexing Set"
] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Union",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Union is ... |
proofwiki-10360 | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union | :$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the cardinality $\size I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcup_{i \mathop = 1}^n ... | :$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ | Let the [[Definition:Cardinality|cardinality]] $\size I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union/Proof | [
"De Morgan's Laws"
] | [] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Union",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Union is ... |
proofwiki-10361 | De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection | :$\ds S \setminus \bigcap \mathbb T = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
where:
:$\ds \bigcap \mathbb T := \set {x: \forall T' \in \mathbb T: x \in T'}$
that is, the intersection of $\mathbb T$ | Suppose:
:$\ds x \in S \setminus \bigcap \mathbb T$
Note that by Set Difference is Subset we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcap \mathbb T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \notin
| r = \big... | :$\ds S \setminus \bigcap \mathbb T = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
where:
:$\ds \bigcap \mathbb T := \set {x: \forall T' \in \mathbb T: x \in T'}$
that is, the [[Definition:Intersection of Set of Sets|intersection]] of $\mathbb T$ | Suppose:
:$\ds x \in S \setminus \bigcap \mathbb T$
Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcap \mathbb T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \notin
| r ... | De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/General_Case/Difference_with_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/General_Case/Difference_with_Intersection/Proof | [
"De Morgan's Laws"
] | [
"Definition:Set Intersection/Set of Sets"
] | [
"Set Difference is Subset",
"De Morgan's Laws (Predicate Logic)/Denial of Universality"
] |
proofwiki-10362 | De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union | :$\ds S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
where:
:$\ds \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$
that is, the union of $\mathbb T$. | Suppose:
:$\ds x \in S \setminus \bigcup \mathbb T$
Note that by Set Difference is Subset we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcup \mathbb T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \notin
| r = \big... | :$\ds S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
where:
:$\ds \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$
that is, the [[Definition:Union of Set of Sets|union]] of $\mathbb T$. | Suppose:
:$\ds x \in S \setminus \bigcup \mathbb T$
Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \setminus \bigcup \mathbb T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \notin
| r ... | De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/General_Case/Difference_with_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/General_Case/Difference_with_Union | [
"De Morgan's Laws"
] | [
"Definition:Set Union/Set of Sets"
] | [
"Set Difference is Subset",
"De Morgan's Laws (Predicate Logic)/Denial of Existence"
] |
proofwiki-10363 | De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection | :$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in S \setminus \paren {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cap T_2} }
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 ... | :$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in S \setminus \paren {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cap T_2} }
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 ... | De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Difference_with_Intersection | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Difference_with_Intersection | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Disjunction of Negations",
"Rule of Distribution",
"Definition:Set Equality/Definition 1"
] |
proofwiki-10364 | De Morgan's Laws (Set Theory)/Set Difference/Difference with Union | :$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in S \setminus \paren {T_1 \cup T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cup T_2} }
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 ... | :$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in S \setminus \paren {T_1 \cup T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cup T_2} }
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 ... | De Morgan's Laws (Set Theory)/Set Difference/Difference with Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Difference_with_Union | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Set_Difference/Difference_with_Union | [
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations",
"Rule of Idempotence",
"Rule of Commutation",
"Rule of Association",
"Definition:Set Equality/Definition 1"
] |
proofwiki-10365 | Isomorphism between Ring of Integers Modulo 2 and Parity Ring | The ring of integers modulo $2$ and the parity ring are isomorphic. | To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$.
Let $f$ be the mapping from the parity ring $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the ring of integers modulo $2$ $\struct {\Z_2, +_2, \times_2}$:
:$f: \struct {\set {\text{... | The [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]] and the [[Definition:Parity Ring|parity ring]] are [[Definition:Isomorphic Algebraic Structures|isomorphic]]. | To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$.
Let $f$ be the [[Definition:Mapping|mapping]] from the [[Definition:Parity Ring|parity ring]] $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the [[Definition:Ring of Integers Modul... | Isomorphism between Ring of Integers Modulo 2 and Parity Ring | https://proofwiki.org/wiki/Isomorphism_between_Ring_of_Integers_Modulo_2_and_Parity_Ring | https://proofwiki.org/wiki/Isomorphism_between_Ring_of_Integers_Modulo_2_and_Parity_Ring | [
"Ring of Integers Modulo m",
"Parity Ring",
"Ring Isomorphisms",
"Field Isomorphisms"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Parity Ring",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Definition:Parity Ring",
"Definition:Ring of Integers Modulo m",
"Definition:Bijection",
"Definition:Cayley Table"
] |
proofwiki-10366 | Parity Addition is Associative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is associative:
:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$ | From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text {even}, \text {odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Addition is Associative
and:
:Isomorphism Preserves Associativity.
{{qed}} | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Associative Operation|associative]]:
:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$ | From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
:$\struct {\set {\text {even}, \text {odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
The result follows from:
:[[Mod... | Parity Addition is Associative/Proof 1 | https://proofwiki.org/wiki/Parity_Addition_is_Associative | https://proofwiki.org/wiki/Parity_Addition_is_Associative/Proof_1 | [
"Parity Ring",
"Parity Addition is Associative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Associative Operation"
] | [
"Isomorphism between Ring of Integers Modulo 2 and Parity Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Ring of Integers Modulo m",
"Modulo Addition is Associative",
"Isomorphism Preserves Associativity"
] |
proofwiki-10367 | Parity Addition is Associative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is associative:
:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$ | Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & ... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Associative Operation|associative]]:
:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$ | Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of [[Definition:Odd Integer|odd]]:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of [[Definition:Even Integer|even]]:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the [[Definition:Ma... | Parity Addition is Associative/Proof 2 | https://proofwiki.org/wiki/Parity_Addition_is_Associative | https://proofwiki.org/wiki/Parity_Addition_is_Associative/Proof_2 | [
"Parity Ring",
"Parity Addition is Associative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Associative Operation"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Mapping",
"Definition:Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Operation/Binary Operation",
"Definition:Parity of Integer",
"Integer Addition is Associative",
"Definition:Oper... |
proofwiki-10368 | Parity Addition is Commutative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is commutative:
:$\forall a, b \in R: a + b = b + a$ | From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Addition is Commutative
and:
:Isomorphism Preserves Commutativity.
{{qed}} | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Commutative Operation|commutative]]:
:$\forall a, b \in R: a + b = b + a$ | From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
The result follows from:
:[[Modul... | Parity Addition is Commutative/Proof 1 | https://proofwiki.org/wiki/Parity_Addition_is_Commutative | https://proofwiki.org/wiki/Parity_Addition_is_Commutative/Proof_1 | [
"Parity Ring",
"Parity Addition is Commutative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation"
] | [
"Isomorphism between Ring of Integers Modulo 2 and Parity Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Ring of Integers Modulo m",
"Modulo Addition is Commutative",
"Isomorphism Preserves Commutativity"
] |
proofwiki-10369 | Parity Addition is Commutative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is commutative:
:$\forall a, b \in R: a + b = b + a$ | Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & ... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Commutative Operation|commutative]]:
:$\forall a, b \in R: a + b = b + a$ | Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of [[Definition:Odd Integer|odd]]:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of [[Definition:Even Integer|even]]:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the [[Definition:Ma... | Parity Addition is Commutative/Proof 2 | https://proofwiki.org/wiki/Parity_Addition_is_Commutative | https://proofwiki.org/wiki/Parity_Addition_is_Commutative/Proof_2 | [
"Parity Ring",
"Parity Addition is Commutative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Mapping",
"Definition:Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Operation/Binary Operation",
"Definition:Parity of Integer",
"Integer Addition is Commutative",
"Definition:Oper... |
proofwiki-10370 | Parity Multiplication is Associative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is associative:
:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$ | From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Multiplication is Associative
and:
:Isomorphism Preserves Associativity.
{{qed}... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Associative Operation|associative]]:
:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$ | From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
The result follows from:
:[[Modul... | Parity Multiplication is Associative/Proof 1 | https://proofwiki.org/wiki/Parity_Multiplication_is_Associative | https://proofwiki.org/wiki/Parity_Multiplication_is_Associative/Proof_1 | [
"Parity Ring",
"Parity Multiplication is Associative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Associative Operation"
] | [
"Isomorphism between Ring of Integers Modulo 2 and Parity Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Ring of Integers Modulo m",
"Modulo Multiplication is Associative",
"Isomorphism Preserves Associativity"
] |
proofwiki-10371 | Parity Multiplication is Associative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is associative:
:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$ | Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & ... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Associative Operation|associative]]:
:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$ | Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of [[Definition:Odd Integer|odd]]:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of [[Definition:Even Integer|even]]:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the [[Definition:Ma... | Parity Multiplication is Associative/Proof 2 | https://proofwiki.org/wiki/Parity_Multiplication_is_Associative | https://proofwiki.org/wiki/Parity_Multiplication_is_Associative/Proof_2 | [
"Parity Ring",
"Parity Multiplication is Associative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Associative Operation"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Mapping",
"Definition:Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Integer Multiplication is Associative"
] |
proofwiki-10372 | Parity Multiplication is Commutative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is commutative:
:$\forall a, b \in R: a \times b = b \times a$ | From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Multiplication is Commutative
and:
:Isomorphism Preserves Associativity.
{{qed}... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Commutative Operation|commutative]]:
:$\forall a, b \in R: a \times b = b \times a$ | From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
The result follows from:
:[[Modul... | Parity Multiplication is Commutative/Proof 1 | https://proofwiki.org/wiki/Parity_Multiplication_is_Commutative | https://proofwiki.org/wiki/Parity_Multiplication_is_Commutative/Proof_1 | [
"Parity Ring",
"Parity Multiplication is Commutative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation"
] | [
"Isomorphism between Ring of Integers Modulo 2 and Parity Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Ring of Integers Modulo m",
"Modulo Multiplication is Commutative",
"Isomorphism Preserves Associativity"
] |
proofwiki-10373 | Parity Multiplication is Commutative | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is commutative:
:$\forall a, b \in R: a \times b = b \times a$ | Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
$\quad \forall x \in R: \map f x := \begin{cases}... | Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Commutative Operation|commutative]]:
:$\forall a, b \in R: a \times b = b \times a$ | Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of [[Definition:Odd Integer|odd]]:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of [[Definition:Even Integer|even]]:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the [[Definition:Ma... | Parity Multiplication is Commutative/Proof 2 | https://proofwiki.org/wiki/Parity_Multiplication_is_Commutative | https://proofwiki.org/wiki/Parity_Multiplication_is_Commutative/Proof_2 | [
"Parity Ring",
"Parity Multiplication is Commutative"
] | [
"Definition:Parity Ring",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Mapping",
"Definition:Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Integer Multiplication is Commutative"
] |
proofwiki-10374 | Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {R_n, \times}$ be the complex $n$th roots of unity under complex multiplication.
Let $\struct {\Z_n, +_n}$ be the integers modulo $n$ under modulo addition.
Then $\struct {R_n, \times}$ and $\struct {\Z_n, +_n}$ are isomorphic algebraic structures. | The set of integers modulo $n$ is the set exemplified by the integers:
:$\Z_n = \set {0, 1, \ldots, n - 1}$
The complex $n$th roots of unity is the set:
:$R_n = \set {z \in \C: z^n = 1}$
From Complex Roots of Unity in Exponential Form:
:$R_n = \set {1, e^{\theta / n}, e^{2 \theta / n}, \ldots, e^{\left({n - 1}\right) \... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\struct {R_n, \times}$ be the [[Definition:Complex Roots of Unity|complex $n$th roots of unity]] under [[Definition:Complex Multiplication|complex multiplication]].
Let $\struct {\Z_n, +_n}$ be the [[Definition:Integer... | The [[Definition:Integers Modulo m|set of integers modulo $n$]] is the [[Definition:Set|set]] exemplified by the [[Definition:Integer|integers]]:
:$\Z_n = \set {0, 1, \ldots, n - 1}$
The [[Definition:Complex Roots of Unity|complex $n$th roots of unity]] is the [[Definition:Set|set]]:
:$R_n = \set {z \in \C: z^n = 1}$
... | Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition | https://proofwiki.org/wiki/Isomorphism_between_Roots_of_Unity_under_Multiplication_and_Integers_under_Modulo_Addition | https://proofwiki.org/wiki/Isomorphism_between_Roots_of_Unity_under_Multiplication_and_Integers_under_Modulo_Addition | [
"Examples of Group Isomorphisms",
"Roots of Unity",
"Additive Groups of Integers Modulo m"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Root of Unity/Complex",
"Definition:Multiplication/Complex Numbers",
"Definition:Integers Modulo m",
"Definition:Modulo Addition",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Integers Modulo m",
"Definition:Set",
"Definition:Integer",
"Definition:Root of Unity/Complex",
"Definition:Set",
"Complex Roots of Unity in Exponential Form",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Isomorp... |
proofwiki-10375 | Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition | Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {\Z_n, +_4}$ be the integers modulo $4$ under modulo addition.
Then $\struct {U_\C, \times}$ and $\struct {\Z_4, +_4}$ are isomorphic algebraic structures. | From Gaussian Integer Units are 4th Roots of Unity:
:$U_\C$ is the set consisting of the (complex) $4$th roots of $1$.
The result follows from Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition. | Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
Let $\struct {\Z_n, +_4}$ be the [[Definition:Integers Modulo m|integers modulo $4$]] under [[Definition:Modulo Addition|modulo addition... | From [[Gaussian Integer Units are 4th Roots of Unity]]:
:$U_\C$ is the [[Definition:Set|set]] consisting of the [[Definition:Complex Roots of Unity|(complex) $4$th roots of $1$]].
The result follows from [[Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition]]. | Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 1 | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Integers_Modulo_4_under_Addition | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Integers_Modulo_4_under_Addition/Proof_1 | [
"Examples of Group Isomorphisms/Order 4",
"Additive Groups of Integers Modulo m",
"Group of Gaussian Integer Units",
"Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition"
] | [
"Definition:Group of Gaussian Integer Units",
"Definition:Multiplication/Complex Numbers",
"Definition:Integers Modulo m",
"Definition:Modulo Addition",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Gaussian Integer Units are 4th Roots of Unity",
"Definition:Set",
"Definition:Root of Unity/Complex",
"Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition"
] |
proofwiki-10376 | Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition | Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {\Z_n, +_4}$ be the integers modulo $4$ under modulo addition.
Then $\struct {U_\C, \times}$ and $\struct {\Z_4, +_4}$ are isomorphic algebraic structures. | Let the mapping $f: \Z_4 \to U_\C$ be defined as:
{{begin-eqn}}
{{eqn | l = \map f 0
| r = 1
}}
{{eqn | l = \map f 1
| r = i
}}
{{eqn | l = \map f 2
| r = -1
}}
{{eqn | l = \map f 3
| r = -i
}}
{{end-eqn}}
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertai... | Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
Let $\struct {\Z_n, +_4}$ be the [[Definition:Integers Modulo m|integers modulo $4$]] under [[Definition:Modulo Addition|modulo addition... | Let the [[Definition:Mapping|mapping]] $f: \Z_4 \to U_\C$ be defined as:
{{begin-eqn}}
{{eqn | l = \map f 0
| r = 1
}}
{{eqn | l = \map f 1
| r = i
}}
{{eqn | l = \map f 2
| r = -1
}}
{{eqn | l = \map f 3
| r = -i
}}
{{end-eqn}}
From [[Isomorphism by Cayley Table]], the two [[Definition:Cayle... | Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 2 | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Integers_Modulo_4_under_Addition | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Integers_Modulo_4_under_Addition/Proof_2 | [
"Examples of Group Isomorphisms/Order 4",
"Additive Groups of Integers Modulo m",
"Group of Gaussian Integer Units",
"Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition"
] | [
"Definition:Group of Gaussian Integer Units",
"Definition:Multiplication/Complex Numbers",
"Definition:Integers Modulo m",
"Definition:Modulo Addition",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Modulo Addition/Cayley Table/Modulo 4",
"Definition:Cayley Table",
"Group of Gaussian Integer Units/Cayley Table",
"Definition:Cayley Table"
] |
proofwiki-10377 | Gaussian Integer Units are 4th Roots of Unity | The units of the ring of Gaussian integers:
:$\set {1, i, -1, -i}$
are the (complex) $4$th roots of $1$. | We have that $i = \sqrt {-1}$ is the imaginary unit.
Thus:
{{begin-eqn}}
{{eqn | l = 1^4
| o =
| rr= = 1
}}
{{eqn | l = i^4
| r = \paren {-1}^2
| rr= = 1
}}
{{eqn | l = \paren {-1}^4
| r = 1^2
| rr= = 1
}}
{{eqn | l = \paren {-i}^4
| r = \paren {-1}^2 \cdot \paren {-1}^2
... | The [[Definition:Unit of Ring|units]] of the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
:$\set {1, i, -1, -i}$
are the [[Definition:Complex Roots of Unity|(complex) $4$th roots of $1$]]. | We have that $i = \sqrt {-1}$ is the [[Definition:Imaginary Unit|imaginary unit]].
Thus:
{{begin-eqn}}
{{eqn | l = 1^4
| o =
| rr= = 1
}}
{{eqn | l = i^4
| r = \paren {-1}^2
| rr= = 1
}}
{{eqn | l = \paren {-1}^4
| r = 1^2
| rr= = 1
}}
{{eqn | l = \paren {-i}^4
| r = \paren ... | Gaussian Integer Units are 4th Roots of Unity | https://proofwiki.org/wiki/Gaussian_Integer_Units_are_4th_Roots_of_Unity | https://proofwiki.org/wiki/Gaussian_Integer_Units_are_4th_Roots_of_Unity | [
"Gaussian Integers",
"Complex Roots of Unity"
] | [
"Definition:Unit of Ring",
"Definition:Ring of Gaussian Integers",
"Definition:Root of Unity/Complex"
] | [
"Definition:Complex Number/Imaginary Unit",
"Definition:Set",
"Definition:Root of Unity/Complex"
] |
proofwiki-10378 | Isomorphism by Cayley Table | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures whose underlying sets are both finite.
Then $\struct {S, \circ}$ and $\struct {T, *}$ are isomorphic {{iff}}:
:a bijection $f: S \to T$ can be found such that:
::the Cayley table of $\struct {T, *}$ can be generated from the Cayley table of $\struct ... | === Necessary Condition ===
Let $S$ and $T$ be isomorphic.
Then by definition there exists an isomorphism $f: S \to T$.
An isomorphism is a bijection by definition.
Thus the existence of the posited bijection has been demonstrated.
By the definition of set equivalence, $S$ and $T$ have the same cardinality.
Let $\card ... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]] whose [[Definition:Underlying Set of Structure|underlying sets]] are both [[Definition:Finite Set|finite]].
Then $\struct {S, \circ}$ and $\struct {T, *}$ are [[Definition:Isomorphic Algebraic Structures|isomorphi... | === Necessary Condition ===
Let $S$ and $T$ be [[Definition:Isomorphic Algebraic Structures|isomorphic]].
Then by definition there exists an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $f: S \to T$.
An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Bijection|bijection]] by ... | Isomorphism by Cayley Table | https://proofwiki.org/wiki/Isomorphism_by_Cayley_Table | https://proofwiki.org/wiki/Isomorphism_by_Cayley_Table | [
"Cayley Tables",
"Isomorphisms (Abstract Algebra)"
] | [
"Definition:Algebraic Structure",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Finite Set",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Cayley Table",
"Definition:Cayley Table",
"Definition:Cayley Table/Entry",
"Definition:Image (Set Theory)/Mapping/... | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Set Equivalence",
"Definition:Cardinality",
"Definition:Cayley Table/Entry",
"Definition:Isomorphism (Abs... |
proofwiki-10379 | Bijection between Integers and Even Integers | Let $\Z$ be the set of integers.
Let $2 \Z$ be the set of even integers.
Then there exists a bijection $f: \Z \to 2 \Z$ between the two. | Let $f: \Z \to 2 \Z$ be the mapping defined as:
:$\forall n \in \Z: \map f n = 2 n$
Let $m, n \in \Z$ such that $\map f m = \map f n$.
{{begin-eqn}}
{{eqn | l = \map f m
| r = \map f n
}}
{{eqn | ll= \leadsto
| l = 2 m
| r = 2 n
| c = Definition of $f$
}}
{{eqn | ll= \leadsto
| l = m
... | Let $\Z$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]].
Let $2 \Z$ be the [[Definition:Set|set]] of [[Definition:Even Integer|even integers]].
Then there exists a [[Definition:Bijection|bijection]] $f: \Z \to 2 \Z$ between the two. | Let $f: \Z \to 2 \Z$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall n \in \Z: \map f n = 2 n$
Let $m, n \in \Z$ such that $\map f m = \map f n$.
{{begin-eqn}}
{{eqn | l = \map f m
| r = \map f n
}}
{{eqn | ll= \leadsto
| l = 2 m
| r = 2 n
| c = Definition of $f$
}}
{{eqn | ll= \le... | Bijection between Integers and Even Integers | https://proofwiki.org/wiki/Bijection_between_Integers_and_Even_Integers | https://proofwiki.org/wiki/Bijection_between_Integers_and_Even_Integers | [
"Examples of Bijections",
"Integers",
"Even Integers"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Set",
"Definition:Even Integer",
"Definition:Bijection"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
"Category:Examples of Bijections",
"Category:Integers",
"Category:Even Integers"
] |
proofwiki-10380 | Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition | Let $\struct {\Z, +}$ be the algebraic structure formed by the set of integers under the operation of addition.
Let $\struct {2 \Z, +}$ be the algebraic structure formed by the set of even integers under the operation of addition.
Then $\struct {\Z, +}$ and $\struct {2 \Z, +}$ are isomorphic. | Let $f: \Z \to 2 \Z$ be the mapping:
:$\forall n \in \Z: \map f n = 2 n$
From Bijection between Integers and Even Integers, $f$ is a bijection.
Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {m + n}
| r = 2 \paren {m + n}
| c = Definition of $f$
}}
{{eqn | r = 2 m + 2 n
| c = Integer Multip... | Let $\struct {\Z, +}$ be the [[Definition:Algebraic Structure with One Operation|algebraic structure]] formed by the [[Definition:Set|set]] of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Addition|addition]].
Let $\struct {2 \Z, +}$ be the [[Definition:Alg... | Let $f: \Z \to 2 \Z$ be the [[Definition:Mapping|mapping]]:
:$\forall n \in \Z: \map f n = 2 n$
From [[Bijection between Integers and Even Integers]], $f$ is a [[Definition:Bijection|bijection]].
Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {m + n}
| r = 2 \paren {m + n}
| c = Definition of ... | Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition | https://proofwiki.org/wiki/Set_of_Integers_under_Addition_is_Isomorphic_to_Set_of_Even_Integers_under_Addition | https://proofwiki.org/wiki/Set_of_Integers_under_Addition_is_Isomorphic_to_Set_of_Even_Integers_under_Addition | [
"Integers",
"Even Integers",
"Additive Group of Integers"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Integer",
"Definition:Operation/Binary Operation",
"Definition:Addition/Integers",
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Even Integer",
"Definition:Operation/Binary Operation",
"De... | [
"Definition:Mapping",
"Bijection between Integers and Even Integers",
"Definition:Bijection",
"Integer Multiplication Distributes over Addition",
"Integer Multiplication Distributes over Addition",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-10381 | Natural Numbers under Addition do not form Group | The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is not a group. | From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ has an identity element $0$.
However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$.
Thus the general element of $\struct {\N, +}$ has no inverse.
Hence the result by definition of group.
{{qed}} | The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\N, +}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Addition|addition]] $+$ is not a [[Definition:Group|group]]. | From [[Natural Numbers under Addition form Commutative Monoid]], $\struct {\N, +}$ has an [[Definition:Identity Element|identity element]] $0$.
However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$.
Thus the general element of $\struct {\N, +}$ has no [[Definition:Inverse El... | Natural Numbers under Addition do not form Group | https://proofwiki.org/wiki/Natural_Numbers_under_Addition_do_not_form_Group | https://proofwiki.org/wiki/Natural_Numbers_under_Addition_do_not_form_Group | [
"Natural Number Addition",
"Examples of Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Group"
] | [
"Natural Numbers under Addition form Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Group"
] |
proofwiki-10382 | Natural Numbers under Multiplication do not form Group | The algebraic structure $\struct {\N, \times}$ consisting of the set of natural numbers $\N$ under multiplication $\times$ is not a group. | {{AimForCont}} that $\struct {\N, \times}$ is a group.
We have that $1 \times 1 = 1$ and so is idempotent.
From Identity is only Idempotent Element in Group it follows that $1$ is the identity of $\struct {\N, \times}$.
Let $x \in \N$ such that $x \ne 0$ and $x \ne 1$.
There exists no $y \in \N$ such that $x \times y =... | The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\N, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]]. | {{AimForCont}} that $\struct {\N, \times}$ is a [[Definition:Group|group]].
We have that $1 \times 1 = 1$ and so is [[Definition:Idempotent Element|idempotent]].
From [[Identity is only Idempotent Element in Group]] it follows that $1$ is the [[Definition:Identity Element|identity]] of $\struct {\N, \times}$.
Let $x... | Natural Numbers under Multiplication do not form Group/Proof 2 | https://proofwiki.org/wiki/Natural_Numbers_under_Multiplication_do_not_form_Group | https://proofwiki.org/wiki/Natural_Numbers_under_Multiplication_do_not_form_Group/Proof_2 | [
"Natural Number Multiplication",
"Examples of Groups",
"Natural Numbers under Multiplication do not form Group"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Idempotence/Element",
"Identity is only Idempotent Element in Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Proof by Contradiction",
"Definition:Group"
] |
proofwiki-10383 | Rational Numbers under Multiplication do not form Group | The algebraic structure $\struct {\Q, \times}$ consisting of the set of rational numbers $\Q$ under multiplication $\times$ is not a group. | {{AimForCont}} that $\struct {\Q, \times}$ is a group.
By the definition of the number $0 \in \Q$:
:$\forall x \in \Q: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\Q, \times}$ is the trivial group.
But $\Q$ contains other elements b... | The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\Q, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ under [[Definition:Rational Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]]. | {{AimForCont}} that $\struct {\Q, \times}$ is a [[Definition:Group|group]].
By the definition of the [[Definition:Zero (Number)|number $0 \in \Q$]]:
:$\forall x \in \Q: x \times 0 = 0 = 0 \times x$
Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
From [[... | Rational Numbers under Multiplication do not form Group | https://proofwiki.org/wiki/Rational_Numbers_under_Multiplication_do_not_form_Group | https://proofwiki.org/wiki/Rational_Numbers_under_Multiplication_do_not_form_Group | [
"Rational Multiplication",
"Examples of Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Rational Number",
"Definition:Multiplication/Rational Numbers",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Zero (Number)",
"Definition:Zero Element",
"Definition:Abstract Algebra",
"Group with Zero Element is Trivial",
"Definition:Trivial Group",
"Definition:Element",
"Proof by Contradiction",
"Definition:Group"
] |
proofwiki-10384 | Real Numbers under Multiplication do not form Group | The algebraic structure $\struct {\R, \times}$ consisting of the set of real numbers $\R$ under multiplication $\times$ is not a group. | {{AimForCont}} that $\struct {\R, \times}$ is a group.
By the definition of the number $0 \in \R$:
:$\forall x \in \R: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\R, \times}$ is the trivial group.
But $\R$ contains other elements b... | The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\R, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ under [[Definition:Real Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]]. | {{AimForCont}} that $\struct {\R, \times}$ is a [[Definition:Group|group]].
By the definition of the [[Definition:Zero (Number)|number $0 \in \R$]]:
:$\forall x \in \R: x \times 0 = 0 = 0 \times x$
Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
From [[... | Real Numbers under Multiplication do not form Group | https://proofwiki.org/wiki/Real_Numbers_under_Multiplication_do_not_form_Group | https://proofwiki.org/wiki/Real_Numbers_under_Multiplication_do_not_form_Group | [
"Real Multiplication",
"Examples of Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Real Number",
"Definition:Multiplication/Real Numbers",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Zero (Number)",
"Definition:Zero Element",
"Definition:Abstract Algebra",
"Group with Zero Element is Trivial",
"Definition:Trivial Group",
"Definition:Element",
"Proof by Contradiction",
"Definition:Group"
] |
proofwiki-10385 | Complex Numbers under Multiplication do not form Group | The algebraic structure $\struct {\C, \times}$ consisting of the set of complex numbers $\C$ under multiplication $\times$ is not a group. | {{AimForCont}} that $\struct {\C, \times}$ is a group.
By the definition of the number $0 \in \C$:
:$\forall x \in \C: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\C, \times}$ is the trivial group.
But $\C$ contains other elements b... | The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\C, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ under [[Definition:Complex Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]]. | {{AimForCont}} that $\struct {\C, \times}$ is a [[Definition:Group|group]].
By the definition of the [[Definition:Zero (Number)|number $0 \in \C$]]:
:$\forall x \in \C: x \times 0 = 0 = 0 \times x$
Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
From [[... | Complex Numbers under Multiplication do not form Group | https://proofwiki.org/wiki/Complex_Numbers_under_Multiplication_do_not_form_Group | https://proofwiki.org/wiki/Complex_Numbers_under_Multiplication_do_not_form_Group | [
"Complex Multiplication",
"Examples of Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Complex Number",
"Definition:Multiplication/Complex Numbers",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Zero (Number)",
"Definition:Zero Element",
"Definition:Abstract Algebra",
"Group with Zero Element is Trivial",
"Definition:Trivial Group",
"Definition:Element",
"Proof by Contradiction",
"Definition:Group"
] |
proofwiki-10386 | Symmetry Group of Square is Group | The symmetry group of the square is a non-abelian group. | Let us refer to this group as $D_4$.
Taking the group axioms in turn: | The [[Definition:Symmetry Group of Square|symmetry group of the square]] is a [[Definition:Abelian Group|non-abelian]] [[Definition:Group|group]]. | Let us refer to this group as $D_4$.
Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Symmetry Group of Square is Group | https://proofwiki.org/wiki/Symmetry_Group_of_Square_is_Group | https://proofwiki.org/wiki/Symmetry_Group_of_Square_is_Group | [
"Symmetry Group of Square"
] | [
"Definition:Symmetry Group of Square",
"Definition:Abelian Group",
"Definition:Group",
"Definition:Symmetry Group of Square",
"Definition:Symmetry Group of Square"
] | [
"Axiom:Group Axioms"
] |
proofwiki-10387 | Dihedral Group is Group | Let $D_n$ be the dihedral group of order $2 n$.
Then $D_n$ is indeed a group. | $D_n$ is by definition the symmetry group of the regular $n$-gon.
The result follows from Symmetry Group is Group.
{{qed}} | Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order $2 n$]].
Then $D_n$ is indeed a [[Definition:Group|group]]. | $D_n$ is by definition the [[Definition:Symmetry Group|symmetry group]] of the [[Definition:Regular Polygon|regular $n$-gon]].
The result follows from [[Symmetry Group is Group]].
{{qed}} | Dihedral Group is Group | https://proofwiki.org/wiki/Dihedral_Group_is_Group | https://proofwiki.org/wiki/Dihedral_Group_is_Group | [
"Dihedral Groups"
] | [
"Definition:Dihedral Group",
"Definition:Order of Structure",
"Definition:Group"
] | [
"Definition:Symmetry Group",
"Definition:Polygon/Regular",
"Symmetry Group is Group"
] |
proofwiki-10388 | Symmetric Group is Subgroup of Monoid of Self-Maps | Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself
Let $\struct {\Gamma \paren S, \circ}$ denote the symmetric group on $S$.
Let $\struct {S^S, \circ}$ be the monoid of self-maps under composition of mappings.
Then $\struct {\Gamma \paren S, \circ}$ is a subgroup of $\struct {S^S, \circ}$. | By Symmetric Group is Group, $\struct {\Gamma \paren S, \circ}$ is a group.
Let $\phi \in \Gamma \paren S$ be a permutation on $S$.
As a permutation is a self-map, it follows that $\phi \in S^S$.
Thus by definition $\Gamma \paren S$ is a subset of $S^S$.
So by definition, $\Gamma \paren S$, is a subgroup of $\struct {S... | Let $S$ be a [[Definition:Set|set]].
Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself
Let $\struct {\Gamma \paren S, \circ}$ denote the [[Definition:Symmetric Group|symmetric group on $S$]].
Let $\struct {S^S, \circ}$ be the [[Definition:Monoid|monoid]] of [[Definition:Self-... | By [[Symmetric Group is Group]], $\struct {\Gamma \paren S, \circ}$ is a [[Definition:Group|group]].
Let $\phi \in \Gamma \paren S$ be a [[Definition:Permutation|permutation]] on $S$.
As a [[Definition:Permutation|permutation]] is a [[Definition:Self-Map|self-map]], it follows that $\phi \in S^S$.
Thus by definition... | Symmetric Group is Subgroup of Monoid of Self-Maps | https://proofwiki.org/wiki/Symmetric_Group_is_Subgroup_of_Monoid_of_Self-Maps | https://proofwiki.org/wiki/Symmetric_Group_is_Subgroup_of_Monoid_of_Self-Maps | [
"Symmetric Groups",
"Subgroups",
"Monoids"
] | [
"Definition:Set",
"Definition:Set of All Mappings",
"Definition:Symmetric Group",
"Definition:Monoid",
"Definition:Self-Map",
"Definition:Composition of Mappings",
"Definition:Subgroup"
] | [
"Symmetric Group is Group",
"Definition:Group",
"Definition:Permutation",
"Definition:Permutation",
"Definition:Self-Map",
"Definition:Subset",
"Definition:Subgroup",
"Category:Symmetric Groups",
"Category:Subgroups",
"Category:Monoids"
] |
proofwiki-10389 | Cancellability by Cayley Table | Let $\struct {S, \circ}$ be a finite algebraic structure.
Let $\TT$ be the Cayley table for $\left({S, \circ}\right)$.
Let $a \in S$ be an element of $S$.
Then $a$ is cancellable for $\circ$ {{iff}}:
:$(1): \quad$ no element of $S$ is repeated in $\TT$ in the row headed by $a$
and:
:$(2): \quad$ no element of $S$ is re... | === Necessary Condition ===
Let $a \in S$ be cancellable for $\circ$.
Suppose there exists $x \in S$ which appears twice in a row in $\TT$ headed by $a$.
Thus by definition of the structure of a Cayley table:
:$\exists y_1, y_2 \in S: a \circ y_1 = x = a \circ y_2$
such that $y_1 \ne y_2$.
That contradicts the stipulat... | Let $\struct {S, \circ}$ be a [[Definition:Finite Algebraic Structure|finite algebraic structure]].
Let $\TT$ be the [[Definition:Cayley Table|Cayley table]] for $\left({S, \circ}\right)$.
Let $a \in S$ be an [[Definition:Element|element]] of $S$.
Then $a$ is [[Definition:Cancellable Element|cancellable]] for $\cir... | === Necessary Condition ===
Let $a \in S$ be [[Definition:Cancellable Element|cancellable]] for $\circ$.
Suppose there exists $x \in S$ which appears twice in a row in $\TT$ headed by $a$.
Thus by definition of the structure of a [[Definition:Cayley Table|Cayley table]]:
:$\exists y_1, y_2 \in S: a \circ y_1 = x = a... | Cancellability by Cayley Table | https://proofwiki.org/wiki/Cancellability_by_Cayley_Table | https://proofwiki.org/wiki/Cancellability_by_Cayley_Table | [
"Cayley Tables"
] | [
"Definition:Finite Algebraic Structure",
"Definition:Cayley Table",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Element",
"Definition:Element"
] | [
"Definition:Cancellable Element",
"Definition:Cayley Table",
"Proof by Contradiction",
"Definition:Cancellable Element",
"Definition:Element",
"Definition:Cayley Table",
"Proof by Contradiction",
"Definition:Cancellable Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
... |
proofwiki-10390 | Subset not necessarily Submagma | Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$.
Then it is not necessarily the case that:
: $\struct {T, \circ} \subseteq \struct {S, \circ}$
That is, it does not always follow that $\struct {T, \circ}$ is a submagma of $\struct {S, \circ}$. | Let $\struct {\Z, -}$ be the magma which is the set of integers under the operation of subtraction.
We have that the natural numbers $\N$ are a subset of the integers.
Consider $\struct {\N, -}$, the natural numbers under subtraction.
We have that Natural Number Subtraction is not Closed.
For example:
: $1 - 2 = -1 \no... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $T \subseteq S$.
Then it is not necessarily the case that:
: $\struct {T, \circ} \subseteq \struct {S, \circ}$
That is, it does not always follow that $\struct {T, \circ}$ is a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$. | Let $\struct {\Z, -}$ be the [[Definition:Magma|magma]] which is the [[Definition:Set|set]] of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Subtraction|subtraction]].
We have that the [[Definition:Natural Numbers|natural numbers]] $\N$ are a [[Definition:S... | Subset not necessarily Submagma | https://proofwiki.org/wiki/Subset_not_necessarily_Submagma | https://proofwiki.org/wiki/Subset_not_necessarily_Submagma | [
"Magmas",
"Submagmas",
"Subsets"
] | [
"Definition:Magma",
"Definition:Submagma"
] | [
"Definition:Magma",
"Definition:Set",
"Definition:Integer",
"Definition:Operation/Binary Operation",
"Definition:Subtraction/Integers",
"Definition:Natural Numbers",
"Definition:Subset",
"Definition:Integer",
"Definition:Natural Numbers",
"Definition:Subtraction/Natural Numbers",
"Natural Number... |
proofwiki-10391 | Closed Subsets of Symmetry Group of Square | Recall the symmetry group of the square: | Recall that a submagma of an algebraic structure $\SS$ is a subsets of $\SS$ which is closed.
Let $\XX$ be the set of all submagmas of $\SS$.
From Empty Set is Submagma of Magma:
:$\O \in \XX$
From Magma is Submagma of Itself:
:$\SS \in \XX$
From Idempotent Magma Element forms Singleton Submagma:
:$\set e \in \XX$
Let ... | Recall the [[Definition:Symmetry Group of Square|symmetry group of the square]]: | Recall that a [[Definition:Submagma|submagma]] of an [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\SS$ is a [[Definition:Subset|subsets]] of $\SS$ which is [[Definition:Closed Algebraic Structure|closed]].
Let $\XX$ be the [[Definition:Set|set]] of all [[Definition:Submagma|submagmas]] o... | Closed Subsets of Symmetry Group of Square | https://proofwiki.org/wiki/Closed_Subsets_of_Symmetry_Group_of_Square | https://proofwiki.org/wiki/Closed_Subsets_of_Symmetry_Group_of_Square | [
"Symmetry Groups",
"Symmetry Group of Square"
] | [
"Definition:Symmetry Group of Square",
"Definition:Symmetry Group of Square"
] | [
"Definition:Submagma",
"Definition:Algebraic Structure/One Operation",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set",
"Definition:Submagma",
"Empty Set is Submagma of Magma",
"Magma is Submagma of Itself",
"Idempotent Magma Element forms Singleton... |
proofwiki-10392 | Piecewise Continuous Function does not necessarily have Improper Integrals | Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.
Let $f$ be a piecewise continuous function:
{{:Definition:Piecewise Continuous Function}}
Then it is not necessarily the case that $f$ is a piecewise continuous function with improper integrals:
{{:Definition:Piecewise Continuous Functi... | Consider the function:
:<nowiki>$\map f x = \begin{cases}
0 & : x = a \\
\dfrac 1 {x - a} & : x \in \hointl a b
\end{cases}$</nowiki>
Since $\dfrac 1 {x - a}$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.
Therefore, $f$ is a piecewise continuous function for the (finite) subdivision $\set {a, b}... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, $a < b$.
Let $f$ be a [[Definition:Piecewise Continuous Function|piecewise continuous function]]:
{{:Definition:Piecewise Continuous Function}}
Then it is not necessarily the ca... | Consider the function:
:<nowiki>$\map f x = \begin{cases}
0 & : x = a \\
\dfrac 1 {x - a} & : x \in \hointl a b
\end{cases}$</nowiki>
Since $\dfrac 1 {x - a}$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$, $f$ is [[Definition:Continuous Real Function on Open Interval|continu... | Piecewise Continuous Function does not necessarily have Improper Integrals | https://proofwiki.org/wiki/Piecewise_Continuous_Function_does_not_necessarily_have_Improper_Integrals | https://proofwiki.org/wiki/Piecewise_Continuous_Function_does_not_necessarily_have_Improper_Integrals | [
"Piecewise Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function",
"Definition:Piecewise Continuous Function/Improper Integrals"
] | [
"Definition:Continuous Real Function/Open Interval",
"Definition:Continuous Real Function/Open Interval",
"Definition:Piecewise Continuous Function",
"Definition:Subdivision of Interval/Finite",
"Definition:Improper Integral/Open Interval",
"Definition:Improper Integral/Open Interval",
"Definition:Piece... |
proofwiki-10393 | Right Cancellable Element is Right Cancellable in Subset | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be right cancellable in $S$.
Then $x$ is also right cancellable in $T$. | Let $x \in T$ be right cancellable in $S$.
That is:
:$\forall a, b \in S: a \circ x = b \circ x \implies a = b$
Therefore:
:$\forall c, d \in T: c \circ x = d \circ x \implies c = d$
Thus $x$ is right cancellable in $T$.
{{qed}}
Category:Abstract Algebra
nrnys061i0ttzgrwzrlslxd3d9q0jel | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be [[Definition:Right Cancellable Element|right cancellable]] in $S$.
Then $x$ is also [[Definition:Right Cancellable Element|right cancellab... | Let $x \in T$ be [[Definition:Right Cancellable Element|right cancellable]] in $S$.
That is:
:$\forall a, b \in S: a \circ x = b \circ x \implies a = b$
Therefore:
:$\forall c, d \in T: c \circ x = d \circ x \implies c = d$
Thus $x$ is [[Definition:Right Cancellable Element|right cancellable]] in $T$.
{{qed}}
[[Cat... | Right Cancellable Element is Right Cancellable in Subset | https://proofwiki.org/wiki/Right_Cancellable_Element_is_Right_Cancellable_in_Subset | https://proofwiki.org/wiki/Right_Cancellable_Element_is_Right_Cancellable_in_Subset | [
"Abstract Algebra"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Cancellable Element/Right Cancellable"
] | [
"Definition:Cancellable Element/Right Cancellable",
"Definition:Cancellable Element/Right Cancellable",
"Category:Abstract Algebra"
] |
proofwiki-10394 | Left Cancellable Element is Left Cancellable in Subset | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be left cancellable in $S$.
Then $x$ is also left cancellable in $T$. | Let $x \in T$ be left cancellable in $S$.
That is:
:$\forall a, b \in S: x \circ a = x \circ b \implies a = b$
Therefore:
:$\forall c, d \in T: x \circ c = x \circ d \implies c = d$
Thus $x$ is left cancellable in $T$.
{{qed}}
Category:Abstract Algebra
qf6y06293gf4xvsea4gad3uofljsg07 | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be [[Definition:Left Cancellable Element|left cancellable]] in $S$.
Then $x$ is also [[Definition:Left Cancellable Element|left cancellable]]... | Let $x \in T$ be [[Definition:Left Cancellable Element|left cancellable]] in $S$.
That is:
:$\forall a, b \in S: x \circ a = x \circ b \implies a = b$
Therefore:
:$\forall c, d \in T: x \circ c = x \circ d \implies c = d$
Thus $x$ is [[Definition:Left Cancellable Element|left cancellable]] in $T$.
{{qed}}
[[Categor... | Left Cancellable Element is Left Cancellable in Subset | https://proofwiki.org/wiki/Left_Cancellable_Element_is_Left_Cancellable_in_Subset | https://proofwiki.org/wiki/Left_Cancellable_Element_is_Left_Cancellable_in_Subset | [
"Abstract Algebra"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Left Cancellable"
] | [
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Left Cancellable",
"Category:Abstract Algebra"
] |
proofwiki-10395 | Left Cancellable Elements of Semigroup form Subsemigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$.
Then $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$. | Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$:
:$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$
Let $x, y \in C_\lambda$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ a
| r = \paren {x \circ y} \circ b
| c =
}}
{{eqn | ll= ... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $C_\lambda$ be the set of [[Definition:Left Cancellable Element|left cancellable elements]] of $\struct {S, \circ}$.
Then $\struct {C_\lambda, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$. | Let $C_\lambda$ be the set of [[Definition:Left Cancellable Element|left cancellable elements]] of $\struct {S, \circ}$:
:$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$
Let $x, y \in C_\lambda$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ a
| r = \paren {x... | Left Cancellable Elements of Semigroup form Subsemigroup | https://proofwiki.org/wiki/Left_Cancellable_Elements_of_Semigroup_form_Subsemigroup | https://proofwiki.org/wiki/Left_Cancellable_Elements_of_Semigroup_form_Subsemigroup | [
"Subsemigroups",
"Cancellability"
] | [
"Definition:Semigroup",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Subsemigroup"
] | [
"Definition:Cancellable Element/Left Cancellable",
"Definition:Associative Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Category:Subsemigroups",
"Category:Cancellability"
] |
proofwiki-10396 | Right Cancellable Elements of Semigroup form Subsemigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$.
Then $\struct {C_\rho, \circ}$ is a subsemigroup of $\struct {S, \circ}$. | Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$:
:$C_\rho = \set {x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}$
Let $x, y \in C_\rho$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ \paren {x \circ y}
| r = b \circ \paren {x \circ y}
| c =
}}
{{eqn | ll= \leadst... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $C_\rho$ be the set of [[Definition:Right Cancellable Element|right cancellable elements]] of $\struct {S, \circ}$.
Then $\struct {C_\rho, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$. | Let $C_\rho$ be the set of [[Definition:Right Cancellable Element|right cancellable elements]] of $\struct {S, \circ}$:
:$C_\rho = \set {x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}$
Let $x, y \in C_\rho$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ \paren {x \circ y}
| r = b \circ \paren {... | Right Cancellable Elements of Semigroup form Subsemigroup | https://proofwiki.org/wiki/Right_Cancellable_Elements_of_Semigroup_form_Subsemigroup | https://proofwiki.org/wiki/Right_Cancellable_Elements_of_Semigroup_form_Subsemigroup | [
"Subsemigroups",
"Cancellability"
] | [
"Definition:Semigroup",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Subsemigroup"
] | [
"Definition:Cancellable Element/Right Cancellable",
"Definition:Associative Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Category:Subsemigroups",
"Category:Cancellability"
] |
proofwiki-10397 | Surjection iff Right Cancellable/Necessary Condition | Let $f$ be a surjection.
Then $f$ is right cancellable. | Let $f: X \to Y$ be surjective.
Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.
As $f$ is a surjection:
:$\Img f = Y$
by definition.
But in order for $h_1 \circ f$ to be defined, it is necessary that $Y = \Dom {h_1}$.
Similarly, for $h_2 \circ f$ to be defined, it is necessary that $Y = \Dom {h_2}$.
So it ... | Let $f$ be a [[Definition:Surjection|surjection]].
Then $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Let $f: X \to Y$ be [[Definition:Surjection|surjective]].
Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.
As $f$ is a [[Definition:Surjection|surjection]]:
:$\Img f = Y$
by definition.
But in order for $h_1 \circ f$ to be [[Definition:Composition of Mappings|defined]], it is necessary that $Y = \Dom {h... | Surjection iff Right Cancellable/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition/Proof_1 | [
"Surjection iff Right Cancellable"
] | [
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Surjection",
"Definition:Surjection",
"Definition:Composition of Mappings",
"Definition:Composition of Mappings",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
... |
proofwiki-10398 | Surjection iff Right Cancellable/Necessary Condition | Let $f$ be a surjection.
Then $f$ is right cancellable. | Let $f: X \to Y$ be surjective.
Then from Surjection iff Right Inverse:
:$\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = h \circ I_Y
| c =
}}
{{eqn | r = h \circ \paren {f \circ g}
| c =
}}
... | Let $f$ be a [[Definition:Surjection|surjection]].
Then $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Let $f: X \to Y$ be [[Definition:Surjection|surjective]].
Then from [[Surjection iff Right Inverse]]:
:$\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = h \circ I_Y
| c =
}}
{{eqn | r = h \circ \... | Surjection iff Right Cancellable/Necessary Condition/Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition/Proof_2 | [
"Surjection iff Right Cancellable"
] | [
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Surjection",
"Surjection iff Right Inverse",
"Composition of Mappings is Associative",
"Composition of Mappings is Associative",
"Definition:Right Cancellable Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] |
proofwiki-10399 | Surjection iff Right Cancellable/Sufficient Condition | Let $f$ be a mapping which is right cancellable.
Then $f$ is a surjection. | Suppose $f$ is a mapping which is not surjective.
Then:
:$\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows.
:$\map {h_1} y = a: y \in Y$
:<nowiki>$\map {h_2} y = \begin {cases}
a & : y \ne y_1 \\
b & : y = y_1
\end {cases}$</nowiki>
Thus we have ... | Let $f$ be a [[Definition:Mapping|mapping]] which is [[Definition:Right Cancellable Mapping|right cancellable]].
Then $f$ is a [[Definition:Surjection|surjection]]. | Suppose $f$ is a [[Definition:Mapping|mapping]] which is not [[Definition:Surjection|surjective]].
Then:
:$\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows.
:$\map {h_1} y = a: y \in Y$
:<nowiki>$\map {h_2} y = \begin {cases}
a & : y \ne y_... | Surjection iff Right Cancellable/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition/Proof_1 | [
"Surjection iff Right Cancellable"
] | [
"Definition:Mapping",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping",
"Rule of Transposition",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] |
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