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-3000 | Sum of Squares of Binomial Coefficients | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | Consider the Binomial Theorem:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
| l = \paren {1 + x}^n
| r = \sum_{j \mathop = 0}^n \dbinom n j x^j
| c =
}}
{{eqn | r = \dbinom n 0 x^0 + \dbinom n 1 x^1 + \dbinom n 2 x^2 + \cdots + \dbinom n {n - 1} x^{n - 1} + \dbinom n n x^n
| c =
}}
{{eqn... | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | Consider the [[Binomial Theorem]]:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
| l = \paren {1 + x}^n
| r = \sum_{j \mathop = 0}^n \dbinom n j x^j
| c =
}}
{{eqn | r = \dbinom n 0 x^0 + \dbinom n 1 x^1 + \dbinom n 2 x^2 + \cdots + \dbinom n {n - 1} x^{n - 1} + \dbinom n n x^n
| c =
}}
... | Sum of Squares of Binomial Coefficients/Algebraic Proof | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients/Algebraic_Proof | [
"Sum of Squares of Binomial Coefficients",
"Binomial Coefficients"
] | [] | [
"Binomial Theorem",
"Symmetry Rule for Binomial Coefficients",
"Definition:Coefficient",
"Definition:Coefficient",
"Binomial Theorem"
] |
proofwiki-3001 | Sum of Squares of Binomial Coefficients | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | Consider the number of paths in the integer lattice from $\tuple {0, 0}$ to $\tuple {n, n}$ using only single steps of the form:
:$\tuple {i, j} \to \tuple {i + 1, j}$
:$\tuple {i, j} \to \tuple {i, j + 1}$
that is, either to the right or up.
This process takes $2 n$ steps, of which $n$ are steps to the right.
Thus the... | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | Consider the number of paths in the integer lattice from $\tuple {0, 0}$ to $\tuple {n, n}$ using only single steps of the form:
:$\tuple {i, j} \to \tuple {i + 1, j}$
:$\tuple {i, j} \to \tuple {i, j + 1}$
that is, either to the right or up.
This process takes $2 n$ steps, of which $n$ are steps to the right.
Thus t... | Sum of Squares of Binomial Coefficients/Combinatorial Proof | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients/Combinatorial_Proof | [
"Sum of Squares of Binomial Coefficients",
"Binomial Coefficients"
] | [] | [] |
proofwiki-3002 | Sum of Squares of Binomial Coefficients | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
$\map P 0$ is true, as this just says:
:$\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$
This holds by definition.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\dbinom 1 0^2 + \dbin... | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
| r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}} | For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
$\map P 0$ is true, as this just says:
:$\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$
This holds by [[Definition:Binomial Coefficient|definition]].
=== Basis for the Induction... | Sum of Squares of Binomial Coefficients/Inductive Proof | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients | https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients/Inductive_Proof | [
"Sum of Squares of Binomial Coefficients",
"Binomial Coefficients"
] | [] | [
"Definition:Proposition",
"Definition:Binomial Coefficient",
"Definition:Binomial Coefficient",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Pascal's Rule",
"Translation of Index Variable of Summation",
"Sum of Squares of Binomial Coefficient... |
proofwiki-3003 | Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces | Let $\mathbb K$ be a subfield of $\C$.
Let $V$ be a semi-inner product space over $\mathbb K$.
Let $x, y$ be vectors in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$ | Let $x, y \in V$.
Let $\lambda \in \mathbb K$.
Then:
{{begin-eqn}}
{{eqn | l = 0
| o = \le
| r = \innerprod {x - \lambda y} {x - \lambda y}
| c = Semi-inner product axioms $\paren {3}$: Non-Negative Definiteness
}}
{{eqn | r = \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \... | Let $\mathbb K$ be a [[Definition:Field (Abstract Algebra)|subfield]] of $\C$.
Let $V$ be a [[Definition:Semi-Inner Product Space|semi-inner product space]] over $\mathbb K$.
Let $x, y$ be [[Definition:Vector (Linear Algebra)|vectors]] in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$ | Let $x, y \in V$.
Let $\lambda \in \mathbb K$.
Then:
{{begin-eqn}}
{{eqn | l = 0
| o = \le
| r = \innerprod {x - \lambda y} {x - \lambda y}
| c = [[Axiom:Complex Semi-Inner Product Axioms|Semi-inner product axioms $\paren {3}$: Non-Negative Definiteness]]
}}
{{eqn | r = \innerprod x x + \innerprod... | Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 1 | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Inner_Product_Spaces | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Inner_Product_Spaces/Proof_1 | [
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Inner Product Spaces",
"Semi-Inner Product Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Semi-Inner Product Space",
"Definition:Vector/Linear Algebra"
] | [
"Axiom:Complex Semi-Inner Product Axioms",
"Axiom:Complex Semi-Inner Product Axioms",
"Axiom:Complex Semi-Inner Product Axioms",
"Definition:Complex Conjugate",
"Definition:Field (Abstract Algebra)",
"Product of Complex Conjugates",
"Modulus in Terms of Conjugate",
"Number Field has Rational Numbers a... |
proofwiki-3004 | Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces | Let $\mathbb K$ be a subfield of $\C$.
Let $V$ be a semi-inner product space over $\mathbb K$.
Let $x, y$ be vectors in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$ | This proof assumes that $V$ is a semi-inner product space over $\R$.
Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of semi-inner products.
Define $f_{x, y}: \R \to \R_{\ge 0}$ by:
:$\map {f_{x, y} } \lambda = \innerprod {x - \lambda y} {x - \lambda y}$
Then by property $(4)$ of... | Let $\mathbb K$ be a [[Definition:Field (Abstract Algebra)|subfield]] of $\C$.
Let $V$ be a [[Definition:Semi-Inner Product Space|semi-inner product space]] over $\mathbb K$.
Let $x, y$ be [[Definition:Vector (Linear Algebra)|vectors]] in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$ | This proof assumes that $V$ is a [[Definition:Semi-Inner Product Space|semi-inner product space]] over $\R$.
Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of [[Definition:Semi-Inner Product|semi-inner products]].
Define $f_{x, y}: \R \to \R_{\ge 0}$ by:
:$\map {f_{x, y} } \l... | Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 2 | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Inner_Product_Spaces | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Inner_Product_Spaces/Proof_2 | [
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Inner Product Spaces",
"Semi-Inner Product Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Semi-Inner Product Space",
"Definition:Vector/Linear Algebra"
] | [
"Definition:Semi-Inner Product Space",
"Definition:Semi-Inner Product",
"Definition:Semi-Inner Product",
"Definition:Semi-Inner Product",
"Definition:Semi-Inner Product",
"Definition:Quadratic_Equation",
"Definition:Real Number",
"Definition:Root of Polynomial",
"Solution to Quadratic Equation",
"... |
proofwiki-3005 | Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals | Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Then:
:$\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map g t}^2 \rd t$ | {{begin-eqn}}
{{eqn | q = \forall x \in \R
| l = 0
| o = \le
| r = \paren {x \map f t + \map g t}^2
}}
{{eqn | l = 0
| o = \le
| r = \int_a^b \paren {x \map f t + \map g t}^2 \rd t
| c = Relative Sizes of Definite Integrals
}}
{{eqn | r = x^2 \int_a^b \paren {\map f t}^2 \rd t + 2 x ... | Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Then:
:$\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map... | {{begin-eqn}}
{{eqn | q = \forall x \in \R
| l = 0
| o = \le
| r = \paren {x \map f t + \map g t}^2
}}
{{eqn | l = 0
| o = \le
| r = \int_a^b \paren {x \map f t + \map g t}^2 \rd t
| c = [[Relative Sizes of Definite Integrals]]
}}
{{eqn | r = x^2 \int_a^b \paren {\map f t}^2 \rd t + ... | Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Definite_Integrals | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Definite_Integrals | [
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Definite Integrals"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function/Interval",
"Definition:Real Interval/Closed"
] | [
"Relative Sizes of Definite Integrals",
"Linear Combination of Integrals/Definite",
"Definition:Quadratic Equation",
"Cauchy's Inequality",
"Definition:Discriminant of Polynomial/Quadratic Equation"
] |
proofwiki-3006 | Cauchy-Schwarz Inequality/Complex Numbers | :$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$
where all of $w_i, z_i \in \C$. | Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers.
Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < ... | :$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$
where all of $w_i, z_i \in \C$. | Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary [[Definition:Complex Number|complex numbers]].
Take the [[Binet-Cauchy Identity]]:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j}... | Cauchy-Schwarz Inequality/Complex Numbers | https://proofwiki.org/wiki/Cauchy-Schwarz_Inequality/Complex_Numbers | https://proofwiki.org/wiki/Cauchy-Schwarz_Inequality/Complex_Numbers | [
"Complex Analysis",
"Cauchy-Bunyakovsky-Schwarz Inequality"
] | [] | [
"Definition:Complex Number",
"Binet-Cauchy Identity",
"Modulus in Terms of Conjugate",
"Complex Modulus is Non-Negative"
] |
proofwiki-3007 | Ceiling of Floor is Floor | :$\ceiling {\floor x} = \floor x$ | Let $y = \floor x$.
By Floor Function is Integer, we have that $y \in \Z$.
Then from Real Number is Integer iff equals Ceiling, we have:
:$\ceiling y = y$
So:
:$\ceiling {\floor x} = \floor x$
{{qed}} | :$\ceiling {\floor x} = \floor x$ | Let $y = \floor x$.
By [[Floor Function is Integer]], we have that $y \in \Z$.
Then from [[Real Number is Integer iff equals Ceiling]], we have:
:$\ceiling y = y$
So:
:$\ceiling {\floor x} = \floor x$
{{qed}} | Ceiling of Floor is Floor | https://proofwiki.org/wiki/Ceiling_of_Floor_is_Floor | https://proofwiki.org/wiki/Ceiling_of_Floor_is_Floor | [
"Floor Function",
"Ceiling Function"
] | [] | [
"Floor Function is Integer",
"Real Number is Integer iff equals Ceiling"
] |
proofwiki-3008 | Floor of Ceiling is Ceiling | :$\floor {\ceiling x} = \ceiling x$ | Let $y = \ceiling x$.
By Ceiling Function is Integer, we have that $y \in \Z$.
From Real Number is Integer iff equals Floor, we have:
:$\floor y = y$
So:
:$\floor {\ceiling x} = \ceiling x$
{{qed}} | :$\floor {\ceiling x} = \ceiling x$ | Let $y = \ceiling x$.
By [[Ceiling Function is Integer]], we have that $y \in \Z$.
From [[Real Number is Integer iff equals Floor]], we have:
:$\floor y = y$
So:
:$\floor {\ceiling x} = \ceiling x$
{{qed}} | Floor of Ceiling is Ceiling | https://proofwiki.org/wiki/Floor_of_Ceiling_is_Ceiling | https://proofwiki.org/wiki/Floor_of_Ceiling_is_Ceiling | [
"Floor Function",
"Ceiling Function"
] | [] | [
"Ceiling Function is Integer",
"Real Number is Integer iff equals Floor"
] |
proofwiki-3009 | Floor equals Ceiling iff Integer | :<nowiki>$\floor x = \begin {cases}
\ceiling x & : x \in \Z \\
\ceiling x - 1 & : x \notin \Z \\
\end {cases}$</nowiki>
or equivalently:
:<nowiki>$\ceiling x = \begin {cases}
\floor x & : x \in \Z \\
\floor x + 1 & : x \notin \Z \\
\end {cases}$</nowiki>
where $\Z$ is the set of integers. | From Real Number is Integer iff equals Floor:
:$x \in \Z \implies x = \floor x$
From Real Number is Integer iff equals Ceiling:
:$x \in \Z \implies x = \ceiling x$
So:
:$x \in \Z \implies \floor x = \ceiling x$
Now let $x \notin \Z$.
From the definition of the floor function:
:$\floor x = \map \sup {\set {m \in \Z: m \... | :<nowiki>$\floor x = \begin {cases}
\ceiling x & : x \in \Z \\
\ceiling x - 1 & : x \notin \Z \\
\end {cases}$</nowiki>
or equivalently:
:<nowiki>$\ceiling x = \begin {cases}
\floor x & : x \in \Z \\
\floor x + 1 & : x \notin \Z \\
\end {cases}$</nowiki>
where $\Z$ is the set of [[Definition:Integer|integers]]. | From [[Real Number is Integer iff equals Floor]]:
:$x \in \Z \implies x = \floor x$
From [[Real Number is Integer iff equals Ceiling]]:
:$x \in \Z \implies x = \ceiling x$
So:
:$x \in \Z \implies \floor x = \ceiling x$
Now let $x \notin \Z$.
From the definition of the [[Definition:Floor Function|floor function]]:
... | Floor equals Ceiling iff Integer | https://proofwiki.org/wiki/Floor_equals_Ceiling_iff_Integer | https://proofwiki.org/wiki/Floor_equals_Ceiling_iff_Integer | [
"Floor Function",
"Ceiling Function"
] | [
"Definition:Integer"
] | [
"Real Number is Integer iff equals Floor",
"Real Number is Integer iff equals Ceiling",
"Definition:Floor Function",
"Definition:Ceiling Function"
] |
proofwiki-3010 | Ceiling of Negative equals Negative of Floor | :$\ceiling {-x} = -\floor x$ | From Integer equals Floor iff between Number and One Less we have:
:$x - 1 < \floor x \le x$
and so, by multiplying both sides by -1:
:$-x + 1 > -\floor x \ge -x$
From Integer equals Ceiling iff between Number and One More we have:
:$\ceiling x = n \iff x \le n < x + 1$
Hence:
:$-x \le -\floor x < -x + 1 \implies \ceil... | :$\ceiling {-x} = -\floor x$ | From [[Integer equals Floor iff between Number and One Less]] we have:
:$x - 1 < \floor x \le x$
and so, by multiplying both sides by -1:
:$-x + 1 > -\floor x \ge -x$
From [[Integer equals Ceiling iff between Number and One More]] we have:
:$\ceiling x = n \iff x \le n < x + 1$
Hence:
:$-x \le -\floor x < -x + 1 \imp... | Ceiling of Negative equals Negative of Floor | https://proofwiki.org/wiki/Ceiling_of_Negative_equals_Negative_of_Floor | https://proofwiki.org/wiki/Ceiling_of_Negative_equals_Negative_of_Floor | [
"Floor Function",
"Ceiling Function"
] | [] | [
"Integer equals Floor iff between Number and One Less",
"Integer equals Ceiling iff between Number and One More"
] |
proofwiki-3011 | Floor of Negative equals Negative of Ceiling | :$\floor {-x} = -\ceiling x$ | From Integer equals Ceiling iff between Number and One More we have:
:$x \le \ceiling x < x + 1$
and so, by multiplying by -1:
:$-x \ge -\ceiling x > -x - 1$
From Integer equals Floor iff between Number and One Less we have:
:$\floor x = n \iff x - 1 < n \le x$
Hence:
:$-x - 1 < -\ceiling x \le -x \implies \floor {-x} ... | :$\floor {-x} = -\ceiling x$ | From [[Integer equals Ceiling iff between Number and One More]] we have:
:$x \le \ceiling x < x + 1$
and so, by multiplying by -1:
:$-x \ge -\ceiling x > -x - 1$
From [[Integer equals Floor iff between Number and One Less]] we have:
:$\floor x = n \iff x - 1 < n \le x$
Hence:
:$-x - 1 < -\ceiling x \le -x \implies \f... | Floor of Negative equals Negative of Ceiling | https://proofwiki.org/wiki/Floor_of_Negative_equals_Negative_of_Ceiling | https://proofwiki.org/wiki/Floor_of_Negative_equals_Negative_of_Ceiling | [
"Floor Function",
"Ceiling Function"
] | [] | [
"Integer equals Ceiling iff between Number and One More",
"Integer equals Floor iff between Number and One Less"
] |
proofwiki-3012 | Integer is Coprime to 1 | Every integer is coprime to $1$.
That is:
:$\forall n \in \Z: n \perp 1$ | Follows from the definitions of coprime and greatest common divisor as follows.
When $n \in \Z: n \ne 0$ we have:
:$\gcd \set {n, 1} = 1$
Then by definition again:
:$\gcd \set {n, 0} = n$
and so when $n = 1$ we have:
:$\gcd \set {1, 0} = 1$
{{qed}} | Every [[Definition:Integer|integer]] is [[Definition:Coprime Integers|coprime]] to $1$.
That is:
:$\forall n \in \Z: n \perp 1$ | Follows from the definitions of [[Definition:Coprime Integers|coprime]] and [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] as follows.
When $n \in \Z: n \ne 0$ we have:
:$\gcd \set {n, 1} = 1$
Then by definition again:
:$\gcd \set {n, 0} = n$
and so when $n = 1$ we have:
:$\gcd \set {1, 0... | Integer is Coprime to 1 | https://proofwiki.org/wiki/Integer_is_Coprime_to_1 | https://proofwiki.org/wiki/Integer_is_Coprime_to_1 | [
"Coprime Integers"
] | [
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Coprime/Integers",
"Definition:Greatest Common Divisor/Integers"
] |
proofwiki-3013 | Integers Coprime to Zero | The only integers which are coprime to zero are $1$ and $-1$.
That is:
:$n \in \Z: n \perp 0 \iff n \in \set {1, -1}$
In particular, note that two integers which are coprime to each other cannot both be $0$. | Let $n \in \Z$ such that $n \perp 0$.
First we note that from Coprime Integers cannot Both be Zero, it cannot be the case that $n = 0$.
From the definition of coprime, we have:
:$m \perp n \iff \gcd \set {m, n} = 1$
From the definition of greatest common divisor:
:$\gcd \set {n, 0} = \size n$
where $\size n$ is the abs... | The only [[Definition:Integer|integers]] which are [[Definition:Coprime Integers|coprime]] to [[Definition:Zero (Number)|zero]] are $1$ and $-1$.
That is:
:$n \in \Z: n \perp 0 \iff n \in \set {1, -1}$
In particular, note that two [[Definition:Integer|integers]] which are [[Definition:Coprime Integers|coprime]] to e... | Let $n \in \Z$ such that $n \perp 0$.
First we note that from [[Coprime Integers cannot Both be Zero]], it cannot be the case that $n = 0$.
From the definition of [[Definition:Coprime Integers|coprime]], we have:
:$m \perp n \iff \gcd \set {m, n} = 1$
From the definition of [[Definition:Greatest Common Divisor of I... | Integers Coprime to Zero | https://proofwiki.org/wiki/Integers_Coprime_to_Zero | https://proofwiki.org/wiki/Integers_Coprime_to_Zero | [
"Coprime Integers"
] | [
"Definition:Integer",
"Definition:Coprime/Integers",
"Definition:Zero (Number)",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Coprime Integers cannot Both be Zero",
"Definition:Coprime/Integers",
"Definition:Greatest Common Divisor/Integers",
"Definition:Absolute Value"
] |
proofwiki-3014 | Sum of Binomial Coefficients over Upper Index | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \binom j m
| r = \binom {n + 1} {m + 1}
| c =
}}
{{eqn | r = \dbinom 0 m + \dbinom 1 m + \dbinom 2 m + \cdots + \dbinom n m = \dbinom {n + 1} {m + 1}
| c =
}}
{{end-eqn}} | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
=== Basis for the Induction ===
$\map P 0$ says:
:$\dbinom 0 m = \dbinom 1 {m + 1}$
When $m = 0$ we have by definition:
:$\dbinom 0 0 = 1 = \dbinom 1 1$
When $m > 0$ we also have ... | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \binom j m
| r = \binom {n + 1} {m + 1}
| c =
}}
{{eqn | r = \dbinom 0 m + \dbinom 1 m + \dbinom 2 m + \cdots + \dbinom n m = \dbinom {n + 1} {m + 1}
| c =
}}
{{end-eqn}} | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
=== Basis for the Induction ===
$\map P 0$ says:
:$\dbinom 0 m = \dbinom 1 {m + 1}$
When $m = 0$ we have by... | Sum of Binomial Coefficients over Upper Index/Proof 1 | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Upper_Index | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Upper_Index/Proof_1 | [
"Binomial Coefficients",
"Sum of Binomial Coefficients over Upper Index"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Binomial Coefficient",
"Definition:Binomial Coefficient",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Sum of Binomial Coefficients over Upper Index",
"Pascal's Rule... |
proofwiki-3015 | Divisibility of Product of Consecutive Integers | The product of $n$ consecutive positive integers is divisible by the product of the ''first'' $n$ consecutive positive integers.
That is:
:$\ds \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$ | {{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \paren {m + k}
| r = \frac {\paren {m + n}!} {m!}
| c =
}}
{{eqn | r = n! \frac {\paren {m + n}!} {m! \, n!}
| c =
}}
{{eqn | r = n! \binom {m + n} m
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \binom {m + n} m \prod_{k \mathop = 1}^n k... | The [[Definition:Integer Multiplication|product]] of $n$ consecutive [[Definition:Positive Integer|positive integers]] is [[Definition:Divisor of Integer|divisible]] by the [[Definition:Integer Multiplication|product]] of the ''first'' $n$ consecutive [[Definition:Positive Integer|positive integers]].
That is:
:$\ds \... | {{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \paren {m + k}
| r = \frac {\paren {m + n}!} {m!}
| c =
}}
{{eqn | r = n! \frac {\paren {m + n}!} {m! \, n!}
| c =
}}
{{eqn | r = n! \binom {m + n} m
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \binom {m + n} m \prod_{k \mathop = 1}^n k... | Divisibility of Product of Consecutive Integers | https://proofwiki.org/wiki/Divisibility_of_Product_of_Consecutive_Integers | https://proofwiki.org/wiki/Divisibility_of_Product_of_Consecutive_Integers | [
"Divisibility of Product of Consecutive Integers",
"Divisibility",
"Factorials"
] | [
"Definition:Multiplication/Integers",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Multiplication/Integers",
"Definition:Positive/Integer"
] | [
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-3016 | Equal Numbers are Congruent | :$\forall x, y, z \in \R: x = y \implies x \equiv y \pmod z$
where $x \equiv y \pmod z$ denotes congruence modulo $z$. | {{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x - y
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x - y
| r = 0 \cdot z
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \equiv
| r = y
| rr= \pmod z
| c = {{Defof|Congruen... | :$\forall x, y, z \in \R: x = y \implies x \equiv y \pmod z$
where $x \equiv y \pmod z$ denotes [[Definition:Congruence (Number Theory)|congruence modulo $z$]]. | {{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x - y
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x - y
| r = 0 \cdot z
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \equiv
| r = y
| rr= \pmod z
| c = {{Defof|Congruen... | Equal Numbers are Congruent | https://proofwiki.org/wiki/Equal_Numbers_are_Congruent | https://proofwiki.org/wiki/Equal_Numbers_are_Congruent | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)"
] | [] |
proofwiki-3017 | Complex Roots of Unity in Exponential Form | Let $n \in \Z$ be an integer such that $n > 0$.
Let $z \in \C$ be a complex number such that $z^n = 1$.
Then:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of $n$th roots of unity.
That is:
:$z \in \set {1, e^{2 i \pi / n}, e^{4 i \pi / n}, \ldots, e^{2 \paren {n - 1} i \pi / n} }$
Thus for every... | Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$.
Then:
:$z^n \in \set {e^{2 i k \pi}: k \in \N_n}$
Hence $z^n = 1$.
Now suppose $z^n = 1$.
We determine the possible values of $z$ using Roots of Complex Number.
Let $z = r e^{i \theta}$, where:
:$r$ is a positive real number
:$\theta$ is a real number.
Then $\cmod {z^n}... | Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Let $z \in \C$ be a [[Definition:Complex Number|complex number]] such that $z^n = 1$.
Then:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the [[Definition:Complex Roots of Unity|set of $n$th roots of unity]].
That is:
:$z \in \set ... | Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$.
Then:
:$z^n \in \set {e^{2 i k \pi}: k \in \N_n}$
Hence $z^n = 1$.
Now suppose $z^n = 1$.
We determine the possible values of $z$ using [[Roots of Complex Number]].
Let $z = r e^{i \theta}$, where:
:$r$ is a [[Definition:Positive Real Number|positive real number]]... | Complex Roots of Unity in Exponential Form | https://proofwiki.org/wiki/Complex_Roots_of_Unity_in_Exponential_Form | https://proofwiki.org/wiki/Complex_Roots_of_Unity_in_Exponential_Form | [
"Complex Roots of Unity"
] | [
"Definition:Integer",
"Definition:Complex Number",
"Definition:Root of Unity/Complex",
"Definition:Integer"
] | [
"Roots of Complex Number",
"Definition:Positive/Real Number",
"Definition:Real Number"
] |
proofwiki-3018 | Roots of Unity under Multiplication form Cyclic Group | Let $n \in \Z$ be an integer such that $n > 0$.
The $n$th complex roots of unity under the operation of multiplication form the cyclic group which is isomorphic to $C_n$. | From Complex Roots of Unity in Exponential Form:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of complex $n$th roots of unity.
Let $\omega = e^{2 i \pi / n}$.
Then we have:
:$U_n = \set {\omega^k: k \in \N_n}$
that is:
:$U_n = \set {\omega^0, \omega^1, \omega^2, \ldots, \omega^{n - 1} }$
Let $\o... | Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
The [[Definition:Complex Roots of Unity|$n$th complex roots of unity]] under the operation of [[Definition:Integer Multiplication|multiplication]] form the [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Group Isomorphism|isomorphic]... | From [[Complex Roots of Unity in Exponential Form]]:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the [[Definition:Complex Roots of Unity|set of complex $n$th roots of unity]].
Let $\omega = e^{2 i \pi / n}$.
Then we have:
:$U_n = \set {\omega^k: k \in \N_n}$
that is:
:$U_n = \set {\omega^0, \omega^1,... | Roots of Unity under Multiplication form Cyclic Group | https://proofwiki.org/wiki/Roots_of_Unity_under_Multiplication_form_Cyclic_Group | https://proofwiki.org/wiki/Roots_of_Unity_under_Multiplication_form_Cyclic_Group | [
"Multiplicative Groups of Complex Roots of Unity",
"Complex Roots of Unity",
"Roots of Unity",
"Examples of Cyclic Groups"
] | [
"Definition:Integer",
"Definition:Root of Unity/Complex",
"Definition:Multiplication/Integers",
"Definition:Cyclic Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Complex Roots of Unity in Exponential Form",
"Definition:Root of Unity/Complex",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group",
"Cyclic Groups of Same Order are Isomorphic"
] |
proofwiki-3019 | Inverse Mapping is Unique | Let $f: S \to T$ be a mapping.
If $f$ has an inverse mapping, then that inverse mapping is unique.
That is, if:
:$f$ and $g$ are inverse mappings of each other
and
:$f$ and $h$ are inverse mappings of each other
then $g = h$. | By the definition of inverse mapping:
{{begin-eqn}}
{{eqn | l = g \circ f
| r = I_S
| c =
}}
{{eqn | r = h \circ f
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f \circ g
| r = I_T
| c =
}}
{{eqn | r = f \circ h
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = h
| r... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
If $f$ has an [[Definition:Inverse Mapping|inverse mapping]], then that inverse mapping is [[Definition:Unique|unique]].
That is, if:
:$f$ and $g$ are [[Definition:Inverse Mapping|inverse mappings]] of each other
and
:$f$ and $h$ are [[Definition:Inverse Mapping|... | By the definition of [[Definition:Inverse Mapping/Definition 2|inverse mapping]]:
{{begin-eqn}}
{{eqn | l = g \circ f
| r = I_S
| c =
}}
{{eqn | r = h \circ f
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f \circ g
| r = I_T
| c =
}}
{{eqn | r = f \circ h
| c =
}}
{{end-e... | Inverse Mapping is Unique/Proof 1 | https://proofwiki.org/wiki/Inverse_Mapping_is_Unique | https://proofwiki.org/wiki/Inverse_Mapping_is_Unique/Proof_1 | [
"Inverse Mappings",
"Inverse Mapping is Unique"
] | [
"Definition:Mapping",
"Definition:Inverse Mapping",
"Definition:Unique",
"Definition:Inverse Mapping",
"Definition:Inverse Mapping"
] | [
"Definition:Inverse Mapping/Definition 2",
"Composition of Mappings is Associative",
"Definition:Inverse Mapping/Definition 2"
] |
proofwiki-3020 | Inverse Mapping is Unique | Let $f: S \to T$ be a mapping.
If $f$ has an inverse mapping, then that inverse mapping is unique.
That is, if:
:$f$ and $g$ are inverse mappings of each other
and
:$f$ and $h$ are inverse mappings of each other
then $g = h$. | We need to show that:
:$\forall t \in T: \map g t = \map h t$
So:
{{begin-eqn}}
{{eqn | l = \map f {\map g t}
| r = t
| c = {{Defof|Inverse Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map h t
| r = \map h {\map f {\map g t} }
| c =
}}
{{eqn | ll= \leadsto
| l = \map h t
| r = \ma... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
If $f$ has an [[Definition:Inverse Mapping|inverse mapping]], then that inverse mapping is [[Definition:Unique|unique]].
That is, if:
:$f$ and $g$ are [[Definition:Inverse Mapping|inverse mappings]] of each other
and
:$f$ and $h$ are [[Definition:Inverse Mapping|... | We need to show that:
:$\forall t \in T: \map g t = \map h t$
So:
{{begin-eqn}}
{{eqn | l = \map f {\map g t}
| r = t
| c = {{Defof|Inverse Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map h t
| r = \map h {\map f {\map g t} }
| c =
}}
{{eqn | ll= \leadsto
| l = \map h t
| r = \... | Inverse Mapping is Unique/Proof 2 | https://proofwiki.org/wiki/Inverse_Mapping_is_Unique | https://proofwiki.org/wiki/Inverse_Mapping_is_Unique/Proof_2 | [
"Inverse Mappings",
"Inverse Mapping is Unique"
] | [
"Definition:Mapping",
"Definition:Inverse Mapping",
"Definition:Unique",
"Definition:Inverse Mapping",
"Definition:Inverse Mapping"
] | [] |
proofwiki-3021 | Special Linear Group is Subgroup of General Linear Group | Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is a subgroup of the general linear group $\GL {n, K}$. | Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, they are nonsingular.
So $\SL {n, K}$ is a subset of $\GL {n, K}$.
Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$.
As $\mathbf A$ is nonsingular we have that it has a... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_K$ and [[Definition:Unity of Field|unity]] is $1_K$.
Let $\SL {n, K}$ be the [[Definition:Special Linear Group|special linear group of order $n$ over $K$]].
Then $\SL {n, K}$ is a [[Definition:Subgroup|subgroup]] of... | Because the [[Definition:Determinant of Matrix|determinants]] of the [[Definition:Element|elements]] of $\SL {n, K}$ are not $0_K$, they are [[Definition:Nonsingular Matrix|nonsingular]].
So $\SL {n, K}$ is a [[Definition:Subset|subset]] of $\GL {n, K}$.
Now we need to show that $\SL {n, K}$ is a [[Definition:Subgrou... | Special Linear Group is Subgroup of General Linear Group | https://proofwiki.org/wiki/Special_Linear_Group_is_Subgroup_of_General_Linear_Group | https://proofwiki.org/wiki/Special_Linear_Group_is_Subgroup_of_General_Linear_Group | [
"Matrix Algebra",
"Special Linear Group",
"General Linear Group"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Special Linear Group",
"Definition:Subgroup",
"Definition:General Linear Group"
] | [
"Definition:Determinant/Matrix",
"Definition:Element",
"Definition:Nonsingular Matrix",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Nonsingular Matrix",
"Definition:Inverse Matrix",
"Determinant of Inverse Matrix",
"Determinant of Matrix Product",
"Two-Step Subgroup Test"
] |
proofwiki-3022 | Determinant of Unit Matrix | Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
The determinant of the unit matrix of order $n$ over $R$ is equal to $1_R$. | Let $\mathbf I_n$ denote the unit matrix of order $n$ over $R$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\map \det {\mathbf I_n} = 1_R$ | Let $R$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
The [[Definition:Determinant of Matrix|determinant]] of the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order $n$]] over $R$ ... | Let $\mathbf I_n$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order $n$]] over $R$.
The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map \det {\mathbf I_n}... | Determinant of Unit Matrix | https://proofwiki.org/wiki/Determinant_of_Unit_Matrix | https://proofwiki.org/wiki/Determinant_of_Unit_Matrix | [
"Determinants",
"Unit Matrices"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Determinant/Matrix",
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order"
] | [
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-3023 | Determinant of Inverse Matrix | Let $K$ be a field whose zero is $0_K$ and whose unity is $1_K$.
Let $\mathbf A$ be a nonsingular matrix of order $n$ over $K$.
Then the determinant of its inverse is given by:
:$\map \det {\mathbf A^{-1} } = \dfrac {1_K} {\map \det {\mathbf A} }$ | By definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf I_n$
where $\mathbf I_n$ is the unit matrix.
By Determinant of Unit Matrix:
:$\map \det {\mathbf I_n} = 1_K$
By Determinant of Matrix Product:
:$\map \det {\mathbf A^{-1} } \map \det {\mathbf A} = \map \det {\mathbf A^{-1} \mathbf A}$
Hence the resul... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_K$ and whose [[Definition:Unity of Field|unity]] is $1_K$.
Let $\mathbf A$ be a [[Definition:Nonsingular Matrix|nonsingular matrix]] of [[Definition:Order of Square Matrix|order $n$]] over $K$.
Then the [[Definition... | By definition of [[Definition:Inverse Matrix|inverse matrix]]:
:$\mathbf A \mathbf A^{-1} = \mathbf I_n$
where $\mathbf I_n$ is the [[Definition:Unit Matrix|unit matrix]].
By [[Determinant of Unit Matrix]]:
:$\map \det {\mathbf I_n} = 1_K$
By [[Determinant of Matrix Product]]:
:$\map \det {\mathbf A^{-1} } \map \det ... | Determinant of Inverse Matrix | https://proofwiki.org/wiki/Determinant_of_Inverse_Matrix | https://proofwiki.org/wiki/Determinant_of_Inverse_Matrix | [
"Determinants",
"Inverse Matrices"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Nonsingular Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Determinant/Matrix",
"Definition:Inverse Matrix"
] | [
"Definition:Inverse Matrix",
"Definition:Unit Matrix",
"Determinant of Unit Matrix",
"Determinant of Matrix Product"
] |
proofwiki-3024 | Transposition is of Odd Parity | Let $S_n$ denote the set of permutations on $n$ letters.
Let $\pi \in S_n$ be a transposition.
Then $\pi$ is of odd parity. | Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition.
Let $\Delta_n$ be defined as Product of Differences.
Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$.
Thus:
:$\begin{pmatrix} 1 & 2 \end{pmatrix... | Let $S_n$ denote the [[Definition:Permutation on n Letters|set of permutations on $n$ letters]].
Let $\pi \in S_n$ be a [[Definition:Transposition|transposition]].
Then $\pi$ is of [[Definition:Parity of Permutation|odd parity]]. | Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a [[Definition:Transposition|transposition]].
Let $\Delta_n$ be defined as [[Definition:Product of Differences|Product of Differences]].
Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the fa... | Transposition is of Odd Parity | https://proofwiki.org/wiki/Transposition_is_of_Odd_Parity | https://proofwiki.org/wiki/Transposition_is_of_Odd_Parity | [
"Transpositions"
] | [
"Definition:Permutation on n Letters",
"Definition:Transposition",
"Definition:Parity of Permutation"
] | [
"Definition:Transposition",
"Definition:Product of Differences",
"Conjugates of Transpositions",
"Transposition is Self-Inverse",
"Parity of Conjugate of Permutation",
"Conjugates of Transpositions",
"Transposition is Self-Inverse",
"Parity of Conjugate of Permutation"
] |
proofwiki-3025 | Stabilizer in Group of Transformations | Let $X$ be any set with $n$ elements (where $n \in \Z_{>0}$).
Consider the symmetric group on $n$ letters $S_n$ as a group of transformations on $X$.
Let $x \in X$.
Then the stabilizer of $x$ is isomorphic to $S_{n - 1}$. | Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$.
By the definition of cardinality, $H$ is equivalent to $\N_n$.
{{WLOG}} we can consider $S_n$ acting directly on $\N_n$.
The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$.
A permut... | Let $X$ be any [[Definition:Set|set]] with $n$ [[Definition:Element|elements]] (where $n \in \Z_{>0}$).
Consider the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]] $S_n$ as a [[Group Action determines Bijection|group of transformations]] on $X$.
Let $x \in X$.
Then the [[Definition:Stabi... | Consider the [[Definition:Initial Segment of Natural Numbers|initial segment of natural numbers]] $\N_n = \set {1, 2, \ldots, n}$.
By the definition of [[Definition:Cardinality|cardinality]], $H$ is [[Definition:Set Equivalence|equivalent]] to $\N_n$.
{{WLOG}} we can consider $S_n$ acting directly on $\N_n$.
The [[... | Stabilizer in Group of Transformations | https://proofwiki.org/wiki/Stabilizer_in_Group_of_Transformations | https://proofwiki.org/wiki/Stabilizer_in_Group_of_Transformations | [
"Stabilizers",
"Symmetric Groups"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Symmetric Group/n Letters",
"Group Action determines Bijection",
"Definition:Stabilizer",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Initial Segment of Natural Numbers",
"Definition:Cardinality",
"Definition:Set Equivalence",
"Definition:Stabilizer",
"Definition:Permutation",
"Definition:Permutation",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] |
proofwiki-3026 | Subset Product Action is Group Action | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
For any $S \in \powerset G$ and for any $g \in G$, the subset product action:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
is a group action. | Let $g \in G$.
First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that:
:$g * S \subseteq G$
Next we note:
:$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s: s \in S} = S$
and so {{GroupActionAxiom|2}} is satisfied.
Now let $g, h \in G$.
We have:
{{begin-... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\powerset G$ be the [[Definition:Power Set|power set]] of $\struct {G, \circ}$.
For any $S \in \powerset G$ and for any $g \in G$, the [[Definition:Subset Product Action|subset product action]]:
:$\fo... | Let $g \in G$.
First we note that since $G$ is [[Definition:Closed Algebraic Structure|closed]], and $g \circ S$ consists of [[Definition:Product Element|products]] of [[Definition:Element|elements]] of $G$, it follows that:
:$g * S \subseteq G$
Next we note:
:$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s... | Subset Product Action is Group Action | https://proofwiki.org/wiki/Subset_Product_Action_is_Group_Action | https://proofwiki.org/wiki/Subset_Product_Action_is_Group_Action | [
"Subset Product Action",
"Subset Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Set",
"Definition:Subset Product Action",
"Definition:Group Action"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Group Product/Product Element",
"Definition:Element"
] |
proofwiki-3027 | Subgroup of Elements whose Order Divides Integer | Let $A$ be an abelian group.
Let $k \in \Z$ and $B$ be a set of the form:
:$\left\{{x \in A : x^k = e}\right\}$
Then $B$ is a subgroup of $A$. | First note that the identity $e$ satisfies $e^k = e$ and so $B$ is non-empty.
Now assume that $a, b \in B$.
Then:
{{begin-eqn}}
{{eqn | l=\left({ab^{-1} }\right)^k
| r=a^k \left({b^{-1} }\right)^k
| c=Power of Product in Abelian Group
}}
{{eqn | r=a^k \left({b^k}\right)^{-1}
| c=Powers of Group Elemen... | Let $A$ be an [[Definition:Abelian Group|abelian group]].
Let $k \in \Z$ and $B$ be a [[Definition:Set|set]] of the form:
:$\left\{{x \in A : x^k = e}\right\}$
Then $B$ is a [[Definition:Subgroup|subgroup]] of $A$. | First note that the [[Definition:Identity Element|identity]] $e$ satisfies $e^k = e$ and so $B$ is non-empty.
Now assume that $a, b \in B$.
Then:
{{begin-eqn}}
{{eqn | l=\left({ab^{-1} }\right)^k
| r=a^k \left({b^{-1} }\right)^k
| c=[[Power of Product in Abelian Group]]
}}
{{eqn | r=a^k \left({b^k}\right)... | Subgroup of Elements whose Order Divides Integer | https://proofwiki.org/wiki/Subgroup_of_Elements_whose_Order_Divides_Integer | https://proofwiki.org/wiki/Subgroup_of_Elements_whose_Order_Divides_Integer | [
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Set",
"Definition:Subgroup"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Power of Product in Abelian Group",
"Powers of Group Elements",
"Inverse of Identity Element is Itself",
"One-Step Subgroup Test",
"Definition:Subgroup",
"Category:Abelian Groups"
] |
proofwiki-3028 | Abelian Group Factored by Prime | Let $G$ be a finite abelian group.
Let $p$ be a prime.
Factor the order of $G$ as:
:$\order G = m p^n$
such that $p$ does not divide $m$.
Then:
:$G = H \times K$
where:
:$H = \set {x \in G : x^{p^n} = e}$
and:
:$K = \set {x \in G : x^m = e}$
Furthermore:
:$\order H = p^n$ | From Subgroup of Elements whose Order Divides Integer, both $H$ and $K$ are subgroups of $G$.
Also, because $G$ is abelian, $H$ and $K$ are normal by Subgroup of Abelian Group is Normal.
In order to prove $G = H \times K$, by the Internal Direct Product Theorem it suffices to show that $G = H K$ and $H \cap K = \set e$... | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
Let $p$ be a [[Definition:Prime Number|prime]].
Factor the [[Definition:Order of Group|order of $G$]] as:
:$\order G = m p^n$
such that $p$ does not [[Definition:Divisor of Integer|divide]] $m$.
Then:
:$G = H \times K$
where:... | From [[Subgroup of Elements whose Order Divides Integer]], both $H$ and $K$ are [[Definition:Subgroup|subgroups]] of $G$.
Also, because $G$ is [[Definition:Abelian Group|abelian]], $H$ and $K$ are [[Definition:Normal Subgroup|normal]] by [[Subgroup of Abelian Group is Normal]].
In order to prove $G = H \times K$, by... | Abelian Group Factored by Prime | https://proofwiki.org/wiki/Abelian_Group_Factored_by_Prime | https://proofwiki.org/wiki/Abelian_Group_Factored_by_Prime | [
"Abelian Groups"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Prime Number",
"Definition:Order of Structure",
"Definition:Divisor (Algebra)/Integer"
] | [
"Subgroup of Elements whose Order Divides Integer",
"Definition:Subgroup",
"Definition:Abelian Group",
"Definition:Normal Subgroup",
"Subgroup of Abelian Group is Normal",
"Internal Direct Product Theorem",
"Definition:Integer",
"Bézout's Identity",
"Definition:Integer",
"Element to Power of Group... |
proofwiki-3029 | Abelian Group of Prime-power Order is Product of Cyclic Groups | Let $G$ be an abelian group of prime-power order.
Let $a$ be an element of maximal order in $G$.
Then $G$ can be written in the form $\gen a \times K$ for some $K \le G$. | Suppose $\order G = p^n$ with $p$ a prime.
We proceed by induction on $n$. | Let $G$ be an [[Definition:Abelian Group|abelian group]] of [[Definition:Prime Number|prime]]-power [[Definition:Order of Group|order]].
Let $a$ be an [[Definition:Element|element]] of maximal [[Definition:Order of Group Element|order]] in $G$.
Then $G$ can be written in the form $\gen a \times K$ for some $K \le G$... | Suppose $\order G = p^n$ with $p$ a [[Definition:Prime Number|prime]].
We proceed by [[Second Principle of Mathematical Induction|induction]] on $n$. | Abelian Group of Prime-power Order is Product of Cyclic Groups | https://proofwiki.org/wiki/Abelian_Group_of_Prime-power_Order_is_Product_of_Cyclic_Groups | https://proofwiki.org/wiki/Abelian_Group_of_Prime-power_Order_is_Product_of_Cyclic_Groups | [
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Prime Number",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Definition:Prime Number",
"Second Principle of Mathematical Induction",
"Second Principle of Mathematical Induction"
] |
proofwiki-3030 | Power of Product in Abelian Group | Let $\struct {G, \circ}$ be an abelian group.
Then:
:$\forall x, y \in G: \forall k \in \Z: \paren {x \circ y}^k = x^k \circ y^k$ | By definition of abelian group, $x$ and $y$ commute.
That is:
:$x \circ y = y \circ x$
The result follows from Power of Product of Commutative Elements in Group.
{{qed}} | Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]].
Then:
:$\forall x, y \in G: \forall k \in \Z: \paren {x \circ y}^k = x^k \circ y^k$ | By definition of [[Definition:Abelian Group|abelian group]], $x$ and $y$ [[Definition:Commute|commute]].
That is:
:$x \circ y = y \circ x$
The result follows from [[Power of Product of Commutative Elements in Group]].
{{qed}} | Power of Product in Abelian Group | https://proofwiki.org/wiki/Power_of_Product_in_Abelian_Group | https://proofwiki.org/wiki/Power_of_Product_in_Abelian_Group | [
"Abelian Groups"
] | [
"Definition:Abelian Group"
] | [
"Definition:Abelian Group",
"Definition:Commutative/Elements",
"Power of Product of Commutative Elements in Group"
] |
proofwiki-3031 | Union of Mappings with Disjoint Domains is Mapping | Let $S_1, S_2, T_1, T_2$ be sets.
Let $f: S_1 \to T_1$ and $g: S_2 \to T_2$ be mappings.
Let $h = f \cup g$ be their union.
If $S_1 \cap S_2 = \O$, then $h: S_1 \cup S_2 \to T_1 \cup T_2$ is a mapping whose domain is $S_1 \cup S_2$. | From the definition of mapping, it is clear that $h$ is a relation.
Suppose $\tuple {x, y_1}, \tuple {x, y_2} \in h$.
Clearly $x \in S_1$ or $x \in S_2$ but as $S_1 \cap S_2 = \O$ it is not in both.
If $x \in S_1$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in f$, and $f$ is a mapping.
Similarly, if $x \in S... | Let $S_1, S_2, T_1, T_2$ be [[Definition:Set|sets]].
Let $f: S_1 \to T_1$ and $g: S_2 \to T_2$ be [[Definition:Mapping|mappings]].
Let $h = f \cup g$ be their [[Definition:Union of Mappings|union]].
If $S_1 \cap S_2 = \O$, then $h: S_1 \cup S_2 \to T_1 \cup T_2$ is a [[Definition:Mapping|mapping]] whose [[Definitio... | From the definition of [[Definition:Mapping|mapping]], it is clear that $h$ is a [[Definition:Relation|relation]].
Suppose $\tuple {x, y_1}, \tuple {x, y_2} \in h$.
Clearly $x \in S_1$ or $x \in S_2$ but as $S_1 \cap S_2 = \O$ it is not in both.
If $x \in S_1$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \... | Union of Mappings with Disjoint Domains is Mapping | https://proofwiki.org/wiki/Union_of_Mappings_with_Disjoint_Domains_is_Mapping | https://proofwiki.org/wiki/Union_of_Mappings_with_Disjoint_Domains_is_Mapping | [
"Mapping Theory",
"Set Union"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Union of Relations",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping"
] |
proofwiki-3032 | Identity Element of Natural Number Addition is Zero | The identity element for the natural numbers under addition is zero ($0$):
:$\forall n \in \N: 0 + n = n$ | Take the definition of natural numbers $\N$ as a naturally ordered semigroup $\struct {\N, +, \le}$.
The result follows from Zero is Identity in Naturally Ordered Semigroup.
{{qed}} | The [[Definition:Identity Element|identity element]] for the [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Addition|addition]] is [[Definition:Zero (Number)|zero ($0$)]]:
:$\forall n \in \N: 0 + n = n$ | Take the definition of [[Definition:Natural Number|natural numbers]] $\N$ as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {\N, +, \le}$.
The result follows from [[Zero is Identity in Naturally Ordered Semigroup]].
{{qed}} | Identity Element of Natural Number Addition is Zero | https://proofwiki.org/wiki/Identity_Element_of_Natural_Number_Addition_is_Zero | https://proofwiki.org/wiki/Identity_Element_of_Natural_Number_Addition_is_Zero | [
"Natural Number Addition",
"Examples of Identity Elements"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Zero (Number)"
] | [
"Definition:Natural Numbers",
"Definition:Naturally Ordered Semigroup",
"Zero is Identity in Naturally Ordered Semigroup"
] |
proofwiki-3033 | Sum of Tangent and Cotangent | :$\tan x + \cot x = \sec x \csc x$ | {{begin-eqn}}
{{eqn | l = \tan x + \cot x
| r = \frac {\sin x} {\cos x} + \cot x
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\sin x} {\cos x} + \frac {\cos x} {\sin x}
| c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\sin^2 x + \cos^2x} {\cos x \sin x}
| c =
}}
... | :$\tan x + \cot x = \sec x \csc x$ | {{begin-eqn}}
{{eqn | l = \tan x + \cot x
| r = \frac {\sin x} {\cos x} + \cot x
| c = [[Tangent is Sine divided by Cosine]]
}}
{{eqn | r = \frac {\sin x} {\cos x} + \frac {\cos x} {\sin x}
| c = [[Cotangent is Cosine divided by Sine]]
}}
{{eqn | r = \frac {\sin^2 x + \cos^2x} {\cos x \sin x}
| ... | Sum of Tangent and Cotangent | https://proofwiki.org/wiki/Sum_of_Tangent_and_Cotangent | https://proofwiki.org/wiki/Sum_of_Tangent_and_Cotangent | [
"Trigonometric Identities"
] | [] | [
"Tangent is Sine divided by Cosine",
"Cotangent is Cosine divided by Sine",
"Sum of Squares of Sine and Cosine",
"Secant is Reciprocal of Cosine",
"Cosecant is Reciprocal of Sine",
"Category:Trigonometric Identities"
] |
proofwiki-3034 | Tangent times Tangent plus Cotangent | :$\tan x \paren {\tan x + \cot x} = \sec^2 x$ | {{begin-eqn}}
{{eqn|l = \tan x \left({\tan x + \cot x}\right)
|r = \tan x \sec x \csc x
|c = Sum of Tangent and Cotangent
}}
{{eqn|r = \frac {\sin x} {\cos x} \sec x \csc x
|c = Tangent is Sine divided by Cosine
}}
{{eqn|r = \frac {\sin x} {\cos^2 x} \csc x
|c = Secant is Reciprocal of Cosine
}}
{{e... | :$\tan x \paren {\tan x + \cot x} = \sec^2 x$ | {{begin-eqn}}
{{eqn|l = \tan x \left({\tan x + \cot x}\right)
|r = \tan x \sec x \csc x
|c = [[Sum of Tangent and Cotangent]]
}}
{{eqn|r = \frac {\sin x} {\cos x} \sec x \csc x
|c = [[Tangent is Sine divided by Cosine]]
}}
{{eqn|r = \frac {\sin x} {\cos^2 x} \csc x
|c = [[Secant is Reciprocal of Cos... | Tangent times Tangent plus Cotangent/Proof 1 | https://proofwiki.org/wiki/Tangent_times_Tangent_plus_Cotangent | https://proofwiki.org/wiki/Tangent_times_Tangent_plus_Cotangent/Proof_1 | [
"Trigonometric Identities",
"Tangent times Tangent plus Cotangent"
] | [] | [
"Sum of Tangent and Cotangent",
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine",
"Cosecant is Reciprocal of Sine",
"Secant is Reciprocal of Cosine"
] |
proofwiki-3035 | Tangent times Tangent plus Cotangent | :$\tan x \paren {\tan x + \cot x} = \sec^2 x$ | {{begin-eqn}}
{{eqn | l = \tan x \paren {\tan x + \cot x}
| r = \frac {\sin x} {\cos x} \paren {\frac {\sin x} {\cos x} + \frac {\cos x} {\sin x} }
| c = {{Defof|Tangent Function}} and {{Defof|Cotangent}}
}}
{{eqn | r = \frac {\sin x} {\cos x} \paren {\frac {\sin^2 x + \cos^2 x} {\cos x \sin x} }
}}
{{eqn |... | :$\tan x \paren {\tan x + \cot x} = \sec^2 x$ | {{begin-eqn}}
{{eqn | l = \tan x \paren {\tan x + \cot x}
| r = \frac {\sin x} {\cos x} \paren {\frac {\sin x} {\cos x} + \frac {\cos x} {\sin x} }
| c = {{Defof|Tangent Function}} and {{Defof|Cotangent}}
}}
{{eqn | r = \frac {\sin x} {\cos x} \paren {\frac {\sin^2 x + \cos^2 x} {\cos x \sin x} }
}}
{{eqn |... | Tangent times Tangent plus Cotangent/Proof 2 | https://proofwiki.org/wiki/Tangent_times_Tangent_plus_Cotangent | https://proofwiki.org/wiki/Tangent_times_Tangent_plus_Cotangent/Proof_2 | [
"Trigonometric Identities",
"Tangent times Tangent plus Cotangent"
] | [] | [
"Sum of Squares of Sine and Cosine"
] |
proofwiki-3036 | Secant Minus Cosine | :$\sec x - \cos x = \sin x \tan x$ | {{begin-eqn}}
{{eqn | l = \sec x - \cos x
| r = \frac 1 {\cos x} - \cos x
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {1 - \cos^2 x} {\cos x}
}}
{{eqn | r = \frac {\sin^2 x} {\cos x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sin x \tan x
| c = Tangent is Sine divided ... | :$\sec x - \cos x = \sin x \tan x$ | {{begin-eqn}}
{{eqn | l = \sec x - \cos x
| r = \frac 1 {\cos x} - \cos x
| c = [[Secant is Reciprocal of Cosine]]
}}
{{eqn | r = \frac {1 - \cos^2 x} {\cos x}
}}
{{eqn | r = \frac {\sin^2 x} {\cos x}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \sin x \tan x
| c = [[Tangent is Sin... | Secant Minus Cosine | https://proofwiki.org/wiki/Secant_Minus_Cosine | https://proofwiki.org/wiki/Secant_Minus_Cosine | [
"Trigonometric Identities"
] | [] | [
"Secant is Reciprocal of Cosine",
"Sum of Squares of Sine and Cosine",
"Tangent is Sine divided by Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3037 | Square of Tangent Minus Square of Sine | :$\tan^2 x - \sin^2 x = \tan^2 x \ \sin^2 x$ | {{begin-eqn}}
{{eqn | l=\tan^2 x - \sin^2 x
| r=\frac {\sin^2 x} {\cos^2x} - \sin^2 x
| c=Tangent is Sine divided by Cosine
}}
{{eqn | r=\frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x}
}}
{{eqn | r=\frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x}
}}
{{eqn | r=\tan^2 x \left({1 - \cos^2 x}\right)
... | :$\tan^2 x - \sin^2 x = \tan^2 x \ \sin^2 x$ | {{begin-eqn}}
{{eqn | l=\tan^2 x - \sin^2 x
| r=\frac {\sin^2 x} {\cos^2x} - \sin^2 x
| c=[[Tangent is Sine divided by Cosine]]
}}
{{eqn | r=\frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x}
}}
{{eqn | r=\frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x}
}}
{{eqn | r=\tan^2 x \left({1 - \cos^2 x}\right)
... | Square of Tangent Minus Square of Sine | https://proofwiki.org/wiki/Square_of_Tangent_Minus_Square_of_Sine | https://proofwiki.org/wiki/Square_of_Tangent_Minus_Square_of_Sine | [
"Trigonometric Identities"
] | [] | [
"Tangent is Sine divided by Cosine",
"Tangent is Sine divided by Cosine",
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3038 | Difference of Fourth Powers of Cosine and Sine | :$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$ | {{begin-eqn}}
{{eqn | l = \sin^4 x - \cos^4 x
| r = \sin^2 x \left({1 - \cos^2 x}\right) - \cos^2 x \left({1 - \sin^2 x}\right)
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sin^2 x - \sin^2 x \ \cos^2 x - \cos^2 x + \sin^2 x \ \cos^2 x
}}
{{eqn | r = \sin^2 x - \cos^2 x
}}
{{end-eqn}}
{{qed}} | :$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$ | {{begin-eqn}}
{{eqn | l = \sin^4 x - \cos^4 x
| r = \sin^2 x \left({1 - \cos^2 x}\right) - \cos^2 x \left({1 - \sin^2 x}\right)
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \sin^2 x - \sin^2 x \ \cos^2 x - \cos^2 x + \sin^2 x \ \cos^2 x
}}
{{eqn | r = \sin^2 x - \cos^2 x
}}
{{end-eqn}}
{{qed}} | Difference of Fourth Powers of Cosine and Sine/Proof 1 | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Cosine_and_Sine | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Cosine_and_Sine/Proof_1 | [
"Difference of Fourth Powers of Cosine and Sine",
"Sine Function",
"Cosine Function"
] | [] | [
"Sum of Squares of Sine and Cosine"
] |
proofwiki-3039 | Difference of Fourth Powers of Cosine and Sine | :$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$ | {{begin-eqn}}
{{eqn | l = \sin^2 x - \cos^2 x
| r = \left({\sin^2 x - \cos^2 x}\right) \left({\sin^2 x + \cos^2 x}\right)
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sin^4 x - \cos^4 x
| c = Difference of Two Squares
}}
{{end-eqn}}
{{qed}} | :$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$ | {{begin-eqn}}
{{eqn | l = \sin^2 x - \cos^2 x
| r = \left({\sin^2 x - \cos^2 x}\right) \left({\sin^2 x + \cos^2 x}\right)
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \sin^4 x - \cos^4 x
| c = [[Difference of Two Squares]]
}}
{{end-eqn}}
{{qed}} | Difference of Fourth Powers of Cosine and Sine/Proof 2 | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Cosine_and_Sine | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Cosine_and_Sine/Proof_2 | [
"Difference of Fourth Powers of Cosine and Sine",
"Sine Function",
"Cosine Function"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Difference of Two Squares"
] |
proofwiki-3040 | Cosecant Minus Sine | :$\csc x - \sin x = \cos x \ \cot x$ | {{begin-eqn}}
{{eqn | l=\csc x - \sin x
| r=\frac 1 {\sin x} - \sin x
| c={{Defof|Cosecant}}
}}
{{eqn | r=\frac {1 - \sin^2 x} {\sin x}
}}
{{eqn | r=\frac {\cos^2 x} {\sin x}
| c=Sum of Squares of Sine and Cosine
}}
{{eqn | r=\cos x \ \cot x
| c={{Defof|Cotangent}}
}}
{{end-eqn}}
{{qed}}
Categor... | :$\csc x - \sin x = \cos x \ \cot x$ | {{begin-eqn}}
{{eqn | l=\csc x - \sin x
| r=\frac 1 {\sin x} - \sin x
| c={{Defof|Cosecant}}
}}
{{eqn | r=\frac {1 - \sin^2 x} {\sin x}
}}
{{eqn | r=\frac {\cos^2 x} {\sin x}
| c=[[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r=\cos x \ \cot x
| c={{Defof|Cotangent}}
}}
{{end-eqn}}
{{qed}}
[[... | Cosecant Minus Sine | https://proofwiki.org/wiki/Cosecant_Minus_Sine | https://proofwiki.org/wiki/Cosecant_Minus_Sine | [
"Trigonometric Identities"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3041 | Sum of Squares of Secant and Cosecant | :$\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$ | {{begin-eqn}}
{{eqn | l = \sec^2 x + \csc^2 x
| r = \frac 1 {\cos^2 x} + \frac 1 {\sin^2 x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac {\sin^2 x + \cos^2 x} {\cos^2 x \sin^2 x}
| c =
}}
{{eqn | r = \frac 1 {\cos^2 x \sin^2 x}
| c = Sum of Squares of Sine and Cos... | :$\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$ | {{begin-eqn}}
{{eqn | l = \sec^2 x + \csc^2 x
| r = \frac 1 {\cos^2 x} + \frac 1 {\sin^2 x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac {\sin^2 x + \cos^2 x} {\cos^2 x \sin^2 x}
| c =
}}
{{eqn | r = \frac 1 {\cos^2 x \sin^2 x}
| c = [[Sum of Squares of Sine and C... | Sum of Squares of Secant and Cosecant | https://proofwiki.org/wiki/Sum_of_Squares_of_Secant_and_Cosecant | https://proofwiki.org/wiki/Sum_of_Squares_of_Secant_and_Cosecant | [
"Trigonometric Identities"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3042 | Difference of Fourth Powers of Secant and Tangent | :$\sec^4 x - \tan^4 x = \sec^2 x + \tan^2 x$ | {{begin-eqn}}
{{eqn | l = \sec^4 x - \tan^4 x
| r = \sec^4 x - \frac {\sin^4 x} {\cos^4 x}
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac 1 {\cos^4 x} - \frac {\sin^4 x} {\cos^4 x}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {1 - \sin^4 x} {\cos^4x}
}}
{{eqn | r = \frac {1 ... | :$\sec^4 x - \tan^4 x = \sec^2 x + \tan^2 x$ | {{begin-eqn}}
{{eqn | l = \sec^4 x - \tan^4 x
| r = \sec^4 x - \frac {\sin^4 x} {\cos^4 x}
| c = [[Tangent is Sine divided by Cosine]]
}}
{{eqn | r = \frac 1 {\cos^4 x} - \frac {\sin^4 x} {\cos^4 x}
| c = [[Secant is Reciprocal of Cosine]]
}}
{{eqn | r = \frac {1 - \sin^4 x} {\cos^4x}
}}
{{eqn | r = \... | Difference of Fourth Powers of Secant and Tangent | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Secant_and_Tangent | https://proofwiki.org/wiki/Difference_of_Fourth_Powers_of_Secant_and_Tangent | [
"Trigonometric Identities"
] | [] | [
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine",
"Sum of Squares of Sine and Cosine",
"Sum of Squares of Sine and Cosine",
"Sum of Squares of Sine and Cosine",
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3043 | Sum of Reciprocals of One Plus and Minus Sine | :$\dfrac 1 {1 - \sin x} + \dfrac 1 {1 + \sin x} = 2 \sec^2 x$ | {{begin-eqn}}
{{eqn | l = \frac 1 {1 - \sin x} + \frac 1 {1 + \sin x}
| r = \frac {1 + \sin x + 1 - \sin x} {1 - \sin^2 x}
| c = Difference of Two Squares
}}
{{eqn | r = \frac 2 {\cos^2 x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 2 \sec^2 x
| c = {{Defof|Secant Function}}
}}
{{end-... | :$\dfrac 1 {1 - \sin x} + \dfrac 1 {1 + \sin x} = 2 \sec^2 x$ | {{begin-eqn}}
{{eqn | l = \frac 1 {1 - \sin x} + \frac 1 {1 + \sin x}
| r = \frac {1 + \sin x + 1 - \sin x} {1 - \sin^2 x}
| c = [[Difference of Two Squares]]
}}
{{eqn | r = \frac 2 {\cos^2 x}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = 2 \sec^2 x
| c = {{Defof|Secant Function}}
}... | Sum of Reciprocals of One Plus and Minus Sine | https://proofwiki.org/wiki/Sum_of_Reciprocals_of_One_Plus_and_Minus_Sine | https://proofwiki.org/wiki/Sum_of_Reciprocals_of_One_Plus_and_Minus_Sine | [
"Trigonometric Identities",
"Sine Function"
] | [] | [
"Difference of Two Squares",
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities",
"Category:Sine Function"
] |
proofwiki-3044 | Difference of Reciprocals of One Plus and Minus Sine | :$\ds \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x} = 2 \tan x \sec x$ | {{begin-eqn}}
{{eqn | l = \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}
| r = \frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}
| c = Difference of Two Squares
}}
{{eqn | r = \frac {2 \sin x} {\cos^2 x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {2 \tan x} {\cos x}
... | :$\ds \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x} = 2 \tan x \sec x$ | {{begin-eqn}}
{{eqn | l = \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}
| r = \frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}
| c = [[Difference of Two Squares]]
}}
{{eqn | r = \frac {2 \sin x} {\cos^2 x}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \frac {2 \tan x} {... | Difference of Reciprocals of One Plus and Minus Sine | https://proofwiki.org/wiki/Difference_of_Reciprocals_of_One_Plus_and_Minus_Sine | https://proofwiki.org/wiki/Difference_of_Reciprocals_of_One_Plus_and_Minus_Sine | [
"Trigonometric Identities",
"Sine Function"
] | [] | [
"Difference of Two Squares",
"Sum of Squares of Sine and Cosine",
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine",
"Category:Trigonometric Identities",
"Category:Sine Function"
] |
proofwiki-3045 | Sum of Secant and Tangent | :$\sec x + \tan x = \dfrac {1 + \sin x} {\cos x}$ | {{begin-eqn}}
{{eqn | l = \sec x + \tan x
| r = \sec x + \frac {\sin x} {\cos x}
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac 1 {\cos x} + \frac {\sin x} {\cos x}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {1 + \sin x} {\cos x}
| c =
}}
{{end-eqn}}
{{qed}} | :$\sec x + \tan x = \dfrac {1 + \sin x} {\cos x}$ | {{begin-eqn}}
{{eqn | l = \sec x + \tan x
| r = \sec x + \frac {\sin x} {\cos x}
| c = [[Tangent is Sine divided by Cosine]]
}}
{{eqn | r = \frac 1 {\cos x} + \frac {\sin x} {\cos x}
| c = [[Secant is Reciprocal of Cosine]]
}}
{{eqn | r = \frac {1 + \sin x} {\cos x}
| c =
}}
{{end-eqn}}
{{qed}} | Sum of Secant and Tangent | https://proofwiki.org/wiki/Sum_of_Secant_and_Tangent | https://proofwiki.org/wiki/Sum_of_Secant_and_Tangent | [
"Trigonometric Identities"
] | [] | [
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine"
] |
proofwiki-3046 | Cosine over Sum of Secant and Tangent | :$\dfrac {\cos x} {\sec x + \tan x} = 1 - \sin x$ | {{begin-eqn}}
{{eqn | l = \frac {\cos x} {\sec x + \tan x}
| r = \frac {\cos^2 x} {1 + \sin x}
| c = Sum of Secant and Tangent
}}
{{eqn | r = \frac {1 - \sin^2 x} {1 + \sin x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {\paren {1 + \sin x} \paren {1 - \sin x} } {1 + \sin x}
| c... | :$\dfrac {\cos x} {\sec x + \tan x} = 1 - \sin x$ | {{begin-eqn}}
{{eqn | l = \frac {\cos x} {\sec x + \tan x}
| r = \frac {\cos^2 x} {1 + \sin x}
| c = [[Sum of Secant and Tangent]]
}}
{{eqn | r = \frac {1 - \sin^2 x} {1 + \sin x}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \frac {\paren {1 + \sin x} \paren {1 - \sin x} } {1 + \sin x}
... | Cosine over Sum of Secant and Tangent | https://proofwiki.org/wiki/Cosine_over_Sum_of_Secant_and_Tangent | https://proofwiki.org/wiki/Cosine_over_Sum_of_Secant_and_Tangent | [
"Trigonometric Identities"
] | [] | [
"Sum of Secant and Tangent",
"Sum of Squares of Sine and Cosine",
"Difference of Two Squares",
"Category:Trigonometric Identities"
] |
proofwiki-3047 | Secant Plus One over Secant Squared | :$\dfrac {\sec x + 1} {\sec^2 x} = \dfrac {\sin^2 x} {\sec x - 1}$ | {{begin-eqn}}
{{eqn | l = \frac {\sec x + 1} {\sec^2 x}
| r = \cos^2 x \paren {\frac 1 {\cos x} + 1}
| c = {{Defof|Secant Function}}
}}
{{eqn | r = \cos x + \cos^2x
}}
{{eqn | r = \cos x \paren {1 + \cos x}
}}
{{eqn | r = \cos x \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {1 - \cos x}
}}
{{eqn | r = \f... | :$\dfrac {\sec x + 1} {\sec^2 x} = \dfrac {\sin^2 x} {\sec x - 1}$ | {{begin-eqn}}
{{eqn | l = \frac {\sec x + 1} {\sec^2 x}
| r = \cos^2 x \paren {\frac 1 {\cos x} + 1}
| c = {{Defof|Secant Function}}
}}
{{eqn | r = \cos x + \cos^2x
}}
{{eqn | r = \cos x \paren {1 + \cos x}
}}
{{eqn | r = \cos x \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {1 - \cos x}
}}
{{eqn | r = \f... | Secant Plus One over Secant Squared | https://proofwiki.org/wiki/Secant_Plus_One_over_Secant_Squared | https://proofwiki.org/wiki/Secant_Plus_One_over_Secant_Squared | [
"Trigonometric Identities"
] | [] | [
"Difference of Two Squares",
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3048 | Sine Plus Cosine times Tangent Plus Cotangent | :$\paren {\sin x + \cos x} \paren {\tan x + \cot x} = \sec x + \csc x$ | {{begin-eqn}}
{{eqn | l = \paren {\sin x + \cos x} \paren {\tan x + \cot x}
| r = \paren {\sin x + \cos x} \paren {\sec x \csc x}
| c = Sum of Tangent and Cotangent
}}
{{eqn | r = \frac {\sin x + \cos x} {\sin x \cos x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac 1 {\co... | :$\paren {\sin x + \cos x} \paren {\tan x + \cot x} = \sec x + \csc x$ | {{begin-eqn}}
{{eqn | l = \paren {\sin x + \cos x} \paren {\tan x + \cot x}
| r = \paren {\sin x + \cos x} \paren {\sec x \csc x}
| c = [[Sum of Tangent and Cotangent]]
}}
{{eqn | r = \frac {\sin x + \cos x} {\sin x \cos x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac 1 ... | Sine Plus Cosine times Tangent Plus Cotangent | https://proofwiki.org/wiki/Sine_Plus_Cosine_times_Tangent_Plus_Cotangent | https://proofwiki.org/wiki/Sine_Plus_Cosine_times_Tangent_Plus_Cotangent | [
"Trigonometric Identities"
] | [] | [
"Sum of Tangent and Cotangent",
"Category:Trigonometric Identities"
] |
proofwiki-3049 | Tangent over Secant Plus One | :$\dfrac {\tan x} {\sec x + 1} = \dfrac {\sec x - 1} {\tan x}$ | {{begin-eqn}}
{{eqn | l = \frac {\tan x} {\sec x + 1}
| r = \frac {\sin x} {\cos x \paren {\sec x + 1} }
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\sin x} {\cos x \paren {\frac 1 {\cos x} + 1} }
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {\sin x} {1 + \cos x}
|... | :$\dfrac {\tan x} {\sec x + 1} = \dfrac {\sec x - 1} {\tan x}$ | {{begin-eqn}}
{{eqn | l = \frac {\tan x} {\sec x + 1}
| r = \frac {\sin x} {\cos x \paren {\sec x + 1} }
| c = [[Tangent is Sine divided by Cosine]]
}}
{{eqn | r = \frac {\sin x} {\cos x \paren {\frac 1 {\cos x} + 1} }
| c = [[Secant is Reciprocal of Cosine]]
}}
{{eqn | r = \frac {\sin x} {1 + \cos x}... | Tangent over Secant Plus One | https://proofwiki.org/wiki/Tangent_over_Secant_Plus_One | https://proofwiki.org/wiki/Tangent_over_Secant_Plus_One | [
"Trigonometric Identities"
] | [] | [
"Tangent is Sine divided by Cosine",
"Secant is Reciprocal of Cosine",
"Sum of Squares of Sine and Cosine",
"Difference of Two Squares",
"Secant is Reciprocal of Cosine",
"Tangent is Sine divided by Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3050 | Squares of Linear Combination of Sine and Cosine | :$\paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2 = a^2 + b^2$ | {{begin-eqn}}
{{eqn | l = \paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2
| r = a^2 \cos^2 x + 2 a b \cos x \ \sin x + b^2 \sin^2 x
}}
{{eqn | o =
| ro= +
| r = b^2 \cos^2 x - 2 a b \sin x \ \cos x + a^2 \sin^2 x
}}
{{eqn | r = \paren {a^2 + b^2} \paren {\sin^2 x + \cos^2 x}
}}
{{eqn |... | :$\paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2 = a^2 + b^2$ | {{begin-eqn}}
{{eqn | l = \paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2
| r = a^2 \cos^2 x + 2 a b \cos x \ \sin x + b^2 \sin^2 x
}}
{{eqn | o =
| ro= +
| r = b^2 \cos^2 x - 2 a b \sin x \ \cos x + a^2 \sin^2 x
}}
{{eqn | r = \paren {a^2 + b^2} \paren {\sin^2 x + \cos^2 x}
}}
{{eqn |... | Squares of Linear Combination of Sine and Cosine | https://proofwiki.org/wiki/Squares_of_Linear_Combination_of_Sine_and_Cosine | https://proofwiki.org/wiki/Squares_of_Linear_Combination_of_Sine_and_Cosine | [
"Trigonometric Identities"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities"
] |
proofwiki-3051 | Reciprocal of One Minus Secant | :$\dfrac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3} = \dfrac 1 {1 - \sec x}$ | {{begin-eqn}}
{{eqn | l = \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3}
| r = \frac {1 - \cos^2 x - 2 \cos x - 1} {1 - \cos^2 x + 3 \cos x - 3}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {\cos^2 x - 2 \cos x} {\cos^2 x - 3 \cos x + 2}
}}
{{eqn | r = \frac {\cos x \paren {\cos x... | :$\dfrac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3} = \dfrac 1 {1 - \sec x}$ | {{begin-eqn}}
{{eqn | l = \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3}
| r = \frac {1 - \cos^2 x - 2 \cos x - 1} {1 - \cos^2 x + 3 \cos x - 3}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \frac {\cos^2 x - 2 \cos x} {\cos^2 x - 3 \cos x + 2}
}}
{{eqn | r = \frac {\cos x \paren {\c... | Reciprocal of One Minus Secant | https://proofwiki.org/wiki/Reciprocal_of_One_Minus_Secant | https://proofwiki.org/wiki/Reciprocal_of_One_Minus_Secant | [
"Trigonometric Identities",
"Reciprocals"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities",
"Category:Reciprocals"
] |
proofwiki-3052 | Reciprocal of One Plus Cosecant | :$\dfrac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2} = \dfrac 1 {1 + \csc x}$ | {{begin-eqn}}
{{eqn | l = \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2}
| r = \frac {1 - \sin^2 x + 3 \sin x - 1} {1 - \sin^2 x + 2 \sin x + 2}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {\sin^2 x - 3 \sin x} {\sin^2 x - 2 \sin x - 3}
}}
{{eqn | r = \frac {\sin x \paren {\sin x... | :$\dfrac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2} = \dfrac 1 {1 + \csc x}$ | {{begin-eqn}}
{{eqn | l = \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2}
| r = \frac {1 - \sin^2 x + 3 \sin x - 1} {1 - \sin^2 x + 2 \sin x + 2}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \frac {\sin^2 x - 3 \sin x} {\sin^2 x - 2 \sin x - 3}
}}
{{eqn | r = \frac {\sin x \paren {\s... | Reciprocal of One Plus Cosecant | https://proofwiki.org/wiki/Reciprocal_of_One_Plus_Cosecant | https://proofwiki.org/wiki/Reciprocal_of_One_Plus_Cosecant | [
"Trigonometric Identities",
"Reciprocals"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Category:Trigonometric Identities",
"Category:Reciprocals"
] |
proofwiki-3053 | Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE/General Result | Let $n \in \Z_{>0}$ be a strictly positive integer.
The ordinary differential equation:
:$a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} c + a_0 \, \map f x = 0$
can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$. | Let $y = \map f x$.
First the following are established:
{{begin-eqn}}
{{eqn | l = x
| r = e^t
| c = Derivative of Exponential Function
}}
{{eqn | ll= \leadsto
| l = \frac {\d x} {\d t}
| r = e^t
| c =
}}
{{eqn | n = 1
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
The [[Definition:Ordinary Differential Equation|ordinary differential equation]]:
:$a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} c + a_0 \, \map f x = 0$
can be transformed to a [[Definition:Linear nth Ord... | Let $y = \map f x$.
First the following are established:
{{begin-eqn}}
{{eqn | l = x
| r = e^t
| c = [[Derivative of Exponential Function]]
}}
{{eqn | ll= \leadsto
| l = \frac {\d x} {\d t}
| r = e^t
| c =
}}
{{eqn | n = 1
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = \... | Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE/General Result | https://proofwiki.org/wiki/Conversion_of_Cauchy-Euler_Equation_to_Constant_Coefficient_Linear_ODE/General_Result | https://proofwiki.org/wiki/Conversion_of_Cauchy-Euler_Equation_to_Constant_Coefficient_Linear_ODE/General_Result | [
"Cauchy-Euler Equation"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Differential Equation/Ordinary",
"Definition:Linear nth Order ODE with Constant Coefficients"
] | [
"Derivative of Exponential Function",
"Derivative of Inverse Function",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Linear nth Order ODE with Constant Coefficients",
"Definition:Linear nth Order ODE with Constant Coefficients",
"Definition:Linear nth Order ODE with Const... |
proofwiki-3054 | Condition for Composite Relation with Inverse to be Identity | Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR \circ \RR^{-1}$ be the composite of $\RR$ with its inverse.
Let $I_T$ be the identity mapping on $T$.
Then:
:$\RR \circ \RR^{-1} = I_T$
{{iff}}:
:$\RR$ is many-to-one
and:
:$\RR$ is right-total. | === Sufficient Condition ===
Let $\RR \circ \RR^{-1} = I_T$.
{{AimForCont}}:
:$\exists t \in T: t \notin \Img \RR$
Then:
:$t \notin \Img {\RR \circ \RR^{-1} }$
But:
:$t \in \Img {I_T}$
by definition of the identity mapping on $T$.
Hence:
:$\RR \circ \RR^{-1} \ne I_T$
From this contradiction we deduce that:
:$\RR \circ ... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Let $\RR \circ \RR^{-1}$ be the [[Definition:Composition of Relations|composite]] of $\RR$ with its [[Definition:Inverse Relation|inverse]].
Let $I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Then:
:$\RR \ci... | === Sufficient Condition ===
Let $\RR \circ \RR^{-1} = I_T$.
{{AimForCont}}:
:$\exists t \in T: t \notin \Img \RR$
Then:
:$t \notin \Img {\RR \circ \RR^{-1} }$
But:
:$t \in \Img {I_T}$
by definition of the [[Definition:Identity Mapping|identity mapping]] on $T$.
Hence:
:$\RR \circ \RR^{-1} \ne I_T$
From this con... | Condition for Composite Relation with Inverse to be Identity | https://proofwiki.org/wiki/Condition_for_Composite_Relation_with_Inverse_to_be_Identity | https://proofwiki.org/wiki/Condition_for_Composite_Relation_with_Inverse_to_be_Identity | [
"Condition for Composite Relation with Inverse to be Identity",
"Inverse Relations",
"Identity Mappings",
"Composite Relations"
] | [
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Identity Mapping",
"Definition:Many-to-One Relation",
"Definition:Right-Total Relation"
] | [
"Definition:Identity Mapping",
"Definition:Set Difference",
"Set Difference with Superset is Empty Set",
"Image is Subset of Codomain",
"Definition:Right-Total Relation",
"Definition:Many-to-One Relation",
"Definition:Inverse Relation",
"Definition:Composition of Relations",
"Definition:Identity Map... |
proofwiki-3055 | Set of All Relations is a Monoid | The set of all relations $\Bbb E = \set {\RR: \RR \subseteq S \times S}$ on a set $S$ forms a monoid in which the operation is composition of relations.
The only invertible elements of $\Bbb E$ are permutations.
If $\RR$ is invertible, its inverse is $\RR^{-1}$. | === Closure ===
A relation followed by another relation is another relation, from the definition of composition of relations.
As the domain and codomain of two relations are the same, then:
:$\forall \RR_1, \RR_2 \subseteq S \times S: \RR_1 \circ \RR_2 \subseteq S \times S$
Therefore composition of relations on a set i... | The set of all [[Definition:Relation|relations]] $\Bbb E = \set {\RR: \RR \subseteq S \times S}$ on a [[Definition:Set|set]] $S$ forms a [[Definition:Monoid|monoid]] in which the [[Definition:Binary Operation|operation]] is [[Definition:Composition of Relations|composition of relations]].
The only [[Definition:Invert... | === Closure ===
A [[Definition:Relation|relation]] followed by another [[Definition:Relation|relation]] is another [[Definition:Relation|relation]], from the definition of [[Definition:Composition of Relations|composition of relations]].
As the [[Definition:Domain of Relation|domain]] and [[Definition:Codomain of Rel... | Set of All Relations is a Monoid | https://proofwiki.org/wiki/Set_of_All_Relations_is_a_Monoid | https://proofwiki.org/wiki/Set_of_All_Relations_is_a_Monoid | [
"Relation Theory",
"Examples of Monoids"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Monoid",
"Definition:Operation/Binary Operation",
"Definition:Composition of Relations",
"Definition:Invertible Element",
"Definition:Permutation",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Relation",
"Definition:Relation",
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Domain (Set Theory)/Relation",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-3056 | Inverse Relation is Left and Right Inverse iff Bijection | Let $\RR \subseteq S \times T$ be a relation on a cartesian product $S \times T$.
Let:
:$I_S$ be the identity mapping on $S$
:$I_T$ be the identity mapping on $T$.
Let $\RR^{-1}$ be the inverse relation of $\RR$.
Then $\RR$ is a bijection {{iff}}:
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$
where $\circ... | === Necessary Condition ===
Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$.
From Left and Right Inverse Relations Implies Bijection, it follows that $\RR$ is a bijection.
{{qed|lemma}} | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on a [[Definition:Cartesian Product|cartesian product]] $S \times T$.
Let:
:$I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
:$I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $\RR^{-1}$ be the [[Definitio... | === Necessary Condition ===
Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$.
From [[Left and Right Inverse Relations Implies Bijection]], it follows that $\RR$ is a [[Definition:Bijection|bijection]].
{{qed|lemma}} | Inverse Relation is Left and Right Inverse iff Bijection | https://proofwiki.org/wiki/Inverse_Relation_is_Left_and_Right_Inverse_iff_Bijection | https://proofwiki.org/wiki/Inverse_Relation_is_Left_and_Right_Inverse_iff_Bijection | [
"Inverse Relations",
"Bijections"
] | [
"Definition:Relation",
"Definition:Cartesian Product",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Relation",
"Definition:Bijection",
"Definition:Composition of Relations"
] | [
"Left and Right Inverse Relations Implies Bijection",
"Definition:Bijection",
"Definition:Bijection"
] |
proofwiki-3057 | Integer Multiples Closed under Addition | Let $n \Z$ be the set of integer multiples of $n$.
Then the algebraic structure $\struct {n \Z, +}$ is closed under addition. | Let $x, y \in n \Z$.
Then $\exists p, q \in \Z: x = n p, y = n q$.
So $x + y = n p + n q = n \paren {p + q}$ where $p + q \in \Z$.
Thus $x + y \in n \Z$ and so $\struct {n \Z, +}$ is closed.
{{qed}} | Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {n \Z, +}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Addition|addition]]. | Let $x, y \in n \Z$.
Then $\exists p, q \in \Z: x = n p, y = n q$.
So $x + y = n p + n q = n \paren {p + q}$ where $p + q \in \Z$.
Thus $x + y \in n \Z$ and so $\struct {n \Z, +}$ is [[Definition:Closed Algebraic Structure|closed]].
{{qed}} | Integer Multiples Closed under Addition | https://proofwiki.org/wiki/Integer_Multiples_Closed_under_Addition | https://proofwiki.org/wiki/Integer_Multiples_Closed_under_Addition | [
"Integer Addition",
"Algebraic Closure"
] | [
"Definition:Set of Integer Multiples",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Integers"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-3058 | Integer Multiples Closed under Multiplication | Let $n \Z$ be the set of integer multiples of $n$.
Then the algebraic structure $\struct {n \Z, \times}$ is closed under multiplication. | Let $x, y \in n \Z$.
Then $\exists p, q \in \Z: x = n p, y = n q$.
So $x y = n p \cdot n q = n \paren {n p q}$ where $n p q \in \Z$.
Thus $x y \in n \Z$ and so $\struct {n \Z, \times}$ is closed.
{{qed}} | Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {n \Z, \times}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]]. | Let $x, y \in n \Z$.
Then $\exists p, q \in \Z: x = n p, y = n q$.
So $x y = n p \cdot n q = n \paren {n p q}$ where $n p q \in \Z$.
Thus $x y \in n \Z$ and so $\struct {n \Z, \times}$ is closed.
{{qed}} | Integer Multiples Closed under Multiplication | https://proofwiki.org/wiki/Integer_Multiples_Closed_under_Multiplication | https://proofwiki.org/wiki/Integer_Multiples_Closed_under_Multiplication | [
"Integer Multiplication",
"Algebraic Closure"
] | [
"Definition:Set of Integer Multiples",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Integers"
] | [] |
proofwiki-3059 | Non-Zero Integers Closed under Multiplication | The set of non-zero integers is closed under multiplication. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ as:
:$\tuple... | The [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]]. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Non-Zero Integers Closed under Multiplication | https://proofwiki.org/wiki/Non-Zero_Integers_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Integers_Closed_under_Multiplication | [
"Integer Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Integer",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Integers"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Definition:Multiplication/Integers",
"Integer Multipl... |
proofwiki-3060 | Two-Step Subgroup Test using Subset Product | Let $G$ be a group.
Let $\O\subset H \subseteq G$ be a non-empty subset of $G$.
Then $H$ is a subgroup of $G$ {{iff}}:
:$H H \subseteq H$
:$H^{-1} \subseteq H$
where:
:$H H$ is the product of $H$ with itself
:$H^{-1}$ is the inverse of $H$. | This is a reformulation of the Two-Step Subgroup Test in terms of subset product. | Let $G$ be a [[Definition:Group|group]].
Let $\O\subset H \subseteq G$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $G$.
Then $H$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}}:
:$H H \subseteq H$
:$H^{-1} \subseteq H$
where:
:$H H$ is the [[Definition:Subset Product|product]] of... | This is a reformulation of the [[Two-Step Subgroup Test]] in terms of [[Definition:Subset Product|subset product]]. | Two-Step Subgroup Test using Subset Product | https://proofwiki.org/wiki/Two-Step_Subgroup_Test_using_Subset_Product | https://proofwiki.org/wiki/Two-Step_Subgroup_Test_using_Subset_Product | [
"Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Subset Product",
"Definition:Inverse of Subset/Group"
] | [
"Two-Step Subgroup Test",
"Definition:Subset Product",
"Definition:Subset Product",
"Two-Step Subgroup Test"
] |
proofwiki-3061 | One-Step Subgroup Test using Subset Product | Let $G$ be a group.
Let $\O \subset H \subseteq G$ be a non-empty subset of $G$.
Then $H$ is a subgroup of $G$ {{iff}}:
:$H H^{-1} \subseteq H$
where:
:$H^{-1}$ is the inverse of $H$
:$H H ^{-1}$ is the product of $H$ with $H^{-1}$. | This is a reformulation of the One-Step Subgroup Test in terms of subset product. | Let $G$ be a [[Definition:Group|group]].
Let $\O \subset H \subseteq G$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $G$.
Then $H$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}}:
:$H H^{-1} \subseteq H$
where:
:$H^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $H$
... | This is a reformulation of the [[One-Step Subgroup Test]] in terms of [[Definition:Subset Product|subset product]]. | One-Step Subgroup Test using Subset Product | https://proofwiki.org/wiki/One-Step_Subgroup_Test_using_Subset_Product | https://proofwiki.org/wiki/One-Step_Subgroup_Test_using_Subset_Product | [
"Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Empty Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Inverse of Subset/Group",
"Definition:Subset Product"
] | [
"One-Step Subgroup Test",
"Definition:Subset Product",
"Definition:Subset Product",
"One-Step Subgroup Test",
"Definition:Subset Product",
"One-Step Subgroup Test"
] |
proofwiki-3062 | Product of Subset with Union | Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$. | Let $x \circ t \in X \circ \paren {Y \cup Z}$.
We have $x \in X, t \in Y \cup Z$ by definition of subset product.
By definition of set union, it follows that $t \in Y$ or $t \in Z$.
So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.
That is:
:$x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z... | Let $\struct {G, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
where $X \circ Y$ denotes th... | Let $x \circ t \in X \circ \paren {Y \cup Z}$.
We have $x \in X, t \in Y \cup Z$ by definition of [[Definition:Subset Product|subset product]].
By definition of [[Definition:Set Union|set union]], it follows that $t \in Y$ or $t \in Z$.
So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.
That is... | Product of Subset with Union/Proof 1 | https://proofwiki.org/wiki/Product_of_Subset_with_Union | https://proofwiki.org/wiki/Product_of_Subset_with_Union/Proof_1 | [
"Subset Products",
"Set Union",
"Product of Subset with Union"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Subset Product"
] | [
"Definition:Subset Product",
"Definition:Set Union",
"Definition:Set Union",
"Definition:Set Union"
] |
proofwiki-3063 | Product of Subset with Union | Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$. | Consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \map \RR S$
Then:
{{begin-eqn}}
{{eqn | l = X \circ \paren {Y \cup Z}
| r = \map \RR {Y \cup Z}
| c =
}}
{{eqn | r = \map \RR y \cup \map... | Let $\struct {G, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
where $X \circ Y$ denotes th... | Consider the [[Definition:Relation|relation]] $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \map \RR S$
Then:
{{begin-eqn}}
{{eqn | l = X \circ \paren {Y \cup Z}
| r = \map \RR {Y \cup Z}
| c =
}}
{{eq... | Product of Subset with Union/Proof 2 | https://proofwiki.org/wiki/Product_of_Subset_with_Union | https://proofwiki.org/wiki/Product_of_Subset_with_Union/Proof_2 | [
"Subset Products",
"Set Union",
"Product of Subset with Union"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Subset Product"
] | [
"Definition:Relation",
"Image of Union under Relation",
"Definition:Relation",
"Image of Union under Relation"
] |
proofwiki-3064 | Quotient Group is Subgroup of Power Structure of Group | Let $\struct {G, \circ}$ be a group.
Let $\struct {N, \circ}$ be a normal subgroup of $\struct {G, \circ}$.
Then $\struct {G / N, \circ_N}$ is a subgroup of $\struct {\powerset G, \circ_\PP}$, where:
:$\struct {G / N, \circ_N}$ is the quotient group of $G$ by $N$
:$\struct {\powerset G, \circ_\PP}$ is the power structu... | Follows directly from:
* Quotient Group is Group
* Cosets of $G$ by $N$ are subsets of $G$ and therefore elements of $\powerset G$
* The operation $\circ_N$ is defined as the subset product of cosets.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {N, \circ}$ be a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {G, \circ}$.
Then $\struct {G / N, \circ_N}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\powerset G, \circ_\PP}$, where:
:$\struct {G / N, \circ_N}$ is the [[Defi... | Follows directly from:
* [[Quotient Group is Group]]
* [[Definition:Coset|Cosets]] of $G$ by $N$ are [[Definition:Subset|subsets]] of $G$ and therefore elements of $\powerset G$
* The operation $\circ_N$ is defined as the [[Definition:Subset Product|subset product]] of [[Definition:Coset|cosets]].
{{qed}} | Quotient Group is Subgroup of Power Structure of Group | https://proofwiki.org/wiki/Quotient_Group_is_Subgroup_of_Power_Structure_of_Group | https://proofwiki.org/wiki/Quotient_Group_is_Subgroup_of_Power_Structure_of_Group | [
"Quotient Groups",
"Power Structures"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Quotient Group",
"Definition:Power Structure"
] | [
"Quotient Group is Group",
"Definition:Coset",
"Definition:Subset",
"Definition:Subset Product",
"Definition:Coset"
] |
proofwiki-3065 | Hilbert's Nullstellensatz | Let $k$ be an algebraically closed field.
Let $n \geq 0$ be an natural number.
Let $k \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Then for every ideal $J \subseteq k \sqbrk {x_1, \ldots, x_n}$, the associated ideal of its zero-locus equals its radical:
:$\map I {\map Z J} = \map \Rad J$ | Note first that the operations $\map I {\, \cdot \,}$ and $\map Z {\, \cdot \,}$ are inclusion reversing by Ideal and Zero Locus are Order Reversing.
That is:
:$X \subseteq Y \subseteq k^n \implies \map I X \supseteq \map I Y$
:$I \subseteq J \implies \map Z I \supseteq \map Z J$
Let $m_a$ be the ideal $\ideal {x_1 - a... | Let $k$ be an [[Definition:Algebraically Closed Field|algebraically closed field]].
Let $n \geq 0$ be an [[Definition:Natural Number|natural number]].
Let $k \sqbrk {x_1, \ldots, x_n}$ be the [[Definition:Polynomial Ring in Multiple Variables|polynomial ring]] in $n$ variables over $k$.
Then for every [[Definition:... | Note first that the operations $\map I {\, \cdot \,}$ and $\map Z {\, \cdot \,}$ are [[Definition:Inclusion-Reversing Mapping|inclusion reversing]] by [[Ideal and Zero Locus are Order Reversing]].
That is:
:$X \subseteq Y \subseteq k^n \implies \map I X \supseteq \map I Y$
:$I \subseteq J \implies \map Z I \supseteq \... | Hilbert's Nullstellensatz | https://proofwiki.org/wiki/Hilbert's_Nullstellensatz | https://proofwiki.org/wiki/Hilbert's_Nullstellensatz | [
"Ideal Theory",
"Algebraic Geometry"
] | [
"Definition:Algebraically Closed Field",
"Definition:Natural Numbers",
"Definition:Polynomial Ring",
"Definition:Ideal of Ring",
"Definition:Vanishing Ideal of Subset of Affine Space",
"Definition:Zero Locus of Set of Polynomials",
"Definition:Radical of Ideal of Ring"
] | [
"Definition:Inclusion-Reversing Mapping",
"Ideal and Zero Locus are Order Reversing"
] |
proofwiki-3066 | Subset of Domain is Subset of Preimage of Image | Let $f: S \to T$ be a mapping.
Then:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
:$f \sqbrk A$ denotes the image of $A$ under $f$
:$f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
:$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
This can be expressed in the language an... | As a mapping is by definition a left-total relation.
Therefore Preimage of Image under Left-Total Relation is Superset applies:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where $\RR$ is a relation.
Hence:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
:$f \sqbrk A$ denotes the [[Definition:Image of Subset under Mapping|image of $A$ under $f$]]
:$f^{-1} \sqbrk A$ denotes the [[Definition:Preimage of Subset under Mapping|preimage... | As a [[Definition:Mapping|mapping]] is by definition a [[Definition:Left-Total Relation|left-total relation]].
Therefore [[Preimage of Image under Left-Total Relation is Superset]] applies:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where $\RR$ is a [[Definition:Relation|relation]].
He... | Subset of Domain is Subset of Preimage of Image | https://proofwiki.org/wiki/Subset_of_Domain_is_Subset_of_Preimage_of_Image | https://proofwiki.org/wiki/Subset_of_Domain_is_Subset_of_Preimage_of_Image | [
"Preimages under Mappings",
"Composite Mappings",
"Subset of Domain is Subset of Preimage of Image"
] | [
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Preimage/Mapping/Subset",
"Definition:Composition of Mappings",
"Definition:Direct Image Mapping/Mapping",
"Definition:Inverse Image Mapping/Mapping"
] | [
"Definition:Mapping",
"Definition:Left-Total Relation",
"Preimage of Image under Left-Total Relation is Superset",
"Definition:Relation"
] |
proofwiki-3067 | Clairaut's Differential Equation | '''Clairaut's differential equation''' is a first order ordinary differential equation which can be put into the form:
:$y = x y' + \map f {y'}$
Its general solution is:
:$y = C x + \map f C$
where $C$ is a constant. | We have:
:$y = x y' + \map f {y'}$
Differentiating the equation {{WRT|Differentiation}} $x$ we have:
{{begin-eqn}}
{{eqn | l = y'
| r = y' + x y'' + y'' \map {f'} {y'}
| c =
}}
{{eqn | ll= \leadsto
| l = 0
| r = \map {y''} {x + \map {f'} {y'} }
| c =
}}
{{end-eqn}} | '''[[Clairaut's Differential Equation|Clairaut's differential equation]]''' is a [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]] which can be put into the form:
:$y = x y' + \map f {y'}$
Its [[Definition:General Solution to Differential Equation|general solution]]... | We have:
:$y = x y' + \map f {y'}$
Differentiating the equation {{WRT|Differentiation}} $x$ we have:
{{begin-eqn}}
{{eqn | l = y'
| r = y' + x y'' + y'' \map {f'} {y'}
| c =
}}
{{eqn | ll= \leadsto
| l = 0
| r = \map {y''} {x + \map {f'} {y'} }
| c =
}}
{{end-eqn}} | Clairaut's Differential Equation | https://proofwiki.org/wiki/Clairaut's_Differential_Equation | https://proofwiki.org/wiki/Clairaut's_Differential_Equation | [
"Clairaut's Differential Equation",
"First Order ODEs"
] | [
"Clairaut's Differential Equation",
"Definition:First Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Constant"
] | [] |
proofwiki-3068 | Noether Normalization Lemma | Let $k$ be a field.
Let $A$ be a non-trivial finitely generated $k$-algebra.
{{explain|the above link is for Definition:Non-Trivial Ring -- we need to define a Definition:Non-Trivial Algebra}}
Then there exists $n \in \N$ and a finite injective morphism of $k$-algebra:
:$k \sqbrk {x_1, \dotsc, x_n} \to A$
{{DefinitionW... | Since $A$ is finitely generated, we prove this by induction on the number $m$ of generators as a $k$-algebra. | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $A$ be a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Finitely Generated R-Algebra|finitely generated]] [[Definition:K-Algebra|$k$-algebra]].
{{explain|the above link is for [[Definition:Non-Trivial Ring]] -- we need to define a [[Definition:... | Since $A$ is [[Definition:Finitely Generated R-Algebra|finitely generated]], we prove this by [[Proof by Mathematical Induction|induction]] on the number $m$ of [[Definition:Generator of Algebra|generators]] as a [[Definition:K-Algebra|$k$-algebra]]. | Noether Normalization Lemma | https://proofwiki.org/wiki/Noether_Normalization_Lemma | https://proofwiki.org/wiki/Noether_Normalization_Lemma | [
"Commutative Algebra"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Non-Trivial Ring",
"Definition:Finitely Generated R-Algebra",
"Definition:K-Algebra",
"Definition:Non-Trivial Ring",
"Definition:Non-Trivial Algebra",
"Definition:Finite Ring Homomorphism",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:K... | [
"Definition:Finitely Generated R-Algebra",
"Principle of Mathematical Induction",
"Definition:Generator of Algebra",
"Definition:K-Algebra"
] |
proofwiki-3069 | Image of Preimage under Mapping | Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B = B \cap f \sqbrk S$
where:
:$f \sqbrk S$ denotes the image of $S$ under $f$
:$f^{-1} \sqbrk B$ denotes the preimage of $B$ under $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$. | As $f$ is a mapping it follows by definition that $f$ is also a relation
Applying Image of Preimage under Relation is Subset directly:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq B$
From Preimage of Subset is Subset of Preimage:
:$B \subseteq T \i... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B = B \cap f \sqbrk S$
where:
:$f \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image of $S$ under $f$]]
:$f^{-1} \sqbrk B$ denotes the [[Definition:Preimage of Subset under Mapping|... | As $f$ is a [[Definition:Mapping|mapping]] it follows by definition that $f$ is also a [[Definition:Relation|relation]]
Applying [[Image of Preimage under Relation is Subset]] directly:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq B$
From [[Preim... | Image of Preimage under Mapping | https://proofwiki.org/wiki/Image_of_Preimage_under_Mapping | https://proofwiki.org/wiki/Image_of_Preimage_under_Mapping | [
"Preimages under Mappings"
] | [
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Preimage/Mapping/Subset",
"Definition:Composition of Relations"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Preimage under Relation is Subset",
"Preimage of Subset is Subset of Preimage",
"Preimage of Mapping equals Domain",
"Image of Subset under Mapping is Subset of Image",
"Intersection is Largest Subset",
"Image of Subset under Mapping is Subset of ... |
proofwiki-3070 | Isomorphism of External Direct Products | Let:
:$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
:$\phi_1$ be an isomorphism fro... | From Homomorphism of External Direct Products, we have that $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ is a homomorphism.
From Cartesian Product of Bijections is Bijection, we have that $\phi_1 \times \phi_2$ is a bijection.
Thus $\phi_1 \times \phi_2$ is a bijective homomor... | Let:
:$\struct {S_1 \times S_2, \circ}$ be the [[Definition:External Direct Product|external direct product]] of two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the [[Definition:External Direct Produc... | From [[Homomorphism of External Direct Products]], we have that $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
From [[Cartesian Product of Bijections is Bijection]], we have that $\phi_1 \times \phi_2$ is a [[Def... | Isomorphism of External Direct Products | https://proofwiki.org/wiki/Isomorphism_of_External_Direct_Products | https://proofwiki.org/wiki/Isomorphism_of_External_Direct_Products | [
"Isomorphism of External Direct Products",
"Isomorphisms (Abstract Algebra)",
"External Direct Products"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Mapping",
"Definition:Isomor... | [
"Homomorphism of External Direct Products",
"Definition:Homomorphism (Abstract Algebra)",
"Cartesian Product of Bijections is Bijection",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-3071 | Equivalence of Definitions of Independent Subgroups | {{TFAE|def = Independent Subgroups}}
Let $G$ be a group whose identity is $e$.
Let $\sequence {H_n}$ be a sequence of independent subgroups of $G$. | === Definition 1 implies Definition 2 ===
Let $H_1, H_2, \ldots, H_n$ be independent by definition 1:
:$\ds \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \closedint 1 n: h_k = e$
where $h_k \in H_k$ for all $k \in \closedint 1 n$.
Let $\ds u \in \paren {\prod_{j \mathop = 1}^{k - 1} H_j} \cap H_k$.
Then:
:$\ds \ex... | {{TFAE|def = Independent Subgroups}}
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\sequence {H_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Subgroups|independent subgroups]] of $G$. | === Definition 1 implies Definition 2 ===
Let $H_1, H_2, \ldots, H_n$ be [[Definition:Independent Subgroups/Definition 1|independent by definition 1]]:
:$\ds \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \closedint 1 n: h_k = e$
where $h_k \in H_k$ for all $k \in \closedint 1 n$.
Let $\ds u \in \paren {\prod_{... | Equivalence of Definitions of Independent Subgroups | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Independent_Subgroups | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Independent_Subgroups | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Sequence",
"Definition:Independent Subgroups"
] | [
"Definition:Independent Subgroups/Definition 1",
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Independent Subgroups/Definition 2",
"Definition:Independent Subgroups/Definition 2",
"Definition:Independent Subgroups/Definition 1"
] |
proofwiki-3072 | Dual of Total Ordering is Total Ordering | Let $\preccurlyeq$ be a total ordering.
Then its dual ordering $\succcurlyeq$ is also a total ordering. | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
From Dual Ordering is Ordering we have that $\succcurlyeq$ is an ordering.
Let $x, y \in S$.
Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total.
By definition of dual ordering, also $y \succcurlyeq x$ or $x \succcurlyeq y$.
Hence the res... | Let $\preccurlyeq$ be a [[Definition:Total Ordering|total ordering]].
Then its [[Definition:Dual Ordering|dual ordering]] $\succcurlyeq$ is also a [[Definition:Total Ordering|total ordering]]. | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
From [[Dual Ordering is Ordering]] we have that $\succcurlyeq$ is an [[Definition:Ordering|ordering]].
Let $x, y \in S$.
Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total.
By definition of [[Defin... | Dual of Total Ordering is Total Ordering | https://proofwiki.org/wiki/Dual_of_Total_Ordering_is_Total_Ordering | https://proofwiki.org/wiki/Dual_of_Total_Ordering_is_Total_Ordering | [
"Total Orderings",
"Dual Orderings"
] | [
"Definition:Total Ordering",
"Definition:Dual Ordering",
"Definition:Total Ordering"
] | [
"Definition:Totally Ordered Set",
"Dual Ordering is Ordering",
"Definition:Ordering",
"Definition:Dual Ordering",
"Definition:Total Ordering"
] |
proofwiki-3073 | Dual of Lattice Ordering is Lattice Ordering | Let $\struct {S, \preccurlyeq}$ be a lattice.
Let $\preccurlyeq$ be the lattice ordering on $\struct {S, \preccurlyeq}$.
Then its dual ordering $\succcurlyeq$ is also a lattice ordering. | Let $\struct {S, \preccurlyeq}$ be a lattice.
It is to be shown that:
:for all $x, y \in S$, the ordered set $\struct {\set {x, y}, \succcurlyeq}$ admits both a supremum and an infimum.
Let $x, y \in S$.
Then $\struct {\set {x, y}, \preccurlyeq}$ admits both a supremum and an infimum.
Let $c = \map \sup {\set {x, y}, ... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Lattice (Ordered Set)|lattice]].
Let $\preccurlyeq$ be the [[Definition:Lattice Ordering|lattice ordering]] on $\struct {S, \preccurlyeq}$.
Then its [[Definition:Dual Ordering|dual ordering]] $\succcurlyeq$ is also a [[Definition:Lattice Ordering|lattice ordering]]. | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Lattice (Ordered Set)|lattice]].
It is to be shown that:
:for all $x, y \in S$, the [[Definition:Ordered Set|ordered set]] $\struct {\set {x, y}, \succcurlyeq}$ admits both a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
Let $... | Dual of Lattice Ordering is Lattice Ordering | https://proofwiki.org/wiki/Dual_of_Lattice_Ordering_is_Lattice_Ordering | https://proofwiki.org/wiki/Dual_of_Lattice_Ordering_is_Lattice_Ordering | [
"Lattice Theory",
"Dual Orderings"
] | [
"Definition:Lattice (Ordered Set)",
"Definition:Lattice Ordering",
"Definition:Dual Ordering",
"Definition:Lattice Ordering"
] | [
"Definition:Lattice (Ordered Set)",
"Definition:Ordered Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Dual Ordering",
"Definition:Upper Bound of ... |
proofwiki-3074 | Order Isomorphism on Totally Ordered Set preserves Total Ordering | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.
Then $\struct {S, \preccurlyeq_1}$ is a totally ordered set {{iff}} $\struct {T, \preccurlyeq_2}$ is also a totally ordered set. | === Necessary Condition ===
Let $\struct {S, \preccurlyeq_1}$ be a totally ordered set
Then by definition $\preccurlyeq_1$ is a total ordering.
Let $x, y \in S$.
Then either $x \preccurlyeq_1 y$ or $y \preccurlyeq_1 x$.
From the definition of order isomorphism, either:
:$\map \phi x \preccurlyeq_2 \map \phi y$
or:
:$\m... | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then $\struct {S, \preccurlyeq_1}$ is a [[Definition:Totally Ordered Set|to... | === Necessary Condition ===
Let $\struct {S, \preccurlyeq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]]
Then by definition $\preccurlyeq_1$ is a [[Definition:Total Ordering|total ordering]].
Let $x, y \in S$.
Then either $x \preccurlyeq_1 y$ or $y \preccurlyeq_1 x$.
From the definition of [[Defin... | Order Isomorphism on Totally Ordered Set preserves Total Ordering | https://proofwiki.org/wiki/Order_Isomorphism_on_Totally_Ordered_Set_preserves_Total_Ordering | https://proofwiki.org/wiki/Order_Isomorphism_on_Totally_Ordered_Set_preserves_Total_Ordering | [
"Order Isomorphisms",
"Total Orderings"
] | [
"Definition:Ordered Set",
"Definition:Order Isomorphism",
"Definition:Totally Ordered Set",
"Definition:Totally Ordered Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Total Ordering",
"Definition:Order Isomorphism",
"Definition:Total Ordering",
"Definition:Totally Ordered Set",
"Definition:Order Isomorphism",
"Definition:Totally Ordered Set"
] |
proofwiki-3075 | Order Isomorphism on Well-Ordered Set preserves Well-Ordering | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_2, \preccurlyeq_2}$ be an order isomorphism.
Then $\struct {S_1, \preccurlyeq_1}$ is a well-ordered set {{iff}} $\struct {S_2, \preccurlyeq_2}$ is also a well-ordered set. | Let $\struct {S_1, \preccurlyeq_1}$ be a well-ordered set.
Then by definition $\preccurlyeq_1$ is a well-ordering.
By Well-Ordering is Total Ordering, $\preccurlyeq_1$ is a total ordering.
From Order Isomorphism on Totally Ordered Set preserves Total Ordering, it follows that $\struct {S_2, \preccurlyeq_2}$ is a totall... | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_2, \preccurlyeq_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then $\struct {S_1, \preccurlyeq_1}$ is a [[Definition:Well-Ordered... | Let $\struct {S_1, \preccurlyeq_1}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Then by definition $\preccurlyeq_1$ is a [[Definition:Well-Ordering|well-ordering]].
By [[Well-Ordering is Total Ordering]], $\preccurlyeq_1$ is a [[Definition:Total Ordering|total ordering]].
From [[Order Isomorphism on Total... | Order Isomorphism on Well-Ordered Set preserves Well-Ordering | https://proofwiki.org/wiki/Order_Isomorphism_on_Well-Ordered_Set_preserves_Well-Ordering | https://proofwiki.org/wiki/Order_Isomorphism_on_Well-Ordered_Set_preserves_Well-Ordering | [
"Order Isomorphisms",
"Well-Orderings"
] | [
"Definition:Ordered Set",
"Definition:Order Isomorphism",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set"
] | [
"Definition:Well-Ordered Set",
"Definition:Well-Ordering",
"Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2",
"Definition:Total Ordering",
"Order Isomorphism on Totally Ordered Set preserves Total Ordering",
"Definition:Totally Ordered Set",
"Definition:Well-Founded Relatio... |
proofwiki-3076 | Order Isomorphism on Lattice preserves Lattice Structure | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.
Then $\struct {S, \preccurlyeq_1}$ is a lattice {{iff}} $\struct {T, \preccurlyeq_2}$ is also a lattice. | Let $\struct {S, \preccurlyeq_1}$ be a lattice
Then by definition $\preccurlyeq_1$ is a lattice ordering.
We need to show that for all $x, y \in S$, the ordered set $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_2}$ admits both a supremum and an infimum.
Let $x, y \in S$.
Then $\struct {\set {x, y}, \preccurl... | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then $\struct {S, \preccurlyeq_1}$ is a [[Definition:Lattice (Ordered Set)|... | Let $\struct {S, \preccurlyeq_1}$ be a [[Definition:Lattice (Ordered Set)|lattice]]
Then by definition $\preccurlyeq_1$ is a [[Definition:Lattice Ordering|lattice ordering]].
We need to show that for all $x, y \in S$, the [[Definition:Ordered Set|ordered set]] $\struct {\set {\map \phi x, \map \phi y}, \preccurlyeq_... | Order Isomorphism on Lattice preserves Lattice Structure | https://proofwiki.org/wiki/Order_Isomorphism_on_Lattice_preserves_Lattice_Structure | https://proofwiki.org/wiki/Order_Isomorphism_on_Lattice_preserves_Lattice_Structure | [
"Lattice Theory",
"Order Isomorphisms"
] | [
"Definition:Ordered Set",
"Definition:Order Isomorphism",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)"
] | [
"Definition:Lattice (Ordered Set)",
"Definition:Lattice Ordering",
"Definition:Ordered Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Image (Set T... |
proofwiki-3077 | Set of Subgroups forms Complete Lattice | Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Then:
:$\struct {\mathbb G, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subgroups of $G$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$. | Let $\O \subset \mathbb H \subseteq \mathbb G$.
By Intersection of Subgroups: General Result, $\bigcap \mathbb H$ is the largest subgroup of $G$ contained in each of the elements of $\mathbb H$.
Thus, not only is $\ds \bigcap \mathbb H$ a lower bound of $\mathbb H$, but also the largest, and therefore an infimum.
The s... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$\struct {\mathbb G, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:S... | Let $\O \subset \mathbb H \subseteq \mathbb G$.
By [[Intersection of Subgroups/General Result|Intersection of Subgroups: General Result]], $\bigcap \mathbb H$ is the largest [[Definition:Subgroup|subgroup]] of $G$ contained in each of the elements of $\mathbb H$.
Thus, not only is $\ds \bigcap \mathbb H$ a [[Definiti... | Set of Subgroups forms Complete Lattice/Proof 1 | https://proofwiki.org/wiki/Set_of_Subgroups_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Subgroups_forms_Complete_Lattice/Proof_1 | [
"Examples of Complete Lattices",
"Subgroups",
"Set of Subgroups forms Complete Lattice"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Subgroup",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subgroup",
"Definition:Infimum of Set"
] | [
"Intersection of Subgroups is Subgroup/General Result",
"Definition:Subgroup",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Subgroup",
"Definition:Complete Lattice"
] |
proofwiki-3078 | Set of Subgroups forms Complete Lattice | Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Then:
:$\struct {\mathbb G, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subgroups of $G$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$. | From Group is Subgroup of Itself:
:$\struct {G, \circ} \in \mathbb G$
Let $\mathbb H$ be a non-empty subset of $\mathbb G$.
From Intersection of Subgroups is Subgroup: General Result:
:$\ds \bigcap \mathbb H \in \mathbb G$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$\struct {\mathbb G, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:S... | From [[Group is Subgroup of Itself]]:
:$\struct {G, \circ} \in \mathbb G$
Let $\mathbb H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\mathbb G$.
From [[Intersection of Subgroups is Subgroup/General Result|Intersection of Subgroups is Subgroup: General Result]]:
:$\ds \bigcap \mathbb ... | Set of Subgroups forms Complete Lattice/Proof 2 | https://proofwiki.org/wiki/Set_of_Subgroups_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Subgroups_forms_Complete_Lattice/Proof_2 | [
"Examples of Complete Lattices",
"Subgroups",
"Set of Subgroups forms Complete Lattice"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Subgroup",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subgroup",
"Definition:Infimum of Set"
] | [
"Group is Subgroup of Itself",
"Definition:Non-Empty Set",
"Definition:Subset",
"Intersection of Subgroups is Subgroup/General Result",
"Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice",
"Definition:Complete Lattice",
"Definition:Infimum of Set"
] |
proofwiki-3079 | Index Laws/Sum of Indices/Monoid | Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ be defined as the $n$th power of $a$:
:<nowiki>$a^n = \begin{cases}
e & : n = 0 \\
a^x \circ a & : n = x + 1
\end{cases}$</nowiki>
That is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ copies of } ... | Because $\struct {S, \circ}$ is a monoid, it is {{afortiori}} also a semigroup.
From Index Laws for Semigroup: Sum of Indices:
:$\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$
It remains to be shown that the result... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is $e$.
For $a \in S$, let $\circ^n a = a^n$ be defined as the [[Definition:Power of Element of Magma with Identity|$n$th power of $a$]]:
:<nowiki>$a^n = \begin{cases}
e & : n = 0 \\
a^x \circ a & : n = x... | Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is {{afortiori}} also a [[Definition:Semigroup|semigroup]].
From [[Index Laws for Semigroup/Sum of Indices|Index Laws for Semigroup: Sum of Indices]]:
:$\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
That is:... | Index Laws/Sum of Indices/Monoid | https://proofwiki.org/wiki/Index_Laws/Sum_of_Indices/Monoid | https://proofwiki.org/wiki/Index_Laws/Sum_of_Indices/Monoid | [
"Index Laws",
"Monoids"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power of Element/Magma with Identity"
] | [
"Definition:Monoid",
"Definition:Semigroup",
"Index Laws/Sum of Indices/Semigroup",
"Integer Addition Identity is Zero",
"Identity Element of Natural Number Addition is Zero",
"Identity Element of Natural Number Addition is Zero"
] |
proofwiki-3080 | Power of Product of Commuting Elements in Semigroup equals Product of Powers | :$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
:$\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | :$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | The proof proceeds by the [[Principle of Mathematical Induction]]:
Let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | Power of Product of Commuting Elements in Semigroup equals Product of Powers | https://proofwiki.org/wiki/Power_of_Product_of_Commuting_Elements_in_Semigroup_equals_Product_of_Powers | https://proofwiki.org/wiki/Power_of_Product_of_Commuting_Elements_in_Semigroup_equals_Product_of_Powers | [
"Semigroups",
"Commutativity",
"Powers (Abstract Algebra)"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-3081 | Number of Compositions | A $k$-composition of a positive integer $n$ is an ordered $k$-tuple: $c = \tuple {c_1, c_2, \ldots, c_k}$ such that $c_1 + c_2 + \cdots + c_k = n$ and $c_i $ are strictly positive integers.
The number of $k$-composition of $n$ is $\dbinom {n - 1} {k - 1}$ and the total number of compositions of $n$ is $2^{n - 1}$ (that... | Consider the following array consisting of $n$ ones and $n - 1$ blanks:
:$\begin{bmatrix} 1 \ \_ \ 1 \ \_ \ \cdots \ \_ \ 1 \ \_ \ 1 \end{bmatrix}$
In each blank we can either put a comma or a plus sign.
Each way of choosing $,$ or $+$ will give a composition of $n$ with the commas separating the individual $c_i$'s.
It... | A [[Definition:Composition (Combinatorics)|$k$-composition]] of a [[Definition:Positive Integer|positive integer]] $n$ is an [[Definition:Ordered Tuple|ordered $k$-tuple]]: $c = \tuple {c_1, c_2, \ldots, c_k}$ such that $c_1 + c_2 + \cdots + c_k = n$ and $c_i $ are [[Definition:Strictly Positive|strictly positive]] [[D... | Consider the following array consisting of $n$ ones and $n - 1$ blanks:
:$\begin{bmatrix} 1 \ \_ \ 1 \ \_ \ \cdots \ \_ \ 1 \ \_ \ 1 \end{bmatrix}$
In each blank we can either put a comma or a plus sign.
Each way of choosing $,$ or $+$ will give a composition of $n$ with the commas separating the individual $c_i$'s.... | Number of Compositions | https://proofwiki.org/wiki/Number_of_Compositions | https://proofwiki.org/wiki/Number_of_Compositions | [
"Combinatorics"
] | [
"Definition:Composition (Combinatorics)",
"Definition:Positive/Integer",
"Definition:Ordered Tuple",
"Definition:Strictly Positive",
"Definition:Integer",
"Definition:Composition (Combinatorics)"
] | [
"Product Rule for Counting",
"Binomial Theorem",
"Category:Combinatorics"
] |
proofwiki-3082 | Sum of Sequences of Fifth and Seventh Powers | :$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ | :$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ | Sum of Sequences of Fifth and Seventh Powers | https://proofwiki.org/wiki/Sum_of_Sequences_of_Fifth_and_Seventh_Powers | https://proofwiki.org/wiki/Sum_of_Sequences_of_Fifth_and_Seventh_Powers | [
"Sums of Sequences",
"Fifth Powers",
"Seventh Powers"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-3083 | Cardinality of Set of Strictly Increasing Mappings | Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets.
Let the cardinality of $S$ and $T$ be:
:$\card S = m, \card T = n$
Then the number of strictly increasing mappings from $S$ to $T$ is:
:$\dbinom n m = \dfrac {n!} {m! \ \paren {n - m}!}$.
where $\dbinom n m$ is a binomial coefficient. | From:
:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
and:
:Strictly Monotone Mapping with Totally Ordered Domain is Injective
a strictly increasing mapping $\phi$ from $S$ to $T$ is an order isomorphism from $S$ to $\map \phi S$.
Let $\mathbb F$ be the set of all strictly increasing mappin... | Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be [[Definition:Totally Ordered Set|tosets]].
Let the [[Definition:Cardinality|cardinality]] of $S$ and $T$ be:
:$\card S = m, \card T = n$
Then the number of [[Definition:Strictly Increasing Mapping|strictly increasing mappings]] from $S$ to $T$ is:
:$\dbin... | From:
:[[Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing]]
and:
:[[Strictly Monotone Mapping with Totally Ordered Domain is Injective]]
a [[Definition:Strictly Increasing Mapping|strictly increasing mapping]] $\phi$ from $S$ to $T$ is an [[Definition:Order Isomorphism|order isomorphism]] fro... | Cardinality of Set of Strictly Increasing Mappings | https://proofwiki.org/wiki/Cardinality_of_Set_of_Strictly_Increasing_Mappings | https://proofwiki.org/wiki/Cardinality_of_Set_of_Strictly_Increasing_Mappings | [
"Combinatorics"
] | [
"Definition:Totally Ordered Set",
"Definition:Cardinality",
"Definition:Strictly Increasing/Mapping",
"Definition:Binomial Coefficient"
] | [
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing",
"Strictly Monotone Mapping with Totally Ordered Domain is Injective",
"Definition:Strictly Increasing/Mapping",
"Definition:Order Isomorphism",
"Definition:Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Set",
"... |
proofwiki-3084 | Construction of Inverse Completion/Cartesian Product with Cancellable Elements | Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.
That is:
:$\forall \tuple {x, y}, \tuple {u, v} \in S \times C: \tuple {x, ... | By Cancellable Elements of Semigroup form Subsemigroup, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {S, \circ}$, where $\circ {\restriction_C}$ is the restriction of $\circ$ to $C$.
By Restriction of Commutative Operation is Commutative, as $\struct {C, \circ {\restriction_C} }$ is a substructu... | Let $\struct {S \times C, \oplus}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where $\oplus$ is the [[Definition:Operation Induced by Direct Product|operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction... | By [[Cancellable Elements of Semigroup form Subsemigroup]], $\struct {C, \circ {\restriction_C} }$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$, where $\circ {\restriction_C}$ is the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $C$.
By [[Restriction of Commutative Operatio... | Construction of Inverse Completion/Cartesian Product with Cancellable Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Cartesian_Product_with_Cancellable_Elements | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Cartesian_Product_with_Cancellable_Elements | [
"Construction of Inverse Completion"
] | [
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Commutative Semigroup"
] | [
"Cancellable Elements of Semigroup form Subsemigroup",
"Definition:Subsemigroup",
"Definition:Restriction/Operation",
"Restriction of Commutative Operation is Commutative",
"Definition:Algebraic Substructure",
"Definition:Commutative/Algebraic Structure",
"Definition:Commutative/Algebraic Structure",
... |
proofwiki-3085 | Construction of Inverse Completion/Equivalence Relation | The cross-relation $\boxtimes$ is an equivalence relation on $\struct {S \times C, \oplus}$. | 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 Equivalence Relation.
{{q... | The [[Definition:Cross-Relation|cross-relation]] $\boxtimes$ is an [[Definition:Equivalence Relation|equivalence relation]] on $\struct {S \times C, \oplus}$. | 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 | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Equivalence_Relation | [
"Construction of Inverse Completion"
] | [
"Definition:Cross-Relation",
"Definition:Equivalence Relation"
] | [
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Semigroup is Subsemigroup of Itself",
"Definition:Subsemigroup",
"Cross-Relation is Equivalence Relation"
] |
proofwiki-3086 | Construction of Inverse Completion/Quotient Structure | Let the quotient structure defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
=== Quotient Structure is Commutative Semigroup ===
{{:Construction of Inverse Com... | From the defined equivalence relation, we have that:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
is a congruence relation on $\struct {S \times C, \oplus}$.
From the definition of the members of the equivalence classes:
:$(1) \quad \forall x, y \in S, a, b \in C: \tuple {x \circ ... | Let the [[Definition:Quotient Structure|quotient structure]] defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the [[Definition:Operation Induced on Quotient Set|operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$]].... | From the defined [[Construction of Inverse Completion/Congruence Relation|equivalence relation]], we have that:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
is a [[Definition:Congruence Relation|congruence relation]] on $\struct {S \times C, \oplus}$.
From the definition of the ... | Construction of Inverse Completion/Quotient Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure | https://proofwiki.org/wiki/Construction_of_Inverse_Completion/Quotient_Structure | [
"Construction of Inverse Completion"
] | [
"Definition:Quotient Structure",
"Definition:Operation Induced on Quotient Set",
"Construction of Inverse Completion/Quotient Structure is Commutative Semigroup",
"Construction of Inverse Completion/Quotient Mapping is Injective",
"Construction of Inverse Completion/Quotient Mapping is Monomorphism",
"Con... | [
"Construction of Inverse Completion/Congruence 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",
"Definition:Equivalence Class",... |
proofwiki-3087 | Integers form Commutative Ring | The set of integers $\Z$ forms a commutative ring under addition and multiplication. | We have that:
:From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is an abelian group.
:From Integers under Multiplication form Countably Infinite Commutative Monoid, the algebraic structure $\struct {\Z, \times}$ is a commutative monoid and therefore a commutative semigroup.
:In... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ forms a [[Definition:Commutative Ring|commutative ring]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]]. | We have that:
:From [[Integers under Addition form Abelian Group]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, +}$ is an [[Definition:Abelian Group|abelian group]].
:From [[Integers under Multiplication form Countably Infinite Commutative Monoid]], the [[Definition:Algebraic Structure|al... | Integers form Commutative Ring | https://proofwiki.org/wiki/Integers_form_Commutative_Ring | https://proofwiki.org/wiki/Integers_form_Commutative_Ring | [
"Integers",
"Commutative Rings"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Commutative Ring",
"Definition:Addition/Integers",
"Definition:Multiplication/Integers"
] | [
"Integers under Addition form Abelian Group",
"Definition:Algebraic Structure",
"Definition:Abelian Group",
"Integers under Multiplication form Countably Infinite Commutative Monoid",
"Definition:Algebraic Structure",
"Definition:Commutative Monoid",
"Definition:Commutative Semigroup",
"Integer Multip... |
proofwiki-3088 | Symmetric Group is Group | Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group. | Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$.
Thus $\struct {\map \Gamma S, \circ}$ is closed.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
From Set of all Self-Maps under Composition forms Monoid, we have that $\stru... | Let $S$ be a [[Definition:Set|set]].
Let $\map \Gamma S$ denote the [[Definition:Set|set]] of all [[Definition:Permutation|permutations]] on $S$.
Then $\struct {\map \Gamma S, \circ}$, the [[Definition:Symmetric Group|symmetric group on $S$]], forms a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn:
=== {{Group-axiom|0|nolink}} ===
By [[Composite of Permutations is Permutation]], $S$ is itself a [[Definition:Permutation|permutation]] on $S$.
Thus $\struct {\map \Gamma S, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
{{qed|lemma}}
=== {{Gr... | Symmetric Group is Group/Proof 1 | https://proofwiki.org/wiki/Symmetric_Group_is_Group | https://proofwiki.org/wiki/Symmetric_Group_is_Group/Proof_1 | [
"Symmetric Group is Group",
"Symmetric Groups"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Permutation",
"Definition:Symmetric Group",
"Definition:Group"
] | [
"Axiom:Group Axioms",
"Composite of Permutations is Permutation",
"Definition:Permutation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Set of all Self-Maps under Composition forms Monoid",
"Definition:Associative Operation",
"Set of all Self-Maps under Composition forms Monoid",
"De... |
proofwiki-3089 | Symmetric Group is Group | Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group. | A direct application of Set of Invertible Mappings forms Symmetric Group.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\map \Gamma S$ denote the [[Definition:Set|set]] of all [[Definition:Permutation|permutations]] on $S$.
Then $\struct {\map \Gamma S, \circ}$, the [[Definition:Symmetric Group|symmetric group on $S$]], forms a [[Definition:Group|group]]. | A direct application of [[Set of Invertible Mappings forms Symmetric Group]].
{{qed}} | Symmetric Group is Group/Proof 2 | https://proofwiki.org/wiki/Symmetric_Group_is_Group | https://proofwiki.org/wiki/Symmetric_Group_is_Group/Proof_2 | [
"Symmetric Group is Group",
"Symmetric Groups"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Permutation",
"Definition:Symmetric Group",
"Definition:Group"
] | [
"Set of Invertible Mappings forms Symmetric Group"
] |
proofwiki-3090 | Quotient Ring Defined by Ring Itself is Null Ring | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {R / R, +, \circ}$ be the quotient ring defined by $R$.
Then $\struct {R / R, +, \circ}$ is a null ring. | From Ring is Ideal of Itself, it is clear we can form the quotient ring $\struct {R / R, +, \circ}$.
But $R / R = 0_R$ and so is the null ring.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $\struct {R / R, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] defined by $R$.
Then $\struct {R / R, +, \circ}$ is a [[Definition:Null Ring|null ring]]. | From [[Ring is Ideal of Itself]], it is clear we can form the [[Definition:Quotient Ring|quotient ring]] $\struct {R / R, +, \circ}$.
But $R / R = 0_R$ and so is the [[Definition:Null Ring|null ring]].
{{qed}} | Quotient Ring Defined by Ring Itself is Null Ring | https://proofwiki.org/wiki/Quotient_Ring_Defined_by_Ring_Itself_is_Null_Ring | https://proofwiki.org/wiki/Quotient_Ring_Defined_by_Ring_Itself_is_Null_Ring | [
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Quotient Ring",
"Definition:Null Ring"
] | [
"Ring is Ideal of Itself",
"Definition:Quotient Ring",
"Definition:Null Ring"
] |
proofwiki-3091 | Set of Subrings forms Complete Lattice | Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice. | Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Subrings is Largest Subring Contained in all Subrings:
:$\bigcap \mathbb S$ is the largest subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Subrings Containing Subset is Smallest:
:The intersection of the set of all su... | Let $\struct {K, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Subring|subrings]] of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]]. | Let $\O \subset \mathbb S \subseteq \mathbb K$.
By [[Intersection of Subrings is Largest Subring Contained in all Subrings]]:
:$\bigcap \mathbb S$ is the largest [[Definition:Subring|subring]] of $K$ contained in each of the elements of $\mathbb S$.
By [[Intersection of Subrings Containing Subset is Smallest]]:
:The... | Set of Subrings forms Complete Lattice | https://proofwiki.org/wiki/Set_of_Subrings_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Subrings_forms_Complete_Lattice | [
"Complete Lattices",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Set",
"Definition:Subring",
"Definition:Complete Lattice"
] | [
"Intersection of Subrings is Largest Subring Contained in all Subrings",
"Definition:Subring",
"Intersection of All Subrings Containing Subset is Smallest",
"Definition:Set Intersection",
"Definition:Subring",
"Definition:Subring",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Defin... |
proofwiki-3092 | Set of Ideals forms Complete Lattice | Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all ideals of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice. | Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Ring Ideals is Largest Ideal Contained in all Ideals:
:$\bigcap \mathbb S$ is the largest ideal of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Ring Ideals Containing Subset is Smallest:
:The intersection of the set of all id... | Let $\struct {K, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]]. | Let $\O \subset \mathbb S \subseteq \mathbb K$.
By [[Intersection of Ring Ideals is Largest Ideal Contained in all Ideals]]:
:$\bigcap \mathbb S$ is the largest [[Definition:Ideal of Ring|ideal]] of $K$ contained in each of the elements of $\mathbb S$.
By [[Intersection of Ring Ideals Containing Subset is Smallest]]... | Set of Ideals forms Complete Lattice | https://proofwiki.org/wiki/Set_of_Ideals_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Ideals_forms_Complete_Lattice | [
"Complete Lattices",
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Set",
"Definition:Ideal of Ring",
"Definition:Complete Lattice"
] | [
"Intersection of Ring Ideals is Largest Ideal Contained in all Ideals",
"Definition:Ideal of Ring",
"Intersection of All Ring Ideals Containing Subset is Smallest",
"Definition:Set Intersection",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Lower Bound of Set",
"Definition:Infim... |
proofwiki-3093 | Subset Product of Abelian Subgroups | Let $\left({G, \circ}\right)$ be an abelian group.
Let $H_1$ and $H_2$ be subgroups of $G$.
Then $H_1 \circ H_2$ is a subgroup of $G$. | From Subgroup of Abelian Group is Normal, $H_1$ and $H_2$ are normal.
The result follows from Subset Product with Normal Subgroup is Subgroup.
{{qed}}
Category:Abelian Groups
Category:Subset Products
42ddr5761fjsy89srlm2c6znwzueywi | Let $\left({G, \circ}\right)$ be an [[Definition:Abelian Group|abelian group]].
Let $H_1$ and $H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
Then $H_1 \circ H_2$ is a [[Definition:Subgroup|subgroup]] of $G$. | From [[Subgroup of Abelian Group is Normal]], $H_1$ and $H_2$ are [[Definition:Normal Subgroup|normal]].
The result follows from [[Subset Product with Normal Subgroup is Subgroup]].
{{qed}}
[[Category:Abelian Groups]]
[[Category:Subset Products]]
42ddr5761fjsy89srlm2c6znwzueywi | Subset Product of Abelian Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Abelian_Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Abelian_Subgroups | [
"Abelian Groups",
"Subset Products"
] | [
"Definition:Abelian Group",
"Definition:Subgroup",
"Definition:Subgroup"
] | [
"Subgroup of Abelian Group is Normal",
"Definition:Normal Subgroup",
"Subset Product with Normal Subgroup is Subgroup",
"Category:Abelian Groups",
"Category:Subset Products"
] |
proofwiki-3094 | Sum of Ideals is Ideal | Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.
Then:
: $J = J_1 + J_2$ is an ideal of $R$
where $J_1 + J_2$ is as defined in subset product with respect to $\struct{R, +}$. | By definition, $\struct {R, +}$ is an abelian group.
So from Subset Product of Abelian Subgroups, we have that:
:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
is itself a subgroup of $R$.
Now consider $a \circ b$ where $a \in J, b \in R$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| r = \paren {a_1 + a_2} \c... | Let $J_1$ and $J_2$ be [[Definition:Ideal of Ring|ideals]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct{R, +, \circ}$.
Then:
: $J = J_1 + J_2$ is an [[Definition:Ideal of Ring|ideal]] of $R$
where $J_1 + J_2$ is as defined in [[Definition:Subset Product|subset product]] with respect to $\struct{R, +}$. | By [[Definition:Ring (Abstract Algebra)|definition]], $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
So from [[Subset Product of Abelian Subgroups]], we have that:
:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
is itself a [[Definition:Subgroup|subgroup]] of $R$.
Now consider $a \circ b$ wh... | Sum of Ideals is Ideal | https://proofwiki.org/wiki/Sum_of_Ideals_is_Ideal | https://proofwiki.org/wiki/Sum_of_Ideals_is_Ideal | [
"Ideal Theory"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Subset Product"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Abelian Group",
"Subset Product of Abelian Subgroups",
"Definition:Subgroup",
"Definition:Distributive Operation",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] |
proofwiki-3095 | Principal Ideal from Element in Center of Ring | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $b \in R$ be in the center of $R$.
Then:
:$\ideal b = R \circ b = \set {x \circ b: x \in R}$
where $\ideal b$ is the principal ideal generated by $b$. | Let $J = R \circ b$
The center of $R$ is defined as:
:$\map Z R = \set {x \in R: \forall s \in R: s \circ x = x \circ s}$
Therefore:
:$R \circ b = b \circ R = \set {x \circ b: x \in R} = \set {b \circ x: x \in R}$
and so:
:$x \in J \implies x \circ b \in J \land b \circ x \in J$
Thus $J$ is an ideal of $R$ and so $J = ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $b \in R$ be in the [[Definition:Center of Ring|center]] of $R$.
Then:
:$\ideal b = R \circ b = \set {x \circ b: x \in R}$
where $\id... | Let $J = R \circ b$
The [[Definition:Center of Ring|center]] of $R$ is defined as:
:$\map Z R = \set {x \in R: \forall s \in R: s \circ x = x \circ s}$
Therefore:
:$R \circ b = b \circ R = \set {x \circ b: x \in R} = \set {b \circ x: x \in R}$
and so:
:$x \in J \implies x \circ b \in J \land b \circ x \in J$
Thus ... | Principal Ideal from Element in Center of Ring | https://proofwiki.org/wiki/Principal_Ideal_from_Element_in_Center_of_Ring | https://proofwiki.org/wiki/Principal_Ideal_from_Element_in_Center_of_Ring | [
"Principal Ideals of Rings"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Principal Ideal of Ring"
] | [
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Ideal of Ring"
] |
proofwiki-3096 | External Direct Sum of Rings is Ring | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Then their (external) direct product:
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
is a ring. | Consider the structures $\struct {R_1, +_1}, \struct {R_2, +_2}, \ldots, \struct {R_n, +_n}$.
By the definition of a ring, these are all groups.
From External Direct Product of Groups is Group we have that the their external direct product:
:$\ds \struct {R, +} = \prod_{k \mathop = 1}^n \struct {R_k, +_k}$
is a group.
... | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be [[Definition:Ring (Abstract Algebra)|rings]].
Then their [[Definition:Ring Direct Product|(external) direct product]]:
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
is a [[Defini... | Consider the structures $\struct {R_1, +_1}, \struct {R_2, +_2}, \ldots, \struct {R_n, +_n}$.
By the definition of a [[Definition:Ring (Abstract Algebra)|ring]], these are all [[Definition:Group|groups]].
From [[External Direct Product of Groups is Group]] we have that the their [[Definition:External Direct Product|e... | External Direct Sum of Rings is Ring | https://proofwiki.org/wiki/External_Direct_Sum_of_Rings_is_Ring | https://proofwiki.org/wiki/External_Direct_Sum_of_Rings_is_Ring | [
"Direct Sums of Rings",
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Direct Product",
"Definition:Ring (Abstract Algebra)"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"External Direct Product of Groups is Group",
"Definition:External Direct Product",
"Definition:Group",
"Definition:Ring (Abstract Algebra)",
"Definition:Semigroup",
"External Direct Product of Semigroups",
"Definition:External Direct Product... |
proofwiki-3097 | External Direct Product of Ringoids is Ringoid | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be ringoids.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
Then $\struct {R, +, \circ}$ is a ringoid. | By definition of ringoid, $\circ_k$ distributes over $+_k$ for all $k = 1, 2, \ldots, n$.
Let $x, y, z \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ \paren {y + z}
| r = \tuple {x_1, x_2, \ldots, x_n} \circ \paren {\tuple {y_1, y_2, \ldots, y_n} + \tuple {z_1, z_2, \ldots, z_n} }
}}
{{eqn | r = \tuple {x_1, x_2... | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be [[Definition:Ringoid (Abstract Algebra)|ringoids]].
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their [[Definition:External Direct Product|external direct product]].
Then... | By definition of [[Definition:Ringoid (Abstract Algebra)|ringoid]], $\circ_k$ distributes over $+_k$ for all $k = 1, 2, \ldots, n$.
Let $x, y, z \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ \paren {y + z}
| r = \tuple {x_1, x_2, \ldots, x_n} \circ \paren {\tuple {y_1, y_2, \ldots, y_n} + \tuple {z_1, z_2, ... | External Direct Product of Ringoids is Ringoid | https://proofwiki.org/wiki/External_Direct_Product_of_Ringoids_is_Ringoid | https://proofwiki.org/wiki/External_Direct_Product_of_Ringoids_is_Ringoid | [
"External Direct Products",
"Distributive Operations"
] | [
"Definition:Ringoid (Abstract Algebra)",
"Definition:External Direct Product",
"Definition:Ringoid (Abstract Algebra)"
] | [
"Definition:Ringoid (Abstract Algebra)",
"Definition:Distributive Operation",
"Definition:Ringoid (Abstract Algebra)",
"Definition:Ringoid (Abstract Algebra)",
"Category:External Direct Products",
"Category:Distributive Operations"
] |
proofwiki-3098 | Ideal of External Direct Sum of Rings | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Let
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
be their direct product.
For each $k \in \closedint 1 n$, let:
:${R_k}' = \set {\tuple {x_1, \ldots, x_n} \in R: \forall j... | Let $y = \tuple {y_1, y_2, \ldots, y_n} \in R$.
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in R_k$
By definition of direct product, we have:
:$x \circ y = \tuple {x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}$
:$y \circ x = \tuple {y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}$
But we have:
:$\forall j \ne ... | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be [[Definition:Ring (Abstract Algebra)|rings]].
Let
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
be their [[Definition:Ring Direct Product|direct product]].
For each $k \in \c... | Let $y = \tuple {y_1, y_2, \ldots, y_n} \in R$.
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in R_k$
By definition of [[Definition:Ring Direct Product|direct product]], we have:
:$x \circ y = \tuple {x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}$
:$y \circ x = \tuple {y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \cir... | Ideal of External Direct Sum of Rings | https://proofwiki.org/wiki/Ideal_of_External_Direct_Sum_of_Rings | https://proofwiki.org/wiki/Ideal_of_External_Direct_Sum_of_Rings | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Direct Product",
"Definition:Ideal of Ring"
] | [
"Definition:Ring Direct Product",
"Definition:Ideal of Ring"
] |
proofwiki-3099 | Canonical Injection from Ideal of External Direct Sum of Rings | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \dotsc, \struct {R_n, +_n, \circ_n}$ be rings.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
For each $k \in \closedint 1 n$, let:
:$R'_k = \set {\tuple {x_1, \dotsc, x_n} \in R: \for... | From Ideal of External Direct Sum of Rings we have that $R'_k$ is an ideal of $R$, and thus a subring of $R$.
The result follows by application of Canonical Injection is Monomorphism.
{{qed}} | Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \dotsc, \struct {R_n, +_n, \circ_n}$ be [[Definition:Ring (Abstract Algebra)|rings]].
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their [[Definition:External Direct Product|external direct product]].
For each ... | From [[Ideal of External Direct Sum of Rings]] we have that $R'_k$ is an [[Definition:Ideal of Ring|ideal]] of $R$, and thus a [[Definition:Subring|subring]] of $R$.
The result follows by application of [[Canonical Injection is Monomorphism]].
{{qed}} | Canonical Injection from Ideal of External Direct Sum of Rings | https://proofwiki.org/wiki/Canonical_Injection_from_Ideal_of_External_Direct_Sum_of_Rings | https://proofwiki.org/wiki/Canonical_Injection_from_Ideal_of_External_Direct_Sum_of_Rings | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:External Direct Product",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Projection (Mapping Theory)",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Inverse Mapping",
"Definition:Restriction/Mapping"
] | [
"Ideal of External Direct Sum of Rings",
"Definition:Ideal of Ring",
"Definition:Subring",
"Canonical Injection is Monomorphism"
] |
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