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-5400 | Canonical Injection is Injection/General Result | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be algebraic structures with identities $e_1, e_2, \ldots, e_j, \ldots, e_n$ respectively.
The canonical injection:
:$\ds \inj_j: \struct {S_j, \circ_j} \to \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}... | Let:
:$x, y \in S_j: \map {\inj_j} x = \map {\inj_j} y$
Then:
:$\tuple {e_1, e_2, \dotsc, e_{j - 1}, x, e_{j + 1}, \dotsc, e_n} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, y, e_{j + 1}, \dotsc, e_n}$
By Equality of Ordered Tuples, it follows directly that:
:$x = y$
Thus the canonical injections are injective.
{{Qed}} | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identities]] $e_1, e_2, \ldots, e_j, \ldots, e_n$ respectively.
The [[Definition:Canonica... | Let:
:$x, y \in S_j: \map {\inj_j} x = \map {\inj_j} y$
Then:
:$\tuple {e_1, e_2, \dotsc, e_{j - 1}, x, e_{j + 1}, \dotsc, e_n} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, y, e_{j + 1}, \dotsc, e_n}$
By [[Equality of Ordered Tuples]], it follows directly that:
:$x = y$
Thus the [[Definition:Canonical Injection (Abstract ... | Canonical Injection is Injection/General Result | https://proofwiki.org/wiki/Canonical_Injection_is_Injection/General_Result | https://proofwiki.org/wiki/Canonical_Injection_is_Injection/General_Result | [
"Canonical Injections"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Injection"
] | [
"Equality of Ordered Tuples",
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Injection"
] |
proofwiki-5401 | Intermediate Value Theorem (Topology) | Let $X$ be a connected topological space.
Let $\struct {Y, \preceq, \tau}$ be a totally ordered set equipped with the order topology.
Let $f: X \to Y$ be a continuous mapping.
Let $a$ and $b$ are two points of $a, b \in X$ such that:
:$\map f a \prec \map f b$
Let:
:$r \in Y: \map f a \prec r \prec \map f b$
Then there... | Let $a, b \in X$, and let $r \in Y$ lie between $\map f a$ and $\map f b$.
Define the sets:
:$A = f \sqbrk X \cap r^\prec$ and $B = f \sqbrk X \cap r^\succ$
where $r^\prec$ and $r^\succ$ denote the strict lower closure and strict upper closure respectively of $r$ in $Y$.
$A$ and $B$ are disjoint by construction.
$A$ an... | Let $X$ be a [[Definition:Connected Topological Space|connected topological space]].
Let $\struct {Y, \preceq, \tau}$ be a [[Definition:Totally Ordered Set|totally ordered set]] equipped with the [[Definition:Order Topology|order topology]].
Let $f: X \to Y$ be a [[Definition:Continuous Mapping (Topology)|continuous ... | Let $a, b \in X$, and let $r \in Y$ lie between $\map f a$ and $\map f b$.
Define the [[Definition:Set|sets]]:
:$A = f \sqbrk X \cap r^\prec$ and $B = f \sqbrk X \cap r^\succ$
where $r^\prec$ and $r^\succ$ denote the [[Definition:Strict Lower Closure of Element|strict lower closure]] and [[Definition:Strict Upper Clos... | Intermediate Value Theorem (Topology) | https://proofwiki.org/wiki/Intermediate_Value_Theorem_(Topology) | https://proofwiki.org/wiki/Intermediate_Value_Theorem_(Topology) | [
"Connected Topological Spaces",
"Continuous Mappings",
"Order Topologies"
] | [
"Definition:Connected Topological Space",
"Definition:Totally Ordered Set",
"Definition:Order Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Element",
"Definition:Element"
] | [
"Definition:Set",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Upper Closure/Element",
"Definition:Disjoint Sets",
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Set Intersection",
"Definition:Open Set/Topology",
"Definition:Element",
"Definition:Separatio... |
proofwiki-5402 | Finite Union of Countable Sets is Countable | The union of a finite number of countable sets is countable. | Let $S_0, \ldots, S_{n - 1}$ be countable sets.
For $i \in \set {0, \ldots, n - 1}$, let $f_i: \N \to S_i$ be a surjection.
These exist by Surjection from Natural Numbers iff Countable.
Now define $f: \N \to \ds \bigcup_{i \mathop = 0}^{n - 1} S_i$ by:
:$\map f m := \map {f_i} {\floor {\dfrac m n} }$
where:
:$i$ is the... | The [[Definition:Set Union|union]] of a [[Definition:Finite Set|finite number]] of [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]]. | Let $S_0, \ldots, S_{n - 1}$ be [[Definition:Countable Set|countable sets]].
For $i \in \set {0, \ldots, n - 1}$, let $f_i: \N \to S_i$ be a [[Definition:Surjection|surjection]].
These exist by [[Surjection from Natural Numbers iff Countable]].
Now define $f: \N \to \ds \bigcup_{i \mathop = 0}^{n - 1} S_i$ by:
:$\... | Finite Union of Countable Sets is Countable | https://proofwiki.org/wiki/Finite_Union_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Finite_Union_of_Countable_Sets_is_Countable | [
"Set Union",
"Countable Sets"
] | [
"Definition:Set Union",
"Definition:Finite Set",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Surjection",
"Surjection from Natural Numbers iff Countable",
"Definition:Unique",
"Definition:Element",
"Definition:Floor Function",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Set Union",
"Definition:Surjection",
"Definition:S... |
proofwiki-5403 | Binomial Coefficient with Zero | :$\forall r \in \R: \dbinom r 0 = 1$ | From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
But when $k = 0$, we have:
:$\ds \prod_{j \mathop = 0}^{-1} \paren {x - j} = 1$... | :$\forall r \in \R: \dbinom r 0 = 1$ | From the [[Definition:Binomial Coefficient/Real Numbers|definition of binomial coefficients]]:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the [[Definition:Falling Factorial|falling factorial]].
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x ... | Binomial Coefficient with Zero | https://proofwiki.org/wiki/Binomial_Coefficient_with_Zero | https://proofwiki.org/wiki/Binomial_Coefficient_with_Zero | [
"Examples of Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient/Real Numbers",
"Definition:Falling Factorial",
"Definition:Continued Product/Vacuous Product",
"Definition:Factorial"
] |
proofwiki-5404 | Binomial Coefficient with One | :$\forall r \in \R: \dbinom r 1 = r$ | From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
But when $k = 1$, we have:
:$\ds \prod_{j \mathop = 0}^0 \paren {x - j} = \pare... | :$\forall r \in \R: \dbinom r 1 = r$ | From the [[Definition:Binomial Coefficient/Real Numbers|definition of binomial coefficients]]:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the [[Definition:Falling Factorial|falling factorial]].
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - ... | Binomial Coefficient with One | https://proofwiki.org/wiki/Binomial_Coefficient_with_One | https://proofwiki.org/wiki/Binomial_Coefficient_with_One | [
"Examples of Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient/Real Numbers",
"Definition:Falling Factorial",
"Definition:Natural Numbers",
"Definition:Factorial"
] |
proofwiki-5405 | Binomial Coefficient with Self | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes Iverson's convention.
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | From the definition of binomial coefficient:
:$\dbinom n n = \dfrac {n!} {n! \ \paren {n - n}!} = \dfrac {n!} {n! \ 0!}$
the result following directly from the definition of the factorial, where $0! = 1$.
From N Choose Negative Number is Zero:
:$\forall k \in \Z_{<0}: \dbinom n k = 0$
So for $n < 0$:
:$\dbinom n n = 0$... | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes [[Definition:Iverson's Convention|Iverson's convention]].
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | From the definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
:$\dbinom n n = \dfrac {n!} {n! \ \paren {n - n}!} = \dfrac {n!} {n! \ 0!}$
the result following directly from the definition of the [[Definition:Factorial|factorial]], where $0! = 1$.
From [[N Choose Negative Number is Zero]]:
:$\forall... | Binomial Coefficient with Self | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self | [
"Examples of Binomial Coefficients"
] | [
"Definition:Iverson's Convention"
] | [
"Definition:Binomial Coefficient",
"Definition:Factorial",
"N Choose Negative Number is Zero"
] |
proofwiki-5406 | Binomial Coefficient with Self | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes Iverson's convention.
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | The case where $n = 1$ can be taken separately.
From Binomial Coefficient with Zero:
:$\dbinom 1 0 = 1$
demonstrating that the result holds for $n = 1$.
Let $n \in \N: n > 1$.
From the definition of binomial coefficients:
:$\dbinom n {n - 1} = \dfrac {n!} {\paren {n - 1}! \paren {n - \paren {n - 1} }!} = \dfrac {n!} {\... | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes [[Definition:Iverson's Convention|Iverson's convention]].
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | The case where $n = 1$ can be taken separately.
From [[Binomial Coefficient with Zero]]:
:$\dbinom 1 0 = 1$
demonstrating that the result holds for $n = 1$.
Let $n \in \N: n > 1$.
From the [[Definition:Binomial Coefficient|definition of binomial coefficients]]:
:$\dbinom n {n - 1} = \dfrac {n!} {\paren {n - 1}! \p... | Binomial Coefficient with Self minus One/Proof 1 | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self_minus_One/Proof_1 | [
"Examples of Binomial Coefficients"
] | [
"Definition:Iverson's Convention"
] | [
"Binomial Coefficient with Zero",
"Definition:Binomial Coefficient",
"Definition:Factorial"
] |
proofwiki-5407 | Binomial Coefficient with Self | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes Iverson's convention.
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | From Cardinality of Set of Subsets, $\dbinom n {n - 1}$ is the number of combination of things taken $n - 1$ at a time.
Choosing $n - 1$ things from $n$ is the same thing as choosing which $1$ of the elements to be left out.
There are $n$ different choices for that $1$ element.
Therefore there are $n$ ways to choose $n... | :$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes [[Definition:Iverson's Convention|Iverson's convention]].
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$ | From [[Cardinality of Set of Subsets]], $\dbinom n {n - 1}$ is the number of combination of things taken $n - 1$ at a time.
Choosing $n - 1$ things from $n$ is the same thing as choosing which $1$ of the elements to be left out.
There are $n$ different choices for that $1$ element.
Therefore there are $n$ ways to ch... | Binomial Coefficient with Self minus One/Proof 2 | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self | https://proofwiki.org/wiki/Binomial_Coefficient_with_Self_minus_One/Proof_2 | [
"Examples of Binomial Coefficients"
] | [
"Definition:Iverson's Convention"
] | [
"Cardinality of Set of Subsets"
] |
proofwiki-5408 | Binomial Coefficient with Two | :$\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$ | From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
When $k = 2$:
:$\ds \prod_{j \mathop = 0}^1 \paren {x - j} = \frac {\paren {x - ... | :$\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$ | From the [[Definition:Binomial Coefficient/Real Numbers|definition of binomial coefficients]]:
:$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the [[Definition:Falling Factorial|falling factorial]].
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x -... | Binomial Coefficient with Two | https://proofwiki.org/wiki/Binomial_Coefficient_with_Two | https://proofwiki.org/wiki/Binomial_Coefficient_with_Two | [
"Examples of Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient/Real Numbers",
"Definition:Falling Factorial"
] |
proofwiki-5409 | Sum of Binomial Coefficients over Lower Index/Corollary | :$\ds \forall n \in \Z_{\ge 0}: \sum_{i \mathop \in \Z} \binom n i = 2^n$ | From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\dbinom n i = 0$.
The result follows directly from Sum of Binomial Coefficients over Lower Index.
{{qed}}
Category:Binomial Coefficients
Category:Sum of Binomial Coefficients over Lower Index
trh9y7tue8se3pozeep849lc8nqh1ss | :$\ds \forall n \in \Z_{\ge 0}: \sum_{i \mathop \in \Z} \binom n i = 2^n$ | From the definition of the [[Definition:Binomial Coefficient|binomial coefficient]], when $i < 0$ and $i > n$ we have $\dbinom n i = 0$.
The result follows directly from [[Sum of Binomial Coefficients over Lower Index]].
{{qed}}
[[Category:Binomial Coefficients]]
[[Category:Sum of Binomial Coefficients over Lower Ind... | Sum of Binomial Coefficients over Lower Index/Corollary | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index/Corollary | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index/Corollary | [
"Binomial Coefficients",
"Sum of Binomial Coefficients over Lower Index"
] | [] | [
"Definition:Binomial Coefficient",
"Sum of Binomial Coefficients over Lower Index",
"Category:Binomial Coefficients",
"Category:Sum of Binomial Coefficients over Lower Index"
] |
proofwiki-5410 | Binomial Theorem/Integral Index | Let $X$ be one of the standard number systems $\N$, $\Z$, $\Q$, $\R$ or $\C$.
Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
| l = \paren {x + y}^n
| r = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k
| c =
}}
{{eqn | r = x^n + \binom n 1 x^{n - 1} y + \binom n 2 x^{n - 2... | === Basis for the Induction ===
For $n = 0$ we have:
:$\ds \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$
This is the basis for the induction. | Let $X$ be one of the [[Definition:Standard Number System|standard number systems]] $\N$, $\Z$, $\Q$, $\R$ or $\C$.
Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
| l = \paren {x + y}^n
| r = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k
| c =
}}
{{eqn | r = x^n + \bi... | === Basis for the Induction ===
For $n = 0$ we have:
:$\ds \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$
This is the [[Definition:Basis for the Induction|basis for the induction]]. | Binomial Theorem/Integral Index | https://proofwiki.org/wiki/Binomial_Theorem/Integral_Index | https://proofwiki.org/wiki/Binomial_Theorem/Integral_Index | [
"Binomial Theorem",
"Proofs by Induction"
] | [
"Definition:Number",
"Definition:Binomial Coefficient"
] | [
"Definition:Basis for the Induction"
] |
proofwiki-5411 | Binomial Theorem/Ring Theory | Let $\struct {R, +, \odot}$ be a ringoid such that $\struct {R, \odot}$ is a commutative semigroup.
Let $n \in \Z: n \ge 2$.
Then:
:$\ds \forall x, y \in R: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$
where $\dbinom n k = \d... | First we establish the result for when $\struct {R, \odot}$ has an identity element $e$.
For $n = 0$ we have:
:$\ds \odot^0 \paren {x + y} = e = {0 \choose 0} \paren {\odot^{0 - 0} x} \odot \paren {\odot^0 y} = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0 - k} \odot y^k$
For $n = 1$ we have:
:$\ds \odot^1 \paren {x + y} =... | Let $\struct {R, +, \odot}$ be a [[Definition:Ringoid (Abstract Algebra)|ringoid]] such that $\struct {R, \odot}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $n \in \Z: n \ge 2$.
Then:
:$\ds \forall x, y \in R: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \binom n k \pa... | First we establish the result for when $\struct {R, \odot}$ has an [[Definition:Identity Element|identity element]] $e$.
For $n = 0$ we have:
:$\ds \odot^0 \paren {x + y} = e = {0 \choose 0} \paren {\odot^{0 - 0} x} \odot \paren {\odot^0 y} = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0 - k} \odot y^k$
For $n = 1$ we ... | Binomial Theorem/Ring Theory | https://proofwiki.org/wiki/Binomial_Theorem/Ring_Theory | https://proofwiki.org/wiki/Binomial_Theorem/Ring_Theory | [
"Binomial Theorem",
"Proofs by Induction",
"Ring Theory"
] | [
"Definition:Ringoid (Abstract Algebra)",
"Definition:Commutative Semigroup",
"Definition:Binomial Coefficient",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-5412 | Derivative of Absolute Value Function | Let $\size x$ be the absolute value of $x$ for real $x$.
Then:
:$\dfrac \d {\d x} \size x = \dfrac x {\size x}$
for $x \ne 0$. | {{begin-eqn}}
{{eqn | l = \frac \d {\d x} \size x
| r = \frac \d {\d x} \sqrt{x^2}
| c = Square of Real Number is Non-Negative
}}
{{eqn | r = \frac \d {\d x} \paren {x^2}^{\frac 1 2}
}}
{{eqn | r = \frac 1 2 \paren {x^2}^{-\frac 1 2} \cdot 2 x
| c = Chain Rule for Derivatives
}}
{{eqn | r = \frac x {\... | Let $\size x$ be the [[Definition:Absolute Value|absolute value]] of $x$ for [[Definition:Real Number|real]] $x$.
Then:
:$\dfrac \d {\d x} \size x = \dfrac x {\size x}$
for $x \ne 0$. | {{begin-eqn}}
{{eqn | l = \frac \d {\d x} \size x
| r = \frac \d {\d x} \sqrt{x^2}
| c = [[Square of Real Number is Non-Negative]]
}}
{{eqn | r = \frac \d {\d x} \paren {x^2}^{\frac 1 2}
}}
{{eqn | r = \frac 1 2 \paren {x^2}^{-\frac 1 2} \cdot 2 x
| c = [[Chain Rule for Derivatives]]
}}
{{eqn | r = \f... | Derivative of Absolute Value Function | https://proofwiki.org/wiki/Derivative_of_Absolute_Value_Function | https://proofwiki.org/wiki/Derivative_of_Absolute_Value_Function | [
"Derivative of Absolute Value Function",
"Absolute Value Function",
"Derivatives"
] | [
"Definition:Absolute Value",
"Definition:Real Number"
] | [
"Square of Real Number is Non-Negative",
"Derivative of Composite Function",
"Definition:Derivative/Real Function/Derivative at Point",
"Limit iff Limits from Left and Right"
] |
proofwiki-5413 | Inverse Element in Inverse Completion of Commutative Monoid | Let $\struct {S, \circ}$ be a commutative monoid.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then the inverse of an element of $S$ which is invertible for $\circ$ i... | Let the identity of $\struct {S, \circ}$ be $e$.
Let $z$ be the inverse of $y$ for $\circ$:
:$z \circ y = e$
:$y \circ z = e$
From Identity of Inverse Completion of Commutative Monoid:
:$z \circ' y = e$
:$y \circ' z = e$
Hence $z$ is the inverse of $y$ for $\circ'$.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Commutative Monoid|commutative monoid]].
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an [[Definition:Inv... | Let the [[Definition:Identity Element|identity]] of $\struct {S, \circ}$ be $e$.
Let $z$ be the [[Definition:Inverse Element|inverse]] of $y$ for $\circ$:
:$z \circ y = e$
:$y \circ z = e$
From [[Identity of Inverse Completion of Commutative Monoid]]:
:$z \circ' y = e$
:$y \circ' z = e$
Hence $z$ is the [[Definition... | Inverse Element in Inverse Completion of Commutative Monoid | https://proofwiki.org/wiki/Inverse_Element_in_Inverse_Completion_of_Commutative_Monoid | https://proofwiki.org/wiki/Inverse_Element_in_Inverse_Completion_of_Commutative_Monoid | [
"Inverse Completions"
] | [
"Definition:Commutative Monoid",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Identity of Inverse Completion of Commutative Monoid",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-5414 | Construction of Inverse Completion/Congruence Relation | The cross-relation $\boxtimes$ is a congruence relation on $\struct {S \times C, \oplus}$.
=== Members of Equivalence Classes ===
{{:Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes}}
=== Equivalence Class of Equal Elements ===
{{:Construction of Inverse Completion/Equivalence Rela... | From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Cross-Relation is Congruence Relation.
{{qe... | The [[Definition:Cross-Relation|cross-relation]] $\boxtimes$ is a [[Definition:Congruence Relation|congruence relation]] on $\struct {S \times C, \oplus}$.
=== [[Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes|Members of Equivalence Classes]] ===
{{:Construction of Inverse Compl... | From [[Semigroup is Subsemigroup of Itself]], $\struct {S, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
Also from [[Semigroup is Subsemigroup of Itself]], $\struct {C, \circ {\restriction_C} }$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {C, \circ {\restriction_C} }$.
T... | Construction of Inverse Completion/Congruence Relation | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Congruence_Relation | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Congruence_Relation | [
"Congruence Relations",
"Construction of Inverse Completion"
] | [
"Definition:Cross-Relation",
"Definition:Congruence Relation",
"Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes",
"Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements"
] | [
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Cross-Relation is Congruence Relation"
] |
proofwiki-5415 | Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes | $\forall x, y \in S, a, b \in C:$
:$(1): \quad \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
:$(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtime... | From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.
Hence the equivalence class of $\tuple {x, y}$ under $\boxtimes$ is defined for all $\tuple {x, y} \in S \times C$.
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
A... | $\forall x, y \in S, a, b \in C:$
:$(1): \quad \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
:$(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
where $\eqclass {\tuple {x, y} } \boxtimes$ is the [[Definition:Equivalence Class|equivalence class]... | From [[Cross-Relation is Equivalence Relation]] we have that $\boxtimes$ is an [[Definition:Equivalence Relation|equivalence relation]].
Hence the [[Definition:Equivalence Class|equivalence class]] of $\tuple {x, y}$ under $\boxtimes$ is defined for all $\tuple {x, y} \in S \times C$.
From [[Semigroup is Subsemigrou... | Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation/Members_of_Equivalence_Classes | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation/Members_of_Equivalence_Classes | [
"Construction of Inverse Completion"
] | [
"Definition:Equivalence Class"
] | [
"Cross-Relation is Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Elements of Cross-Relation Equivalence Class"
] |
proofwiki-5416 | Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements | :$\forall c, d \in C: \tuple {c, c} \boxtimes \tuple {d, d}$ | From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Equivalence Class of Equal Elements of Cros... | :$\forall c, d \in C: \tuple {c, c} \boxtimes \tuple {d, d}$ | From [[Semigroup is Subsemigroup of Itself]], $\struct {S, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
Also from [[Semigroup is Subsemigroup of Itself]], $\struct {C, \circ {\restriction_C} }$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {C, \circ {\restriction_C} }$.
T... | Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation/Equivalence_Class_of_Equal_Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation/Equivalence_Class_of_Equal_Elements | [
"Construction of Inverse Completion"
] | [] | [
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Equivalence Class of Equal Elements of Cross-Relation"
] |
proofwiki-5417 | Construction of Inverse Completion/Quotient Structure is Commutative Semigroup | :$\struct {T', \oplus'}$ is a commutative semigroup. | The quotient epimorphism from $\struct {S \times C, \oplus}$ onto $\struct {T', \oplus'}$ is given by:
:$q_\boxtimes: \struct {S \times C, \oplus} \to \struct {T', \oplus'}: \map {q_\boxtimes} {x, y} = \eqclass {\tuple {x, y} } \boxtimes$
where, by definition:
{{begin-eqn}}
{{eqn | q = \forall \tuple {x_1, y_1}, \tuple... | :$\struct {T', \oplus'}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]. | The [[Definition:Quotient Epimorphism|quotient epimorphism]] from $\struct {S \times C, \oplus}$ onto $\struct {T', \oplus'}$ is given by:
:$q_\boxtimes: \struct {S \times C, \oplus} \to \struct {T', \oplus'}: \map {q_\boxtimes} {x, y} = \eqclass {\tuple {x, y} } \boxtimes$
where, by definition:
{{begin-eqn}}
{{eqn ... | Construction of Inverse Completion/Quotient Structure is Commutative Semigroup | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure_is_Commutative_Semigroup | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure_is_Commutative_Semigroup | [
"Construction of Inverse Completion"
] | [
"Definition:Commutative Semigroup"
] | [
"Definition:Quotient Epimorphism",
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Epimorphism Preserves Associativity",
"Definition:Associative Operation",
"Epimorphism Preserves Commutativity",
"Definition:Commutative/Operation",
"Definition:Commu... |
proofwiki-5418 | Construction of Inverse Completion/Quotient Mapping is Injective | Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Then $\psi: S \to T'$ is an injection, and does not depend on the particular element $a$ chosen. | {{begin-eqn}}
{{eqn | l = \map \psi x
| r = \map \psi y
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a \in C
| l = \eqclass {\tuple {x \circ a, a} } \boxtimes
| r = \eqclass {\tuple {y \circ a, a} } \boxtimes
| c = Definition of $\eqclass {\tuple {x, y} } \boxtimes$
}}
{{eqn | ll= \l... | Let the [[Definition:Mapping|mapping]] $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Then $\psi: S \to T'$ is an [[Definition:Injection|injection]], and does not depend on the particular element $a$ chosen. | {{begin-eqn}}
{{eqn | l = \map \psi x
| r = \map \psi y
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a \in C
| l = \eqclass {\tuple {x \circ a, a} } \boxtimes
| r = \eqclass {\tuple {y \circ a, a} } \boxtimes
| c = Definition of $\eqclass {\tuple {x, y} } \boxtimes$
}}
{{eqn | ll= \l... | Construction of Inverse Completion/Quotient Mapping is Injective | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_is_Injective | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_is_Injective | [
"Examples of Injections",
"Construction of Inverse Completion"
] | [
"Definition:Mapping",
"Definition:Injection"
] | [
"Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes",
"Definition:Injection"
] |
proofwiki-5419 | Construction of Inverse Completion/Quotient Mapping is Monomorphism | The mapping $\psi: S \to T'$ is a monomorphism. | We have that this quotient mapping $\psi: S \to T'$ is an injection.
Let $x, y \in S$. Then:
{{begin-eqn}}
{{eqn | l = \map \psi x \oplus' \map \psi y
| r = \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {y \circ a, a} } \boxtimes
| c = Definition of $\psi$
}}
{{eqn | r = \eqclass {\tu... | The [[Definition:Mapping|mapping]] $\psi: S \to T'$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]]. | We have that this [[Construction of Inverse Completion/Quotient Mapping is Injective|quotient mapping $\psi: S \to T'$ is an injection]].
Let $x, y \in S$. Then:
{{begin-eqn}}
{{eqn | l = \map \psi x \oplus' \map \psi y
| r = \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {y \circ a, a} }... | Construction of Inverse Completion/Quotient Mapping is Monomorphism | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_is_Monomorphism | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_is_Monomorphism | [
"Construction of Inverse Completion",
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Mapping",
"Definition:Monomorphism (Abstract Algebra)"
] | [
"Construction of Inverse Completion/Quotient Mapping is Injective",
"Definition:Commutative/Operation",
"Definition:Morphism Property",
"Definition:Injection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Monomorphism (Abstract Algebra)"
] |
proofwiki-5420 | Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism | Let $S'$ be the image $\psi \sqbrk S$ of $S$.
Then $\psi$ is an isomorphism from $S$ onto $S'$. | From Quotient Mapping is Monomorphism, $\psi: \struct {S, \circ} \to \struct {S', \oplus'}$ is a monomorphism.
Therefore by definition:
:$\psi$ is a homomorphism
:$\psi$ is an injection.
Because $S'$ is the image of $\psi \sqbrk S$, by Surjection by Restriction of Codomain $\psi$ is a surjection.
Therefore by definitio... | Let $S'$ be the [[Definition:Image of Subset under Mapping|image]] $\psi \sqbrk S$ of $S$.
Then $\psi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $S$ onto $S'$. | From [[Construction of Inverse Completion/Quotient Mapping is Monomorphism|Quotient Mapping is Monomorphism]], $\psi: \struct {S, \circ} \to \struct {S', \oplus'}$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]].
Therefore by definition:
:$\psi$ is a [[Definition:Homomorphism (Abstract Algebra)|homomo... | Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_to_Image_is_Isomorphism | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping_to_Image_is_Isomorphism | [
"Construction of Inverse Completion",
"Isomorphisms (Abstract Algebra)"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Construction of Inverse Completion/Quotient Mapping is Monomorphism",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Injection",
"Restriction of Mapping to Image is Surjection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Isomo... |
proofwiki-5421 | Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements | The set $C'$ of cancellable elements of the semigroup $S'$ is $\psi \sqbrk C$. | We have Morphism Property Preserves Cancellability.
Thus:
:$c \in C \implies \map \psi c \in C'$
So by Image of Subset under Mapping is Subset of Image:
:$\psi \sqbrk C \subseteq C'$
From above, $\psi$ is an isomorphism.
{{explain|Above where?}}
Hence, also from Morphism Property Preserves Cancellability:
:$c' \in C' \... | The [[Definition:Set|set]] $C'$ of [[Definition:Cancellable Element|cancellable elements]] of the [[Definition:Semigroup|semigroup]] $S'$ is $\psi \sqbrk C$. | We have [[Morphism Property Preserves Cancellability]].
Thus:
:$c \in C \implies \map \psi c \in C'$
So by [[Image of Subset under Mapping is Subset of Image]]:
:$\psi \sqbrk C \subseteq C'$
From above, $\psi$ is an [[Definition:Semigroup Isomorphism|isomorphism]].
{{explain|Above where?}}
Hence, also from [[Morph... | Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping/Image_of_Cancellable_Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Mapping/Image_of_Cancellable_Elements | [
"Construction of Inverse Completion"
] | [
"Definition:Set",
"Definition:Cancellable Element",
"Definition:Semigroup"
] | [
"Morphism Property Preserves Cancellability",
"Image of Subset under Mapping is Subset of Image",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomorphism",
"Morphism Property Preserves Cancellability",
"Preimage of Subset is Subset of Preimage",
"Definition:Set Equality"
] |
proofwiki-5422 | Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup | Let $S'$ be the image $\psi \sqbrk S$ of $S$.
Then $\struct {S', \oplus'}$ is a subsemigroup of $\struct {T', \oplus'}$. | We have that $S'$ is the image $\psi \sqbrk S$ of $S$.
For $\struct {S', \oplus'}$ to be a subsemigroup of $\struct {T', \oplus'}$, by Subsemigroup Closure Test we need to show that $\struct {S', \oplus'}$ is closed.
Let $x, y \in S'$.
Then $x = \map \phi {x'}, y = \map \phi {y'}$ for some $x', y' \in S$.
But as $\phi$... | Let $S'$ be the [[Definition:Image of Subset under Mapping|image]] $\psi \sqbrk S$ of $S$.
Then $\struct {S', \oplus'}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {T', \oplus'}$. | We have that $S'$ is the [[Definition:Image of Subset under Mapping|image]] $\psi \sqbrk S$ of $S$.
For $\struct {S', \oplus'}$ to be a [[Definition:Subsemigroup|subsemigroup]] of $\struct {T', \oplus'}$, by [[Subsemigroup Closure Test]] we need to show that $\struct {S', \oplus'}$ is [[Definition:Closed Algebraic Str... | Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Image_of_Quotient_Mapping_is_Subsemigroup | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Image_of_Quotient_Mapping_is_Subsemigroup | [
"Construction of Inverse Completion"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Subsemigroup"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Subsemigroup",
"Subsemigroup Closure Test",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Morphism Property",
"Subsemigroup Closure Test",
"Definition:Subsemigroup"
] |
proofwiki-5423 | Construction of Inverse Completion/Identity of Quotient Structure | Let $c \in C$ be arbitrary.
Then:
:$\eqclass {\tuple {c, c} } \boxtimes$
is the identity of $T'$. | {{begin-eqn}}
{{eqn | l = \paren {x \circ c} \circ y
| r = x \circ \paren {c \circ y}
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x \circ \paren {y \circ c}
| c = {{Defof|Commutative Semigroup}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {\tuple {x, y} } \boxtimes \oplus' \eqclass {\tuple {c, c} } \b... | Let $c \in C$ be arbitrary.
Then:
:$\eqclass {\tuple {c, c} } \boxtimes$
is the [[Definition:Identity Element|identity]] of $T'$. | {{begin-eqn}}
{{eqn | l = \paren {x \circ c} \circ y
| r = x \circ \paren {c \circ y}
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x \circ \paren {y \circ c}
| c = {{Defof|Commutative Semigroup}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {\tuple {x, y} } \boxtimes \oplus' \eqclass {\tuple {c, c} } \b... | Construction of Inverse Completion/Identity of Quotient Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Identity_of_Quotient_Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Identity_of_Quotient_Structure | [
"Construction of Inverse Completion"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Cancellable Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-5424 | Construction of Inverse Completion/Invertible Elements in Quotient Structure | Every cancellable element of $S'$ is invertible in $T'$. | From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.
Call this identity $e_{T'}$.
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
From Image of Cancel... | Every [[Definition:Cancellable Element|cancellable element]] of $S'$ is [[Definition:Invertible Element|invertible]] in $T'$. | From [[Construction of Inverse Completion/Identity of Quotient Structure|Identity of Quotient Structure]], $\struct {T', \oplus'}$ has an [[Definition:Identity Element|identity]], and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.
Call this identity $e_{T'}$.
Let the [[Definition:Mapping|mapping]] $\... | Construction of Inverse Completion/Invertible Elements in Quotient Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Invertible_Elements_in_Quotient_Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Invertible_Elements_in_Quotient_Structure | [
"Construction of Inverse Completion"
] | [
"Definition:Cancellable Element",
"Definition:Invertible Element"
] | [
"Construction of Inverse Completion/Identity of Quotient Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Mapping",
"Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements",
"Definition:Surjection",
"Cancellable Elements of Semigroup form Subs... |
proofwiki-5425 | Construction of Inverse Completion/Generator for Quotient Structure | $T' = S' \cup \paren {C'}^{-1}$ is a generator for the semigroup $T'$. | Let $\tuple {x, y} \in S \times C$. Then:
{{begin-eqn}}
{{eqn | l = \map \psi x \oplus' \paren {\map \psi y}^{-1}
| r = \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ y} } \boxtimes
| c = Invertible Elements in Quotient Structure above
}}
{{eqn | r = \eqclass {\tuple {x \ci... | $T' = S' \cup \paren {C'}^{-1}$ is a [[Definition:Generator of Semigroup|generator]] for the [[Definition:Semigroup|semigroup]] $T'$. | Let $\tuple {x, y} \in S \times C$. Then:
{{begin-eqn}}
{{eqn | l = \map \psi x \oplus' \paren {\map \psi y}^{-1}
| r = \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ y} } \boxtimes
| c = [[Construction of Inverse Completion/Invertible Elements in Quotient Structure|Invert... | Construction of Inverse Completion/Generator for Quotient Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Generator_for_Quotient_Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Generator_for_Quotient_Structure | [
"Construction of Inverse Completion"
] | [
"Definition:Generator of Subsemigroup",
"Definition:Semigroup"
] | [
"Construction of Inverse Completion/Invertible Elements in Quotient Structure",
"Definition:Commutative/Operation",
"Definition:Cancellable Element"
] |
proofwiki-5426 | Construction of Inverse Completion/Quotient Structure is Inverse Completion | $T'$ is an inverse completion of its subsemigroup $S'$. | Every cancellable element of $S'$ is invertible in $T'$, from Invertible Elements in Quotient Structure.
$T' = S' \cup \paren {C'}^{-1}$ is a generator for the semigroup $T'$, from Generator for Quotient Structure.
Hence the result, by definition of inverse completion
{{Qed}} | $T'$ is an [[Definition:Inverse Completion|inverse completion]] of its [[Definition:Subsemigroup|subsemigroup]] $S'$. | Every [[Definition:Cancellable Element|cancellable element]] of $S'$ is [[Definition:Invertible Element|invertible]] in $T'$, from [[Construction of Inverse Completion/Invertible Elements in Quotient Structure|Invertible Elements in Quotient Structure]].
$T' = S' \cup \paren {C'}^{-1}$ is a generator for the [[Definit... | Construction of Inverse Completion/Quotient Structure is Inverse Completion | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure_is_Inverse_Completion | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure_is_Inverse_Completion | [
"Construction of Inverse Completion"
] | [
"Definition:Inverse Completion",
"Definition:Subsemigroup"
] | [
"Definition:Cancellable Element",
"Definition:Invertible Element",
"Construction of Inverse Completion/Invertible Elements in Quotient Structure",
"Definition:Semigroup",
"Construction of Inverse Completion/Generator for Quotient Structure",
"Definition:Inverse Completion"
] |
proofwiki-5427 | Ring of Integers Modulo Composite is not Integral Domain | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Let $m$ be a composite number.
Then $\struct {\Z_m, +, \times}$ is not an integral domain. | Let $m \in \Z: m \ge 2$ be composite.
Then:
:$\exists k, l \in \N_{> 0}: 1 < k < m, 1 < l < m: m = k \times l$
Thus:
{{begin-eqn}}
{{eqn | l = \eqclass 0 m
| r = \eqclass m m
| c =
}}
{{eqn | r = \eqclass {k l} m
| c =
}}
{{eqn | r = \eqclass k m \times \eqclass l m
| c =
}}
{{end-eqn}}
So $\... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Let $m$ be a [[Definition:Composite Number|composite number]].
Then $\struct {\Z_m, +, \times}$ is not an [[Definition:Integral Domain|integral domain]]. | Let $m \in \Z: m \ge 2$ be [[Definition:Composite Number|composite]].
Then:
:$\exists k, l \in \N_{> 0}: 1 < k < m, 1 < l < m: m = k \times l$
Thus:
{{begin-eqn}}
{{eqn | l = \eqclass 0 m
| r = \eqclass m m
| c =
}}
{{eqn | r = \eqclass {k l} m
| c =
}}
{{eqn | r = \eqclass k m \times \eqclass l m... | Ring of Integers Modulo Composite is not Integral Domain | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Composite_is_not_Integral_Domain | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Composite_is_not_Integral_Domain | [
"Ring of Integers Modulo m",
"Integral Domains"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Composite Number",
"Definition:Integral Domain"
] | [
"Definition:Composite Number",
"Definition:Ring (Abstract Algebra)",
"Definition:Zero Divisor/Ring",
"Definition:Integral Domain"
] |
proofwiki-5428 | Ring Zero is not Cancellable | Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $0$ be the ring zero of $R$.
Then $0$ is not a cancellable element for the ring product $\circ$. | {{AimForCont}} $0$ is cancellable.
Let $a, b \in R$ such that $a \ne b$.
By definition of ring zero:
:$0 \circ a = 0 = 0 \circ b$
By our supposition that $0$ is cancellable:
:$a = b$
The result follows by Proof by Contradiction.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] which is [[Definition:Non-Null Ring|not null]].
Let $0$ be the [[Definition:Ring Zero|ring zero]] of $R$.
Then $0$ is not a [[Definition:Cancellable Element|cancellable element]] for the [[Definition:Ring Product|ring product]] $\circ$. | {{AimForCont}} $0$ is [[Definition:Cancellable Element|cancellable]].
Let $a, b \in R$ such that $a \ne b$.
By definition of [[Definition:Ring Zero|ring zero]]:
:$0 \circ a = 0 = 0 \circ b$
By our supposition that $0$ is [[Definition:Cancellable Element|cancellable]]:
:$a = b$
The result follows by [[Proof by Contr... | Ring Zero is not Cancellable | https://proofwiki.org/wiki/Ring_Zero_is_not_Cancellable | https://proofwiki.org/wiki/Ring_Zero_is_not_Cancellable | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Null Ring",
"Definition:Ring Zero",
"Definition:Cancellable Element",
"Definition:Ring (Abstract Algebra)/Product"
] | [
"Definition:Cancellable Element",
"Definition:Ring Zero",
"Definition:Cancellable Element",
"Proof by Contradiction"
] |
proofwiki-5429 | Congruence Relation on Ring induces Ideal | Let $\struct {R, +, \circ}$ be a ring.
Let $\EE$ be a congruence relation on $R$.
Let $J = \eqclass {0_R} \EE$ be the equivalence class of $0_R$ under $\EE$.
Then $J$ is an ideal of $R$. | Let $J = \eqclass {0_R} \EE$.
By Congruence Relation induces Normal Subgroup, $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$.
Thus the elements of $\struct {R, +} / \struct {J, +}$ are the cosets of $\eqclass {0_R} \EE$ by $+$.
We have that $\EE$ is also compatible with $\circ$.
Thus from Quotient Structure ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\EE$ be a [[Definition:Congruence Relation|congruence relation]] on $R$.
Let $J = \eqclass {0_R} \EE$ be the [[Definition:Equivalence Class|equivalence class]] of $0_R$ under $\EE$.
Then $J$ is an [[Definition:Ideal of Ring|ideal]] ... | Let $J = \eqclass {0_R} \EE$.
By [[Congruence Relation induces Normal Subgroup]], $\struct {J, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {R, +}$.
Thus the elements of $\struct {R, +} / \struct {J, +}$ are the [[Definition:Coset|cosets]] of $\eqclass {0_R} \EE$ by $+$.
We have that $\EE$ is... | Congruence Relation on Ring induces Ideal | https://proofwiki.org/wiki/Congruence_Relation_on_Ring_induces_Ideal | https://proofwiki.org/wiki/Congruence_Relation_on_Ring_induces_Ideal | [
"Congruence Relations",
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Congruence Relation",
"Definition:Equivalence Class",
"Definition:Ideal of Ring"
] | [
"Congruence Relation induces Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Coset",
"Definition:Relation Compatible with Operation",
"Quotient Structure is Well-Defined",
"Definition:Ideal of Ring"
] |
proofwiki-5430 | Ideal induces Congruence Relation on Ring | Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$
Then $J$ induces a congruence relation $\EE_J$ on $R$ such that $\struct {R / J, +, \circ}$ is a quotient ring. | From Ideal is Additive Normal Subgroup, we have that $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$.
Let $x \mathop {\EE_J} y$ denote that $x$ and $y$ are in the same coset, that is:
:$x \mathop {\EE_J} y \iff x + N = y + N$
From Congruence Modulo Normal Subgroup is Congruence Relation, $\EE_J$ is a congruen... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$
Then $J$ induces a [[Definition:Congruence Relation|congruence relation]] $\EE_J$ on $R$ such that $\struct {R / J, +, \circ}$ is a [[Definition:Quotient Ring|quotient ring]]. | From [[Ideal is Additive Normal Subgroup]], we have that $\struct {J, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {R, +}$.
Let $x \mathop {\EE_J} y$ denote that $x$ and $y$ are in the same [[Definition:Coset|coset]], that is:
:$x \mathop {\EE_J} y \iff x + N = y + N$
From [[Congruence Modulo N... | Ideal induces Congruence Relation on Ring | https://proofwiki.org/wiki/Ideal_induces_Congruence_Relation_on_Ring | https://proofwiki.org/wiki/Ideal_induces_Congruence_Relation_on_Ring | [
"Congruence Relations",
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Congruence Relation",
"Definition:Quotient Ring"
] | [
"Ideal is Additive Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Coset",
"Congruence Modulo Normal Subgroup is Congruence Relation",
"Definition:Congruence Relation",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence Relation",
"Definition:Quotient Ring"
] |
proofwiki-5431 | Vector Space has Basis between Linearly Independent Set and Spanning Set | Let $V$ be a vector space over a field $F$.
Let $L$ be a linearly independent subset of $V$.
Let $S$ be a set that spans $V$.
Suppose that $L \subseteq S$.
Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$. | Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion.
Note that $L \in \mathscr I$, so $\mathscr I \ne \O$.
Let $\mathscr C$ be a chain in $\mathscr I$.
Let $C = \bigcup \mathscr C$.
{{AimForCont}} that $C$ is linearly dependent.
Then there exist $v_1, v_2, \ldots, v... | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
Let $L$ be a [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subset]] of $V$.
Let $S$ be a [[Definition:Spanning Set of Vector Space|set that spans $V$]].
Suppose that $L ... | Let $\mathscr I$ be the set of [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subsets]] of $S$ that contain $L$, [[Subset Relation is Ordering|ordered by inclusion]].
Note that $L \in \mathscr I$, so $\mathscr I \ne \O$.
Let $\mathscr C$ be a [[Definition:Chain of Subsets|chain]] in ... | Vector Space has Basis between Linearly Independent Set and Spanning Set | https://proofwiki.org/wiki/Vector_Space_has_Basis_between_Linearly_Independent_Set_and_Spanning_Set | https://proofwiki.org/wiki/Vector_Space_has_Basis_between_Linearly_Independent_Set_and_Spanning_Set | [
"Generators of Vector Spaces",
"Bases of Vector Spaces"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Linearly Independent/Set",
"Definition:Subset",
"Definition:Generator of Vector Space",
"Definition:Basis of Vector Space"
] | [
"Definition:Linearly Independent/Set",
"Definition:Subset",
"Subset Relation is Ordering",
"Definition:Chain of Subsets",
"Definition:Linearly Dependent/Set",
"Definition:Chain of Subsets",
"Definition:Linearly Dependent/Set",
"Definition:Contradiction",
"Definition:Linearly Independent/Set",
"Zor... |
proofwiki-5432 | Null Ring is Trivial Ring | Let $R$ be the null ring.
Then $R$ is a trivial ring. | We have that $R$ is the null ring.
That is, by definition it has a single element, which can be denoted $0_R$, such that:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and the ring product are defined as:
{{begin-eqn}}
{{eqn | l = 0_R + 0_R
| r = 0_R
}}
{{eqn | l = 0_R \circ 0_R
| r = 0_R
}}
{{... | Let $R$ be the [[Definition:Null Ring|null ring]].
Then $R$ is a [[Definition:Trivial Ring|trivial ring]]. | We have that $R$ is the [[Definition:Null Ring|null ring]].
That is, by definition it has a single [[Definition:Element|element]], which can be denoted $0_R$, such that:
:$R := \struct {\set {0_R}, +, \circ}$
where [[Definition:Ring Addition|ring addition]] and the [[Definition:Ring Product|ring product]] are defined ... | Null Ring is Trivial Ring | https://proofwiki.org/wiki/Null_Ring_is_Trivial_Ring | https://proofwiki.org/wiki/Null_Ring_is_Trivial_Ring | [
"Trivial Rings",
"Null Ring"
] | [
"Definition:Null Ring",
"Definition:Trivial Ring"
] | [
"Definition:Null Ring",
"Definition:Element",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure/One Operation",
"Definition:Trivial Group",
"Definition:Trivial Ring"
] |
proofwiki-5433 | Null Ring is Ring | Let $R$ be the null ring.
That is, let:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and ring product are defined as:
{{begin-eqn}}
{{eqn | l = 0_R + 0_R
| r = 0_R
}}
{{eqn | l = 0_R \circ 0_R
| r = 0_R
}}
{{end-eqn}}
Then $R$ is a ring. | A null ring is a trivial ring.
So, by Trivial Ring is Commutative Ring, the result follows.
{{qed}}
Category:Null Ring
74kr1nos4nmsxq3pbnyepbpfj63dpjm | Let $R$ be the [[Definition:Null Ring|null ring]].
That is, let:
:$R := \struct {\set {0_R}, +, \circ}$
where [[Definition:Ring Addition|ring addition]] and [[Definition:Ring Product|ring product]] are defined as:
{{begin-eqn}}
{{eqn | l = 0_R + 0_R
| r = 0_R
}}
{{eqn | l = 0_R \circ 0_R
| r = 0_R
}}
{{en... | A [[Null Ring is Trivial Ring|null ring is a trivial ring]].
So, by [[Trivial Ring is Commutative Ring]], the result follows.
{{qed}}
[[Category:Null Ring]]
74kr1nos4nmsxq3pbnyepbpfj63dpjm | Null Ring is Ring | https://proofwiki.org/wiki/Null_Ring_is_Ring | https://proofwiki.org/wiki/Null_Ring_is_Ring | [
"Null Ring"
] | [
"Definition:Null Ring",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Ring (Abstract Algebra)"
] | [
"Null Ring is Trivial Ring",
"Trivial Ring is Commutative Ring",
"Category:Null Ring"
] |
proofwiki-5434 | Trivial Group is Cyclic Group | The trivial group is a cyclic group. | In Trivial Group is Group it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = e$ is in fact a group.
It remains to be shown that it is cyclic.
In order for $G$ to be a cyclic group, every element $x$ of $G$ has to be expressible in the form $x = g^n$ for some $g \in G$ and some $n ... | The [[Definition:Trivial Group|trivial group]] is a [[Definition:Cyclic Group|cyclic group]]. | In [[Trivial Group is Group]] it is shown that the [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\set e, \circ}$ such that $e \circ e = e$ is in fact a [[Definition:Group|group]].
It remains to be shown that it is [[Definition:Cyclic Group|cyclic]].
In order for $G$ to be a [[De... | Trivial Group is Cyclic Group | https://proofwiki.org/wiki/Trivial_Group_is_Cyclic_Group | https://proofwiki.org/wiki/Trivial_Group_is_Cyclic_Group | [
"Trivial Group",
"Cyclic Groups"
] | [
"Definition:Trivial Group",
"Definition:Cyclic Group"
] | [
"Trivial Group is Group",
"Definition:Algebraic Structure/One Operation",
"Definition:Group",
"Definition:Cyclic Group",
"Definition:Cyclic Group",
"Definition:Element",
"Definition:Integer",
"Definition:Element",
"Definition:Cyclic Group"
] |
proofwiki-5435 | Trivial Ring is Commutative Ring | Let $\struct {R, +, \circ}$ be a trivial ring.
Then $\struct {R, +, \circ}$ is a commutative ring. | First we need to show that a trivial ring is actually a ring in the first place.
Taking the ring axioms in turn: | Let $\struct {R, +, \circ}$ be a [[Definition:Trivial Ring|trivial ring]].
Then $\struct {R, +, \circ}$ is a [[Definition:Commutative Ring|commutative ring]]. | First we need to show that a [[Definition:Trivial Ring|trivial ring]] is actually a [[Definition:Ring (Abstract Algebra)|ring]] in the first place.
Taking the [[Axiom:Ring Axioms|ring axioms]] in turn: | Trivial Ring is Commutative Ring | https://proofwiki.org/wiki/Trivial_Ring_is_Commutative_Ring | https://proofwiki.org/wiki/Trivial_Ring_is_Commutative_Ring | [
"Ring Theory",
"Commutative Rings",
"Trivial Rings"
] | [
"Definition:Trivial Ring",
"Definition:Commutative Ring"
] | [
"Definition:Trivial Ring",
"Definition:Ring (Abstract Algebra)",
"Axiom:Ring Axioms",
"Definition:Trivial Ring"
] |
proofwiki-5436 | Congruence Relation and Ideal are Equivalent | Let $\struct {R, +, \circ}$ be a ring.
Let $\EE$ be an equivalence relation on $R$ compatible with both $\circ$ and $+$, that is, a congruence relation on $R$.
Let $J = \eqclass {0_R} \EE$ be the equivalence class of $0_R$ under $\EE$.
Then:
: $(1a): \quad J = \eqclass {0_R} \EE$ is an ideal of $R$
: $(2a): \quad$ The ... | === Part $(1a)$ ===
This is shown on Congruence Relation on Ring induces Ideal.
{{qed|lemma}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\EE$ be an [[Definition:Equivalence Relation|equivalence relation]] on $R$ [[Definition:Relation Compatible with Operation|compatible]] with both $\circ$ and $+$, that is, a [[Definition:Congruence Relation|congruence relation]] on $R$... | === Part $(1a)$ ===
This is shown on [[Congruence Relation on Ring induces Ideal]].
{{qed|lemma}} | Congruence Relation and Ideal are Equivalent | https://proofwiki.org/wiki/Congruence_Relation_and_Ideal_are_Equivalent | https://proofwiki.org/wiki/Congruence_Relation_and_Ideal_are_Equivalent | [
"Ideal Theory",
"Congruence Relations"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Relation Compatible with Operation",
"Definition:Congruence Relation",
"Definition:Equivalence Class",
"Definition:Ideal of Ring",
"Definition:Equivalence Relation",
"Definition:Quotient Ring",
"Definition:Ideal of ... | [
"Congruence Relation on Ring induces Ideal"
] |
proofwiki-5437 | Quotient Ring is Ring | Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $R / J$ is also a ring. | First, it is to be shown that $+$ and $\circ$ are in fact well-defined operations on $R / J$. | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] of $R$ by $J$.
Then $R / J$ is also a [[Definition:Ring (Abstract Algebra)|ring]]. | First, it is to be shown that $+$ and $\circ$ are in fact [[Definition:Well-Defined Operation|well-defined operations]] on $R / J$. | Quotient Ring is Ring | https://proofwiki.org/wiki/Quotient_Ring_is_Ring | https://proofwiki.org/wiki/Quotient_Ring_is_Ring | [
"Quotient Rings",
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Ring (Abstract Algebra)"
] | [
"Definition:Well-Defined/Operation"
] |
proofwiki-5438 | Ideal is Additive Normal Subgroup | Let $\left({R, +, \circ}\right)$ be a ring.
Let $J$ be an ideal of $R$.
Then $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$. | As $J$ is an ideal, $\left({J, +}\right)$ is a subgroup of $\left({R, +}\right)$.
By definition of a ring, $\left({R, +}\right)$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
{{qed}}
Category:Ideal Theory
eas96ddq9bmt95eqmyswfp6se6v6le4 | Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Then $\left({J, +}\right)$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\left({R, +}\right)$. | As $J$ is an [[Definition:Ideal of Ring|ideal]], $\left({J, +}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({R, +}\right)$.
By definition of a [[Definition:Ring (Abstract Algebra)|ring]], $\left({R, +}\right)$ is [[Definition:Abelian Group|abelian]].
The result follows from [[Subgroup of Abelian Group is ... | Ideal is Additive Normal Subgroup | https://proofwiki.org/wiki/Ideal_is_Additive_Normal_Subgroup | https://proofwiki.org/wiki/Ideal_is_Additive_Normal_Subgroup | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Normal Subgroup"
] | [
"Definition:Ideal of Ring",
"Definition:Subgroup",
"Definition:Ring (Abstract Algebra)",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Category:Ideal Theory"
] |
proofwiki-5439 | Congruence Relation on Ring induces Ring | Let $\struct {R, +, \circ}$ be a ring.
Let $\EE$ be a congruence relation on $R$ for both $+$ and $\circ$.
Let $R / \EE$ be the quotient set of $R$ by $\EE$.
Let $+_\EE$ and $\circ_\EE$ be the operations induced on $R / \EE$ by $+$ and $\circ$ respectively.
Then $\struct {R / \EE, +_\EE, \circ_\EE}$ is a ring. | Let $q_\EE$ be the quotient mapping from $\struct {R, +, \circ}$ to $\struct {R / \EE, +_\EE, \circ_\EE}$.
From Quotient Mapping on Structure is Epimorphism:
:$q_\EE: \struct {R, +} \to \struct {R / \EE, +_\EE}$ is an epimorphism
:$q_\EE: \struct {R, \circ} \to \struct {R / \EE, \circ _\EE}$ is an epimorphism.
As the m... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\EE$ be a [[Definition:Congruence Relation|congruence relation]] on $R$ for both $+$ and $\circ$.
Let $R / \EE$ be the [[Definition:Quotient Set|quotient set of $R$ by $\EE$]].
Let $+_\EE$ and $\circ_\EE$ be the [[Definition:Operatio... | Let $q_\EE$ be the [[Definition:Quotient Mapping|quotient mapping]] from $\struct {R, +, \circ}$ to $\struct {R / \EE, +_\EE, \circ_\EE}$.
From [[Quotient Mapping on Structure is Epimorphism]]:
:$q_\EE: \struct {R, +} \to \struct {R / \EE, +_\EE}$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]
:$q_\E... | Congruence Relation on Ring induces Ring | https://proofwiki.org/wiki/Congruence_Relation_on_Ring_induces_Ring | https://proofwiki.org/wiki/Congruence_Relation_on_Ring_induces_Ring | [
"Congruence Relations",
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Congruence Relation",
"Definition:Quotient Set",
"Definition:Operation Induced on Quotient Set",
"Definition:Ring (Abstract Algebra)"
] | [
"Definition:Quotient Mapping",
"Quotient Mapping on Structure is Epimorphism",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Morphism Property",
"Definition:Epimorphism (Abstract Algebra)",
"Epimorphism Preserves Rings",
"Definition:Ring (Abstrac... |
proofwiki-5440 | Subring is not necessarily Ideal | Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +_S, \circ_S}$ be a subring of $R$.
Then it is not necessarily the case that $S$ is also an ideal of $R$. | Consider the field of real numbers $\struct {\R, +, \times}$.
We have that a field is by definition a ring, hence so is $\struct {\R, +, \times}$.
From Rational Numbers form Subfield of Real Numbers and Integers form Subdomain of Rationals, it follows that the integers $\struct {\Z, +, \times}$ are a subring of $\struc... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {S, +_S, \circ_S}$ be a [[Definition:Subring|subring]] of $R$.
Then it is not necessarily the case that $S$ is also an [[Definition:Ideal of Ring|ideal]] of $R$. | Consider the [[Definition:Field of Real Numbers|field of real numbers]] $\struct {\R, +, \times}$.
We have that a [[Definition:Field (Abstract Algebra)|field]] is by definition a [[Definition:Ring (Abstract Algebra)|ring]], hence so is $\struct {\R, +, \times}$.
From [[Rational Numbers form Subfield of Real Numbers]]... | Subring is not necessarily Ideal | https://proofwiki.org/wiki/Subring_is_not_necessarily_Ideal | https://proofwiki.org/wiki/Subring_is_not_necessarily_Ideal | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring",
"Definition:Ideal of Ring"
] | [
"Definition:Field of Real Numbers",
"Definition:Field (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Rational Numbers form Subfield of Real Numbers",
"Integers form Subdomain of Rationals",
"Definition:Integer",
"Definition:Subring",
"Proof by Counterexample",
"Definition:Ideal of Ring"... |
proofwiki-5441 | Characterization of Integrable Functions | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R, f \in \MM_{\overline \R}$ be a $\Sigma$-measurable function.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \in \map {\LL_{\overline \R} } \mu$, that is, $f$ is $\mu$-integrable}}
{{item|(2):|The positive and negative parts $f^+$ and $f^-$ are $\... | We prove the whole cycle of implications:
:$(1) \implies (2) \quad$ by definition of $(1)$
:$(2) \implies (3)\quad$ because $\size f = f^+ + f^-$ and Integral of Positive Measurable Function is Additive
:$(3) \implies (4)\quad$ because $g:= \size f$ exists
It remains to demonstrate $(4) \implies (1)$.
Let $f \in \MM_... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R, f \in \MM_{\overline \R}$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \in \map {\LL_{\overline \R} } \mu$, that is, $f$ is [[Definition:Measu... | We prove the whole cycle of implications:
:$(1) \implies (2) \quad$ by definition of $(1)$
:$(2) \implies (3)\quad$ because $\size f = f^+ + f^-$ and [[Integral of Positive Measurable Function is Additive]]
:$(3) \implies (4)\quad$ because $g:= \size f$ exists
It remains to demonstrate $(4) \implies (1)$.
Let $... | Characterization of Integrable Functions | https://proofwiki.org/wiki/Characterization_of_Integrable_Functions | https://proofwiki.org/wiki/Characterization_of_Integrable_Functions | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Integrable Function/Measure Space",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Integrable Function/Measure Space",
"Definition:Absolute Value of Mapping/Extended Real-Valued Function",
"Definition:Integr... | [
"Integral of Positive Measurable Function is Additive",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Integral of Positive Measurable Function",
"Definition:Integral of Positive Measurable Function",
"Definition:Integrable Function/Measure Space"
] |
proofwiki-5442 | Quotient Ring is Ring/Quotient Ring Addition is Well-Defined | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $+$ is well-defined on $R / J$, that is:
:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies \paren {x_1 + y_1} + J = \paren {x_2 + y_2} + ... | From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$.
Thus, the quotient group $\struct {R / J, +}$ is defined, and as a Quotient Group is Group, $+$ is well-defined. | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] of... | From [[Ideal is Additive Normal Subgroup]] that $J$ is a [[Definition:Normal Subgroup|normal subgroup]] of $R$ under $+$.
Thus, the [[Definition:Quotient Group|quotient group]] $\struct {R / J, +}$ is defined, and as a [[Quotient Group is Group]], $+$ is [[Definition:Well-Defined Operation|well-defined]]. | Quotient Ring is Ring/Quotient Ring Addition is Well-Defined | https://proofwiki.org/wiki/Quotient_Ring_is_Ring/Quotient_Ring_Addition_is_Well-Defined | https://proofwiki.org/wiki/Quotient_Ring_is_Ring/Quotient_Ring_Addition_is_Well-Defined | [
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Well-Defined/Operation"
] | [
"Ideal is Additive Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Quotient Group is Group",
"Definition:Well-Defined/Operation"
] |
proofwiki-5443 | Implicitly Defined Real-Valued Function | Let $F: \struct {\mathbf X' \subseteq \R^{n + 1} } \to \struct {\mathbb I' \subseteq \R}$ have continuous partial derivatives.
{{explain|Can the language of this be brought into line with existing definitions of implicit functions?}}
Let $\tuple {\mathbf x, z}$ denote an element of $\R^{n + 1}$, where $\mathbf x \in \... | This is a special case of the Implicit Function Theorem.
{{qed}} | Let $F: \struct {\mathbf X' \subseteq \R^{n + 1} } \to \struct {\mathbb I' \subseteq \R}$ have [[Definition:Continuous Mapping (Metric Spaces)|continuous]] [[Definition:Partial Derivative|partial derivatives]].
{{explain|Can the language of this be brought into line with existing definitions of implicit functions?}}
... | This is a special case of the [[Implicit Function Theorem]].
{{qed}} | Implicitly Defined Real-Valued Function | https://proofwiki.org/wiki/Implicitly_Defined_Real-Valued_Function | https://proofwiki.org/wiki/Implicitly_Defined_Real-Valued_Function | [
"Calculus"
] | [
"Definition:Continuous Mapping (Metric Space)",
"Definition:Partial Derivative",
"Definition:Element",
"Definition:Unique",
"Definition:Mapping",
"Definition:Real Interval/Open",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Partial Derivative"
] | [
"Implicit Function Theorem"
] |
proofwiki-5444 | Characteristic Function of Subset | Let $A \subseteq B \subseteq S$.
Then:
:$\forall s \in S: \map {\chi_A} s \le \map {\chi_B} s$
where $\chi$ denotes characteristic function. | Both $\chi_A$ and $\chi_B$ take values in $\set {0, 1}$ as they are characteristic functions.
So if $\map {\chi_A} s = 0$, then the statement of the theorem is automatically satisfied (since both $0 \le 0$ and $0 \le 1$).
Now assume $\map {\chi_A} s = 1$.
By definition of $\chi_A$, this happens {{iff}} $s \in A$.
But a... | Let $A \subseteq B \subseteq S$.
Then:
:$\forall s \in S: \map {\chi_A} s \le \map {\chi_B} s$
where $\chi$ denotes [[Definition:Characteristic Function of Set|characteristic function]]. | Both $\chi_A$ and $\chi_B$ take values in $\set {0, 1}$ as they are [[Definition:Characteristic Function of Set|characteristic functions]].
So if $\map {\chi_A} s = 0$, then the statement of the theorem is automatically satisfied (since both $0 \le 0$ and $0 \le 1$).
Now assume $\map {\chi_A} s = 1$.
By [[Definitio... | Characteristic Function of Subset | https://proofwiki.org/wiki/Characteristic_Function_of_Subset | https://proofwiki.org/wiki/Characteristic_Function_of_Subset | [
"Characteristic Functions"
] | [
"Definition:Characteristic Function (Set Theory)/Set"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Proof by Cases",
"Category:Characteristic Functions"
] |
proofwiki-5445 | Characteristic Function Determined by 0-Fiber | Let $A \subseteq S$.
Let $f: S \to \set {0, 1}$ be a mapping.
Denote by $\chi_A$ the characteristic function on $A$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} \chi_A$}}
{{item|(2):|$\forall s \in S: \map f s {{=}} 0 \iff s \notin A$}}
{{end-itemize}}
Using the notion of a fiber, $(2)$ may also be expressed as:
:$... | === $(1)$ implies $(2)$ ===
Follows directly from the definition of characteristic function.
{{qed|lemma}} | Let $A \subseteq S$.
Let $f: S \to \set {0, 1}$ be a [[Definition:Mapping|mapping]].
Denote by $\chi_A$ the [[Definition:Characteristic Function of Set|characteristic function]] on $A$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} \chi_A$}}
{{item|(2):|$\forall s \in S: \map f s {{=}} 0 \iff s \notin A$}}
{{end-... | === $(1)$ implies $(2)$ ===
Follows directly from the definition of [[Definition:Characteristic Function of Set|characteristic function]].
{{qed|lemma}} | Characteristic Function Determined by 0-Fiber | https://proofwiki.org/wiki/Characteristic_Function_Determined_by_0-Fiber | https://proofwiki.org/wiki/Characteristic_Function_Determined_by_0-Fiber | [
"Characteristic Functions"
] | [
"Definition:Mapping",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Preimage/Mapping/Element"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set"
] |
proofwiki-5446 | Ideal induced by Congruence Relation defines that Congruence | Let $\struct {R, +, \circ}$ be a ring.
Let $\EE$ be a congruence relation on $R$.
Let $J = \eqclass {0_R} \EE$ be the ideal induced by $\EE$.
Then the equivalence defined by the coset space $\struct {R, +} / \struct {J, +}$ is $\EE$ itself. | Let $J = \eqclass {0_R} \EE$.
From Congruence Relation on Ring induces Ideal, we have that $J$ is an ideal of $R$.
From Ideal is Additive Normal Subgroup, we have that $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$.
From Normal Subgroup induced by Congruence Relation defines that Congruence, the equivalence ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\EE$ be a [[Definition:Congruence Relation|congruence relation]] on $R$.
Let $J = \eqclass {0_R} \EE$ be the [[Congruence Relation on Ring induces Ideal|ideal induced by $\EE$]].
Then the [[Definition:Equivalence Relation|equivalence... | Let $J = \eqclass {0_R} \EE$.
From [[Congruence Relation on Ring induces Ideal]], we have that $J$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
From [[Ideal is Additive Normal Subgroup]], we have that $\struct {J, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {R, +}$.
From [[Normal Subgrou... | Ideal induced by Congruence Relation defines that Congruence | https://proofwiki.org/wiki/Ideal_induced_by_Congruence_Relation_defines_that_Congruence | https://proofwiki.org/wiki/Ideal_induced_by_Congruence_Relation_defines_that_Congruence | [
"Congruence Relations",
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Congruence Relation",
"Congruence Relation on Ring induces Ideal",
"Definition:Equivalence Relation",
"Definition:Coset Space"
] | [
"Congruence Relation on Ring induces Ideal",
"Definition:Ideal of Ring",
"Ideal is Additive Normal Subgroup",
"Definition:Normal Subgroup",
"Normal Subgroup induced by Congruence Relation defines that Congruence",
"Definition:Equivalence Relation",
"Definition:Set Partition",
"Definition:Congruence Re... |
proofwiki-5447 | Rational Numbers form Null Set under Lebesgue Measure | Let $\lambda$ be $1$-dimensional Lebesgue measure on $\R$.
Let $\Q$ be the set of rational numbers.
Then:
:$\map \lambda \Q = 0$
that is, $\Q$ is a $\lambda$-null set. | We have that the Rational Numbers are Countably Infinite.
The result follows from Countable Set is Null Set under Lebesgue Measure.
{{qed}} | Let $\lambda$ be [[Definition:Lebesgue Measure|$1$-dimensional Lebesgue measure]] on $\R$.
Let $\Q$ be the set of [[Definition:Rational Number|rational numbers]].
Then:
:$\map \lambda \Q = 0$
that is, $\Q$ is a [[Definition:Null Set|$\lambda$-null set]]. | We have that the [[Rational Numbers are Countably Infinite]].
The result follows from [[Countable Set is Null Set under Lebesgue Measure]].
{{qed}} | Rational Numbers form Null Set under Lebesgue Measure | https://proofwiki.org/wiki/Rational_Numbers_form_Null_Set_under_Lebesgue_Measure | https://proofwiki.org/wiki/Rational_Numbers_form_Null_Set_under_Lebesgue_Measure | [
"Lebesgue Measure",
"Rational Numbers",
"Examples of Null Sets"
] | [
"Definition:Lebesgue Measure",
"Definition:Rational Number",
"Definition:Null Set"
] | [
"Rational Numbers are Countably Infinite",
"Countable Set is Null Set under Lebesgue Measure"
] |
proofwiki-5448 | Cartesian Product of Intersections/General Case | :$\ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i} = \prod_{i \mathop \in I} \paren {S_i \cap T_i}$ | Let $\family {x_i}_{i \mathop \in I} \in \ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i}$.
By definition of intersection, this is equivalent to the conjunction of:
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i \mathop \in I} S_i$
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i... | :$\ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i} = \prod_{i \mathop \in I} \paren {S_i \cap T_i}$ | Let $\family {x_i}_{i \mathop \in I} \in \ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i}$.
By definition of [[Definition:Set Intersection|intersection]], this is equivalent to the [[Definition:Conjunction|conjunction]] of:
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i \mathop \... | Cartesian Product of Intersections/General Case | https://proofwiki.org/wiki/Cartesian_Product_of_Intersections/General_Case | https://proofwiki.org/wiki/Cartesian_Product_of_Intersections/General_Case | [
"Cartesian Product of Intersections"
] | [] | [
"Definition:Set Intersection",
"Definition:Conjunction",
"Definition:Cartesian Product/Finite",
"Definition:Set Intersection",
"Definition:Cartesian Product/Finite",
"Definition:Set Equality",
"Category:Cartesian Product of Intersections"
] |
proofwiki-5449 | Cartesian Product of Semirings of Sets | Let $\SS$ and $\TT$ be semirings of sets.
Then $\SS \times \TT$ is also a semiring of sets.
Here, $\times$ denotes Cartesian product. | Recall the conditions for $\SS \times \TT$ to be a semiring of sets:
:$(1): \quad \O \in \SS \times \TT$
:$(2): \quad \SS \times \TT$ is $\cap$-stable
:$(3'):\quad$ If $A, B \in \SS \times \TT$, then there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS \times \TT$ such that $\ds A \se... | Let $\SS$ and $\TT$ be [[Definition:Semiring of Sets|semirings of sets]].
Then $\SS \times \TT$ is also a [[Definition:Semiring of Sets|semiring of sets]].
Here, $\times$ denotes [[Definition:Cartesian Product|Cartesian product]]. | Recall the conditions for $\SS \times \TT$ to be a [[Definition:Semiring of Sets|semiring of sets]]:
:$(1): \quad \O \in \SS \times \TT$
:$(2): \quad \SS \times \TT$ is [[Definition:Stable under Intersection|$\cap$-stable]]
:$(3'):\quad$ If $A, B \in \SS \times \TT$, then there exists a [[Definition:Finite Sequence|fi... | Cartesian Product of Semirings of Sets | https://proofwiki.org/wiki/Cartesian_Product_of_Semirings_of_Sets | https://proofwiki.org/wiki/Cartesian_Product_of_Semirings_of_Sets | [
"Cartesian Product",
"Semirings of Sets"
] | [
"Definition:Semiring of Sets",
"Definition:Semiring of Sets",
"Definition:Cartesian Product"
] | [
"Definition:Semiring of Sets",
"Definition:Stable under Intersection",
"Definition:Finite Sequence",
"Definition:Pairwise Disjoint",
"Definition:Set",
"Definition:Stable under Intersection",
"Definition:Semiring of Sets",
"Definition:Finite",
"Definition:Finite"
] |
proofwiki-5450 | Congruence Relation on Group induces Normal Subgroup | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the equivalence class of $e$ under $\RR$.
Then:
:$\struct {H, \circ \restriction_H}$ is a normal subgroup of $G$
where $\circ \restriction_H$ denotes the restric... | We are given that $\RR$ is a congruence relation for $\circ$.
From Equivalence Relation is Congruence iff Compatible with Operation, we have:
:$\forall u \in G: x \mathrel \RR y \implies \paren {x \circ u} \mathrel \RR \paren {y \circ u}, \paren {u \circ x} \mathrel \RR \paren {u \circ y}$
==== Proof of being a Subgrou... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the [[Definition:Equivalence Class|equivalence class]] of $e$ under $\R... | We are given that $\RR$ is a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
From [[Equivalence Relation is Congruence iff Compatible with Operation]], we have:
:$\forall u \in G: x \mathrel \RR y \implies \paren {x \circ u} \mathrel \RR \paren {y \circ u}, \paren {u \circ x} \mathrel \RR \paren {... | Congruence Relation on Group induces Normal Subgroup | https://proofwiki.org/wiki/Congruence_Relation_on_Group_induces_Normal_Subgroup | https://proofwiki.org/wiki/Congruence_Relation_on_Group_induces_Normal_Subgroup | [
"Normal Subgroups",
"Congruence Relations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Relation",
"Definition:Equivalence Class",
"Definition:Normal Subgroup",
"Definition:Restriction/Operation"
] | [
"Definition:Congruence Relation",
"Equivalence Relation is Congruence iff Compatible with Operation",
"Definition:Subgroup",
"Definition:Empty Set",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Definition:Group",
"Two-Step Subgroup Test",
"Defini... |
proofwiki-5451 | Normal Subgroup induced by Congruence Relation defines that Congruence | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $\eqclass e \RR$ be the equivalence class of $e$ under $\RR$.
Let $N = \eqclass e \RR$ be the normal subgroup induced by $\RR$.
Then $\RR$ is the equivalence relation $\RR_N$ defined by $N$. | Let $\RR_N$ be the equivalence defined by $N$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \RR
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = e
| o = \RR
| r = \paren {x^{-1} \circ y}
| c = $\RR$ is compatible with $\circ$
}}
{{eqn | ll= \leadsto
| l = \paren {e \circ e}
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
Let $\eqclass e \RR$ be the [[Definition:Equivalence Class|equivalence class]] of $e$ under $\RR$.
Let $N = \eqclass e \RR... | Let $\RR_N$ be the [[Congruence Modulo Subgroup is Equivalence Relation|equivalence defined by $N$]].
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \RR
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = e
| o = \RR
| r = \paren {x^{-1} \circ y}
| c = $\RR$ is [[Definition:Relation Compati... | Normal Subgroup induced by Congruence Relation defines that Congruence | https://proofwiki.org/wiki/Normal_Subgroup_induced_by_Congruence_Relation_defines_that_Congruence | https://proofwiki.org/wiki/Normal_Subgroup_induced_by_Congruence_Relation_defines_that_Congruence | [
"Normal Subgroups",
"Congruence Relations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Relation",
"Definition:Equivalence Class",
"Congruence Relation on Group induces Normal Subgroup",
"Congruence Modulo Subgroup is Equivalence Relation"
] | [
"Congruence Modulo Subgroup is Equivalence Relation",
"Definition:Relation Compatible with Operation",
"Definition:Group",
"Congruence Class Modulo Subgroup is Coset"
] |
proofwiki-5452 | Quotient Structure on Group defined by Congruence equals Quotient Group | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $\struct {G / \RR, \circ_\RR}$ be the quotient structure defined by $\RR$.
Let $N = \eqclass e \RR$ be the normal subgroup induced by $\RR$.
Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$.
... | Let $\eqclass x \RR \in G / \RR$.
By Congruence Relation on Group induces Normal Subgroup:
:$\eqclass x \RR = x N$
where $x N$ is the (left) coset of $N$ in $G$.
Similarly, let:
:$y N \in G / N$
Then from Normal Subgroup induced by Congruence Relation defines that Congruence:
:$y N = \eqclass x \RR$
where:
:$\eqclass x... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
Let $\struct {G / \RR, \circ_\RR}$ be the [[Definition:Quotient Structure|quotient structure defined by $\RR$]].
Let $N = ... | Let $\eqclass x \RR \in G / \RR$.
By [[Congruence Relation on Group induces Normal Subgroup]]:
:$\eqclass x \RR = x N$
where $x N$ is the [[Definition:Left Coset|(left) coset]] of $N$ in $G$.
Similarly, let:
:$y N \in G / N$
Then from [[Normal Subgroup induced by Congruence Relation defines that Congruence]]:
:$y N... | Quotient Structure on Group defined by Congruence equals Quotient Group | https://proofwiki.org/wiki/Quotient_Structure_on_Group_defined_by_Congruence_equals_Quotient_Group | https://proofwiki.org/wiki/Quotient_Structure_on_Group_defined_by_Congruence_equals_Quotient_Group | [
"Normal Subgroups",
"Congruence Relations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Relation",
"Definition:Quotient Structure",
"Congruence Relation on Group induces Normal Subgroup",
"Definition:Quotient Group",
"Definition:Subgroup",
"Definition:Semigroup"
] | [
"Congruence Relation on Group induces Normal Subgroup",
"Definition:Coset/Left Coset",
"Normal Subgroup induced by Congruence Relation defines that Congruence",
"Definition:Equivalence Class",
"Left Congruence Modulo Subgroup is Equivalence Relation"
] |
proofwiki-5453 | Vector Scaled by Zero is Zero Vector | Let $F$ be a field whose zero is $0_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \mathbf v \in \mathbf V: 0_F \ast \mathbf v = \bszero$ | {{begin-eqn}}
{{eqn | l = 0_F \circ \mathbf v
| r = \paren {0_F + 0_F} \ast \mathbf v
| c = {{Field-axiom|A3}}
}}
{{eqn | r = 0_F \ast \mathbf v + 0_F \ast \mathbf v
| c = {{Vector-space-axiom|2}}
}}
{{eqn | ll= \leadsto
| l = 0_F \ast \mathbf v + \paren {-0_F \ast \mathbf v}
| r = \paren... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then:
:$\forall \mathbf v \in \mathbf V: 0_F \as... | {{begin-eqn}}
{{eqn | l = 0_F \circ \mathbf v
| r = \paren {0_F + 0_F} \ast \mathbf v
| c = {{Field-axiom|A3}}
}}
{{eqn | r = 0_F \ast \mathbf v + 0_F \ast \mathbf v
| c = {{Vector-space-axiom|2}}
}}
{{eqn | ll= \leadsto
| l = 0_F \ast \mathbf v + \paren {-0_F \ast \mathbf v}
| r = \paren... | Vector Scaled by Zero is Zero Vector | https://proofwiki.org/wiki/Vector_Scaled_by_Zero_is_Zero_Vector | https://proofwiki.org/wiki/Vector_Scaled_by_Zero_is_Zero_Vector | [
"Vector Algebra"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Vector Space",
"Axiom:Vector Space Axioms"
] | [] |
proofwiki-5454 | Zero Vector Scaled is Zero Vector | Let $\struct {\mathbf V, +, \ast}_K$ be a vector space over a division ring $K$, as defined by the vector space axioms.
Then:
:$\forall \lambda \in K: \lambda \ast \bszero = \bszero$
where $\bszero \in \mathbf V$ is the zero vector. | {{begin-eqn}}
{{eqn | l = \lambda \ast \bszero
| r = \lambda \ast \paren {\bszero + \bszero}
| c = {{Abelian-group-axiom|2||underlying}}
}}
{{eqn | r = \lambda \ast \bszero + \lambda \ast \bszero
| c = {{Vector-space-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = \lambda \ast \bszero + \paren {-\lambda... | Let $\struct {\mathbf V, +, \ast}_K$ be a [[Definition:Vector Space over Division Ring|vector space]] over a [[Definition:Division Ring|division ring]] $K$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then:
:$\forall \lambda \in K: \lambda \ast \bszero = \bszero$
where $\bszero \in \mathbf ... | {{begin-eqn}}
{{eqn | l = \lambda \ast \bszero
| r = \lambda \ast \paren {\bszero + \bszero}
| c = {{Abelian-group-axiom|2||underlying}}
}}
{{eqn | r = \lambda \ast \bszero + \lambda \ast \bszero
| c = {{Vector-space-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = \lambda \ast \bszero + \paren {-\lambda... | Zero Vector Scaled is Zero Vector | https://proofwiki.org/wiki/Zero_Vector_Scaled_is_Zero_Vector | https://proofwiki.org/wiki/Zero_Vector_Scaled_is_Zero_Vector | [
"Zero Vectors"
] | [
"Definition:Vector Space/Division Ring",
"Definition:Division Ring",
"Axiom:Vector Space Axioms",
"Definition:Zero Vector"
] | [] |
proofwiki-5455 | Vector Product is Zero only if Factor is Zero | Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \lambda \in F: \forall \mathbf v \in \mathbf V: \lambda \ast \mathbf v = \bszero \implies \paren {\lambda = 0_F \lor \mathbf v = \mathb... | {{AimForCont}} that:
:$\exists \lambda \in F: \exists \mathbf v \in \mathbf V: \lambda \ast \mathbf v = \bszero \land \lambda \ne 0_F \land \mathbf v \ne \bszero$
which is the negation of the exposition of the theorem.
Utilizing the vector space axioms:
{{begin-eqn}}
{{eqn | l = \lambda \ast \mathbf v
| r = \bsze... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms... | {{AimForCont}} that:
:$\exists \lambda \in F: \exists \mathbf v \in \mathbf V: \lambda \ast \mathbf v = \bszero \land \lambda \ne 0_F \land \mathbf v \ne \bszero$
which is the [[Definition:Logical Not|negation]] of the exposition of the theorem.
Utilizing the [[Axiom:Vector Space Axioms|vector space axioms]]:
{{be... | Vector Product is Zero only if Factor is Zero | https://proofwiki.org/wiki/Vector_Product_is_Zero_only_if_Factor_is_Zero | https://proofwiki.org/wiki/Vector_Product_is_Zero_only_if_Factor_is_Zero | [
"Vector Algebra"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Vector Space",
"Axiom:Vector Space Axioms",
"Definition:Zero Vector"
] | [
"Definition:Logical Not",
"Axiom:Vector Space Axioms",
"Zero Vector Scaled is Zero Vector",
"Proof by Contradiction"
] |
proofwiki-5456 | Baire-Osgood Theorem | Let $X$ be a Baire space.
Let $Y$ be a metrizable topological space
Let $f: X \to Y$ be a mapping which is the pointwise limit of a sequence $\sequence {f_n}$ in $\map C {X, Y}$.
{{explain|$\map C {X, Y}$}}
Let $\map D f$ be the set of points where $f$ is discontinuous.
Then $\map D f$ is a meager subset of $X$. | Let $d$ be a metric on $Y$ generating its topology.
Let $\map {\omega_f} x$ denote the oscillation of $f$ at $x$.
We have:
:$\ds \map D f = \bigcup_{n \mathop = 1}^\infty \set {x \in X: \map {\omega_f} x \ge \frac 1 n}$
which is a countable union of closed sets.
Since we have this expression for $\map D f$, the claim f... | Let $X$ be a [[Definition:Baire Space (Topology)|Baire space]].
Let $Y$ be a [[Definition:Metrizable Topology|metrizable topological space]]
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]] which is the [[Definition:Pointwise Limit|pointwise limit]] of a [[Definition:Sequence|sequence]] $\sequence {f_n}$ in $\map... | Let $d$ be a [[Definition:Metric|metric]] on $Y$ generating its [[Definition:Topology|topology]].
Let $\map {\omega_f} x$ denote the [[Definition:Oscillation at Point on Metric Space|oscillation]] of $f$ at $x$.
We have:
:$\ds \map D f = \bigcup_{n \mathop = 1}^\infty \set {x \in X: \map {\omega_f} x \ge \frac 1 n}$... | Baire-Osgood Theorem | https://proofwiki.org/wiki/Baire-Osgood_Theorem | https://proofwiki.org/wiki/Baire-Osgood_Theorem | [
"Topology",
"Meager Spaces",
"Baire Spaces"
] | [
"Definition:Baire Space (Topology)",
"Definition:Metrizable Space",
"Definition:Mapping",
"Definition:Pointwise Limit",
"Definition:Sequence",
"Definition:Discontinuous (Topology)",
"Definition:Meager Space"
] | [
"Definition:Metric Space/Metric",
"Definition:Topology",
"Definition:Oscillation/Metric Space/Point",
"Definition:Set Union/Countable Union",
"Definition:Closed Set/Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Nowhere Dense",
"Definition:Closed Set/Metric Space"
] |
proofwiki-5457 | Null Ring is Ideal | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Then the null ring $\struct {\set {0_R}, +, \circ}$ is an ideal of $\struct {R, +, \circ}$. | From Null Ring is Subring of Ring, $\struct {\set {0_R}, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
Also, from Ring Product with Zero:
:$\forall x \in R: x \circ 0_R = 0_R = 0_R \circ x \in \set {0_R}$
thus fulfilling the condition for $\struct {\set {0_R}, +, \circ}$ to be an ideal of $\struct {R, +, \circ}$.... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Then the [[Definition:Null Ring|null ring]] $\struct {\set {0_R}, +, \circ}$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$. | From [[Null Ring is Subring of Ring]], $\struct {\set {0_R}, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$.
Also, from [[Ring Product with Zero]]:
:$\forall x \in R: x \circ 0_R = 0_R = 0_R \circ x \in \set {0_R}$
thus fulfilling the condition for $\struct {\set {0_R}, +, \circ}$ to be an ... | Null Ring is Ideal | https://proofwiki.org/wiki/Null_Ring_is_Ideal | https://proofwiki.org/wiki/Null_Ring_is_Ideal | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Null Ring",
"Definition:Ideal of Ring"
] | [
"Null Ring is Subring of Ring",
"Definition:Subring",
"Ring Product with Zero",
"Definition:Ideal of Ring"
] |
proofwiki-5458 | Ring Epimorphism Preserves Subrings | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$. | A direct application of Ring Homomorphism Preserves Subrings.
{{qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
Then $\phi \sqbrk S$ is a [[Definition:Subring|subring]] of $R_2$. | A direct application of [[Ring Homomorphism Preserves Subrings]].
{{qed}} | Ring Epimorphism Preserves Subrings | https://proofwiki.org/wiki/Ring_Epimorphism_Preserves_Subrings | https://proofwiki.org/wiki/Ring_Epimorphism_Preserves_Subrings | [
"Ring Epimorphisms",
"Subrings"
] | [
"Definition:Ring Epimorphism",
"Definition:Subring",
"Definition:Subring"
] | [
"Ring Homomorphism Preserves Subrings"
] |
proofwiki-5459 | Preimage of Image of Ideal under Ring Homomorphism | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $K = \map \ker \phi$, be the kernel of $\phi$.
Let $J$ be an ideal of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$ | As an ideal is a subring, the result Preimage of Image of Subring under Ring Homomorphism applies directly.
{{Qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $K = \map \ker \phi$, be the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \... | As an [[Ideal is Subring|ideal is a subring]], the result [[Preimage of Image of Subring under Ring Homomorphism]] applies directly.
{{Qed}} | Preimage of Image of Ideal under Ring Homomorphism | https://proofwiki.org/wiki/Preimage_of_Image_of_Ideal_under_Ring_Homomorphism | https://proofwiki.org/wiki/Preimage_of_Image_of_Ideal_under_Ring_Homomorphism | [
"Ring Homomorphisms",
"Ideal Theory"
] | [
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism",
"Definition:Ideal of Ring"
] | [
"Ideal is Subring",
"Preimage of Image of Subring under Ring Homomorphism"
] |
proofwiki-5460 | Image of Preimage of Ideal under Ring Epimorphism | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S_2$ be an ideal of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$ | As an ideal is a subring, the result Image of Preimage of Subring under Ring Epimorphism applies directly.
{{Qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
Let $S_2$ be an [[Definition:Ideal of Ring|ideal]] of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$ | As an [[Ideal is Subring|ideal is a subring]], the result [[Image of Preimage of Subring under Ring Epimorphism]] applies directly.
{{Qed}} | Image of Preimage of Ideal under Ring Epimorphism | https://proofwiki.org/wiki/Image_of_Preimage_of_Ideal_under_Ring_Epimorphism | https://proofwiki.org/wiki/Image_of_Preimage_of_Ideal_under_Ring_Epimorphism | [
"Ring Epimorphisms",
"Subrings"
] | [
"Definition:Ring Epimorphism",
"Definition:Ideal of Ring"
] | [
"Ideal is Subring",
"Image of Preimage of Subring under Ring Epimorphism"
] |
proofwiki-5461 | Vectors are Right Cancellable | Let $\struct {\mathbf V, +, \circ}$ be a vector space over $\GF$, as defined by the vector space axioms.
Then every $\mathbf v \in \struct {\mathbf V, +}$ is right cancellable:
:$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf a + \mathbf c = \mathbf b + \mathbf c \implies \mathbf a = \mathbf b$ | Utilizing the vector space axioms:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf c
| r = \mathbf b + \mathbf c
}}
{{eqn | ll= \leadsto
| l = \paren {\mathbf a + \mathbf c} - \mathbf c
| r = \paren {\mathbf b + \mathbf c} - \mathbf c
}}
{{eqn | ll= \leadsto
| l = \mathbf a + \paren {\mathbf c - \... | Let $\struct {\mathbf V, +, \circ}$ be a [[Definition:Vector Space|vector space]] over $\GF$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then every $\mathbf v \in \struct {\mathbf V, +}$ is [[Definition:Right Cancellable Element|right cancellable]]:
:$\forall \mathbf a, \mathbf b, \mathbf c ... | Utilizing the [[Axiom:Vector Space Axioms|vector space axioms]]:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf c
| r = \mathbf b + \mathbf c
}}
{{eqn | ll= \leadsto
| l = \paren {\mathbf a + \mathbf c} - \mathbf c
| r = \paren {\mathbf b + \mathbf c} - \mathbf c
}}
{{eqn | ll= \leadsto
| l = \m... | Vectors are Right Cancellable | https://proofwiki.org/wiki/Vectors_are_Right_Cancellable | https://proofwiki.org/wiki/Vectors_are_Right_Cancellable | [
"Vector Algebra"
] | [
"Definition:Vector Space",
"Axiom:Vector Space Axioms",
"Definition:Cancellable Element/Right Cancellable"
] | [
"Axiom:Vector Space Axioms"
] |
proofwiki-5462 | Vectors are Left Cancellable | Let $\struct {\mathbf V, +, \circ}$ be a vector space over $\GF$, as defined by the vector space axioms.
Then every $\mathbf v \in \struct {\mathbf V, +}$ is left cancellable:
:$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf c + \mathbf a = \mathbf c + \mathbf b \implies \mathbf a = \mathbf b$ | Utilizing the vector space axioms:
{{begin-eqn}}
{{eqn | l = \mathbf c + \mathbf a
| r = \mathbf c + \mathbf b
}}
{{eqn | ll= \leadsto
| l = \mathbf a + \mathbf c
| r = \mathbf b + \mathbf c
}}
{{eqn | ll= \leadsto
| l = \mathbf a
| r = \mathbf b
| c = Vectors are Right Cancellable
}... | Let $\struct {\mathbf V, +, \circ}$ be a [[Definition:Vector Space|vector space]] over $\GF$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then every $\mathbf v \in \struct {\mathbf V, +}$ is [[Definition:Left Cancellable Element|left cancellable]]:
:$\forall \mathbf a, \mathbf b, \mathbf c \i... | Utilizing the [[Axiom:Vector Space Axioms|vector space axioms]]:
{{begin-eqn}}
{{eqn | l = \mathbf c + \mathbf a
| r = \mathbf c + \mathbf b
}}
{{eqn | ll= \leadsto
| l = \mathbf a + \mathbf c
| r = \mathbf b + \mathbf c
}}
{{eqn | ll= \leadsto
| l = \mathbf a
| r = \mathbf b
| c = ... | Vectors are Left Cancellable | https://proofwiki.org/wiki/Vectors_are_Left_Cancellable | https://proofwiki.org/wiki/Vectors_are_Left_Cancellable | [
"Vector Algebra"
] | [
"Definition:Vector Space",
"Axiom:Vector Space Axioms",
"Definition:Cancellable Element/Left Cancellable"
] | [
"Axiom:Vector Space Axioms",
"Vectors are Right Cancellable",
"Category:Vector Algebra"
] |
proofwiki-5463 | Zero Vector is Unique | Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms.
Then the zero vector in $\mathbf V$ is unique:
:$\exists! \mathbf 0 \in \mathbf V: \forall \mathbf x \in \mathbf V: \mathbf x + \mathbf 0 = \mathbf x$ | === Proof of Existence ===
Follows from the vector space axioms.
{{qed|lemma}} | Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a [[Definition:Vector Space|vector space]] over $\mathbb F$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then the [[Definition:Zero Vector|zero vector]] in $\mathbf V$ is [[Definition:Unique|unique]]:
:$\exists! \mathbf 0 \in \mathbf V: \fora... | === Proof of Existence ===
Follows from the [[Axiom:Vector Space Axioms|vector space axioms]].
{{qed|lemma}} | Zero Vector is Unique | https://proofwiki.org/wiki/Zero_Vector_is_Unique | https://proofwiki.org/wiki/Zero_Vector_is_Unique | [
"Zero Vectors"
] | [
"Definition:Vector Space",
"Axiom:Vector Space Axioms",
"Definition:Zero Vector",
"Definition:Unique"
] | [
"Axiom:Vector Space Axioms",
"Axiom:Vector Space Axioms"
] |
proofwiki-5464 | Additive Inverse in Vector Space is Unique | Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms.
Then for every $\mathbf v \in \mathbf V$, the additive inverse of $\mathbf v$ is unique:
:$\forall \mathbf v \in \mathbf V: \exists! \paren {-\mathbf v} \in \mathbf V: \mathbf v + \paren {-\mathbf v} = \math... | === Proof of Existence ===
Follows from the vector space axioms.
{{qed|lemma}} | Let $\struct {\mathbf V, +, \circ}_F$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Then for every $\mathbf v \in \mathbf V$, the [[Definition:Inverse Element|additive inverse]] of $\mathbf v... | === Proof of Existence ===
Follows from the [[Axiom:Vector Space Axioms|vector space axioms]].
{{qed|lemma}} | Additive Inverse in Vector Space is Unique | https://proofwiki.org/wiki/Additive_Inverse_in_Vector_Space_is_Unique | https://proofwiki.org/wiki/Additive_Inverse_in_Vector_Space_is_Unique | [
"Vector Algebra",
"Linear Algebra",
"Vector Spaces"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Axiom:Vector Space Axioms",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Unique"
] | [
"Axiom:Vector Space Axioms"
] |
proofwiki-5465 | Vector Inverse is Negative Vector | Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \mathbf v \in \mathbf V: -\mathbf v = -1_F \ast \mathbf v$ | {{begin-eqn}}
{{eqn | l = \mathbf v + \paren {-1_F \ast \mathbf v}
| r = \paren {1_F \ast \mathbf v} + \paren {-1_F \ast \mathbf v}
| c = {{Field-axiom|M3}}
}}
{{eqn | r = \paren {1_F + \paren {-1_F} } \ast \mathbf v
| c = {{Vector-space-axiom|2}}
}}
{{eqn | r = 0_F \ast \mathbf v
| c = {{Field-... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\struct {\mathbf V, +, \ast}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms... | {{begin-eqn}}
{{eqn | l = \mathbf v + \paren {-1_F \ast \mathbf v}
| r = \paren {1_F \ast \mathbf v} + \paren {-1_F \ast \mathbf v}
| c = {{Field-axiom|M3}}
}}
{{eqn | r = \paren {1_F + \paren {-1_F} } \ast \mathbf v
| c = {{Vector-space-axiom|2}}
}}
{{eqn | r = 0_F \ast \mathbf v
| c = {{Field-... | Vector Inverse is Negative Vector | https://proofwiki.org/wiki/Vector_Inverse_is_Negative_Vector | https://proofwiki.org/wiki/Vector_Inverse_is_Negative_Vector | [
"Vector Algebra"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Vector Space",
"Axiom:Vector Space Axioms"
] | [
"Vector Scaled by Zero is Zero Vector",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Additive Inverse in Vector Space is Unique"
] |
proofwiki-5466 | Non-Trivial Commutative Division Ring is Field | Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative.
Then $\struct {R, +, \circ}$ is a field.
Similarly, let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative. | Suppose $\struct {R, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
By definition $\struct {R, +}$ is an abelian group.
Thus {{Field-axiom|A}} are satisfied.
Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative.
Thus {{Field-axiom|M0}} to {{Field-axiom|M2... | Let $\struct {R, +, \circ}$ be a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Division Ring|division ring]] such that $\circ$ is [[Definition:Commutative Operation|commutative]].
Then $\struct {R, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
Similarly, let $\struct {F, +, \circ}$ be a [... | Suppose $\struct {R, +, \circ}$ is a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Division Ring|division ring]] such that $\circ$ is [[Definition:Commutative Operation|commutative]].
By definition $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
Thus {{Field-axiom|A}} are satisfied.
Al... | Non-Trivial Commutative Division Ring is Field | https://proofwiki.org/wiki/Non-Trivial_Commutative_Division_Ring_is_Field | https://proofwiki.org/wiki/Non-Trivial_Commutative_Division_Ring_is_Field | [
"Field Theory",
"Division Rings"
] | [
"Definition:Non-Trivial Ring",
"Definition:Division Ring",
"Definition:Commutative/Operation",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Non-Trivial Ring",
"Definition:Division Ring",
"Definition:Commutative/Operation"
] | [
"Definition:Non-Trivial Ring",
"Definition:Division Ring",
"Definition:Commutative/Operation",
"Definition:Abelian Group",
"Definition:Semigroup",
"Definition:Commutative/Operation",
"Definition:Ring with Unity",
"Definition:Ring (Abstract Algebra)",
"Definition:Division Ring",
"Axiom:Field Axioms... |
proofwiki-5467 | Division Ring has No Proper Zero Divisors | Let $\struct {R, +, \circ}$ be a division ring.
Then $\struct {R, +, \circ}$ has no proper zero divisors. | Let $\struct {R, +, \circ}$ be a division ring whose zero is $0_R$ and whose unity is $1_R$.
By definition of division ring, every element $x$ of $R^* = R \setminus \set {0_R}$ has an element $y$ such that:
:$y \circ x = x \circ y = 1_R$
That is, by definition, every element of $R^*$ is a unit of $R$.
The result follow... | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Then $\struct {R, +, \circ}$ has no [[Definition:Proper Zero Divisor|proper zero divisors]]. | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
By definition of [[Definition:Division Ring|division ring]], every element $x$ of $R^* = R \setminus \set {0_R}$ has an [[Definition:Element|el... | Division Ring has No Proper Zero Divisors | https://proofwiki.org/wiki/Division_Ring_has_No_Proper_Zero_Divisors | https://proofwiki.org/wiki/Division_Ring_has_No_Proper_Zero_Divisors | [
"Division Rings"
] | [
"Definition:Division Ring",
"Definition:Proper Zero Divisor"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Division Ring",
"Definition:Element",
"Definition:Unit of Ring",
"Unit of Ring is not Zero Divisor"
] |
proofwiki-5468 | Maximal Element need not be Greatest Element | Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $M \in S$ be a maximal element of $S$.
Then $M$ is not necessarily the greatest element of $S$. | Proof by Counterexample:
Let $S = \set {a, b, c}$.
Let $\preccurlyeq$ be defined as:
:$x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$
A straightforward but laborious process determines that $\preccurlyeq$ is a partial ordering on $S$.
We have t... | Let $\struct {S, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $M \in S$ be a [[Definition:Maximal Element|maximal element]] of $S$.
Then $M$ is not necessarily the [[Definition:Greatest Element|greatest element]] of $S$. | [[Proof by Counterexample]]:
Let $S = \set {a, b, c}$.
Let $\preccurlyeq$ be defined as:
:$x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$
A straightforward but laborious process determines that $\preccurlyeq$ is a [[Definition:Partial Orderi... | Maximal Element need not be Greatest Element | https://proofwiki.org/wiki/Maximal_Element_need_not_be_Greatest_Element | https://proofwiki.org/wiki/Maximal_Element_need_not_be_Greatest_Element | [
"Maximal Elements",
"Greatest Elements"
] | [
"Definition:Ordered Set",
"Definition:Maximal/Element",
"Definition:Greatest Element"
] | [
"Proof by Counterexample",
"Definition:Partial Ordering",
"Definition:Maximal/Element",
"Definition:Greatest Element",
"Greatest Element is Unique",
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Greatest Element"
] |
proofwiki-5469 | Maximal Ideal of Division Ring | Let $\struct {D, +, \circ}$ be a Division Ring whose zero is $0$.
Let $\struct {J, +, \circ}$ be a maximal ideal of $D$.
Then:
:$J = \set 0$ | From Ideals of Division Ring, the only ideals of a Division Ring $\struct {D, +, \circ}$ are $\struct {D, +, \circ}$ and $\struct {\set 0, +, \circ}$.
Hence the result by definition of maximal ideal.
{{qed}} | Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|Division Ring]] whose [[Definition:Ring Zero|zero]] is $0$.
Let $\struct {J, +, \circ}$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $D$.
Then:
:$J = \set 0$ | From [[Ideals of Division Ring]], the only [[Definition:Ideal of Ring|ideals]] of a [[Definition:Division Ring|Division Ring]] $\struct {D, +, \circ}$ are $\struct {D, +, \circ}$ and $\struct {\set 0, +, \circ}$.
Hence the result by definition of [[Definition:Maximal Ideal of Ring|maximal ideal]].
{{qed}} | Maximal Ideal of Division Ring | https://proofwiki.org/wiki/Maximal_Ideal_of_Division_Ring | https://proofwiki.org/wiki/Maximal_Ideal_of_Division_Ring | [
"Ideal Theory",
"Division Rings"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Maximal Ideal of Ring"
] | [
"Ideals of Division Ring",
"Definition:Ideal of Ring",
"Definition:Division Ring",
"Definition:Maximal Ideal of Ring"
] |
proofwiki-5470 | Linear Transformation is Injective iff Kernel Contains Only Zero | Let $\mathbf V, \mathbf V'$ be vector spaces, with respective zeroes $\mathbf 0, \mathbf 0'$.
Let $T: \mathbf V \to \mathbf V'$ be a linear transformation.
Then:
:$T$ is injective {{iff}} $\map \ker T = \set {\mathbf 0}$
where:
:$\mathbf 0$ is the zero of the domain of $T$
:$\map \ker T$ is the kernel of $T$. | === Sufficient Condition ===
That $\mathbf 0 \in \map \ker T$ follows from Kernel of Linear Transformation contains Zero Vector.
That $\map \ker T$ is a singleton follows from the definition of injection.
{{qed|lemma}} | Let $\mathbf V, \mathbf V'$ be [[Definition:Vector Space|vector spaces]], with respective [[Definition:Zero Vector|zeroes]] $\mathbf 0, \mathbf 0'$.
Let $T: \mathbf V \to \mathbf V'$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
Then:
:$T$ is [[Definition:Injection|injective]] {{i... | === Sufficient Condition ===
That $\mathbf 0 \in \map \ker T$ follows from [[Kernel of Linear Transformation contains Zero Vector]].
That $\map \ker T$ is a [[Definition:Singleton|singleton]] follows from the definition of [[Definition:Injection|injection]].
{{qed|lemma}} | Linear Transformation is Injective iff Kernel Contains Only Zero | https://proofwiki.org/wiki/Linear_Transformation_is_Injective_iff_Kernel_Contains_Only_Zero | https://proofwiki.org/wiki/Linear_Transformation_is_Injective_iff_Kernel_Contains_Only_Zero | [
"Linear Transformations"
] | [
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Linear Transformation/Vector Space",
"Definition:Injection",
"Definition:Zero Vector",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Kernel of Linear Transformation"
] | [
"Kernel of Linear Transformation contains Zero Vector",
"Definition:Singleton",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-5471 | Diagonal Relation on Ring is Ordering Compatible with Ring Structure | Let $\struct {R, +, \circ, \preceq}$ be a ring whose zero is $0_R$.
Then the diagonal relation $\Delta_R$ on $R$ is an ordering compatible with the ring structure of $R$. | From Diagonal Relation is Ordering and Equivalence, we have that $\Delta_R$ is actually an ordering on $R$.
From the definition of the diagonal relation:
:$\tuple {x, y} \in \Delta_R \iff x = y$
Thus:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \Delta_R
| c =
}}
{{eqn | ll= \leadsto
... | Let $\struct {R, +, \circ, \preceq}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Then the [[Definition:Diagonal Relation|diagonal relation]] $\Delta_R$ on $R$ is an [[Definition:Ordering Compatible with Ring Structure|ordering compatible with the ring structure]] of ... | From [[Diagonal Relation is Ordering and Equivalence]], we have that $\Delta_R$ is actually an [[Definition:Ordering|ordering]] on $R$.
From the definition of the [[Definition:Diagonal Relation|diagonal relation]]:
:$\tuple {x, y} \in \Delta_R \iff x = y$
Thus:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in... | Diagonal Relation on Ring is Ordering Compatible with Ring Structure | https://proofwiki.org/wiki/Diagonal_Relation_on_Ring_is_Ordering_Compatible_with_Ring_Structure | https://proofwiki.org/wiki/Diagonal_Relation_on_Ring_is_Ordering_Compatible_with_Ring_Structure | [
"Ring Theory",
"Ordered Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Diagonal Relation",
"Definition:Ordering Compatible with Ring Structure"
] | [
"Diagonal Relation is Ordering and Equivalence",
"Definition:Ordering",
"Definition:Diagonal Relation",
"Definition:Relation Compatible with Operation",
"Definition:Ordering Compatible with Ring Structure"
] |
proofwiki-5472 | Nagata-Smirnov Metrization Theorem | A topological space $T = \struct {S, \tau}$ is metrizable {{iff}} $T$ is regular and has a basis that is $\sigma$-locally finite. | === Necessary Condition ===
Let $T$ be metrizable.
{{:Nagata-Smirnov Metrization Theorem/Necessary Condition}}{{qed|lemma}} | A [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ is [[Definition:Metrizable Space|metrizable]] {{iff}} $T$ is [[Definition:Regular Space|regular]] and has a [[Definition:Basis (Topology)|basis]] that is [[Definition:Sigma-Locally Finite Basis|$\sigma$-locally finite]]. | === [[Nagata-Smirnov Metrization Theorem/Necessary Condition|Necessary Condition]] ===
Let $T$ be [[Definition:Metrizable Space|metrizable]].
{{:Nagata-Smirnov Metrization Theorem/Necessary Condition}}{{qed|lemma}} | Nagata-Smirnov Metrization Theorem | https://proofwiki.org/wiki/Nagata-Smirnov_Metrization_Theorem | https://proofwiki.org/wiki/Nagata-Smirnov_Metrization_Theorem | [
"Nagata-Smirnov Metrization Theorem",
"Metrization Theorems",
"Regular Spaces",
"Metrizable Spaces"
] | [
"Definition:Topological Space",
"Definition:Metrizable Space",
"Definition:Regular Space",
"Definition:Basis (Topology)",
"Definition:Sigma-Locally Finite Basis"
] | [
"Nagata-Smirnov Metrization Theorem/Necessary Condition",
"Definition:Metrizable Space"
] |
proofwiki-5473 | Smirnov Metrization Theorem | Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is metrizable {{iff}} $T$ is paracompact and locally metrizable. | {{proof wanted}}
{{Namedfor|Yurii Mikhailovich Smirnov|cat = Smirnov Y M}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then:
:$T$ is [[Definition:Metrizable Space|metrizable]] {{iff}} $T$ is [[Definition:Paracompact Space|paracompact]] and [[Definition:Locally Metrizable Space|locally metrizable]]. | {{proof wanted}}
{{Namedfor|Yurii Mikhailovich Smirnov|cat = Smirnov Y M}} | Smirnov Metrization Theorem | https://proofwiki.org/wiki/Smirnov_Metrization_Theorem | https://proofwiki.org/wiki/Smirnov_Metrization_Theorem | [
"Metrization Theorems",
"Metrizable Spaces"
] | [
"Definition:Topological Space",
"Definition:Metrizable Space",
"Definition:Paracompact Space",
"Definition:Locally Metrizable Space"
] | [] |
proofwiki-5474 | Identity Mapping is Automorphism/Semigroups | Let $\struct {S, \circ}$ be a semigroup.
Then $I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism. | The main result Identity Mapping is Automorphism holds directly.
{{qed}}
Category:Semigroup Automorphisms
Category:Identity Mappings
7spug7di6v6w2vjjaft8wwmd7e58u1f | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Then $I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a [[Definition:Semigroup Automorphism|semigroup automorphism]]. | The main result [[Identity Mapping is Automorphism]] holds directly.
{{qed}}
[[Category:Semigroup Automorphisms]]
[[Category:Identity Mappings]]
7spug7di6v6w2vjjaft8wwmd7e58u1f | Identity Mapping is Automorphism/Semigroups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Semigroups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Semigroups | [
"Semigroup Automorphisms",
"Identity Mappings"
] | [
"Definition:Semigroup",
"Definition:Semigroup Automorphism"
] | [
"Identity Mapping is Automorphism",
"Category:Semigroup Automorphisms",
"Category:Identity Mappings"
] |
proofwiki-5475 | Identity Mapping is Ordered Ring Automorphism | Let $\struct {S, +, \circ, \preceq}$ be an ordered ring.
Then the identity mapping $I_S: S \to S$ is an ordered ring automorphism. | We have that:
:an identity mapping is an order isomorphism
:an identity mapping is a group automorphism
:an identity mapping is a semigroup automorphism
Hence the result by definition of ordered ring automorphism.
{{qed}} | Let $\struct {S, +, \circ, \preceq}$ be an [[Definition:Ordered Ring|ordered ring]].
Then the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is an [[Definition:Ordered Ring Automorphism|ordered ring automorphism]]. | We have that:
:an [[Identity Mapping is Order Isomorphism|identity mapping is an order isomorphism]]
:an [[Identity Mapping is Group Automorphism|identity mapping is a group automorphism]]
:an [[Identity Mapping is Semigroup Automorphism|identity mapping is a semigroup automorphism]]
Hence the result by definition of ... | Identity Mapping is Ordered Ring Automorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Ordered_Ring_Automorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Ordered_Ring_Automorphism | [
"Ordered Rings",
"Identity Mappings"
] | [
"Definition:Ordered Ring",
"Definition:Identity Mapping",
"Definition:Ordered Ring Automorphism"
] | [
"Identity Mapping is Order Isomorphism",
"Identity Mapping is Automorphism/Groups",
"Identity Mapping is Automorphism/Semigroups",
"Definition:Ordered Ring Automorphism"
] |
proofwiki-5476 | Inverse of Reflexive Relation is Reflexive | Let $\RR$ be a relation on a set $S$.
If $\RR$ is reflexive, then so is $\RR^{-1}$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = {{Defof|Reflexive Relation}}
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \in
| r = \RR^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{end-eqn}... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Reflexive Relation|reflexive]], then so is $\RR^{-1}$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = {{Defof|Reflexive Relation}}
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \in
| r = \RR^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{end-eqn}... | Inverse of Reflexive Relation is Reflexive | https://proofwiki.org/wiki/Inverse_of_Reflexive_Relation_is_Reflexive | https://proofwiki.org/wiki/Inverse_of_Reflexive_Relation_is_Reflexive | [
"Reflexive Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Relation"
] |
proofwiki-5477 | Inverse of Antireflexive Relation is Antireflexive | Let $\RR$ be a relation on a set $S$.
If $\RR$ is antireflexive, then so is $\RR^{-1}$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \notin
| r = \RR
| c = {{Defof|Antireflexive Relation}}
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \notin
| r = \RR^{-1}
| c = {{Defof|Inverse R... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Antireflexive Relation|antireflexive]], then so is $\RR^{-1}$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \notin
| r = \RR
| c = {{Defof|Antireflexive Relation}}
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \notin
| r = \RR^{-1}
| c = {{Defof|Inverse R... | Inverse of Antireflexive Relation is Antireflexive | https://proofwiki.org/wiki/Inverse_of_Antireflexive_Relation_is_Antireflexive | https://proofwiki.org/wiki/Inverse_of_Antireflexive_Relation_is_Antireflexive | [
"Antireflexive Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Antireflexive Relation"
] |
proofwiki-5478 | Inverse of Non-Reflexive Relation is Non-Reflexive | Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-reflexive, then so is $\RR^{-1}$. | Let $\RR$ be non-reflexive.
Then:
{{begin-eqn}}
{{eqn | q = \exists x \in S
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = as $\RR$ is not antireflexive
}}
{{eqn | ll= \leadsto
| q = \exists x \in S
| l = \tuple {x, x}
| o = \in
| r = \RR^{-1}
| c = {{Defof|Inverse R... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Non-Reflexive Relation|non-reflexive]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Non-Reflexive Relation|non-reflexive]].
Then:
{{begin-eqn}}
{{eqn | q = \exists x \in S
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = as $\RR$ is not [[Definition:Antireflexive Relation|antireflexive]]
}}
{{eqn | ll= \leadsto
| q = \exists x \in S
| l = \t... | Inverse of Non-Reflexive Relation is Non-Reflexive | https://proofwiki.org/wiki/Inverse_of_Non-Reflexive_Relation_is_Non-Reflexive | https://proofwiki.org/wiki/Inverse_of_Non-Reflexive_Relation_is_Non-Reflexive | [
"Non-Reflexive Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Non-Reflexive Relation"
] | [
"Definition:Non-Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Non-Reflexive Relation"
] |
proofwiki-5479 | Inverse of Symmetric Relation is Symmetric | Let $\RR$ be a relation on a set $S$.
If $\RR$ is symmetric, then so is $\RR^{-1}$. | Let $\RR$ be symmetric.
Then from Relation equals Inverse iff Symmetric it follows that $\RR^{-1}$ is also symmetric.
{{qed}} | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Symmetric Relation|symmetric]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Symmetric Relation|symmetric]].
Then from [[Relation equals Inverse iff Symmetric]] it follows that $\RR^{-1}$ is also symmetric.
{{qed}} | Inverse of Symmetric Relation is Symmetric | https://proofwiki.org/wiki/Inverse_of_Symmetric_Relation_is_Symmetric | https://proofwiki.org/wiki/Inverse_of_Symmetric_Relation_is_Symmetric | [
"Symmetric Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation",
"Equivalence of Definitions of Symmetric Relation"
] |
proofwiki-5480 | Inverse of Asymmetric Relation is Asymmetric | Let $\RR$ be a relation on a set $S$.
If $\RR$ is asymmetric, then so is $\RR^{-1}$. | Let $\RR$ be asymmetric.
Let $\tuple {x, y} \in \RR^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR^{-1}
}}
{{eqn | ll= \leadsto
| l = \tuple {y, x}
| o = \in
| r = \RR
| c = Inverse of Inverse Relation
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Asymmetric Relation|asymmetric]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]].
Let $\tuple {x, y} \in \RR^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR^{-1}
}}
{{eqn | ll= \leadsto
| l = \tuple {y, x}
| o = \in
| r = \RR
| c = [[Inverse of Inverse Relation]]
}}
{{eqn | ll= ... | Inverse of Asymmetric Relation is Asymmetric | https://proofwiki.org/wiki/Inverse_of_Asymmetric_Relation_is_Asymmetric | https://proofwiki.org/wiki/Inverse_of_Asymmetric_Relation_is_Asymmetric | [
"Asymmetric Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Asymmetric Relation"
] | [
"Definition:Asymmetric Relation",
"Inverse of Inverse Relation",
"Definition:Asymmetric Relation"
] |
proofwiki-5481 | Inverse of Antisymmetric Relation is Antisymmetric | Let $\RR$ be a relation on a set $S$.
If $\RR$ is antisymmetric, then so is $\RR^{-1}$. | Let $\RR$ be antisymmetric.
Then:
:$\tuple {x, y} \land \tuple {y, x} \in \RR \implies x = y$
It follows that:
:$\tuple {y, x} \land \tuple {x, y} \in \RR^{-1} \implies x = y$
Thus it follows that $\RR^{-1}$ is also antisymmetric.
{{qed}} | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Antisymmetric Relation|antisymmetric]].
Then:
:$\tuple {x, y} \land \tuple {y, x} \in \RR \implies x = y$
It follows that:
:$\tuple {y, x} \land \tuple {x, y} \in \RR^{-1} \implies x = y$
Thus it follows that $\RR^{-1}$ is also [[Definition:Antisymmetric Relation|antisymmetric]].
{{qed}} | Inverse of Antisymmetric Relation is Antisymmetric | https://proofwiki.org/wiki/Inverse_of_Antisymmetric_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Inverse_of_Antisymmetric_Relation_is_Antisymmetric | [
"Antisymmetric Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-5482 | Inverse of Non-Symmetric Relation is Non-Symmetric | Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-symmetric, then so is $\RR^{-1}$. | Let $\RR$ be non-symmetric.
Then:
:$\exists \tuple {x_1, y_1} \in \RR \implies \tuple {y_1, x_1} \in \RR$
and also:
:$\exists \tuple {x_2, y_2} \in \RR \implies \tuple {y_2, x_2} \notin \RR$
Thus:
:$\exists \tuple {y_1, x_1} \in \RR^{-1} \implies \tuple {x_1, y_1} \in \RR^{-1}$
and also:
:$\exists \tuple {y_2, x_2} \in... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Non-Symmetric Relation|non-symmetric]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Non-Symmetric Relation|non-symmetric]].
Then:
:$\exists \tuple {x_1, y_1} \in \RR \implies \tuple {y_1, x_1} \in \RR$
and also:
:$\exists \tuple {x_2, y_2} \in \RR \implies \tuple {y_2, x_2} \notin \RR$
Thus:
:$\exists \tuple {y_1, x_1} \in \RR^{-1} \implies \tuple {x_1, y_1} \in \RR^{-1}$
a... | Inverse of Non-Symmetric Relation is Non-Symmetric | https://proofwiki.org/wiki/Inverse_of_Non-Symmetric_Relation_is_Non-Symmetric | https://proofwiki.org/wiki/Inverse_of_Non-Symmetric_Relation_is_Non-Symmetric | [
"Non-Symmetric Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Non-Symmetric Relation"
] | [
"Definition:Non-Symmetric Relation",
"Definition:Non-Symmetric Relation"
] |
proofwiki-5483 | Inverse of Transitive Relation is Transitive | Let $\RR$ be a relation on a set $S$.
Let $\RR$ be transitive.
Then its inverse $\RR^{-1}$ is also transitive. | Let $\RR$ be transitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \in \RR^{-1}$
and so $\RR^{-1}$ is transitive.
{{qed}} | Let $\RR$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
Let $\RR$ be [[Definition:Transitive Relation|transitive]].
Then its [[Definition:Inverse Relation|inverse]] $\RR^{-1}$ is also [[Definition:Transitive Relation|transitive]]. | Let $\RR$ be [[Definition:Transitive Relation|transitive]].
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \in \RR^{-1}$
and so $\RR^{-1}$ is [[Definition:Transitive Relation|transitive]].
{{qed}} | Inverse of Transitive Relation is Transitive/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Transitive_Relation_is_Transitive | https://proofwiki.org/wiki/Inverse_of_Transitive_Relation_is_Transitive/Proof_1 | [
"Transitive Relations",
"Inverse Relations",
"Inverse of Transitive Relation is Transitive"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Inverse Relation",
"Definition:Transitive Relation"
] | [
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] |
proofwiki-5484 | Inverse of Transitive Relation is Transitive | Let $\RR$ be a relation on a set $S$.
Let $\RR$ be transitive.
Then its inverse $\RR^{-1}$ is also transitive. | Let $\RR$ be transitive.
Thus by definition:
:$\RR \circ \RR \subseteq \RR$
Thus:
{{begin-eqn}}
{{eqn | l = \RR^{-1} \circ \RR^{-1}
| r = \paren {\RR \circ \RR}^{-1}
| c = Inverse of Composite Relation
}}
{{eqn | o = \subseteq
| r = \RR^{-1}
| c = Inverse of Subset of Relation is Subset of Inver... | Let $\RR$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
Let $\RR$ be [[Definition:Transitive Relation|transitive]].
Then its [[Definition:Inverse Relation|inverse]] $\RR^{-1}$ is also [[Definition:Transitive Relation|transitive]]. | Let $\RR$ be [[Definition:Transitive Relation|transitive]].
Thus by definition:
:$\RR \circ \RR \subseteq \RR$
Thus:
{{begin-eqn}}
{{eqn | l = \RR^{-1} \circ \RR^{-1}
| r = \paren {\RR \circ \RR}^{-1}
| c = [[Inverse of Composite Relation]]
}}
{{eqn | o = \subseteq
| r = \RR^{-1}
| c = [[Inver... | Inverse of Transitive Relation is Transitive/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Transitive_Relation_is_Transitive | https://proofwiki.org/wiki/Inverse_of_Transitive_Relation_is_Transitive/Proof_2 | [
"Transitive Relations",
"Inverse Relations",
"Inverse of Transitive Relation is Transitive"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Inverse Relation",
"Definition:Transitive Relation"
] | [
"Definition:Transitive Relation",
"Inverse of Composite Relation",
"Inverse of Subset of Relation is Subset of Inverse",
"Definition:Transitive Relation"
] |
proofwiki-5485 | Inverse of Antitransitive Relation is Antitransitive | Let $\RR$ be a relation on a set $S$.
If $\RR$ is antitransitive, then so is $\RR^{-1}$. | Let $\RR$ be antitransitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \notin \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \notin \RR^{-1}$
and so $\RR^{-1}$ is antitransitive.
{{qed}} | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Antitransitive Relation|antitransitive]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Antitransitive Relation|antitransitive]].
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \notin \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \notin \RR^{-1}$
and so $\RR^{-1}$ is [[Definition:Antitransitive Relation|antitransitive]].
{{qe... | Inverse of Antitransitive Relation is Antitransitive | https://proofwiki.org/wiki/Inverse_of_Antitransitive_Relation_is_Antitransitive | https://proofwiki.org/wiki/Inverse_of_Antitransitive_Relation_is_Antitransitive | [
"Antitransitive Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Antitransitive Relation"
] | [
"Definition:Antitransitive Relation",
"Definition:Antitransitive Relation"
] |
proofwiki-5486 | Inverse of Non-Transitive Relation is Non-Transitive | Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-transitive, then so is $\RR^{-1}$. | Let $\RR$ be non-transitive.
Then:
:$\exists x_1, y_1, z_1 \in S: \tuple {x_1, y_1}, \tuple {y_1, z_1} \in \RR, \tuple {x_1, z_1} \in \RR$
:$\exists x_2, y_2, z_2 \in S: \tuple {x_2, y_2}, \tuple {y_2, z_2} \in \RR, \tuple {x_2, z_2} \notin \RR$
So:
:$\exists x_1, y_1, z_1 \in S: \tuple {y_1, x_1}, \tuple {z_1, y_1} \i... | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ is [[Definition:Non-Transitive Relation|non-transitive]], then so is $\RR^{-1}$. | Let $\RR$ be [[Definition:Non-Transitive Relation|non-transitive]].
Then:
:$\exists x_1, y_1, z_1 \in S: \tuple {x_1, y_1}, \tuple {y_1, z_1} \in \RR, \tuple {x_1, z_1} \in \RR$
:$\exists x_2, y_2, z_2 \in S: \tuple {x_2, y_2}, \tuple {y_2, z_2} \in \RR, \tuple {x_2, z_2} \notin \RR$
So:
:$\exists x_1, y_1, z_1 \in ... | Inverse of Non-Transitive Relation is Non-Transitive | https://proofwiki.org/wiki/Inverse_of_Non-Transitive_Relation_is_Non-Transitive | https://proofwiki.org/wiki/Inverse_of_Non-Transitive_Relation_is_Non-Transitive | [
"Non-Transitive Relations",
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Non-Transitive Relation"
] | [
"Definition:Non-Transitive Relation",
"Definition:Non-Transitive Relation"
] |
proofwiki-5487 | Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism | Let $\struct {S, +, \circ, \preceq}$ and $\struct {T, \oplus, *, \preccurlyeq}$ be ordered rings.
Let $\phi: S \to T$ be an ordered ring isomorphism.
Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism. | By definition, $\phi$ is a bijection.
By Bijection iff Inverse is Bijection, $\phi^{-1}$ is also a bijection.
By definition, an ordered ring isomorphism from $\phi: \struct {S, +, \circ, \preceq} \to \struct {T, \oplus, *, \preccurlyeq}$ is:
:an order isomorphism from the ordered set $\struct {S, \preceq}$ to the order... | Let $\struct {S, +, \circ, \preceq}$ and $\struct {T, \oplus, *, \preccurlyeq}$ be [[Definition:Ordered Ring|ordered rings]].
Let $\phi: S \to T$ be an [[Definition:Ordered Ring Isomorphism|ordered ring isomorphism]].
Then $\phi^{-1}: T \to S$ is also an [[Definition:Ordered Ring Isomorphism|ordered ring isomorphis... | By definition, $\phi$ is a [[Definition:Bijection|bijection]].
By [[Bijection iff Inverse is Bijection]], $\phi^{-1}$ is also a [[Definition:Bijection|bijection]].
By definition, an [[Definition:Ordered Ring Isomorphism|ordered ring isomorphism]] from $\phi: \struct {S, +, \circ, \preceq} \to \struct {T, \oplus, *, ... | Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism | https://proofwiki.org/wiki/Inverse_of_Ordered_Ring_Isomorphism_is_Ordered_Ring_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Ordered_Ring_Isomorphism_is_Ordered_Ring_Isomorphism | [
"Ordered Rings"
] | [
"Definition:Ordered Ring",
"Definition:Ordered Ring Isomorphism",
"Definition:Ordered Ring Isomorphism"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Ordered Ring Isomorphism",
"Definition:Order Isomorphism",
"Definition:Ordered Set",
"Definition:Ordered Set",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group",
"Defini... |
proofwiki-5488 | Orthogonal Group is Subgroup of General Linear Group | Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\GL {n, k}$ be the $n$th general linear group on $k$.
Then $\map {\operatorname O} {n, k}$ is a subgroup of $\GL {n, k}$. | From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal.
Let $\mathbf A, \mathbf B \in \map {\operatorname O} {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are orthogonal.
Then by Inverse of Orthogonal Matrix is Orthogonal:
:$\mathbf B^{-1}$ is a orthogonal matrix.
By Product of Orthogon... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\map {\operatorname O} {n, k}$ be the $n$th [[Definition:Orthogonal Group|orthogonal group]] on $k$.
Let $\GL {n, k}$ be the $n$th [[Definition:General Linear Group|general linear group]] on $k$.
Then $\map {\operatorname O} {n, k}$ is a [[Definition:... | From [[Unit Matrix is Orthogonal]], the [[Definition:Unit Matrix|unit matrix $\mathbf I_n$]] is [[Definition:Orthogonal Matrix|orthogonal]].
Let $\mathbf A, \mathbf B \in \map {\operatorname O} {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are [[Definition:Orthogonal Matrix|orthogonal]].
Then by [[Invers... | Orthogonal Group is Subgroup of General Linear Group | https://proofwiki.org/wiki/Orthogonal_Group_is_Subgroup_of_General_Linear_Group | https://proofwiki.org/wiki/Orthogonal_Group_is_Subgroup_of_General_Linear_Group | [
"Orthogonal Groups",
"General Linear Group"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Orthogonal Group",
"Definition:General Linear Group",
"Definition:Subgroup"
] | [
"Unit Matrix is Orthogonal",
"Definition:Unit Matrix",
"Definition:Orthogonal Matrix",
"Definition:Orthogonal Matrix",
"Inverse of Orthogonal Matrix is Orthogonal",
"Definition:Orthogonal Matrix",
"Product of Orthogonal Matrices is Orthogonal Matrix",
"Definition:Orthogonal Matrix",
"Definition:Orth... |
proofwiki-5489 | Orthogonal Group is Group | Let $k$ be a field.
The $n$th orthogonal group on $k$ is a group. | A direct corollary of Orthogonal Group is Subgroup of General Linear Group.
{{qed}} | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
The $n$th [[Definition:Orthogonal Group|orthogonal group]] on $k$ is a [[Definition:Group|group]]. | A direct corollary of [[Orthogonal Group is Subgroup of General Linear Group]].
{{qed}} | Orthogonal Group is Group | https://proofwiki.org/wiki/Orthogonal_Group_is_Group | https://proofwiki.org/wiki/Orthogonal_Group_is_Group | [
"Orthogonal Groups"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Orthogonal Group",
"Definition:Group"
] | [
"Orthogonal Group is Subgroup of General Linear Group"
] |
proofwiki-5490 | Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism | Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be ordered rings.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring isomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring is... | From Composite of Order Isomorphisms is Order Isomorphism, $\psi \circ \phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_3, \preccurlyeq_3}$ is an order isomorphism.
From Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\psi \circ \phi$ is an algebraic structure isomorphism.
From Isomorphism Preserves... | Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be [[Definition:Ordered Ring|ordered rings]].
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be [[Definition:Ordered Ring Isomorphism|ordered ring isomorphisms]].
Then the [[De... | From [[Composite of Order Isomorphisms is Order Isomorphism]], $\psi \circ \phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_3, \preccurlyeq_3}$ is an [[Definition:Order Isomorphism|order isomorphism]].
From [[Composite of Isomorphisms in Algebraic Structure is Isomorphism]], $\psi \circ \phi$ is an [[Definition:Isom... | Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Ring_Isomorphisms_is_Ordered_Ring_Isomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Ring_Isomorphisms_is_Ordered_Ring_Isomorphism | [
"Ring Isomorphisms",
"Order Isomorphisms"
] | [
"Definition:Ordered Ring",
"Definition:Ordered Ring Isomorphism",
"Definition:Composition of Mappings",
"Definition:Ordered Ring Isomorphism"
] | [
"Composite of Order Isomorphisms is Order Isomorphism",
"Definition:Order Isomorphism",
"Composite of Isomorphisms is Isomorphism/Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Isomorphism Preserves Groups",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Isomorphism... |
proofwiki-5491 | Composite of Ordered Ring Monomorphisms is Ordered Ring Monomorphism | Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be ordered rings.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring monomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring m... | From Composite of Order Embeddings is Order Embedding, $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an order embedding.
From Composite of Monomorphisms is Monomorphism, $\psi \circ \phi$ is a monomorphism.
From Group Monomorphism preserves Group, it follows that $\psi \circ \phi$ is a gro... | Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be [[Definition:Ordered Ring|ordered rings]].
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be [[Definition:Ordered Ring Monomorphism|ordered ring monomorphisms]].
Then the [[... | From [[Composite of Order Embeddings is Order Embedding]], $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an [[Definition:Order Embedding|order embedding]].
From [[Composite of Monomorphisms is Monomorphism]], $\psi \circ \phi$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism... | Composite of Ordered Ring Monomorphisms is Ordered Ring Monomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Ring_Monomorphisms_is_Ordered_Ring_Monomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Ring_Monomorphisms_is_Ordered_Ring_Monomorphism | [
"Ring Monomorphisms",
"Order Embeddings",
"Ordered Rings"
] | [
"Definition:Ordered Ring",
"Definition:Ordered Ring Monomorphism",
"Definition:Composition of Mappings",
"Definition:Ordered Ring Monomorphism"
] | [
"Composite of Order Embeddings is Order Embedding",
"Definition:Order Embedding",
"Composite of Monomorphisms is Monomorphism",
"Definition:Monomorphism (Abstract Algebra)",
"Group Monomorphism preserves Group",
"Definition:Group Monomorphism",
"Semigroup Monomorphism preserves Semigroup",
"Definition... |
proofwiki-5492 | Rescaling is Linear Transformation | Let $\struct {R, +, \times}$ be a commutative ring.
Let $\struct {V, +, \ast}_R$ be an $R$-module.
Then for any $r \in R$, the rescaling operator:
:$m_r: V \to V, v \mapsto r \ast v$
is a linear transformation. | Let $v \in V$ and $s \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map {m_r} {s \ast v}
| r = r \ast \paren {s \ast v}
| c = {{Defof|Rescaling Operator}}
}}
{{eqn | r = \paren {r \times s} \ast v
| c = $V$ is an $R$-module
}}
{{eqn | r = \paren {s \times r} \ast v
| c = $R$ is a commutative ring
}}
{... | Let $\struct {R, +, \times}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {V, +, \ast}_R$ be an [[Definition:Module over Ring|$R$-module]].
Then for any $r \in R$, the [[Definition:Rescaling Operator|rescaling operator]]:
:$m_r: V \to V, v \mapsto r \ast v$
is a [[Definition:Linear Transform... | Let $v \in V$ and $s \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map {m_r} {s \ast v}
| r = r \ast \paren {s \ast v}
| c = {{Defof|Rescaling Operator}}
}}
{{eqn | r = \paren {r \times s} \ast v
| c = $V$ is an [[Definition:Module over Ring|$R$-module]]
}}
{{eqn | r = \paren {s \times r} \ast v
| ... | Rescaling is Linear Transformation | https://proofwiki.org/wiki/Rescaling_is_Linear_Transformation | https://proofwiki.org/wiki/Rescaling_is_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Commutative Ring",
"Definition:Module over Ring",
"Definition:Rescaling Operator",
"Definition:Linear Transformation"
] | [
"Definition:Module over Ring",
"Definition:Commutative Ring",
"Definition:Module over Ring",
"Definition:Module over Ring",
"Definition:Linear Transformation",
"Category:Linear Transformations"
] |
proofwiki-5493 | Determinant of Rescaling Matrix | Let $R$ be a commutative ring.
Let $r \in R$.
Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by:
:$\sqbrk {r \, \mathbf I_n}_{i j} = \begin {cases} r & : i = j \\ 0 & : i \ne j \end {cases}$
Then:
:$\map \det {r \, \mathbf I_n} = r^n$
where $\det$ denotes determinant. | From Determinant of Diagonal Matrix, it follows directly that:
:$\map \det {r \, \mathbf I_n} = \ds \prod_{i \mathop = 1}^n r = r^n$
{{qed}} | Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
Let $r \in R$.
Let $r \, \mathbf I_n$ be the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]] defined by:
:$\sqbrk {r \, \mathbf I_n}_{i j} = \begin {cases} r & : i = j \\ 0 & : i \ne j \end {cases}$
Then:
... | From [[Determinant of Diagonal Matrix]], it follows directly that:
:$\map \det {r \, \mathbf I_n} = \ds \prod_{i \mathop = 1}^n r = r^n$
{{qed}} | Determinant of Rescaling Matrix | https://proofwiki.org/wiki/Determinant_of_Rescaling_Matrix | https://proofwiki.org/wiki/Determinant_of_Rescaling_Matrix | [
"Determinant of Rescaling Matrix",
"Determinants"
] | [
"Definition:Commutative Ring",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Determinant/Matrix"
] | [
"Determinant of Diagonal Matrix"
] |
proofwiki-5494 | Inverse of Rescaling Matrix | Let $R$ be a commutative ring with unity.
Let $r \in R$ be a unit in $R$.
Let $r \, \mathbf I_n$ be the $n \times n$ rescaling matrix of $r$.
Then:
:$\paren {r \, \mathbf I_n}^{-1} = r^{-1} \, \mathbf I_n$ | By definition, a rescaling matrix is also a diagonal matrix.
Hence Inverse of Diagonal Matrix applies, and since $r$ is a unit, it gives the desired result.
{{qed}}
Category:Matrix Algebra
fntdpuxeksnrvrzyedfpvwv41vz2cis | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $r \in R$ be a [[Definition:Unit of Ring|unit]] in $R$.
Let $r \, \mathbf I_n$ be the $n \times n$ [[Definition:Rescaling Matrix|rescaling matrix]] of $r$.
Then:
:$\paren {r \, \mathbf I_n}^{-1} = r^{-1} \, \mathbf I_n$ | By definition, a [[Definition:Rescaling Matrix|rescaling matrix]] is also a [[Definition:Diagonal Matrix|diagonal matrix]].
Hence [[Inverse of Diagonal Matrix]] applies, and since $r$ is a [[Definition:Unit of Ring|unit]], it gives the desired result.
{{qed}}
[[Category:Matrix Algebra]]
fntdpuxeksnrvrzyedfpvwv41vz2ci... | Inverse of Rescaling Matrix | https://proofwiki.org/wiki/Inverse_of_Rescaling_Matrix | https://proofwiki.org/wiki/Inverse_of_Rescaling_Matrix | [
"Matrix Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Unit of Ring",
"Definition:Rescaling Matrix"
] | [
"Definition:Rescaling Matrix",
"Definition:Diagonal Matrix",
"Inverse of Diagonal Matrix",
"Definition:Unit of Ring",
"Category:Matrix Algebra"
] |
proofwiki-5495 | Integral Multiple of Ring Element/General Result | :$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$. | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$
In what follows, we make extensive use of Integral Multiple of Ring Element:
:$\forall n \in \Z: \forall x \in R: \paren {m \cdot x} \circ x = m \cdot \paren ... | :$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$
In what follows, we make extensive use of [[Integral Multiple of Ring Element]]:
:$\for... | Integral Multiple of Ring Element/General Result | https://proofwiki.org/wiki/Integral_Multiple_of_Ring_Element/General_Result | https://proofwiki.org/wiki/Integral_Multiple_of_Ring_Element/General_Result | [
"Ring Theory",
"Proofs by Induction"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Integral Multiple of Ring Element",
"Principle of Mathematical Induction"
] |
proofwiki-5496 | Characteristic of Division Ring is Zero or Prime | Let $\struct {D, +, \circ}$ be a division ring.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number. | By definition, a division ring has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
Thus... | Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $\Char D$ be the [[Definition:Characteristic of Ring|characteristic]] of $D$.
Then $\Char D$ is either $0$ or a [[Definition:Prime Number|prime number]]. | By definition, a [[Definition:Division Ring|division ring]] has no [[Definition:Proper Zero Divisor|proper zero divisors]].
If $\struct {D, +, \circ}$ is [[Definition:Finite Ring|finite]], then from [[Characteristic of Finite Ring with No Zero Divisors]], $\Char D$ is [[Definition:Prime Number|prime]].
On the other ... | Characteristic of Division Ring is Zero or Prime | https://proofwiki.org/wiki/Characteristic_of_Division_Ring_is_Zero_or_Prime | https://proofwiki.org/wiki/Characteristic_of_Division_Ring_is_Zero_or_Prime | [
"Division Rings"
] | [
"Definition:Division Ring",
"Definition:Characteristic of Ring",
"Definition:Prime Number"
] | [
"Definition:Division Ring",
"Definition:Proper Zero Divisor",
"Definition:Finite Ring",
"Characteristic of Finite Ring with No Zero Divisors",
"Definition:Prime Number",
"Definition:Finite Ring"
] |
proofwiki-5497 | Characteristic of Integral Domain is Zero or Prime | Let $\struct {D, +, \circ}$ be an integral domain.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number. | By definition, an integral domain has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
T... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $\Char D$ be the [[Definition:Characteristic of Ring|characteristic]] of $D$.
Then $\Char D$ is either $0$ or a [[Definition:Prime Number|prime number]]. | By definition, an [[Definition:Integral Domain|integral domain]] has no [[Definition:Proper Zero Divisor|proper zero divisors]].
If $\struct {D, +, \circ}$ is [[Definition:Finite Ring|finite]], then from [[Characteristic of Finite Ring with No Zero Divisors]], $\Char D$ is [[Definition:Prime Number|prime]].
On the o... | Characteristic of Integral Domain is Zero or Prime | https://proofwiki.org/wiki/Characteristic_of_Integral_Domain_is_Zero_or_Prime | https://proofwiki.org/wiki/Characteristic_of_Integral_Domain_is_Zero_or_Prime | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Characteristic of Ring",
"Definition:Prime Number"
] | [
"Definition:Integral Domain",
"Definition:Proper Zero Divisor",
"Definition:Finite Ring",
"Characteristic of Finite Ring with No Zero Divisors",
"Definition:Prime Number",
"Definition:Finite Ring"
] |
proofwiki-5498 | Sigma-Algebra Generated by Complements of Generators | Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\GG$ be a generator for $\Sigma$.
Then:
:$\GG' := \set {X \setminus G: G \in \GG}$
the set of relative complements of $\GG$, is also a generator for $\Sigma$. | {{begin-eqn}}
{{eqn | q = \forall G \in \GG
| l = G
| o = \in
| r = \Sigma
| c = {{Defof|Sigma-Algebra Generated by Collection of Subsets/Generator|Generator of Sigma-Algebra}}
}}
{{eqn | ll= \leadsto
| q = \forall G \in \GG
| l = X \setminus G
| o = \in
| r = \Sigma
... | Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a set $X$.
Let $\GG$ be a [[Definition:Sigma-Algebra Generated by Collection of Subsets|generator]] for $\Sigma$.
Then:
:$\GG' := \set {X \setminus G: G \in \GG}$
the set of [[Definition:Relative Complement|relative complements]] of $\GG$, is also ... | {{begin-eqn}}
{{eqn | q = \forall G \in \GG
| l = G
| o = \in
| r = \Sigma
| c = {{Defof|Sigma-Algebra Generated by Collection of Subsets/Generator|Generator of Sigma-Algebra}}
}}
{{eqn | ll= \leadsto
| q = \forall G \in \GG
| l = X \setminus G
| o = \in
| r = \Sigma
... | Sigma-Algebra Generated by Complements of Generators | https://proofwiki.org/wiki/Sigma-Algebra_Generated_by_Complements_of_Generators | https://proofwiki.org/wiki/Sigma-Algebra_Generated_by_Complements_of_Generators | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets",
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Relative Complement",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Set Difference with Set Difference",
"Intersection with Subset is Subset",
"Category:Sigma-Algebras Generated by Collection of Subsets"
] |
proofwiki-5499 | Sigma-Algebra Extended by Single Set | Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $S \subseteq X$ be a subset of $X$.
For subsets $T \subseteq X$ of $X$, denote $T^\complement$ for the set difference $X \setminus T$.
Then:
:$\map \sigma {\Sigma \cup \set S} = \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
whe... | Define $\Sigma'$ as follows:
:$\Sigma' := \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
Picking $E_1 = X$ and $E_2 = \O$ (allowed by Sigma-Algebra Contains Empty Set), it follows that $S \in \Sigma'$.
On the other hand, for any $E_1 \in \Sigma$, have by Intersection Distributes o... | Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a set $X$.
Let $S \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
For [[Definition:Subset|subsets]] $T \subseteq X$ of $X$, denote $T^\complement$ for the [[Definition:Set Difference|set difference]] $X \setminus T$.
Then:
:$\map \sigma {\Si... | Define $\Sigma'$ as follows:
:$\Sigma' := \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
Picking $E_1 = X$ and $E_2 = \O$ (allowed by [[Sigma-Algebra Contains Empty Set]]), it follows that $S \in \Sigma'$.
On the other hand, for any $E_1 \in \Sigma$, have by [[Intersection Dis... | Sigma-Algebra Extended by Single Set | https://proofwiki.org/wiki/Sigma-Algebra_Extended_by_Single_Set | https://proofwiki.org/wiki/Sigma-Algebra_Extended_by_Single_Set | [
"Sigma-Algebras"
] | [
"Definition:Sigma-Algebra",
"Definition:Subset",
"Definition:Subset",
"Definition:Set Difference",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Sigma-Algebra Contains Empty Set",
"Intersection Distributes over Union",
"Union with Relative Complement",
"Sigma-Algebra Closed under Union",
"Sigma-Algebra Closed under Finite Intersection",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"De Morgan's Laws (Set Theory)/Set Difference/Diffe... |
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