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-15700 | Field Norm of Complex Number is Multiplicative Function | Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is a multiplicative function on $\C$. | {{begin-eqn}}
{{eqn | l = \map N {z_1 z_2}
| r = \cmod {z_1 z_2}^2
| c = Definition of $N$
}}
{{eqn | r = \paren {\cmod {z_1} \cmod {z_2} }^2
| c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | r = \cmod {z_1}^2 \cmod {z_2}^2
| c =
}}
{{eqn | r = \map N {z_1} \map N {z_2}
| c =... | Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
Then ... | {{begin-eqn}}
{{eqn | l = \map N {z_1 z_2}
| r = \cmod {z_1 z_2}^2
| c = Definition of $N$
}}
{{eqn | r = \paren {\cmod {z_1} \cmod {z_2} }^2
| c = [[Complex Modulus of Product of Complex Numbers]]
}}
{{eqn | r = \cmod {z_1}^2 \cmod {z_2}^2
| c =
}}
{{eqn | r = \map N {z_1} \map N {z_2}
|... | Field Norm of Complex Number is Multiplicative Function | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_Multiplicative_Function | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_Multiplicative_Function | [
"Field Norm of Complex Number"
] | [
"Definition:Complex Number",
"Definition:Field Norm of Complex Number",
"Definition:Complex Modulus",
"Definition:Multiplicative Function on Ring"
] | [
"Complex Modulus of Product of Complex Numbers",
"Definition:Multiplicative Function on Ring"
] |
proofwiki-15701 | Field Norm of Complex Number is not Norm | Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is not a norm on $\C$. | Proof by Counterexample:
Let $z_1 = z_2 = 1$.
Then:
{{begin-eqn}}
{{eqn | l = \map N {z_1 + z_2}
| r = \cmod {z_1 + z_2}^2
| c = Definition of $N$
}}
{{eqn | r = 2^2
| c =
}}
{{eqn | r = 4
| c =
}}
{{end-eqn}}
But:
{{begin-eqn}}
{{eqn | l = \map N {z_1} + \map N {z_2}
| r = \cmod {z_1}^2... | Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
Then ... | [[Proof by Counterexample]]:
Let $z_1 = z_2 = 1$.
Then:
{{begin-eqn}}
{{eqn | l = \map N {z_1 + z_2}
| r = \cmod {z_1 + z_2}^2
| c = Definition of $N$
}}
{{eqn | r = 2^2
| c =
}}
{{eqn | r = 4
| c =
}}
{{end-eqn}}
But:
{{begin-eqn}}
{{eqn | l = \map N {z_1} + \map N {z_2}
| r = \cm... | Field Norm of Complex Number is not Norm | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_not_Norm | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_not_Norm | [
"Field Norm of Complex Number"
] | [
"Definition:Complex Number",
"Definition:Field Norm of Complex Number",
"Definition:Complex Modulus",
"Definition:Norm/Ring"
] | [
"Proof by Counterexample",
"Definition:Triangle Inequality",
"Definition:Norm/Ring",
"Category:Field Norm of Complex Number"
] |
proofwiki-15702 | Units of 5th Cyclotomic Ring | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$. | Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Thus by definition:
:$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eq... | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
The [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$. | Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Thus by definition:
:$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$... | Units of 5th Cyclotomic Ring | https://proofwiki.org/wiki/Units_of_5th_Cyclotomic_Ring | https://proofwiki.org/wiki/Units_of_5th_Cyclotomic_Ring | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Unit of Ring"
] | [
"Definition:Field Norm of Complex Number",
"Definition:Unit of Ring",
"Field Norm on 5th Cyclotomic Ring",
"Field Norm on 5th Cyclotomic Ring"
] |
proofwiki-15703 | Field Norm on 5th Cyclotomic Ring | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
Let $\alpha = a + i b \sqrt 5$ be an arbitrary element of $\Z \sqbrk {i \sqrt 5}$.
The field norm of $\alpha$ is given by:
:$\map N \alpha = a^2 + 5 b^2$ | {{begin-eqn}}
{{eqn | l = \map N \alpha
| r = \cmod \alpha^2
| c = {{Defof|Field Norm of Complex Number}}
}}
{{eqn | r = \paren {\sqrt {a^2 + \paren {b \sqrt 5}^2} }^2
| c = {{Defof|Complex Modulus}}
}}
{{eqn | r = a^2 + 5 b^2
| c =
}}
{{end-eqn}}
{{Qed}} | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
Let $\alpha = a + i b \sqrt 5$ be an arbitrary [[Definition:Element|element]] of $\Z \sqbrk {i \sqrt 5}$.
The [[Definition:Field Norm of Complex Number|field norm]] of $\alpha$ is given by:
:$\map N \a... | {{begin-eqn}}
{{eqn | l = \map N \alpha
| r = \cmod \alpha^2
| c = {{Defof|Field Norm of Complex Number}}
}}
{{eqn | r = \paren {\sqrt {a^2 + \paren {b \sqrt 5}^2} }^2
| c = {{Defof|Complex Modulus}}
}}
{{eqn | r = a^2 + 5 b^2
| c =
}}
{{end-eqn}}
{{Qed}} | Field Norm on 5th Cyclotomic Ring | https://proofwiki.org/wiki/Field_Norm_on_5th_Cyclotomic_Ring | https://proofwiki.org/wiki/Field_Norm_on_5th_Cyclotomic_Ring | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Element",
"Definition:Field Norm of Complex Number"
] | [] |
proofwiki-15704 | 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3 | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
There are no elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose field norm is either $2$ or $3$. | Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \map N z
| r = 2
| c =
}}
{{eqn | ll= \leadsto
| l = x^2 + 5 y^2
| r = 2
| c = Field Norm on 5th Cyclotomic Ring
}}
{{eqn | ll= \leadsto
| l = x^2
| r = 2
... | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
There are no [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose [[Definition:Field Norm of Complex Number|field norm]] is either $2$ or $3$. | Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \map N z
| r = 2
| c =
}}
{{eqn | ll= \leadsto
| l = x^2 + 5 y^2
| r = 2
| c = [[Field Norm on 5th Cyclotomic Ring]]
}}
... | 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3 | https://proofwiki.org/wiki/5th_Cyclotomic_Ring_has_no_Elements_with_Field_Norm_of_2_or_3 | https://proofwiki.org/wiki/5th_Cyclotomic_Ring_has_no_Elements_with_Field_Norm_of_2_or_3 | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Element",
"Definition:Field Norm of Complex Number"
] | [
"Definition:Field Norm of Complex Number",
"Field Norm on 5th Cyclotomic Ring",
"Definition:Integer",
"Square Root of Prime is Irrational"
] |
proofwiki-15705 | Irreducible Elements of 5th Cyclotomic Ring | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The following elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are irreducible:
:$2$
:$3$
:$1 + i \sqrt 5$
:$1 - i \sqrt 5$ | {{TheoremWanted|For the concept of irreducibility to be defined, it needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an integral domain.}}
Let $z = x + i y$ be an element of $\Z \sqbrk {i \sqrt 5}$ in the set $S$, where:
:$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$
Let $z$ hav... | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
The following [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are [[Definition:Irreducible Element of Ring|irreducible]]:
:$2$
:$3$
:$1 + i \sqrt 5$
:$1 - i \sqrt 5$ | {{TheoremWanted|For the concept of [[Definition:Irreducible Element of Ring|irreducibility]] to be defined, it needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an [[Definition:Integral Domain|integral domain]].}}
Let $z = x + i y$ be an [[Definition:Element|element]] of $\Z \sqbr... | Irreducible Elements of 5th Cyclotomic Ring | https://proofwiki.org/wiki/Irreducible_Elements_of_5th_Cyclotomic_Ring | https://proofwiki.org/wiki/Irreducible_Elements_of_5th_Cyclotomic_Ring | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Element",
"Definition:Irreducible Element of Ring"
] | [
"Definition:Irreducible Element of Ring",
"Definition:Integral Domain",
"Definition:Element",
"Definition:Set",
"Definition:Trivial Factorization/Non-Trivial Factorization",
"Definition:Unit of Ring",
"Definition:Field Norm of Complex Number",
"Field Norm on 5th Cyclotomic Ring",
"Field Norm on 5th ... |
proofwiki-15706 | Value of Field Norm on 5th Cyclotomic Ring is Integer | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
Let $\alpha = a + i b \sqrt 5$ be an arbitrary element of $\Z \sqbrk {i \sqrt 5}$.
Let $\map N \alpha$ denoted the field norm of $\alpha$.
Then $\map N \alpha$ is an integer. | From Field Norm on 5th Cyclotomic Ring:
:$\map N \alpha = a^2 + 5 b^2$
From the definition of the $5$th cyclotomic ring:
:$\Z \sqbrk {i \sqrt 5} = \set {a + i \sqrt 5 b: a, b \in \Z}$
That is, both $a$ and $b$ are integers.
Hence $a^2 + 5 b^2$ is also an integer.
{{Qed}} | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
Let $\alpha = a + i b \sqrt 5$ be an arbitrary [[Definition:Element|element]] of $\Z \sqbrk {i \sqrt 5}$.
Let $\map N \alpha$ denoted the [[Definition:Field Norm of Complex Number|field norm]] of $\alph... | From [[Field Norm on 5th Cyclotomic Ring]]:
:$\map N \alpha = a^2 + 5 b^2$
From the definition of the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]]:
:$\Z \sqbrk {i \sqrt 5} = \set {a + i \sqrt 5 b: a, b \in \Z}$
That is, both $a$ and $b$ are [[Definition:Integer|integers]].
Hence $a^2 + 5 b^2$ is also an [[... | Value of Field Norm on 5th Cyclotomic Ring is Integer | https://proofwiki.org/wiki/Value_of_Field_Norm_on_5th_Cyclotomic_Ring_is_Integer | https://proofwiki.org/wiki/Value_of_Field_Norm_on_5th_Cyclotomic_Ring_is_Integer | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Element",
"Definition:Field Norm of Complex Number",
"Definition:Integer"
] | [
"Field Norm on 5th Cyclotomic Ring",
"Cyclotomic Ring/Examples/5th",
"Definition:Integer",
"Definition:Integer"
] |
proofwiki-15707 | Elements of 5th Cyclotomic Ring with Field Norm 1 | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The only elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose field norm equals $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$. | From Units of 5th Cyclotomic Ring, $1$ and $-1$ are the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z = 1$.
Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \map N z
| r = 1... | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
The only [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose [[Definition:Field Norm of Complex Number|field norm]] equals $1$ are the [[Definition:Unit of Ring|units]]... | From [[Units of 5th Cyclotomic Ring]], $1$ and $-1$ are the only [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z ... | Elements of 5th Cyclotomic Ring with Field Norm 1 | https://proofwiki.org/wiki/Elements_of_5th_Cyclotomic_Ring_with_Field_Norm_1 | https://proofwiki.org/wiki/Elements_of_5th_Cyclotomic_Ring_with_Field_Norm_1 | [
"Cyclotomic Rings"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Element",
"Definition:Field Norm of Complex Number",
"Definition:Unit of Ring"
] | [
"Units of 5th Cyclotomic Ring",
"Definition:Unit of Ring",
"Definition:Field Norm of Complex Number",
"Field Norm on 5th Cyclotomic Ring"
] |
proofwiki-15708 | 5th Cyclotomic Ring is not a Unique Factorization Domain | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
Then $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is not a unique factorization domain.
The following elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are irreducible:
:$2$
:$3$
:$1 + i \sqrt 5$
:$1 - i \sqrt 5$ | By definition, a unique factorization domain $D$ is an integral domain with the properties that:
:For all $x \in D$ such that $x$ is non-zero and not a unit of $D$:
::$(1): \quad x$ possesses a complete factorization in $D$
::$(2): \quad$ Any two complete factorizations of $x$ are equivalent.
A complete factorization i... | Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
Then $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is not a [[Definition:Unique Factorization Domain|unique factorization domain]].
The following [[Definition:Element|elements]] of $\struct {\Z \sqbrk {... | By definition, a [[Definition:Unique Factorization Domain|unique factorization domain]] $D$ is an [[Definition:Integral Domain|integral domain]] with the properties that:
:For all $x \in D$ such that $x$ is non-[[Definition:Ring Zero|zero]] and not a [[Definition:Unit of Ring|unit]] of $D$:
::$(1): \quad x$ possesses... | 5th Cyclotomic Ring is not a Unique Factorization Domain | https://proofwiki.org/wiki/5th_Cyclotomic_Ring_is_not_a_Unique_Factorization_Domain | https://proofwiki.org/wiki/5th_Cyclotomic_Ring_is_not_a_Unique_Factorization_Domain | [
"Cyclotomic Rings",
"Unique Factorization Domains"
] | [
"Cyclotomic Ring/Examples/5th",
"Definition:Unique Factorization Domain",
"Definition:Element",
"Definition:Irreducible Element of Ring"
] | [
"Definition:Unique Factorization Domain",
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unit of Ring",
"Definition:Complete Factorization",
"Definition:Complete Factorization",
"Definition:Equivalent Factorizations",
"Definition:Complete Factorization",
"Definition:Tidy Factoriza... |
proofwiki-15709 | Quotient of Ring of Polynomials in Ring Element on Integral Domain by that Polynomial is that Domain | Let $\struct {D, +, \times}$ be an integral domain.
Let $X \in R$ be transcencental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Let $D \sqbrk X / \ideal X$ denote the quotient ring of $D \sqbrk X$ by the ideal of $D$ generated by $X$.
Then:
:$D \sqbrk X / \ideal X \cong D$ | Let $n \in \Z_{> 0}$ be arbitrary.
Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a polynomial over $D$ in $X$.
Consider the mapping $\phi: D \sqbrk X \to D$ defined as:
:$\forall P \in D \sqbrk X: \map \phi P = a_0$
Let:
:$P_1 = a_m X^m + a_{m - 1} X^{m - 1} + \dotsb + a_1 X + a_0$
:$P_2 = b_m X^m +... | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcencental over $D$]].
Let $D \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $D$.
Let $D \sqbrk X / \ideal X$ d... | Let $n \in \Z_{> 0}$ be arbitrary.
Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a [[Definition:Polynomial over Ring in One Variable|polynomial over $D$ in $X$]].
Consider the [[Definition:Mapping|mapping]] $\phi: D \sqbrk X \to D$ defined as:
:$\forall P \in D \sqbrk X: \map \phi P = a_0$
Let:... | Quotient of Ring of Polynomials in Ring Element on Integral Domain by that Polynomial is that Domain | https://proofwiki.org/wiki/Quotient_of_Ring_of_Polynomials_in_Ring_Element_on_Integral_Domain_by_that_Polynomial_is_that_Domain | https://proofwiki.org/wiki/Quotient_of_Ring_of_Polynomials_in_Ring_Element_on_Integral_Domain_by_that_Polynomial_is_that_Domain | [
"Polynomial Rings",
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Quotient Ring",
"Definition:Ideal of Ring"
] | [
"Definition:Polynomial over Ring/One Variable",
"Definition:Mapping",
"Definition:Ring Homomorphism",
"First Isomorphism Theorem/Rings"
] |
proofwiki-15710 | Polynomials in Integers with Even Constant Term forms Ideal | Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Let $S \subseteq \Z \sqbrk X$ be the set of polynomials over $\Z$ in $X$ which have a constant term which is even.
Then $S$ is an ideal of $\Z \sqbrk X$. | For example, $X + 2$ is a polynomials over $\Z$ in $X$ with an even constant term.
So $S$ is not empty.
Let $P_1 = \ds \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \ds \sum_{k \mathop = 0}^n b_k X^k$ be elements of $S$.
We have:
{{begin-eqn}}
{{eqn | l = P_1 - P_2
| r = \sum_{k \mathop = 0}^n a_k X^k + \sum_{k \mat... | Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
Let $S \subseteq \Z \sqbrk X$ be the [[Definition:Set|set]] of [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] which have a [[Definition:Constant Term of Polynomial|constan... | For example, $X + 2$ is a [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] with an [[Definition:Even Integer|even]] [[Definition:Constant Term of Polynomial|constant term]].
So $S$ is not [[Definition:Empty Set|empty]].
Let $P_1 = \ds \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \ds \su... | Polynomials in Integers with Even Constant Term forms Ideal | https://proofwiki.org/wiki/Polynomials_in_Integers_with_Even_Constant_Term_forms_Ideal | https://proofwiki.org/wiki/Polynomials_in_Integers_with_Even_Constant_Term_forms_Ideal | [
"Polynomial Rings",
"Integers"
] | [
"Definition:Ring of Polynomials in Ring Element",
"Definition:Set",
"Definition:Polynomial over Ring/One Variable",
"Definition:Constant Term of Polynomial",
"Definition:Even Integer",
"Definition:Ideal of Ring"
] | [
"Definition:Polynomial over Ring/One Variable",
"Definition:Even Integer",
"Definition:Constant Term of Polynomial",
"Definition:Empty Set",
"Definition:Element",
"Definition:Constant Term of Polynomial",
"Definition:Even Integer",
"Definition:Constant Term of Polynomial",
"Definition:Even Integer",... |
proofwiki-15711 | Polynomials in Integers is not Principal Ideal Domain | Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Then $\Z \sqbrk X$ is not a principal ideal domain. | Let $J$ be the ideal formed from the set of polynomials over $\Z$ in $X$ which have a constant term which is even.
From Polynomials in Integers with Even Constant Term forms Ideal, $J$ is indeed an ideal.
{{AimForCont}} $J$ is a principal ideal of $\Z \sqbrk X$ such that $J = \ideal f$.
But $2 \in J$, and so $2$ is a m... | Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
Then $\Z \sqbrk X$ is not a [[Definition:Principal Ideal Domain|principal ideal domain]]. | Let $J$ be the [[Definition:Ideal of Ring|ideal]] formed from the [[Definition:Set|set]] of [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] which have a [[Definition:Constant Term of Polynomial|constant term]] which is [[Definition:Even Integer|even]].
From [[Polynomials in Integers wi... | Polynomials in Integers is not Principal Ideal Domain | https://proofwiki.org/wiki/Polynomials_in_Integers_is_not_Principal_Ideal_Domain | https://proofwiki.org/wiki/Polynomials_in_Integers_is_not_Principal_Ideal_Domain | [
"Polynomial Rings",
"Integers",
"Principal Ideal Domains"
] | [
"Definition:Ring of Polynomials in Ring Element",
"Definition:Principal Ideal Domain"
] | [
"Definition:Ideal of Ring",
"Definition:Set",
"Definition:Polynomial over Ring/One Variable",
"Definition:Constant Term of Polynomial",
"Definition:Even Integer",
"Polynomials in Integers with Even Constant Term forms Ideal",
"Definition:Ideal of Ring",
"Definition:Principal Ideal of Ring",
"Definit... |
proofwiki-15712 | Polynomials in Integers is Unique Factorization Domain | Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Then $\Z \sqbrk X$ is a unique factorization domain. | We have that Integers form Unique Factorization Domain.
The result follows from Gauss's Lemma on Unique Factorization Domains.
{{qed}} | Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
Then $\Z \sqbrk X$ is a [[Definition:Unique Factorization Domain|unique factorization domain]]. | We have that [[Integers form Unique Factorization Domain]].
The result follows from [[Gauss's Lemma on Unique Factorization Domains]].
{{qed}} | Polynomials in Integers is Unique Factorization Domain | https://proofwiki.org/wiki/Polynomials_in_Integers_is_Unique_Factorization_Domain | https://proofwiki.org/wiki/Polynomials_in_Integers_is_Unique_Factorization_Domain | [
"Polynomial Rings",
"Integers",
"Unique Factorization Domains"
] | [
"Definition:Ring of Polynomials in Ring Element",
"Definition:Unique Factorization Domain"
] | [
"Integers form Unique Factorization Domain",
"Gauss's Lemma on Unique Factorization Domains"
] |
proofwiki-15713 | Ideal of Ring of Polynomials over Field has Unique Monic Polynomial forming Principal Ideal | Let $F$ be a field.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $J$ be a non-null ideal of $F \sqbrk X$.
Then there exists exactly one monic polynomial $f \in F \sqbrk X$ such that:
:$J = \ideal f$
where $\ideal f$ is the principal ideal generated by $f$ in $F \sqbrk X$. | {{MissingLinks}}
{{Proofread}} | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $F$.
Let $J$ be a non-[[Definition:Null Ideal|null]] [[Definition:Ideal of Ring|ideal]] of $F \sqbrk X$.
Then there exists [[Definition:Unique|exactl... | {{MissingLinks}}
{{Proofread}} | Ideal of Ring of Polynomials over Field has Unique Monic Polynomial forming Principal Ideal | https://proofwiki.org/wiki/Ideal_of_Ring_of_Polynomials_over_Field_has_Unique_Monic_Polynomial_forming_Principal_Ideal | https://proofwiki.org/wiki/Ideal_of_Ring_of_Polynomials_over_Field_has_Unique_Monic_Polynomial_forming_Principal_Ideal | [
"Principal Ideals of Rings",
"Polynomial Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Null Ideal",
"Definition:Ideal of Ring",
"Definition:Unique",
"Definition:Monic Polynomial",
"Definition:Principal Ideal of Ring"
] | [] |
proofwiki-15714 | Gaussian Integers form Principal Ideal Domain | The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
forms a principal ideal domain. | Follows immediately from:
:Gaussian Integers form Euclidean Domain
:Euclidean Domain is Principal Ideal Domain.
{{qed}} | The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
:$\struct {\Z \sqbrk i, +, \times}$
forms a [[Definition:Principal Ideal Domain|principal ideal domain]]. | Follows immediately from:
:[[Gaussian Integers form Euclidean Domain]]
:[[Euclidean Domain is Principal Ideal Domain]].
{{qed}} | Gaussian Integers form Principal Ideal Domain/Proof 2 | https://proofwiki.org/wiki/Gaussian_Integers_form_Principal_Ideal_Domain | https://proofwiki.org/wiki/Gaussian_Integers_form_Principal_Ideal_Domain/Proof_2 | [
"Gaussian Integers form Principal Ideal Domain",
"Gaussian Integers",
"Examples of Principal Ideal Domains"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Principal Ideal Domain"
] | [
"Gaussian Integers form Euclidean Domain",
"Euclidean Domain is Principal Ideal Domain"
] |
proofwiki-15715 | Polynomial X^2 + 1 is Irreducible in Ring of Real Polynomials | Let $\R \sqbrk X$ be the ring of polynomials in $X$ over the real numbers $\R$.
Then the polynomial $X^2 + 1$ is an irreducible element of $\R \sqbrk X$. | {{AimForCont}} $x^2 + 1$ has a non-trivial factorization in $\R \sqbrk X$.
Then:
:$\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$
and from the Polynomial Factor Theorem:
:$\alpha^2 + 1 = 0$
But that means:
:$\alpha^2 = -1$
and such an $\alpha$ does not exist in $\R$.
Hence the result by Proof by ... | Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over the [[Definition:Real Number|real numbers]] $\R$.
Then the [[Definition:Polynomial in Ring Element|polynomial]] $X^2 + 1$ is an [[Definition:Irreducible Element of Ring|irreducible element]] of $\R \sqbrk X$. | {{AimForCont}} $x^2 + 1$ has a [[Definition:Non-Trivial Factorization|non-trivial factorization]] in $\R \sqbrk X$.
Then:
:$\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$
and from the [[Polynomial Factor Theorem]]:
:$\alpha^2 + 1 = 0$
But that means:
:$\alpha^2 = -1$
and such an $\alpha$ doe... | Polynomial X^2 + 1 is Irreducible in Ring of Real Polynomials | https://proofwiki.org/wiki/Polynomial_X^2_+_1_is_Irreducible_in_Ring_of_Real_Polynomials | https://proofwiki.org/wiki/Polynomial_X^2_+_1_is_Irreducible_in_Ring_of_Real_Polynomials | [
"Irreducible Elements of Rings",
"Polynomial Theory"
] | [
"Definition:Ring of Polynomials in Ring Element",
"Definition:Real Number",
"Definition:Polynomial in Ring Element",
"Definition:Irreducible Element of Ring"
] | [
"Definition:Trivial Factorization/Non-Trivial Factorization",
"Polynomial Factor Theorem",
"Proof by Contradiction"
] |
proofwiki-15716 | Matrix Multiplication is not Commutative | Let $R$ be a ring with unity.
Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n \ne 1$.
Let $\map {\MM_R} n$ denote the $n \times n$ matrix space over $R$.
Then (conventional) matrix multiplication over $\map {\MM_R} n$ is not commutative:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B}... | The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$ | Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $n \ne 1$.
Let $\map {\MM_R} n$ denote the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
Then [[Definition:Matrix Product (Convention... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$ | Matrix Multiplication is not Commutative | https://proofwiki.org/wiki/Matrix_Multiplication_is_not_Commutative | https://proofwiki.org/wiki/Matrix_Multiplication_is_not_Commutative | [
"Conventional Matrix Multiplication",
"Examples of Commutative Operations",
"Proofs by Induction",
"Matrix Multiplication is not Commutative"
] | [
"Definition:Ring with Unity",
"Definition:Strictly Positive/Integer",
"Definition:Matrix Space",
"Definition:Matrix Product (Conventional)",
"Definition:Commutative/Operation",
"Definition:Commutative Ring"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-15717 | Matrix Multiplication on Square Matrices over Trivial Ring is Commutative | Let $\struct {R, +, \circ}$ be the trivial ring over an underlying set.
Let $\map {\MM_R} n$ denote the $n \times n$ matrix space over $R$.
Then (conventional) matrix multiplication is commutative over $\map {\MM_R} n$:
:$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$ | Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be order $n$ square matrices over $R$.
By definition of matrix multiplication, $\mathbf A \mathbf B = \mathbf C = \sqbrk c_n$ where:
:$\ds \forall i \in \closedint 1 n, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
But by definiti... | Let $\struct {R, +, \circ}$ be the [[Definition:Trivial Ring|trivial ring]] over an [[Definition:Underlying Set of Structure|underlying set]].
Let $\map {\MM_R} n$ denote the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
Then [[Definition:Matrix Product (Conventional)|(conventional) matrix multipli... | Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be [[Definition:Square Matrix|order $n$ square matrices]] over $R$.
By definition of [[Definition:Matrix Product (Conventional)|matrix multiplication]], $\mathbf A \mathbf B = \mathbf C = \sqbrk c_n$ where:
:$\ds \forall i \in \closedint 1 n, j \in \closedint... | Matrix Multiplication on Square Matrices over Trivial Ring is Commutative | https://proofwiki.org/wiki/Matrix_Multiplication_on_Square_Matrices_over_Trivial_Ring_is_Commutative | https://proofwiki.org/wiki/Matrix_Multiplication_on_Square_Matrices_over_Trivial_Ring_is_Commutative | [
"Conventional Matrix Multiplication",
"Trivial Rings"
] | [
"Definition:Trivial Ring",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Matrix Space",
"Definition:Matrix Product (Conventional)",
"Definition:Commutative/Operation"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Matrix Product (Conventional)",
"Definition:Trivial Ring",
"Definition:Ring Zero",
"Definition:Zero Matrix",
"Definition:Zero Matrix",
"Definition:Commutative/Operation",
"Category:Conventional Matrix Multiplication",
"Category:Trivial Rings"
] |
proofwiki-15718 | Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle | Let $y$ be a smooth curve, embedded in $2$-dimensional Euclidean space.
Let $y$ have a total length of $l$.
Let it be contained in the upper halfplane with an exception of endpoints, which are on the x-axis and are given.
Suppose, $y$, together with a line segment connecting $y$'s endpoints, maximizes the enclosed area... | {{WLOG}}, we choose our point of reference such that $y$ intersect x-axis at points $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.
The area below the curve $y$ is a functional of the following form:
:$\ds A \sqbrk y = \int_{-a}^a y \rd x$
Furthermore, $y$ has to satisfy the following conditions:
:$\map y {-a} ... | Let $y$ be a [[Definition:Smooth Curve|smooth curve]], embedded in [[Definition:Dimension of Vector Space|$2$-dimensional]] [[Definition:Real Euclidean Space|Euclidean space]].
Let $y$ have a total [[Definition:Length of Curve|length]] of $l$.
Let it be contained in the upper halfplane with an exception of [[Definiti... | {{WLOG}}, we choose our [[Definition:Point of Reference|point of reference]] such that $y$ intersect x-[[Definition:Coordinate Axis|axis]] at [[Definition:Point|points]] $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.
The [[Definition:Area|area]] below the [[Definition:Curve|curve]] $y$ is a [[Definition:Real F... | Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle | https://proofwiki.org/wiki/Area_between_Smooth_Curve_and_Line_with_Fixed_Endpoints_is_Maximized_by_Arc_of_Circle | https://proofwiki.org/wiki/Area_between_Smooth_Curve_and_Line_with_Fixed_Endpoints_is_Maximized_by_Arc_of_Circle | [
"Calculus of Variations",
"Isoperimetrical Problems"
] | [
"Definition:Smooth Curve",
"Definition:Dimension of Vector Space",
"Definition:Euclidean Space/Real",
"Definition:Arc Length",
"Definition:Directed Smooth Curve/Endpoints",
"Definition:Axis/Coordinate Axes",
"Definition:Line/Segment",
"Definition:Directed Smooth Curve/Endpoints",
"Definition:Area",
... | [
"Definition:Frame of Reference/Point of Reference",
"Definition:Axis/Coordinate Axes",
"Definition:Point",
"Definition:Area",
"Definition:Line/Curve",
"Definition:Functional/Real",
"Simplest Variational Problem with Subsidiary Conditions",
"Definition:Constant Mapping",
"Definition:Functional/Real",... |
proofwiki-15719 | Ring Subtraction equals Zero iff Elements are Equal | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$
Then:
:$\forall a, b \in R: a - b = 0_R \iff a = b$
where $a - b$ denotes ring subtraction. | {{begin-eqn}}
{{eqn | l = a - b
| r = 0_R
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a + \paren {-b}
| r = 0_R
| c = {{Defof|Ring Subtraction}}
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {a + \paren {-b} } + b
| r = 0_R + b
| c = Cancellation Laws
}}
{{eqn | ll= \leadst... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$
Then:
:$\forall a, b \in R: a - b = 0_R \iff a = b$
where $a - b$ denotes [[Definition:Ring Subtraction|ring subtraction]]. | {{begin-eqn}}
{{eqn | l = a - b
| r = 0_R
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a + \paren {-b}
| r = 0_R
| c = {{Defof|Ring Subtraction}}
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {a + \paren {-b} } + b
| r = 0_R + b
| c = [[Cancellation Laws]]
}}
{{eqn | ll= \le... | Ring Subtraction equals Zero iff Elements are Equal | https://proofwiki.org/wiki/Ring_Subtraction_equals_Zero_iff_Elements_are_Equal | https://proofwiki.org/wiki/Ring_Subtraction_equals_Zero_iff_Elements_are_Equal | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Subtraction/Ring"
] | [
"Cancellation Laws"
] |
proofwiki-15720 | Preordering of Products under Operation Compatible with Preordering | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\precsim$ be a preordering on $S$.
Then $\precsim$ is compatible with $\circ$ {{iff}}:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$ | By definition, $\precsim$ is compatible with $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| l = x \precsim y
| o = \implies
| r = \paren {x \circ z} \precsim \paren {y \circ z}
}}
{{eqn | l = x \precsim y
| o = \implies
| r = \paren {z \circ x} \precsim \paren {z \circ ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\precsim$ be a [[Definition:Preordering|preordering]] on $S$.
Then $\precsim$ is [[Definition:Relation Compatible with Operation|compatible with $\circ$]] {{iff}}:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \... | By definition, $\precsim$ is [[Definition:Relation Compatible with Operation|compatible with $\circ$]] {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| l = x \precsim y
| o = \implies
| r = \paren {x \circ z} \precsim \paren {y \circ z}
}}
{{eqn | l = x \precsim y
| o = \implies
... | Preordering of Products under Operation Compatible with Preordering | https://proofwiki.org/wiki/Preordering_of_Products_under_Operation_Compatible_with_Preordering | https://proofwiki.org/wiki/Preordering_of_Products_under_Operation_Compatible_with_Preordering | [
"Preorder Theory",
"Compatible Relations"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Preordering",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation"
] |
proofwiki-15721 | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1 | :$\exists x \in \Z_{>0}: p \nmid x, x \ge \dfrac {p + 1} 2$ | Let $x = p + 1$.
Then $p \nmid x$ and:
:$x = p + 1 > p > 0$
{{qed}}
Category:P-adic Norm not Complete on Rational Numbers
klt8vnjgaediwd7qt6bk8eururfgx8p | :$\exists x \in \Z_{>0}: p \nmid x, x \ge \dfrac {p + 1} 2$ | Let $x = p + 1$.
Then $p \nmid x$ and:
:$x = p + 1 > p > 0$
{{qed}}
[[Category:P-adic Norm not Complete on Rational Numbers]]
klt8vnjgaediwd7qt6bk8eururfgx8p | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_1 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_1 | [
"P-adic Norm not Complete on Rational Numbers"
] | [] | [
"Category:P-adic Norm not Complete on Rational Numbers"
] |
proofwiki-15722 | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 5 | :$\ds \lim_{n \mathop \to \infty} {x_n}^k = a$ in $\struct {\Q, \norm {\,\cdot\,}_p}$ | By assumption:
:$\forall n \in \N: p^n \divides \paren { {x_n}^k - a}$
By the definition of the $p$-adic norm:
:$\forall n \in \N: \norm { {x_n}^k - a}_p \le \dfrac 1 {p^n}$
By Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
By Squeeze Theorem for Real Sequences:
:$\ds ... | :$\ds \lim_{n \mathop \to \infty} {x_n}^k = a$ in $\struct {\Q, \norm {\,\cdot\,}_p}$ | By assumption:
:$\forall n \in \N: p^n \divides \paren { {x_n}^k - a}$
By the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
:$\forall n \in \N: \norm { {x_n}^k - a}_p \le \dfrac 1 {p^n}$
By [[Sequence of Powers of Number less than One]]:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
By [[Squeez... | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 5 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_5 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_5 | [
"P-adic Norm not Complete on Rational Numbers"
] | [] | [
"Definition:P-adic Norm",
"Sequence of Powers of Number less than One",
"Squeeze Theorem/Sequences/Real Numbers",
"Definition:Convergent Sequence/Normed Division Ring",
"Category:P-adic Norm not Complete on Rational Numbers"
] |
proofwiki-15723 | Characterisation of Cauchy Sequence in Non-Archimedean Norm/Corollary 1 | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\sequence {x_n}$ be a sequence of integers such that:
:$\forall n: x_{n + 1} \equiv x_n \pmod {p^n}$
Then:
:$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$. | By hypothesis:
:$\forall n \in \N: p^n \divides \paren {x_{n + 1} - x_n}$
By the definition of the $p$-adic norm:
:$\forall n \in \N: \norm {x_{n + 1} - x_n}_p \le \dfrac 1 {p^n}$
From Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
From the Squeeze Theorem for Real Seq... | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Integer|integers]] such that:
:$\forall n: x_{n + 1} \equiv x_n \pm... | By [[Definition:Hypothesis|hypothesis]]:
:$\forall n \in \N: p^n \divides \paren {x_{n + 1} - x_n}$
By the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
:$\forall n \in \N: \norm {x_{n + 1} - x_n}_p \le \dfrac 1 {p^n}$
From [[Sequence of Powers of Number less than One]]:
:$\ds \lim_{n \mathop \to \infty... | Characterisation of Cauchy Sequence in Non-Archimedean Norm/Corollary 1 | https://proofwiki.org/wiki/Characterisation_of_Cauchy_Sequence_in_Non-Archimedean_Norm/Corollary_1 | https://proofwiki.org/wiki/Characterisation_of_Cauchy_Sequence_in_Non-Archimedean_Norm/Corollary_1 | [
"Characterisation of Cauchy Sequence in Non-Archimedean Norm"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Sequence",
"Definition:Integer",
"Definition:Cauchy Sequence/Normed Division Ring"
] | [
"Definition:Hypothesis",
"Definition:P-adic Norm",
"Sequence of Powers of Number less than One",
"Squeeze Theorem/Sequences/Real Numbers",
"P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers",
"Definition:P-adic Norm",
"Definition:Non-Archimedean/Norm (Division Ring)",
"Characterisation o... |
proofwiki-15724 | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2 | :$a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$ | Since $x, p > 0$ then $a > 0$.
{{AimForCont}} for some $c \in \Z:c^k = a$.
Since $c^k \in \Z$, by Nth Root of Integer is Integer or Irrational then:
:$c \in \Z$
Suppose $k$ is odd.
Since $a > 0$, by Odd Power Function is Strictly Increasing then $c > 0$
Hence $a = \size c^k$
On the other hand, suppose $k$ is even, that... | :$a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$ | Since $x, p > 0$ then $a > 0$.
{{AimForCont}} for some $c \in \Z:c^k = a$.
Since $c^k \in \Z$, by [[Nth Root of Integer is Integer or Irrational]] then:
:$c \in \Z$
Suppose $k$ is [[Definition:Odd Integer|odd]].
Since $a > 0$, by [[Odd Power Function is Strictly Increasing]] then $c > 0$
Hence $a = \size c^k$
On... | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_2 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_2 | [
"P-adic Norm not Complete on Rational Numbers"
] | [] | [
"Nth Root of Integer is Integer or Irrational",
"Definition:Odd Integer",
"Odd Power Function is Strictly Increasing",
"Definition:Even Integer",
"Equivalence of Definitions of Absolute Value Function",
"Difference of Two Powers",
"Definition:Factor",
"Definition:Factor",
"Definition:Contradiction",... |
proofwiki-15725 | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 3 | :$\map f {x_1} \equiv 0 \pmod p$ | {{begin-eqn}}
{{eqn | l = \map f {x_1}
| r = x_1^k - \paren {x_1^k + p}
}}
{{eqn | r = \paren {x_1^k - x_1^k} - p
}}
{{eqn | r = -p
}}
{{eqn | o = \equiv
| r = 0 \pmod p
}}
{{end-eqn}}
{{qed}}
Category:P-adic Norm not Complete on Rational Numbers
4y313yewjfwjz6avent2oo6zfqueega | :$\map f {x_1} \equiv 0 \pmod p$ | {{begin-eqn}}
{{eqn | l = \map f {x_1}
| r = x_1^k - \paren {x_1^k + p}
}}
{{eqn | r = \paren {x_1^k - x_1^k} - p
}}
{{eqn | r = -p
}}
{{eqn | o = \equiv
| r = 0 \pmod p
}}
{{end-eqn}}
{{qed}}
[[Category:P-adic Norm not Complete on Rational Numbers]]
4y313yewjfwjz6avent2oo6zfqueega | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 3 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_3 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_3 | [
"P-adic Norm not Complete on Rational Numbers"
] | [] | [
"Category:P-adic Norm not Complete on Rational Numbers"
] |
proofwiki-15726 | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 4 | :$\map {f'} {x_1} \not \equiv 0 \pmod p$ | By Euclid's Lemma for Prime Divisors:
:$p \nmid k x_1^{k - 1}$
Hence:
:$k x_1^{k - 1} \not \equiv 0 \mod p$
The formal derivative $\map {f'} X \in \Z \sqbrk X$ of $\map f X$ is by definition:
:$k X^{k - 1}$
Then:
:$\map {f'} {x_1} = k x_1^{k - 1} \not \equiv 0 \pmod p$
{{qed}}
Category:P-adic Norm not Complete on Ratio... | :$\map {f'} {x_1} \not \equiv 0 \pmod p$ | By [[Euclid's Lemma for Prime Divisors]]:
:$p \nmid k x_1^{k - 1}$
Hence:
:$k x_1^{k - 1} \not \equiv 0 \mod p$
The [[Definition:Formal Derivative of Polynomial|formal derivative]] $\map {f'} X \in \Z \sqbrk X$ of $\map f X$ is by definition:
:$k X^{k - 1}$
Then:
:$\map {f'} {x_1} = k x_1^{k - 1} \not \equiv 0 \pmod... | P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 4 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_4 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2/Lemma_4 | [
"P-adic Norm not Complete on Rational Numbers"
] | [] | [
"Euclid's Lemma for Prime Divisors",
"Definition:Formal Derivative of Polynomial",
"Category:P-adic Norm not Complete on Rational Numbers"
] |
proofwiki-15727 | Surjection from Natural Numbers iff Countable/Corollary 1 | Let $T$ be a countably infinite set.
Let $S$ be a non-empty set.
Then $S$ is countable {{iff}} there exists a surjection $f: T \to S$. | Let $g: T \to \N$ be a bijection from $T$ to $\N$.
By Inverse of Bijection is Bijection, $g^{-1}: \N \to T$ is a bijection from $\N$ to $T$. | Let $T$ be a [[Definition:Countably Infinite Set|countably infinite set]].
Let $S$ be a [[Definition:Non-Empty Set|non-empty set]].
Then $S$ is [[Definition:Countable Set|countable]] {{iff}} [[Definition:Existential Quantifier|there exists]] a [[Definition:Surjection|surjection]] $f: T \to S$. | Let $g: T \to \N$ be a [[Definition:Bijection|bijection]] from $T$ to $\N$.
By [[Inverse of Bijection is Bijection]], $g^{-1}: \N \to T$ is a [[Definition:Bijection|bijection]] from $\N$ to $T$. | Surjection from Natural Numbers iff Countable/Corollary 1 | https://proofwiki.org/wiki/Surjection_from_Natural_Numbers_iff_Countable/Corollary_1 | https://proofwiki.org/wiki/Surjection_from_Natural_Numbers_iff_Countable/Corollary_1 | [
"Countable Sets",
"Surjections"
] | [
"Definition:Countably Infinite/Set",
"Definition:Non-Empty Set",
"Definition:Countable Set",
"Definition:Existential Quantifier",
"Definition:Surjection"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection"
] |
proofwiki-15728 | Surjection from Natural Numbers iff Countable/Corollary 2 | Let $T$ be a countably infinite set.
Let $S$ be an uncountable set.
Let $f:T \to S$ be a mapping.
Then $f$ is not a surjection. | By Corollary 1 no mapping from $T$ to $S$ is a surjection.
{{qed}}
Category:Countable Sets
Category:Surjections
220auysxelec96dq78xe8ymn2ex26jk | Let $T$ be a [[Definition:Countably Infinite Set|countably infinite set]].
Let $S$ be an [[Definition:Uncountable Set|uncountable set]].
Let $f:T \to S$ be a [[Definition:Mapping|mapping]].
Then $f$ is not a [[Definition:Surjection|surjection]]. | By [[Surjection from Natural Numbers iff Countable/Corollary 1|Corollary 1]] no [[Definition:Mapping|mapping]] from $T$ to $S$ is a [[Definition:Surjection|surjection]].
{{qed}}
[[Category:Countable Sets]]
[[Category:Surjections]]
220auysxelec96dq78xe8ymn2ex26jk | Surjection from Natural Numbers iff Countable/Corollary 2 | https://proofwiki.org/wiki/Surjection_from_Natural_Numbers_iff_Countable/Corollary_2 | https://proofwiki.org/wiki/Surjection_from_Natural_Numbers_iff_Countable/Corollary_2 | [
"Countable Sets",
"Surjections"
] | [
"Definition:Countably Infinite/Set",
"Definition:Uncountable/Set",
"Definition:Mapping",
"Definition:Surjection"
] | [
"Surjection from Natural Numbers iff Countable/Corollary 1",
"Definition:Mapping",
"Definition:Surjection",
"Category:Countable Sets",
"Category:Surjections"
] |
proofwiki-15729 | P-adic Numbers are Uncountable | Let $p$ be any prime number.
The set of $p$-adic numbers $\Q_p$ is an uncountable set. | Let $P$ be the set of sequences on $\set{i : i \in \N : 0 \le i < p}$.
That is:
:$P = \set{\sequence{d_n} : d_n \in \N : 0 \le d_n < p}$
From Cantor's Diagonal Argument:
:$P$ is an uncountable set
Let $f: P \to \Q_p$ be the mapping from $P$ to $\Z_p$ defined by:
:$\forall \sequence{d_n} \in P : \map f {\sequence{d_n}} ... | Let $p$ be any [[Definition:Prime Number|prime number]].
The [[Definition:Set|set]] of [[Definition:P-adic Numbers|$p$-adic numbers]] $\Q_p$ is an [[Definition:Uncountable Set|uncountable set]]. | Let $P$ be the [[Definition:Set|set]] of [[Definition:Sequence|sequences]] on $\set{i : i \in \N : 0 \le i < p}$.
That is:
:$P = \set{\sequence{d_n} : d_n \in \N : 0 \le d_n < p}$
From [[Cantor's Diagonal Argument]]:
:$P$ is an [[Definition:Uncountable Set|uncountable set]]
Let $f: P \to \Q_p$ be the [[Definition:M... | P-adic Numbers are Uncountable | https://proofwiki.org/wiki/P-adic_Numbers_are_Uncountable | https://proofwiki.org/wiki/P-adic_Numbers_are_Uncountable | [
"P-adic Number Theory"
] | [
"Definition:Prime Number",
"Definition:Set",
"Definition:P-adic Number",
"Definition:Uncountable/Set"
] | [
"Definition:Set",
"Definition:Sequence",
"Cantor's Diagonal Argument",
"Definition:Uncountable/Set",
"Definition:Mapping",
"Definition:P-adic Integer",
"Definition:P-adic Expansion",
"P-adic Integer has Unique Coherent Sequence Representative/P-adic Expansion",
"Definition:Bijection",
"Definition:... |
proofwiki-15730 | Ordered Integral Domain is Totally Ordered Ring | Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
Then $\struct {D, +, \times, \le}$ is a totally ordered ring. | By definition, $\struct {D, +, \times, \le}$ is an integral domain endowed with a strict positivity property.
From Strict Positivity Property induces Total Ordering, the ordering $\le$ on $\struct {D, +, \times, \le}$ is a total ordering.
Hence the result by definition of totally ordered ring.
{{qed}}
Category:Ordered ... | Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
Then $\struct {D, +, \times, \le}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]]. | By definition, $\struct {D, +, \times, \le}$ is an [[Definition:Integral Domain|integral domain]] endowed with a [[Definition:Strict Positivity Property|strict positivity property]].
From [[Strict Positivity Property induces Total Ordering]], the [[Definition:Ordering|ordering]] $\le$ on $\struct {D, +, \times, \le}$ ... | Ordered Integral Domain is Totally Ordered Ring | https://proofwiki.org/wiki/Ordered_Integral_Domain_is_Totally_Ordered_Ring | https://proofwiki.org/wiki/Ordered_Integral_Domain_is_Totally_Ordered_Ring | [
"Ordered Integral Domains",
"Totally Ordered Rings"
] | [
"Definition:Ordered Integral Domain",
"Definition:Totally Ordered Ring"
] | [
"Definition:Integral Domain",
"Definition:Strict Positivity Property",
"Strict Positivity Property induces Total Ordering",
"Definition:Ordering",
"Definition:Total Ordering",
"Definition:Totally Ordered Ring",
"Category:Ordered Integral Domains",
"Category:Totally Ordered Rings"
] |
proofwiki-15731 | Strict Negativity is equivalent to Strictly Preceding Zero | :$\map N a \iff a < 0$ | {{begin-eqn}}
{{eqn | l = \map N a
| o = \leadstoandfrom
| r = \map P {-a}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadstoandfrom
| r = \map P {-a + 0}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = a < 0
| c = Strict Positivity Property induces Total Ordering
}... | :$\map N a \iff a < 0$ | {{begin-eqn}}
{{eqn | l = \map N a
| o = \leadstoandfrom
| r = \map P {-a}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadstoandfrom
| r = \map P {-a + 0}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = a < 0
| c = [[Strict Positivity Property induces Total Ordering... | Strict Negativity is equivalent to Strictly Preceding Zero | https://proofwiki.org/wiki/Strict_Negativity_is_equivalent_to_Strictly_Preceding_Zero | https://proofwiki.org/wiki/Strict_Negativity_is_equivalent_to_Strictly_Preceding_Zero | [
"Ordered Integral Domains"
] | [] | [
"Strict Positivity Property induces Total Ordering"
] |
proofwiki-15732 | Strict Negativity is equivalent to Strict Positivity of Negative | :$\map P a \iff \map N {-a}$ | {{begin-eqn}}
{{eqn | l = \map P a
| o = \leadstoandfrom
| r = \map P {-\paren {-a} }
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \map N {-a}
| c = {{Defof|Strict Negativity Property}}
}}
{{end-eqn}}
{{qed}} | :$\map P a \iff \map N {-a}$ | {{begin-eqn}}
{{eqn | l = \map P a
| o = \leadstoandfrom
| r = \map P {-\paren {-a} }
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \map N {-a}
| c = {{Defof|Strict Negativity Property}}
}}
{{end-eqn}}
{{qed}} | Strict Negativity is equivalent to Strict Positivity of Negative | https://proofwiki.org/wiki/Strict_Negativity_is_equivalent_to_Strict_Positivity_of_Negative | https://proofwiki.org/wiki/Strict_Negativity_is_equivalent_to_Strict_Positivity_of_Negative | [
"Ordered Integral Domains"
] | [] | [] |
proofwiki-15733 | Sum of Strictly Negative Elements is Strictly Negative | :$\map N a, \map N b \implies \map N {a + b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map N b
| o = \leadsto
| r = \map P {-a}, \map P {-b}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} + \paren {-b} }
| c = Strict Positivity Property: $(P \, 1)$
}}
{{eqn | o = \leadsto
| r = \map P {-... | :$\map N a, \map N b \implies \map N {a + b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map N b
| o = \leadsto
| r = \map P {-a}, \map P {-b}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} + \paren {-b} }
| c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 1)$]]
}}
... | Sum of Strictly Negative Elements is Strictly Negative | https://proofwiki.org/wiki/Sum_of_Strictly_Negative_Elements_is_Strictly_Negative | https://proofwiki.org/wiki/Sum_of_Strictly_Negative_Elements_is_Strictly_Negative | [
"Ordered Integral Domains"
] | [] | [
"Definition:Strict Positivity Property"
] |
proofwiki-15734 | Product of Two Strictly Negative Elements is Strictly Positive | :$\map N a, \map N b \implies \map P {a \times b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map N b
| o = \leadsto
| r = \map P {-a}, \map P {-b}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} \times \paren {-b} }
| c = Strict Positivity Property: $(P \, 2)$
}}
{{eqn | o = \leadsto
| r = \map... | :$\map N a, \map N b \implies \map P {a \times b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map N b
| o = \leadsto
| r = \map P {-a}, \map P {-b}
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} \times \paren {-b} }
| c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 2)$]... | Product of Two Strictly Negative Elements is Strictly Positive | https://proofwiki.org/wiki/Product_of_Two_Strictly_Negative_Elements_is_Strictly_Positive | https://proofwiki.org/wiki/Product_of_Two_Strictly_Negative_Elements_is_Strictly_Positive | [
"Ordered Integral Domains"
] | [] | [
"Definition:Strict Positivity Property",
"Product of Ring Negatives"
] |
proofwiki-15735 | Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative | :$\map N a, \map P b \implies \map N {a \times b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map P b
| o = \leadsto
| r = \map P {-a}, \map P b
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} \times b}
| c = Strict Positivity Property: $(P \, 2)$
}}
{{eqn | o = \leadsto
| r = \map P {-\paren {a... | :$\map N a, \map P b \implies \map N {a \times b}$ | {{begin-eqn}}
{{eqn | l = \map N a, \map P b
| o = \leadsto
| r = \map P {-a}, \map P b
| c = {{Defof|Strict Negativity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a} \times b}
| c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 2)$]]
}}
{{eqn | o... | Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative | https://proofwiki.org/wiki/Product_of_Strictly_Negative_Element_with_Strictly_Positive_Element_is_Strictly_Negative | https://proofwiki.org/wiki/Product_of_Strictly_Negative_Element_with_Strictly_Positive_Element_is_Strictly_Negative | [
"Ordered Integral Domains"
] | [] | [
"Definition:Strict Positivity Property",
"Product with Ring Negative"
] |
proofwiki-15736 | Equivalence of Definitions of Well-Ordered Integral Domain | {{TFAE|def = Well-Ordered Integral Domain}}
Let $\struct {D, +, \times \le}$ be an ordered integral domain whose zero is $0_D$. | === $(1)$ implies $(2)$ ===
Let $\struct {D, +, \times \le}$ be a well-ordered integral domain by definition 1.
Then by definition the ordering $\le$ is a well-ordering on the set $P$ of (strictly) positive elements of $D$.
Let $S \subseteq P$.
Thus by definition of well-ordering, $S$ has a minimal element.
Thus $\stru... | {{TFAE|def = Well-Ordered Integral Domain}}
Let $\struct {D, +, \times \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$. | === $(1)$ implies $(2)$ ===
Let $\struct {D, +, \times \le}$ be a [[Definition:Well-Ordered Integral Domain/Definition 1|well-ordered integral domain by definition 1]].
Then by definition the [[Definition:Total Ordering Induced by Strict Positivity Property|ordering $\le$]] is a [[Definition:Well-Ordering|well-orderi... | Equivalence of Definitions of Well-Ordered Integral Domain | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordered_Integral_Domain | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordered_Integral_Domain | [
"Well-Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Ring Zero"
] | [
"Definition:Well-Ordered Integral Domain/Definition 1",
"Definition:Total Ordering Induced by Strict Positivity Property",
"Definition:Well-Ordering",
"Definition:Set",
"Definition:Strictly Positive",
"Definition:Well-Ordering",
"Definition:Minimal/Element",
"Definition:Well-Ordered Integral Domain/De... |
proofwiki-15737 | Principle of Mathematical Induction/Zero-Based | Let $\map P n$ be a propositional function depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is true
:$(2): \quad \forall k \in \N: k \ge 0 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N$. | Consider $\N$ defined as a Peano structure.
The result follows from Principle of Mathematical Induction for Peano Structure.
{{qed}} | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N: k \ge 0 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \in \N$. | Consider $\N$ defined as a [[Definition:Peano Structure|Peano structure]].
The result follows from [[Principle of Mathematical Induction for Peano Structure]].
{{qed}} | Principle of Mathematical Induction/Zero-Based | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/Zero-Based | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/Zero-Based | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:Peano Structure",
"Principle of Mathematical Induction/Peano Structure"
] |
proofwiki-15738 | Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | Let $S$ be the set defined as:
:$S := \set {n \in \N_{>0}: \map P n \text { is false} }$
{{AimForCont}} $S \ne \O$.
From the Well-Ordering Principle it follows that $S$ has a minimal element $m$.
From $(1)$ we have that $\map P 1$ holds.
Hence $1 \notin S$.
Therefore $m \ne 1$.
Therefore $m - 1 \in \N_{>0}$.
But $m$ is... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \... | Let $S$ be the [[Definition:Set|set]] defined as:
:$S := \set {n \in \N_{>0}: \map P n \text { is false} }$
{{AimForCont}} $S \ne \O$.
From the [[Well-Ordering Principle]] it follows that $S$ has a [[Definition:Minimal Element|minimal element]] $m$.
From $(1)$ we have that $\map P 1$ holds.
Hence $1 \notin S$.
Th... | Principle of Mathematical Induction/One-Based/Proof 1 | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based/Proof_1 | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:Set",
"Well-Ordering Principle",
"Definition:Minimal/Element",
"Definition:Minimal/Element"
] |
proofwiki-15739 | Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | Let $M$ be the set of all $n \in \N_{>0}$ for which $\map P n$ holds.
By $(1)$ we have that $1 \in M$.
By $(2)$ we have that if $k \in M$ then $k + 1 \in M$.
From the Axiomatization of $1$-Based Natural Numbers, Axiom $(\text F)$, it follows that $M = \N_{>0}$.
{{qed}} | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \... | Let $M$ be the [[Definition:Set|set]] of all $n \in \N_{>0}$ for which $\map P n$ holds.
By $(1)$ we have that $1 \in M$.
By $(2)$ we have that if $k \in M$ then $k + 1 \in M$.
From the [[Axiom:Axiomatization of 1-Based Natural Numbers|Axiomatization of $1$-Based Natural Numbers]], Axiom $(\text F)$, it follows that... | Principle of Mathematical Induction/One-Based/Proof 2 | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based/Proof_2 | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:Set",
"Axiom:Axiomatization of 1-Based Natural Numbers"
] |
proofwiki-15740 | Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | We have that Natural Numbers are Non-Negative Integers.
Then we have that Integers form Well-Ordered Integral Domain.
The result follows from Induction on Well-Ordered Integral Domain.
{{qed}} | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \... | We have that [[Natural Numbers are Non-Negative Integers]].
Then we have that [[Integers form Well-Ordered Integral Domain]].
The result follows from [[Induction on Well-Ordered Integral Domain]].
{{qed}} | Principle of Mathematical Induction/One-Based/Proof 3 | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based/Proof_3 | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Natural Numbers are Non-Negative Integers",
"Integers form Well-Ordered Integral Domain",
"Principle of Mathematical Induction/Well-Ordered Integral Domain"
] |
proofwiki-15741 | Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \... | For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is [[Definition:True|true]].
Now suppose that $\map {P'} n$ hold... | Second Principle of Mathematical Induction/One-Based/Proof 1 | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based/Proof_1 | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:True",
"Principle of Mathematical Induction"
] |
proofwiki-15742 | Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the Well-Ordering Principle $S$ contains a minimal element $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is [[Definition:True|true]] for all $n \... | Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the [[Well-Ordering Principle]] $S$ contains a [[Definition:Minimal Element|minimal element]] $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist som... | Second Principle of Mathematical Induction/One-Based/Proof 2 | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based/Proof_2 | [
"Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Well-Ordering Principle",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Contradiction",
"Proof by Contradiction"
] |
proofwiki-15743 | Principle of Finite Induction/Zero-Based | Let $S \subseteq \N$ be a subset of the natural numbers.
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N : n \in S \implies n + 1 \in S$
Then:
:$S = \N$ | Consider $\N$ defined as a naturally ordered semigroup.
The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup.
{{qed}} | Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]].
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N : n \in S \implies n + 1 \in S$
Then:
:$S = \N$ | Consider $\N$ defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
The result follows directly from [[Principle of Mathematical Induction for Naturally Ordered Semigroup]].
{{qed}} | Principle of Finite Induction/Zero-Based | https://proofwiki.org/wiki/Principle_of_Finite_Induction/Zero-Based | https://proofwiki.org/wiki/Principle_of_Finite_Induction/Zero-Based | [
"Principle of Finite Induction"
] | [
"Definition:Subset",
"Definition:Natural Numbers"
] | [
"Definition:Naturally Ordered Semigroup",
"Principle of Mathematical Induction/Naturally Ordered Semigroup"
] |
proofwiki-15744 | Principle of Finite Induction/One-Based | Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Consider $\N$ defined as a naturally ordered semigroup.
The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result.
{{qed}} | Let $S \subseteq \N_{>0}$ be a [[Definition:Subset|subset]] of the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Consider $\N$ defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
The result follows directly from [[Principle of Mathematical Induction for Naturally Ordered Semigroup/General Result|Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result]].
{{qed}} | Principle of Finite Induction/One-Based/Proof 1 | https://proofwiki.org/wiki/Principle_of_Finite_Induction/One-Based | https://proofwiki.org/wiki/Principle_of_Finite_Induction/One-Based/Proof_1 | [
"Principle of Finite Induction"
] | [
"Definition:Subset",
"Axiom:Axiomatization of 1-Based Natural Numbers"
] | [
"Definition:Naturally Ordered Semigroup",
"Principle of Mathematical Induction/Naturally Ordered Semigroup/General Result"
] |
proofwiki-15745 | Principle of Finite Induction/One-Based | Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Let $T$ be the set of all $1$-based natural numbers not in $S$:
:$T = \N_{>0} \setminus S$
{{AimForCont}} $T$ is non-empty.
From the Well-Ordering Principle, $T$ has a smallest element.
Let this smallest element be denoted $a$.
We have been given that $1 \in S$.
So:
:$a > 1$
and so:
:$0 < a - 1 < a$
As $a$ is the small... | Let $S \subseteq \N_{>0}$ be a [[Definition:Subset|subset]] of the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Let $T$ be the [[Definition:Set|set]] of all [[Definition:1-Based Natural Numbers|$1$-based natural numbers]] not in $S$:
:$T = \N_{>0} \setminus S$
{{AimForCont}} $T$ is [[Definition:Non-Empty Set|non-empty]].
From the [[Well-Ordering Principle]], $T$ has a [[Definition:Smallest Element|smallest element]].
Let this... | Principle of Finite Induction/One-Based/Proof 2 | https://proofwiki.org/wiki/Principle_of_Finite_Induction/One-Based | https://proofwiki.org/wiki/Principle_of_Finite_Induction/One-Based/Proof_2 | [
"Principle of Finite Induction"
] | [
"Definition:Subset",
"Axiom:Axiomatization of 1-Based Natural Numbers"
] | [
"Definition:Set",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Non-Empty Set",
"Well-Ordering Principle",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:By Hypothesis",
"Definition:Contradiction",
"Proof by Contradiction",... |
proofwiki-15746 | Second Principle of Finite Induction/Zero-Based | Let $S \subseteq \N$ be a subset of the natural numbers.
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N: \paren {\forall k: 0 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N$ | Define $T$ as:
:$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 0: n \in T$
Firstly, we have that $0 \in T$ {{iff}} the following condition holds:
:$\forall k: 0 \le k \le 0 \implies k \in S$
Since $0 \i... | Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]].
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N: \paren {\forall k: 0 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N$ | Define $T$ as:
:$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 0: n \in T$
Firstly, we have that $0 \in T$ {{iff}} the following condition holds:
:$\forall k: 0 \le k \le 0 \implies k \in S$
Sin... | Second Principle of Finite Induction/Zero-Based | https://proofwiki.org/wiki/Second_Principle_of_Finite_Induction/Zero-Based | https://proofwiki.org/wiki/Second_Principle_of_Finite_Induction/Zero-Based | [
"Second Principle of Finite Induction"
] | [
"Definition:Subset",
"Definition:Natural Numbers"
] | [
"Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor",
"Principle of Finite Induction",
"Category:Second Principle of Finite Induction"
] |
proofwiki-15747 | Second Principle of Finite Induction/One-Based | Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Define $T$ as:
:$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 1: n \in T$
Firstly, we have that $1 \in T$ {{iff}} the following condition holds:
:$\forall k: 1 \le k \le 1 \implies k \in S$
Since $... | Let $S \subseteq \N_{>0}$ be a [[Definition:Subset|subset]] of the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N_{>0}$ | Define $T$ as:
:$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 1: n \in T$
Firstly, we have that $1 \in T$ {{iff}} the following condition holds:
:$\forall k: 1 \le k \le 1 \implies k \in S$
... | Second Principle of Finite Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Finite_Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Finite_Induction/One-Based | [
"Second Principle of Finite Induction"
] | [
"Definition:Subset",
"Axiom:Axiomatization of 1-Based Natural Numbers"
] | [
"Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor",
"Principle of Finite Induction"
] |
proofwiki-15748 | Second Principle of Mathematical Induction/Zero-Based | Let $\map P n$ be a propositional function depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is true
:$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N$. | For each $n \in \N$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 0 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N$.
It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\ma... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P... | For each $n \in \N$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 0 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N$.
It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is [[Definition:True|true]].
Now suppose that $\map {P'} n$ holds.
By $(2... | Second Principle of Mathematical Induction/Zero-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/Zero-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/Zero-Based | [
"Second Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:True",
"Principle of Mathematical Induction"
] |
proofwiki-15749 | Second Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:... | For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is [[Definition:True|true]].
Now suppose that $\map {P'} n$ hold... | Second Principle of Mathematical Induction/One-Based/Proof 1 | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based/Proof_1 | [
"Second Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Definition:True",
"Principle of Mathematical Induction"
] |
proofwiki-15750 | Second Principle of Mathematical Induction/One-Based | Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$. | Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the Well-Ordering Principle $S$ contains a minimal element $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is [[Definition:True|true]]
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:... | Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the [[Well-Ordering Principle]] $S$ contains a [[Definition:Minimal Element|minimal element]] $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist som... | Second Principle of Mathematical Induction/One-Based/Proof 2 | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based | https://proofwiki.org/wiki/Second_Principle_of_Mathematical_Induction/One-Based/Proof_2 | [
"Second Principle of Mathematical Induction"
] | [
"Definition:Propositional Function",
"Definition:True",
"Definition:True"
] | [
"Well-Ordering Principle",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Contradiction",
"Proof by Contradiction"
] |
proofwiki-15751 | Null Sequences form Maximal Left and Right Ideal/Corollary 1 | Then $\NN$ is a maximal ring ideal of $\CC$. | By Null Sequences form Maximal Left and Right Ideal then $\NN$ is a maximal left ideal of $\CC$.
A field is by definition a commutative ring.
In a commutative ring, a maximal left ideal is by definition a maximal ideal.
{{qed}} | Then $\NN$ is a [[Definition:Maximal Ideal of Ring|maximal ring ideal]] of $\CC$. | By [[Null Sequences form Maximal Left and Right Ideal]] then $\NN$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] of $\CC$.
A [[Definition:Field (Abstract Algebra)|field]] is by definition a [[Definition:Commutative Ring|commutative ring]].
In a [[Definition:Commutative Ring|commutative ring]], a ... | Null Sequences form Maximal Left and Right Ideal/Corollary 1 | https://proofwiki.org/wiki/Null_Sequences_form_Maximal_Left_and_Right_Ideal/Corollary_1 | https://proofwiki.org/wiki/Null_Sequences_form_Maximal_Left_and_Right_Ideal/Corollary_1 | [
"Null Sequences form Maximal Left and Right Ideal"
] | [
"Definition:Maximal Ideal of Ring"
] | [
"Null Sequences form Maximal Left and Right Ideal",
"Definition:Maximal Ideal of Ring/Left",
"Definition:Field (Abstract Algebra)",
"Definition:Commutative Ring",
"Definition:Commutative Ring",
"Definition:Maximal Ideal of Ring/Left",
"Definition:Maximal Ideal of Ring"
] |
proofwiki-15752 | Non-Zero Integer has Finite Number of Divisors | Let $n \in \Z_{\ne 0}$ be a non-zero integer.
Then $n$ has a finite number of divisors. | Let $S$ be the set of all divisors of $n$.
Then from Absolute Value of Integer is not less than Divisors:
:$\forall m \in S: -n \le m \le n$
Thus $S$ is finite.
{{qed}}
Category:Number Theory
Category:Divisors
dhwii31ikzevtdctqmctmo6ixevhs9s | Let $n \in \Z_{\ne 0}$ be a non-[[Definition:Zero (Number)|zero]] [[Definition:Integer|integer]].
Then $n$ has a [[Definition:Finite Set|finite number]] of [[Definition:Divisor of Integer|divisors]]. | Let $S$ be the [[Definition:Set|set]] of all [[Definition:Divisor of Integer|divisors]] of $n$.
Then from [[Absolute Value of Integer is not less than Divisors]]:
:$\forall m \in S: -n \le m \le n$
Thus $S$ is [[Definition:Finite Set|finite]].
{{qed}}
[[Category:Number Theory]]
[[Category:Divisors]]
dhwii31ikzevtdct... | Non-Zero Integer has Finite Number of Divisors | https://proofwiki.org/wiki/Non-Zero_Integer_has_Finite_Number_of_Divisors | https://proofwiki.org/wiki/Non-Zero_Integer_has_Finite_Number_of_Divisors | [
"Number Theory",
"Divisors"
] | [
"Definition:Zero (Number)",
"Definition:Integer",
"Definition:Finite Set",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Set",
"Definition:Divisor (Algebra)/Integer",
"Absolute Value of Integer is not less than Divisors",
"Definition:Finite Set",
"Category:Number Theory",
"Category:Divisors"
] |
proofwiki-15753 | Coprimality Relation is Non-Reflexive | :$\perp$ is non-reflexive. | Proof by Counterexample:
We have from GCD of Integer and Divisor:
:$\gcd \set {n, n} = n$
and so, for example:
:$\gcd \set {2, 2} = 2$
and so:
:$2 \not \perp 2$
Hence $\perp$ is not reflexive.
But we also note that:
:$\gcd \set {1, 1} = 1$
and so:
:$1 \perp 1$
demonstrating that $\perp$ is not antireflexive either.
The... | :$\perp$ is [[Definition:Non-Reflexive Relation|non-reflexive]]. | [[Proof by Counterexample]]:
We have from [[GCD of Integer and Divisor]]:
:$\gcd \set {n, n} = n$
and so, for example:
:$\gcd \set {2, 2} = 2$
and so:
:$2 \not \perp 2$
Hence $\perp$ is not [[Definition:Reflexive Relation|reflexive]].
But we also note that:
:$\gcd \set {1, 1} = 1$
and so:
:$1 \perp 1$
demonstrat... | Coprimality Relation is Non-Reflexive | https://proofwiki.org/wiki/Coprimality_Relation_is_Non-Reflexive | https://proofwiki.org/wiki/Coprimality_Relation_is_Non-Reflexive | [
"Coprime Integers"
] | [
"Definition:Non-Reflexive Relation"
] | [
"Proof by Counterexample",
"GCD of Integer and Divisor",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation"
] |
proofwiki-15754 | Coprimality Relation is Symmetric | :$\perp$ is symmetric. | {{begin-eqn}}
{{eqn | l = x
| o = \perp
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {x, y}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {y, x}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = y
| o = \perp
| r = x
| c =
}}
{... | :$\perp$ is [[Definition:Symmetric Relation|symmetric]]. | {{begin-eqn}}
{{eqn | l = x
| o = \perp
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {x, y}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {y, x}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = y
| o = \perp
| r = x
| c =
}}
{... | Coprimality Relation is Symmetric | https://proofwiki.org/wiki/Coprimality_Relation_is_Symmetric | https://proofwiki.org/wiki/Coprimality_Relation_is_Symmetric | [
"Coprime Integers"
] | [
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation"
] |
proofwiki-15755 | Coprimality Relation is not Antisymmetric | :$\perp$ is not antisymmetric. | Proof by Counterexample:
We have:
:$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$
and so:
:$3 \perp 5$ and $5 \perp 3$
However, it is not the case that $3 = 5$.
The result follows by definition of antisymmetric relation.
{{qed}}
Category:Coprime Integers
1u7e7pm5fi56pokyvmkmvxole4btnun | :$\perp$ is not [[Definition:Antisymmetric Relation|antisymmetric]]. | [[Proof by Counterexample]]:
We have:
:$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$
and so:
:$3 \perp 5$ and $5 \perp 3$
However, it is not the case that $3 = 5$.
The result follows by definition of [[Definition:Antisymmetric Relation|antisymmetric relation]].
{{qed}}
[[Category:Coprime Integers]]
1u7e7pm5fi56pokyvmk... | Coprimality Relation is not Antisymmetric | https://proofwiki.org/wiki/Coprimality_Relation_is_not_Antisymmetric | https://proofwiki.org/wiki/Coprimality_Relation_is_not_Antisymmetric | [
"Coprime Integers"
] | [
"Definition:Antisymmetric Relation"
] | [
"Proof by Counterexample",
"Definition:Antisymmetric Relation",
"Category:Coprime Integers"
] |
proofwiki-15756 | Coprimality Relation is Non-Transitive | :$\perp$ is non-transitive. | Proof by Counterexample:
We have:
{{begin-eqn}}
{{eqn | l = \gcd \set {2, 3}
| r = 1
| c =
}}
{{eqn | l = \gcd \set {3, 4}
| r = 1
| c =
}}
{{eqn | l = \gcd \set {2, 4}
| r = 2
| c =
}}
{{end-eqn}}
Hence we have:
:$2 \perp 3$ and $3 \perp 4$
However, it is not the case that $2 \pe... | :$\perp$ is [[Definition:Non-Transitive Relation|non-transitive]]. | [[Proof by Counterexample]]:
We have:
{{begin-eqn}}
{{eqn | l = \gcd \set {2, 3}
| r = 1
| c =
}}
{{eqn | l = \gcd \set {3, 4}
| r = 1
| c =
}}
{{eqn | l = \gcd \set {2, 4}
| r = 2
| c =
}}
{{end-eqn}}
Hence we have:
:$2 \perp 3$ and $3 \perp 4$
However, it is not the case th... | Coprimality Relation is Non-Transitive | https://proofwiki.org/wiki/Coprimality_Relation_is_Non-Transitive | https://proofwiki.org/wiki/Coprimality_Relation_is_Non-Transitive | [
"Coprime Integers",
"Examples of Non-Transitive Relations"
] | [
"Definition:Non-Transitive Relation"
] | [
"Proof by Counterexample",
"Definition:Transitive Relation",
"Definition:Antitransitive Relation",
"Definition:Non-Transitive Relation"
] |
proofwiki-15757 | Ring of Polynomials over Reals is not Field | Let $\R \sqbrk X$ be the ring of polynomials in an indeterminate $X$ over $\R$.
Then $\R \sqbrk X$ is not a field. | Consider the polynomial $x + 1$ in $\R \sqbrk X$.
There exists no polynomial $\map f x$ such that:
:$\paren {x + 1} \map f x = 1$
This is because the {{LHS}} has degree $1$, and the {{RHS}} has degree $0$.
{{qed}} | Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in an [[Definition:Indeterminate|indeterminate]] $X$ over $\R$.
Then $\R \sqbrk X$ is not a [[Definition:Field (Abstract Algebra)|field]]. | Consider the [[Definition:Polynomial over Real Numbers|polynomial]] $x + 1$ in $\R \sqbrk X$.
There exists no [[Definition:Polynomial over Real Numbers|polynomial]] $\map f x$ such that:
:$\paren {x + 1} \map f x = 1$
This is because the {{LHS}} has [[Definition:Degree of Polynomial|degree]] $1$, and the {{RHS}} has... | Ring of Polynomials over Reals is not Field | https://proofwiki.org/wiki/Ring_of_Polynomials_over_Reals_is_not_Field | https://proofwiki.org/wiki/Ring_of_Polynomials_over_Reals_is_not_Field | [
"Polynomial Rings",
"Field Theory"
] | [
"Definition:Ring of Polynomials in Ring Element",
"Definition:Indeterminate",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Degree of Polynomial"
] |
proofwiki-15758 | Even Integers form Commutative Ring | Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a commutative ring.
However, $\struct {2 \Z, +, \times}$ is not an integral domain. | From Integer Multiples form Commutative Ring, $\struct {2 \Z, +, \times}$ is a commutative ring.
As $2 \ne 1$, we also have from Integer Multiples form Commutative Ring that $\struct {2 \Z, +, \times}$ has no unity.
Hence by definition it is not an integral domain.
{{qed}} | Let $2 \Z$ be the [[Definition:Even Integer|set of even integers]].
Then $\struct {2 \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
However, $\struct {2 \Z, +, \times}$ is not an [[Definition:Integral Domain|integral domain]]. | From [[Integer Multiples form Commutative Ring]], $\struct {2 \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
As $2 \ne 1$, we also have from [[Integer Multiples form Commutative Ring]] that $\struct {2 \Z, +, \times}$ has no [[Definition:Unity of Ring|unity]].
Hence by definition it is not an ... | Even Integers form Commutative Ring | https://proofwiki.org/wiki/Even_Integers_form_Commutative_Ring | https://proofwiki.org/wiki/Even_Integers_form_Commutative_Ring | [
"Integers",
"Commutative Rings"
] | [
"Definition:Even Integer",
"Definition:Commutative Ring",
"Definition:Integral Domain"
] | [
"Integer Multiples form Commutative Ring",
"Definition:Commutative Ring",
"Integer Multiples form Commutative Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain"
] |
proofwiki-15759 | Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers | Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$. | From Subrings of Integers are Sets of Integer Multiples, a ring of the form $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
$\struct {2 \Z, +, \times}$ is such an example.
{{qed}} | Let $2 \Z$ be the [[Definition:Even Integer|set of even integers]].
Then $\struct {2 \Z, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$. | From [[Subrings of Integers are Sets of Integer Multiples]], a [[Definition:Ring (Abstract Algebra)|ring]] of the form $\struct {n \Z, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$ when $n \ge 1$.
$\struct {2 \Z, +, \times}$ is such an example.
{{qed}} | Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers | https://proofwiki.org/wiki/Subrings_of_Integers_are_Sets_of_Integer_Multiples/Examples/Even_Integers | https://proofwiki.org/wiki/Subrings_of_Integers_are_Sets_of_Integer_Multiples/Examples/Even_Integers | [
"Subrings of Integers are Sets of Integer Multiples"
] | [
"Definition:Even Integer",
"Definition:Subring"
] | [
"Subrings of Integers are Sets of Integer Multiples",
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] |
proofwiki-15760 | Gaussian Integers does not form Subfield of Complex Numbers | The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
is not a subfield of $\C$. | Proof by Counterexample:
We have that:
:$2 + 0 i \in \Z \sqbrk i$
However there is no $z \in \Z \sqbrk i$ such that:
:$x \paren {2 + 0 i} = 1 + 0 i$
So, by definition, $\Z \sqbrk i$ is not a field.
Thus $\Z \sqbrk i$ is not a subfield of $\C$.
{{qed}} | The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
:$\struct {\Z \sqbrk i, +, \times}$
is not a [[Definition:Subfield|subfield]] of $\C$. | [[Proof by Counterexample]]:
We have that:
:$2 + 0 i \in \Z \sqbrk i$
However there is no $z \in \Z \sqbrk i$ such that:
:$x \paren {2 + 0 i} = 1 + 0 i$
So, by definition, $\Z \sqbrk i$ is not a [[Definition:Field (Abstract Algebra)|field]].
Thus $\Z \sqbrk i$ is not a [[Definition:Subfield|subfield]] of $\C$.
{{qe... | Gaussian Integers does not form Subfield of Complex Numbers | https://proofwiki.org/wiki/Gaussian_Integers_does_not_form_Subfield_of_Complex_Numbers | https://proofwiki.org/wiki/Gaussian_Integers_does_not_form_Subfield_of_Complex_Numbers | [
"Subfields",
"Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Subfield"
] | [
"Proof by Counterexample",
"Definition:Field (Abstract Algebra)",
"Definition:Subfield"
] |
proofwiki-15761 | Ideals of Ring of Integers Modulo m | Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +, \times}$ denote the ring of integers modulo $m$.
The ideals of $\struct {\Z_m, +, \times}$ are of the form:
:$d \Z / m \Z$
where $d$ is a divisor of $m$. | Let $J$ be an ideal of $\struct {\Z_m, +, \times}$.
Recall from Ring of Integers Modulo m is Ring that $\struct {\Z_m, +}$ is the additive group of integers modulo $m$.
From Integers Modulo $m$ under Addition form Cyclic Group, $\struct {\Z_m, +}$ is a cyclic group.
By definition of ideal, $\struct {J, +}$ is a subgrou... | Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\struct {\Z_m, +, \times}$ denote the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
The [[Definition:Ideal of Ring|ideals]] of $\struct {\Z_m, +, \times}$ are of the form:
:$d \Z / m \Z$
where $... | Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\struct {\Z_m, +, \times}$.
Recall from [[Ring of Integers Modulo m is Ring]] that $\struct {\Z_m, +}$ is the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
From [[Integers Modulo m under Addition form Cyclic Group|Integer... | Ideals of Ring of Integers Modulo m | https://proofwiki.org/wiki/Ideals_of_Ring_of_Integers_Modulo_m | https://proofwiki.org/wiki/Ideals_of_Ring_of_Integers_Modulo_m | [
"Ring of Integers Modulo m",
"Ideal Theory",
"Ideals of Ring of Integers Modulo m"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Ring of Integers Modulo m",
"Definition:Ideal of Ring",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Ideal of Ring",
"Ring of Integers Modulo m is Ring",
"Definition:Additive Group of Integers Modulo m",
"Integers Modulo m under Addition form Cyclic Group",
"Definition:Cyclic Group",
"Definition:Ideal of Ring",
"Definition:Subgroup",
"Definition:Subgroup",
"Subgroup of Cyclic Group is C... |
proofwiki-15762 | Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1 | Then the quotient ring $\CC \,\big / \NN$ is a field. | By Quotient Ring of Cauchy Sequences is Division Ring, $\CC \,\big / \NN$ is a division ring.
By {{Corollary|Cauchy Sequences form Ring with Unity}}, $\CC$ is a commutative ring with unity.
By Quotient Ring of Commutative Ring is Commutative, $\CC \,\big / \NN$ is a commutative division ring, that is, a field.
{{qed}} | Then the [[Definition:Quotient Ring|quotient ring]] $\CC \,\big / \NN$ is a [[Definition:Field (Abstract Algebra)|field]]. | By [[Quotient Ring of Cauchy Sequences is Division Ring]], $\CC \,\big / \NN$ is a [[Definition:Division Ring|division ring]].
By {{Corollary|Cauchy Sequences form Ring with Unity}}, $\CC$ is a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
By [[Quotient Ring of Commutative Ring is Commutativ... | Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1 | https://proofwiki.org/wiki/Quotient_Ring_of_Cauchy_Sequences_is_Division_Ring/Corollary_1 | https://proofwiki.org/wiki/Quotient_Ring_of_Cauchy_Sequences_is_Division_Ring/Corollary_1 | [
"Completion of Normed Division Ring"
] | [
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)"
] | [
"Quotient Ring of Cauchy Sequences is Division Ring",
"Definition:Division Ring",
"Definition:Commutative and Unitary Ring",
"Quotient Ring of Commutative Ring is Commutative",
"Definition:Commutative Ring",
"Definition:Division Ring",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-15763 | Quotient Ring of Cauchy Sequences is Normed Division Ring/Corollary 1 | Then $\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a valued field. | By Quotient Ring of Cauchy Sequences is Normed Division Ring, $\CC \,\big / \NN$ is a normed division ring.
By {{Corollary|Quotient Ring of Cauchy Sequences is Division Ring|1}}, $\CC \,\big / \NN$ is a field.
The result follows.
{{qed}} | Then $\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a [[Definition:Valued Field|valued field]]. | By [[Quotient Ring of Cauchy Sequences is Normed Division Ring]], $\CC \,\big / \NN$ is a [[Definition:Normed Division Ring|normed division ring]].
By {{Corollary|Quotient Ring of Cauchy Sequences is Division Ring|1}}, $\CC \,\big / \NN$ is a [[Definition:Field (Abstract Algebra)|field]].
The result follows.
{{qed}} | Quotient Ring of Cauchy Sequences is Normed Division Ring/Corollary 1 | https://proofwiki.org/wiki/Quotient_Ring_of_Cauchy_Sequences_is_Normed_Division_Ring/Corollary_1 | https://proofwiki.org/wiki/Quotient_Ring_of_Cauchy_Sequences_is_Normed_Division_Ring/Corollary_1 | [
"Completion of Normed Division Ring"
] | [
"Definition:Valued Field"
] | [
"Quotient Ring of Cauchy Sequences is Normed Division Ring",
"Definition:Normed Division Ring",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-15764 | Cauchy Sequence Converges Iff Equivalent to Constant Sequence | Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $\\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\sequence {x_n} \in \CC$.
Then $\sequence {x_n}$ converge... | By definition, $\sequence {x_n}$ converges to $a \in R$ {{iff}} $\ds \lim_{n \mathop \to \infty} \norm {x_n - a} = 0$
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n - a}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \sequence {x_n - a}
| o = \in
| r = \NN
| c = {{Def... | Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
Let $\CC$ be the [[Definition:Ring of Cauchy Sequences|ring of Cauchy sequences over $R$]]
Let $\NN$ be the [[Definition:Set|set]] of [[Definition:Null Sequence in Normed Division Ring|null sequences]].
Let $\\CC \,\... | By definition, $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $a \in R$ {{iff}} $\ds \lim_{n \mathop \to \infty} \norm {x_n - a} = 0$
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n - a}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \sequence... | Cauchy Sequence Converges Iff Equivalent to Constant Sequence | https://proofwiki.org/wiki/Cauchy_Sequence_Converges_Iff_Equivalent_to_Constant_Sequence | https://proofwiki.org/wiki/Cauchy_Sequence_Converges_Iff_Equivalent_to_Constant_Sequence | [
"Normed Division Rings"
] | [
"Definition:Normed Division Ring",
"Definition:Ring of Cauchy Sequences",
"Definition:Set",
"Definition:Null Sequence/Normed Division Ring",
"Quotient Ring of Cauchy Sequences is Division Ring",
"Null Sequences form Maximal Left and Right Ideal",
"Definition:Convergent Sequence/Normed Division Ring",
... | [
"Definition:Convergent Sequence/Normed Division Ring",
"Element in Left Coset iff Product with Inverse in Subgroup",
"Category:Normed Division Rings"
] |
proofwiki-15765 | Homomorphism of Ring Subtraction | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall a, b \in R_1: \map \phi {a -_1 b} = \map \phi a -_2 \map \phi b$
where $a -_1 b$ denotes subtraction of $b$ from $a$. | {{begin-eqn}}
{{eqn | l = \map \phi {a -_1 b}
| r = \map \phi {a +_1 \paren {-b} }
| c = {{Defof|Ring Subtraction}}
}}
{{eqn | r = \map \phi a +_2 \map \phi {-b}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \phi a +_2 \paren {-\map \phi b}
| c = Ring Homomorphism Preserves Negatives
}}
... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Then:
:$\forall a, b \in R_1: \map \phi {a -_1 b} = \map \phi a -_2 \map \phi b$
where $a -_1 b$ denotes [[Definition:Ring Subtraction|subtraction]] of $b$ from $a$. | {{begin-eqn}}
{{eqn | l = \map \phi {a -_1 b}
| r = \map \phi {a +_1 \paren {-b} }
| c = {{Defof|Ring Subtraction}}
}}
{{eqn | r = \map \phi a +_2 \map \phi {-b}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \phi a +_2 \paren {-\map \phi b}
| c = [[Ring Homomorphism Preserves Negatives]]... | Homomorphism of Ring Subtraction | https://proofwiki.org/wiki/Homomorphism_of_Ring_Subtraction | https://proofwiki.org/wiki/Homomorphism_of_Ring_Subtraction | [
"Ring Homomorphisms"
] | [
"Definition:Ring Homomorphism",
"Definition:Subtraction/Ring"
] | [
"Ring Homomorphism Preserves Negatives"
] |
proofwiki-15766 | Limit of Modulo Operation/Limit 1 | Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to 0} x \bmod y = 0$. | By Range of Modulo Operation for Positive Modulus and Range of Modulo Operation for Negative Modulus we have:
:$-\size y < x \bmod y < \size y$
The result follows from the Squeeze Theorem for Functions.
{{qed}} | Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
Then $\ds \lim_{y \mathop \to 0} x \bmod y = 0$. | By [[Range of Modulo Operation for Positive Modulus]] and [[Range of Modulo Operation for Negative Modulus]] we have:
:$-\size y < x \bmod y < \size y$
The result follows from the [[Squeeze Theorem for Functions]].
{{qed}} | Limit of Modulo Operation/Limit 1 | https://proofwiki.org/wiki/Limit_of_Modulo_Operation/Limit_1 | https://proofwiki.org/wiki/Limit_of_Modulo_Operation/Limit_1 | [
"Limit of Modulo Operation"
] | [
"Definition:Real Number",
"Definition:Modulo Operation"
] | [
"Range of Modulo Operation for Positive Modulus",
"Range of Modulo Operation for Negative Modulus",
"Squeeze Theorem/Functions"
] |
proofwiki-15767 | Limit of Modulo Operation/Limit 2 | Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$. | As $y \to \infty$:
{{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = x
| mo= <
| r = y
}}
{{eqn | ll= \leadsto
| l = 0
| o = \le
| m = \frac x y
| mo= <
| r = 1
}}
{{eqn | ll= \leadsto
| o =
| m = \floor {\frac x y}
| mo= =
| r = 0
}}
{{end-eqn}}
T... | Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
Then $\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$. | As $y \to \infty$:
{{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = x
| mo= <
| r = y
}}
{{eqn | ll= \leadsto
| l = 0
| o = \le
| m = \frac x y
| mo= <
| r = 1
}}
{{eqn | ll= \leadsto
| o =
| m = \floor {\frac x y}
| mo= =
| r = 0
}}
{{end-eqn}}
... | Limit of Modulo Operation/Limit 2 | https://proofwiki.org/wiki/Limit_of_Modulo_Operation/Limit_2 | https://proofwiki.org/wiki/Limit_of_Modulo_Operation/Limit_2 | [
"Limit of Modulo Operation"
] | [
"Definition:Real Number",
"Definition:Modulo Operation"
] | [
"Definition:Modulo Operation"
] |
proofwiki-15768 | Limit of Modulo Operation | Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then holding $x$ fixed gives:
:$\ds \lim_{y \mathop \to 0} x \bmod y = 0$
:$\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$ | === Limit 1 ===
{{:Limit of Modulo Operation/Limit 1}} | Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
Then holding $x$ fixed gives:
:$\ds \lim_{y \mathop \to 0} x \bmod y = 0$
:$\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$ | === [[Limit of Modulo Operation/Limit 1|Limit 1]] ===
{{:Limit of Modulo Operation/Limit 1}} | Limit of Modulo Operation | https://proofwiki.org/wiki/Limit_of_Modulo_Operation | https://proofwiki.org/wiki/Limit_of_Modulo_Operation | [
"Limits of Real Functions",
"Limit of Modulo Operation"
] | [
"Definition:Real Number",
"Definition:Modulo Operation"
] | [
"Limit of Modulo Operation/Limit 1"
] |
proofwiki-15769 | Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary | Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0_R$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n} = \norm {x_m}$
=== Corollary ===
{{:Non-Null Cauchy Sequence in P-ad... | By Cauchy Sequence Is Eventually Bounded Away From Non-Limit then:
:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$
Let $N = \max \set {N_1, N_2}$.
Let $n, m \ge N$.
Th... | Let $\struct {R, \norm{\,\cdot\,} }$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]] with [[Definition:Ring Zero|zero]] $0_R$
Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] such that $\sequence {x_n}$ does not [[Definition:C... | By [[Cauchy Sequence Is Eventually Bounded Away From Non-Limit]] then:
:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
Since $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] then:
:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_... | Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary | https://proofwiki.org/wiki/Non-Null_Cauchy_Sequence_in_Non-Archimedean_Norm_is_Eventually_Stationary | https://proofwiki.org/wiki/Non-Null_Cauchy_Sequence_in_Non-Archimedean_Norm_is_Eventually_Stationary | [
"Normed Division Rings",
"Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary"
] | [
"Definition:Non-Archimedean/Norm (Division Ring)",
"Definition:Ring Zero",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary/P-adic Norm"
] | [
"Cauchy Sequence Is Eventually Bounded Away From Non-Limit",
"Definition:Cauchy Sequence/Normed Division Ring"
] |
proofwiki-15770 | Equivalence of Definitions of Unit of Ring | Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
{{TFAE|def = Unit of Ring}} | Let $\struct {R, +, \circ}$ be a ring with unity. | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
{{TFAE|def = Unit of Ring}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]]. | Equivalence of Definitions of Unit of Ring | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Unit_of_Ring | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Unit_of_Ring | [
"Units of Rings"
] | [
"Definition:Ring with Unity",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Ring with Unity"
] |
proofwiki-15771 | Bézout's Identity/Euclidean Domain | Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$ and whose unity is $1$.
Let $\nu: D \setminus \set 0 \to \N$ be the Euclidean valuation on $D$.
Let $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.
Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.
Then:
:$\exists x, ... | {{Proofread}}
We are given that $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.
{{WLOG}}, suppose specifically that $b \ne 0$.
Let $S \subseteq D$ be the set defined as:
:$S = \set {x \in D_{\ne 0}: x = m \times a + n \times b: m, n \in D}$
where $D_{\ne 0}$ denotes $D \setminus 0$.
Setting $m = 0$ and $n... | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$ and whose [[Definition:Unity of Ring|unity]] is $1$.
Let $\nu: D \setminus \set 0 \to \N$ be the [[Definition:Euclidean Valuation|Euclidean valuation]] on $D$.
Let $a, b \in D$ such that $a$ a... | {{Proofread}}
We are given that $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.
{{WLOG}}, suppose specifically that $b \ne 0$.
Let $S \subseteq D$ be the [[Definition:Set|set]] defined as:
:$S = \set {x \in D_{\ne 0}: x = m \times a + n \times b: m, n \in D}$
where $D_{\ne 0}$ denotes $D \setminus 0$... | Bézout's Identity/Euclidean Domain | https://proofwiki.org/wiki/Bézout's_Identity/Euclidean_Domain | https://proofwiki.org/wiki/Bézout's_Identity/Euclidean_Domain | [
"Euclidean Domains",
"Greatest Common Divisor",
"Bézout's Identity"
] | [
"Definition:Euclidean Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Euclidean Domain/Valuation",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Element"
] | [
"Definition:Set",
"Definition:Image (Set Theory)/Mapping/Subset",
"Well-Ordering Principle",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Contradiction",
"Definition:Element",
"Definition:Minimal/Element",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Eleme... |
proofwiki-15772 | Euclidean Valuation of Non-Unit is less than that of Product | Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $b, c \in D_{\ne 0}$.
Then:
:If $c$ is not a unit of $D$ then $\map \nu b < \map \nu {b c}$ | By principal ideal domain, $D$ is a principal ideal domain
Consider the principal ideal $U = \ideal b$ of $D$ generated by $b$.
As $\nu$ is a valuation function, we have:
:$\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$
As $U$ is a principal ideal:
:$\forall a \in \ideal b: \exists x \in D: a = x \time... | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$, and [[Definition:Unity of Ring|unity]] is $1$.
Let the [[Definition:Euclidean Valuation|valuation function]] of $D$ be $\nu$.
Let $b, c \in D_{\ne 0}$.
Then:
:If $c$ is not a [[Definition:U... | By [[Euclidean Domain is Principal Ideal Domain|principal ideal domain]], $D$ is a [[Definition:Principal Ideal Domain|principal ideal domain]]
Consider the [[Definition:Principal Ideal of Ring|principal ideal]] $U = \ideal b$ of $D$ generated by $b$.
As $\nu$ is a [[Definition:Euclidean Valuation|valuation function]... | Euclidean Valuation of Non-Unit is less than that of Product | https://proofwiki.org/wiki/Euclidean_Valuation_of_Non-Unit_is_less_than_that_of_Product | https://proofwiki.org/wiki/Euclidean_Valuation_of_Non-Unit_is_less_than_that_of_Product | [
"Euclidean Domains"
] | [
"Definition:Euclidean Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Euclidean Domain/Valuation",
"Definition:Unit of Ring"
] | [
"Euclidean Domain is Principal Ideal Domain",
"Definition:Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Definition:Euclidean Domain/Valuation",
"Definition:Principal Ideal of Ring",
"Euclidean Domain is Principal Ideal Domain",
"Definition:Generator of Ideal of Ring",
"Definition:Div... |
proofwiki-15773 | Element is Unit iff its Euclidean Valuation equals that of 1 | Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $a \in D$.
Then:
:$a$ is a unit of $D$ {{iff}} $\map \nu a = \map \nu 1$ | For $a \in D$ we have that:
:$\map \nu 1 \le \map \nu {1 a} = \map \nu a$
by definition of Euclidean valuation.
Let $a$ be a unit of $D$.
Then:
:$\exists b \in D: a b = 1$
Then:
:$\map \nu a \le \map \nu {a b} = \map \nu 1$
and so:
:$\map \nu a = \map \nu 1$
{{qed|lemma}}
Let $\map \nu a = \map \nu 1$.
We can write thi... | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$, and [[Definition:Unity of Ring|unity]] is $1$.
Let the [[Definition:Euclidean Valuation|valuation function]] of $D$ be $\nu$.
Let $a \in D$.
Then:
:$a$ is a [[Definition:Unit of Ring|unit]]... | For $a \in D$ we have that:
:$\map \nu 1 \le \map \nu {1 a} = \map \nu a$
by definition of [[Definition:Euclidean Valuation|Euclidean valuation]].
Let $a$ be a [[Definition:Unit of Ring|unit]] of $D$.
Then:
:$\exists b \in D: a b = 1$
Then:
:$\map \nu a \le \map \nu {a b} = \map \nu 1$
and so:
:$\map \nu a = \map \... | Element is Unit iff its Euclidean Valuation equals that of 1 | https://proofwiki.org/wiki/Element_is_Unit_iff_its_Euclidean_Valuation_equals_that_of_1 | https://proofwiki.org/wiki/Element_is_Unit_iff_its_Euclidean_Valuation_equals_that_of_1 | [
"Euclidean Domains"
] | [
"Definition:Euclidean Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Euclidean Domain/Valuation",
"Definition:Unit of Ring"
] | [
"Definition:Euclidean Domain/Valuation",
"Definition:Unit of Ring",
"Euclidean Valuation of Non-Unit is less than that of Product",
"Definition:Unit of Ring"
] |
proofwiki-15774 | Gauss's Lemma on Primitive Polynomials over Ring | Let $R$ be a commutative ring with unity.
Let $f, g \in R \sqbrk X$ be primitive polynomials.
{{explain|Definition:Primitive Polynomial over Ring}}
Then $f g$ is primitive. | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
Category:Polynomial Theory
Category:Gauss's Lemma (Polynomial Theory)
flfxswga8xeuuwnzrf368tz9ymcj507 | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $f, g \in R \sqbrk X$ be [[Definition:Primitive Polynomial over Ring|primitive polynomials]].
{{explain|[[Definition:Primitive Polynomial over Ring]]}}
Then $f g$ is [[Definition:Primitive Polynomial over Ring|primitive]]. | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
[[Category:Polynomial Theory]]
[[Category:Gauss's Lemma (Polynomial Theory)]]
flfxswga8xeuuwnzrf368tz9ymcj507 | Gauss's Lemma on Primitive Polynomials over Ring | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Polynomials_over_Ring | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Polynomials_over_Ring | [
"Polynomial Theory",
"Gauss's Lemma (Polynomial Theory)"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Primitive Polynomial (Ring Theory)"
] | [
"Category:Polynomial Theory",
"Category:Gauss's Lemma (Polynomial Theory)"
] |
proofwiki-15775 | Rational Polynomial is Content Times Primitive Polynomial/Existence | Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial. | Consider the coefficients of $f$ expressed as fractions.
Let $k$ be any positive integer that is divisible by the denominators of all the coefficients of $f$.
Such a number is bound to exist: just multiply all those denominators together, for example.
Then $\map f X$ is a polynomial equal to $\dfrac 1 k$ multiplied by ... | Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
w... | Consider the [[Definition:Polynomial Coefficient|coefficients]] of $f$ expressed as [[Definition:Rational Number|fractions]].
Let $k$ be any [[Definition:Positive Integer|positive integer]] that is [[Definition:Divisor of Integer|divisible]] by the [[Definition:Denominator|denominators]] of all the [[Definition:Polyno... | Rational Polynomial is Content Times Primitive Polynomial/Existence | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial/Existence | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial/Existence | [
"Rational Polynomial is Content Times Primitive Polynomial"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)"
] | [
"Definition:Coefficient of Polynomial",
"Definition:Rational Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Fraction/Denominator",
"Definition:Coefficient of Polynomial",
"Definition:Greatest Common Divisor/Integers",
"Definition:Primitive Polynomial (Ring ... |
proofwiki-15776 | Rational Polynomial is Content Times Primitive Polynomial/Uniqueness | Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$ be given.
Then there exist unique content $\cont f$ of $\map f X$ and primitive polynomial $\map {f^*} X$ such that:
:$\map f X = \cont f \, \map {f^*} X$ | Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are primitive.
Then:
:$\map g X = \dfrac a b \map f X$
where $\dfrac a b$ is some rational number which can be expressed as $\dfrac m n$ where $m$ and $n$ are coprime.
Then:
:$\map g X = \dfrac m n \map f X$
that is:
:$m \cdot \map f X = ... | Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X \in \Q \sqbrk X$ be given.
Then there exist [[Definition:Unique... | Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are [[Definition:Primitive Polynomial (Ring Theory)|primitive]].
Then:
:$\map g X = \dfrac a b \map f X$
where $\dfrac a b$ is some [[Definition:Rational Number|rational number]] which can be expressed as $\dfrac m n$ where $m$ and $n$ a... | Rational Polynomial is Content Times Primitive Polynomial/Uniqueness | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial/Uniqueness | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial/Uniqueness | [
"Rational Polynomial is Content Times Primitive Polynomial"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Unique",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)"
] | [
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Rational Number",
"Definition:Coprime/Integers",
"Euclid's Lemma",
"Definition:Divisor (Algebra)/Integer",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Unique"
] |
proofwiki-15777 | Content of Polynomial in Dedekind Domain is Multiplicative | Let $R$ be a Dedekind domain.
Let $f, g \in R \sqbrk X$ be polynomials.
Let $\cont f$ denote the content of $f$.
Then $\cont {f g} = \cont f \cont g$ is the product of $\cont f$ and $\cont g$. | {{ProofWanted}}
Category:Content of Polynomial
Category:Gauss's Lemma (Polynomial Theory)
6r0v780lamw3ass7o2x8pu1nvetx5f8 | Let $R$ be a [[Definition:Dedekind Domain|Dedekind domain]].
Let $f, g \in R \sqbrk X$ be [[Definition:Polynomial over Ring in One Variable|polynomials]].
Let $\cont f$ denote the [[Definition:Content of Polynomial|content]] of $f$.
Then $\cont {f g} = \cont f \cont g$ is the [[Definition:Product of Ideals of Ring|... | {{ProofWanted}}
[[Category:Content of Polynomial]]
[[Category:Gauss's Lemma (Polynomial Theory)]]
6r0v780lamw3ass7o2x8pu1nvetx5f8 | Content of Polynomial in Dedekind Domain is Multiplicative | https://proofwiki.org/wiki/Content_of_Polynomial_in_Dedekind_Domain_is_Multiplicative | https://proofwiki.org/wiki/Content_of_Polynomial_in_Dedekind_Domain_is_Multiplicative | [
"Content of Polynomial",
"Gauss's Lemma (Polynomial Theory)"
] | [
"Definition:Dedekind Domain",
"Definition:Polynomial over Ring/One Variable",
"Definition:Content of Polynomial",
"Definition:Product of Ideals of Ring"
] | [
"Category:Content of Polynomial",
"Category:Gauss's Lemma (Polynomial Theory)"
] |
proofwiki-15778 | Factors of Polynomial with Integer Coefficients have Integer Coefficients | Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map h X \in \Q \sqbrk X$ have coefficients all of which are integers.
Let it be possible to express $\map h X$ as:
:$\map h X = \map f X \, \map g X$
where $\map f X, \map g X \in \Q \sqbrk X$.
Then it i... | Let $\cont h$ denote the content of $\map h X$.
From Polynomial has Integer Coefficients iff Content is Integer:
:$\cont h \in \Z$
Let $\map h X = \map f X \, \map g X$ as suggested.
Then from Rational Polynomial is Content Times Primitive Polynomial:
{{begin-eqn}}
{{eqn | l = \map h X
| r = \cont f \cont g \cdot... | Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map h X \in \Q \sqbrk X$ have [[Definition:Polynomial Coefficient|coeffi... | Let $\cont h$ denote the [[Definition:Content of Rational Polynomial|content]] of $\map h X$.
From [[Polynomial has Integer Coefficients iff Content is Integer]]:
:$\cont h \in \Z$
Let $\map h X = \map f X \, \map g X$ as suggested.
Then from [[Rational Polynomial is Content Times Primitive Polynomial]]:
{{begin-eq... | Factors of Polynomial with Integer Coefficients have Integer Coefficients | https://proofwiki.org/wiki/Factors_of_Polynomial_with_Integer_Coefficients_have_Integer_Coefficients | https://proofwiki.org/wiki/Factors_of_Polynomial_with_Integer_Coefficients_have_Integer_Coefficients | [
"Polynomial Theory"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Coefficient of Polynomial",
"Definition:Integer",
"Definition:Coefficient of Polynomial",
"Definition:Integer"
] | [
"Definition:Content of Polynomial/Rational",
"Polynomial has Integer Coefficients iff Content is Integer",
"Rational Polynomial is Content Times Primitive Polynomial",
"Rational Polynomial is Content Times Primitive Polynomial",
"Content of Rational Polynomial is Multiplicative",
"Definition:Primitive Pol... |
proofwiki-15779 | Polynomial which is Irreducible over Integers is Irreducible over Rationals | Let $\Z \sqbrk X$ be the ring of polynomial forms over the integers in the indeterminate $X$.
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Z \sqbrk X$ be irreducible in $\Z \sqbrk X$.
Then $\map f X$ is also irreducible in $\Q \sqbrk X... | {{AimForCont}} $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.
By hypothesis:
:$\map f X \in \Z \sqbrk X$
and so by definition has coefficients all of which are integers.
But from Factors of Polynomial with Integer Coefficients have Integer Coefficients it follows that $\map f X$ can be... | Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Integer|integers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:F... | {{AimForCont}} $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.
[[Definition:By Hypothesis|By hypothesis]]:
:$\map f X \in \Z \sqbrk X$
and so by definition has [[Definition:Polynomial Coefficient|coefficients]] all of which are [[Definition:Integer|integers]].
But from [[Factors of Po... | Polynomial which is Irreducible over Integers is Irreducible over Rationals | https://proofwiki.org/wiki/Polynomial_which_is_Irreducible_over_Integers_is_Irreducible_over_Rationals | https://proofwiki.org/wiki/Polynomial_which_is_Irreducible_over_Integers_is_Irreducible_over_Rationals | [
"Polynomial Theory"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Integer",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Irreducible Polynomial",
"Definition:Irreducible Polynomial"
] | [
"Definition:By Hypothesis",
"Definition:Coefficient of Polynomial",
"Definition:Integer",
"Factors of Polynomial with Integer Coefficients have Integer Coefficients",
"Definition:Element",
"Definition:Coefficient of Polynomial",
"Definition:Integer",
"Definition:Contradiction",
"Definition:Irreducib... |
proofwiki-15780 | Dirichlet Function is Periodic | Let $D: \R \to \R$ be a Dirichlet function:
:$\forall x \in \R: \map D x = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$
Then $D$ is periodic.
Namely, every non-zero rational number is a periodic element of $D$. | Let $x \in \R$.
Let $L \in \Q$.
If $x \in \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
| r = c
| c = Rational Addition is Closed
}}
{{eqn | r = \map D x
}}
{{end-eqn}}
If $x \notin \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
| r = d
| c = Rational Number plus Irrational Number is ... | Let $D: \R \to \R$ be a [[Definition:Dirichlet Function|Dirichlet function]]:
:$\forall x \in \R: \map D x = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$
Then $D$ is [[Definition:Real Periodic Function|periodic]].
Namely, every non-[[Definition:Zero (Number)|zero]] [[Definition:Rational Number|rati... | Let $x \in \R$.
Let $L \in \Q$.
If $x \in \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
| r = c
| c = [[Rational Addition is Closed]]
}}
{{eqn | r = \map D x
}}
{{end-eqn}}
If $x \notin \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
| r = d
| c = [[Rational Number plus Irrational ... | Dirichlet Function is Periodic | https://proofwiki.org/wiki/Dirichlet_Function_is_Periodic | https://proofwiki.org/wiki/Dirichlet_Function_is_Periodic | [
"Periodic Functions",
"Dirichlet Functions"
] | [
"Definition:Dirichlet Function",
"Definition:Periodic Function/Real",
"Definition:Zero (Number)",
"Definition:Rational Number",
"Definition:Periodic Function/Periodic Element"
] | [
"Rational Addition is Closed",
"Rational Number plus Irrational Number is Irrational",
"Category:Periodic Functions",
"Category:Dirichlet Functions"
] |
proofwiki-15781 | Dirichlet Function has no Period | The Dirichlet functions are periodic by Dirichlet Function is Periodic.
However, they do not admit a period.
That is, there does not exist a smallest value $L \in \R_{> 0}$ such that:
:$\forall x \in \R: \map D x = \map D {x + L}$ | Let $D: \R \to \R$ be a Dirichlet function.
In proving that the Dirichlet Function is Periodic, it was shown that every non-zero rational number is a periodic element of $D$.
Therefore, the period of $D$ must be the smallest element of $\Q_{> 0}$.
But from Rational Numbers are not Well-Ordered under Conventional Orderi... | The [[Definition:Dirichlet Function|Dirichlet functions]] are [[Definition:Real Periodic Function|periodic]] by [[Dirichlet Function is Periodic]].
However, they do not admit a [[Definition:Period of Periodic Real Function|period]].
That is, there does not exist a [[Definition:Smallest Element|smallest]] value $L \i... | Let $D: \R \to \R$ be a [[Definition:Dirichlet Function|Dirichlet function]].
In proving that the [[Dirichlet Function is Periodic]], it was shown that every [[Definition:Nonzero|non-zero]] [[Definition:Rational Number|rational number]] is a [[Definition:Periodic Element|periodic element]] of $D$.
Therefore, the [[D... | Dirichlet Function has no Period | https://proofwiki.org/wiki/Dirichlet_Function_has_no_Period | https://proofwiki.org/wiki/Dirichlet_Function_has_no_Period | [
"Periodic Functions",
"Dirichlet Functions"
] | [
"Definition:Dirichlet Function",
"Definition:Periodic Function/Real",
"Dirichlet Function is Periodic",
"Definition:Periodic Real Function/Period",
"Definition:Smallest Element"
] | [
"Definition:Dirichlet Function",
"Dirichlet Function is Periodic",
"Definition:Nonzero",
"Definition:Rational Number",
"Definition:Periodic Function/Periodic Element",
"Definition:Periodic Real Function/Period",
"Definition:Smallest Element",
"Rational Numbers are not Well-Ordered under Conventional O... |
proofwiki-15782 | Existence of Nonconstant Periodic Function with no Period | There exists a real, non-constant function $f$ such that:
:$(1): \quad f$ is periodic.
:$(2): \quad f$ does '''not''' have a period. | By Dirichlet Function is Periodic and Dirichlet Function has no Period, it is seen that the Dirichlet functions are such an example.
{{qed}}
Category:Periodic Functions
ouf4sp5a3l3br6w4gpr3o3zkq48s3a0 | There exists a [[Definition:Real Function|real]], [[Definition:Nonconstant Function|non-constant function]] $f$ such that:
:$(1): \quad f$ is [[Definition:Real Periodic Function|periodic]].
:$(2): \quad f$ does '''not''' have a [[Definition:Period of Periodic Real Function|period]]. | By [[Dirichlet Function is Periodic]] and [[Dirichlet Function has no Period]], it is seen that the [[Definition:Dirichlet Function|Dirichlet functions]] are such an example.
{{qed}}
[[Category:Periodic Functions]]
ouf4sp5a3l3br6w4gpr3o3zkq48s3a0 | Existence of Nonconstant Periodic Function with no Period | https://proofwiki.org/wiki/Existence_of_Nonconstant_Periodic_Function_with_no_Period | https://proofwiki.org/wiki/Existence_of_Nonconstant_Periodic_Function_with_no_Period | [
"Periodic Functions"
] | [
"Definition:Real Function",
"Definition:Nonconstant Function",
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period"
] | [
"Dirichlet Function is Periodic",
"Dirichlet Function has no Period",
"Definition:Dirichlet Function",
"Category:Periodic Functions"
] |
proofwiki-15783 | Vector Augend plus Addend equals Augend implies Addend is Zero | Let $V$ be a vector space over a field $F$.
Let $\mathbf a, \mathbf b \in V$.
Let $\mathbf a + \mathbf b = \mathbf a$.
Then:
:$\mathbf b = \bszero$
where $\bszero$ is the zero vector of $V$. | {{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
| r = \mathbf a
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}
| r = \paren {-\mathbf a} + \mathbf a
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b
... | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
Let $\mathbf a, \mathbf b \in V$.
Let $\mathbf a + \mathbf b = \mathbf a$.
Then:
:$\mathbf b = \bszero$
where $\bszero$ is the [[Definition:Zero Vector|zero vector]] of $V$. | {{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
| r = \mathbf a
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}
| r = \paren {-\mathbf a} + \mathbf a
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b
... | Vector Augend plus Addend equals Augend implies Addend is Zero | https://proofwiki.org/wiki/Vector_Augend_plus_Addend_equals_Augend_implies_Addend_is_Zero | https://proofwiki.org/wiki/Vector_Augend_plus_Addend_equals_Augend_implies_Addend_is_Zero | [
"Vector Spaces"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Zero Vector"
] | [] |
proofwiki-15784 | Vector Space of All Mappings is Vector Space | Let $\struct {K, +, \circ}$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the vector space of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_K$ is a $K$-vector space. | Follows directly from Module of All Mappings is Module and the definition of vector space. | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $S$ be a [[Definition:Set|set]].
Let $\struct {G^S, +_G', \circ}_R$ be the [[Definition:Vector Space of All Mappings|vector space of all mappings]] from ... | Follows directly from [[Module of All Mappings is Module]] and the definition of [[Definition:Vector Space|vector space]]. | Vector Space of All Mappings is Vector Space | https://proofwiki.org/wiki/Vector_Space_of_All_Mappings_is_Vector_Space | https://proofwiki.org/wiki/Vector_Space_of_All_Mappings_is_Vector_Space | [
"Examples of Vector Spaces"
] | [
"Definition:Division Ring",
"Definition:Vector Space",
"Definition:Set",
"Definition:Vector Space of All Mappings",
"Definition:Vector Space"
] | [
"Module of All Mappings is Module",
"Definition:Vector Space"
] |
proofwiki-15785 | Unitary Module of All Mappings is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is a unitary module. | From Module of All Mappings is Module, we have that $\struct {G^S, +_G', \circ}_R$ is an $R$-module.
To show that $\struct {G^S, +_G', \circ}_R$ is a unitary $R$-module, we verify the following:
:$\forall f \in G^S: 1_R \circ f = f$
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then:
{{begin-eqn}}
{{eqn | q ... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Module over Ring|unitary $R$-module]].
Let $S$ be a [[Definition:Set|set]].
Let $\struct {G^S, +_G', \circ}_R$ be the [[Def... | From [[Module of All Mappings is Module]], we have that $\struct {G^S, +_G', \circ}_R$ is an [[Definition:Module over Ring|$R$-module]].
To show that $\struct {G^S, +_G', \circ}_R$ is a [[Definition:Unitary Module over Ring|unitary $R$-module]], we verify the following:
:$\forall f \in G^S: 1_R \circ f = f$
Let $\... | Unitary Module of All Mappings is Unitary Module | https://proofwiki.org/wiki/Unitary_Module_of_All_Mappings_is_Unitary_Module | https://proofwiki.org/wiki/Unitary_Module_of_All_Mappings_is_Unitary_Module | [
"Examples of Unitary Modules"
] | [
"Definition:Ring with Unity",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unitary Module over Ring",
"Definition:Set",
"Definition:Module of All Mappings",
"Definition:Unitary Module over Ring"
] | [
"Module of All Mappings is Module",
"Definition:Module over Ring",
"Definition:Unitary Module over Ring",
"Definition:Unitary Module over Ring"
] |
proofwiki-15786 | Finite Direct Product of Unitary Modules is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be unitary $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a unitary module. | This is a special case of Direct Product of Unitary Modules is Unitary Module. | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Unitary Module over Ring|unitary $R$-modules]].
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their [[Definition... | This is a special case of [[Direct Product of Unitary Modules is Unitary Module]]. | Finite Direct Product of Unitary Modules is Unitary Module/Proof 1 | https://proofwiki.org/wiki/Finite_Direct_Product_of_Unitary_Modules_is_Unitary_Module | https://proofwiki.org/wiki/Finite_Direct_Product_of_Unitary_Modules_is_Unitary_Module/Proof_1 | [
"Finite Direct Product of Unitary Modules is Unitary Module",
"Unitary Modules",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Unitary Module over Ring",
"Definition:Module Direct Product",
"Definition:Unitary Module over Ring"
] | [
"Direct Product of Unitary Modules is Unitary Module"
] |
proofwiki-15787 | Finite Direct Product of Unitary Modules is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be unitary $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a unitary module. | From Finite Direct Product of Modules is Module we have that $G$ is a module.
It remains to be shown that:
:$\forall x \in G: 1_R \circ x = x$
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
Then:
{{begin-eqn}}
{{eqn | l = 1_R \circ x
| r = 1_R \circ \tuple {x_1, x_2, \ldots, x_n}
| c =
}}
{{eqn | r = \tup... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Unitary Module over Ring|unitary $R$-modules]].
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their [[Definition... | From [[Finite Direct Product of Modules is Module]] we have that $G$ is a [[Definition:Module over Ring|module]].
It remains to be shown that:
:$\forall x \in G: 1_R \circ x = x$
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
Then:
{{begin-eqn}}
{{eqn | l = 1_R \circ x
| r = 1_R \circ \tuple {x_1, x_2, \ld... | Finite Direct Product of Unitary Modules is Unitary Module/Proof 2 | https://proofwiki.org/wiki/Finite_Direct_Product_of_Unitary_Modules_is_Unitary_Module | https://proofwiki.org/wiki/Finite_Direct_Product_of_Unitary_Modules_is_Unitary_Module/Proof_2 | [
"Finite Direct Product of Unitary Modules is Unitary Module",
"Unitary Modules",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Unitary Module over Ring",
"Definition:Module Direct Product",
"Definition:Unitary Module over Ring"
] | [
"Finite Direct Product of Modules is Module",
"Definition:Module over Ring"
] |
proofwiki-15788 | Nonconstant Periodic Function with no Period is Discontinuous Everywhere | Let $f$ be a real periodic function that does not have a period.
Then $f$ is either constant or discontinuous everywhere. | Let $f$ be a real periodic function that does not have a period.
Let $x \in \R$.
If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by Constant Function has no Period.
If $f$ is non-constant, then let $y$ be such a value.
Let $\LL_{f > 0}$ be the set of all positive periodic elemen... | Let $f$ be a [[Definition:Real Periodic Function|real periodic function]] that does not have a [[Definition:Period of Periodic Real Function|period]].
Then $f$ is either [[Definition:Constant Function|constant]] or [[Definition:Discontinuous Real Function/Everywhere|discontinuous everywhere]]. | Let $f$ be a [[Definition:Real Periodic Function|real periodic function]] that does not have a [[Definition:Period of Periodic Real Function|period]].
Let $x \in \R$.
If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by [[Constant Function has no Period]].
If $f$ is [[Definitio... | Nonconstant Periodic Function with no Period is Discontinuous Everywhere | https://proofwiki.org/wiki/Nonconstant_Periodic_Function_with_no_Period_is_Discontinuous_Everywhere | https://proofwiki.org/wiki/Nonconstant_Periodic_Function_with_no_Period_is_Discontinuous_Everywhere | [
"Periodic Functions"
] | [
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period",
"Definition:Constant Mapping",
"Definition:Discontinuous Real Function/Everywhere"
] | [
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period",
"Constant Function has no Period",
"Definition:Nonconstant Function",
"Definition:Set",
"Definition:Positive/Real Number",
"Definition:Periodic Function/Periodic Element",
"Definition:Non-Empty Set",
"Absolute Value of ... |
proofwiki-15789 | Module on Cartesian Product is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module. | This is a special case of Direct Product of Modules is Module.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is an [[Definition:Module over Ring|$R$-module]]. | This is a special case of [[Direct Product of Modules is Module]].
{{qed}} | Module on Cartesian Product is Module/Proof 1 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module/Proof_1 | [
"Module on Cartesian Product is Module",
"Module on Cartesian Product",
"Examples of Modules",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module on Cartesian Product",
"Definition:Module over Ring"
] | [
"Direct Product of Modules is Module"
] |
proofwiki-15790 | Module on Cartesian Product is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module. | This is a special case of the Module of All Mappings, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is an [[Definition:Module over Ring|$R$-module]]. | This is a special case of the [[Definition:Module of All Mappings|Module of All Mappings]], where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}} | Module on Cartesian Product is Module/Proof 2 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module/Proof_2 | [
"Module on Cartesian Product is Module",
"Module on Cartesian Product",
"Examples of Modules",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module on Cartesian Product",
"Definition:Module over Ring"
] | [
"Definition:Module of All Mappings"
] |
proofwiki-15791 | Module on Cartesian Product is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module. | This is a special case of a Finite Direct Product of Modules is Module where each of the $G_k$ is the $R$-module $R$.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is an [[Definition:Module over Ring|$R$-module]]. | This is a special case of a [[Finite Direct Product of Modules is Module]] where each of the $G_k$ is the [[Definition:Module over Ring|$R$-module]] $R$.
{{qed}} | Module on Cartesian Product is Module/Proof 3 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_is_Module/Proof_3 | [
"Module on Cartesian Product is Module",
"Module on Cartesian Product",
"Examples of Modules",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module on Cartesian Product",
"Definition:Module over Ring"
] | [
"Finite Direct Product of Modules is Module",
"Definition:Module over Ring"
] |
proofwiki-15792 | Module on Cartesian Product of Ring with Unity is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module. | This is a special case of Direct Product of Unitary Modules is Unitary Module.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is a [[Definition:Unitary Module over Ring|unitary $R$-module]]. | This is a special case of [[Direct Product of Unitary Modules is Unitary Module]].
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 1 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_1 | [
"Module on Cartesian Product of Ring with Unity is Unitary Module",
"Examples of Unitary Modules",
"Direct Products",
"Module on Cartesian Product"
] | [
"Definition:Ring with Unity",
"Definition:Module on Cartesian Product",
"Definition:Unitary Module over Ring"
] | [
"Direct Product of Unitary Modules is Unitary Module"
] |
proofwiki-15793 | Module on Cartesian Product of Ring with Unity is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module. | This is a special case of the Unitary Module of All Mappings is Unitary Module, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is a [[Definition:Unitary Module over Ring|unitary $R$-module]]. | This is a special case of the [[Unitary Module of All Mappings is Unitary Module]], where $S$ is the [[Definition:Set|set]] $\closedint 1 n \subset \N_{>0}$.
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 2 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_2 | [
"Module on Cartesian Product of Ring with Unity is Unitary Module",
"Examples of Unitary Modules",
"Direct Products",
"Module on Cartesian Product"
] | [
"Definition:Ring with Unity",
"Definition:Module on Cartesian Product",
"Definition:Unitary Module over Ring"
] | [
"Unitary Module of All Mappings is Unitary Module",
"Definition:Set"
] |
proofwiki-15794 | Module on Cartesian Product of Ring with Unity is Unitary Module | Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module. | This is a special case of a Finite Direct Product of Unitary Modules is Unitary Module where each of the $G_k$ is the $R$-module $R$.
{{qed}} | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
Then $\struct {R^n, +, \times}_R$ is a [[Definition:Unitary Module over Ring|unitary $R$-module]]. | This is a special case of a [[Finite Direct Product of Unitary Modules is Unitary Module]] where each of the $G_k$ is the [[Definition:Module over Ring|$R$-module]] $R$.
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 3 | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_3 | [
"Module on Cartesian Product of Ring with Unity is Unitary Module",
"Examples of Unitary Modules",
"Direct Products",
"Module on Cartesian Product"
] | [
"Definition:Ring with Unity",
"Definition:Module on Cartesian Product",
"Definition:Unitary Module over Ring"
] | [
"Finite Direct Product of Unitary Modules is Unitary Module",
"Definition:Module over Ring"
] |
proofwiki-15795 | Vector Space on Cartesian Product is Vector Space | Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''$K$-vector space $K^n$'''.
Then $\struct {K^n, +, \times}_K$ is a $K$-vector space. | This is a special case of the Vector Space of All Mappings, where $S$ is the set $\closedint 1 n \subset \N^*$.
{{qed}} | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''[[Definition:Vector Space on Cartesian Product|$K$-vector space $K^n$]]'''.
Then $\struct {K^n, +, \times}_K$ is a [[Definition:Vector Space|$K$-vector space]]. | This is a special case of the [[Definition:Vector Space of All Mappings|Vector Space of All Mappings]], where $S$ is the set $\closedint 1 n \subset \N^*$.
{{qed}} | Vector Space on Cartesian Product is Vector Space/Proof 1 | https://proofwiki.org/wiki/Vector_Space_on_Cartesian_Product_is_Vector_Space | https://proofwiki.org/wiki/Vector_Space_on_Cartesian_Product_is_Vector_Space/Proof_1 | [
"Examples of Vector Spaces",
"Vector Space on Cartesian Product"
] | [
"Definition:Division Ring",
"Definition:Vector Space on Cartesian Product",
"Definition:Vector Space"
] | [
"Definition:Vector Space of All Mappings"
] |
proofwiki-15796 | Vector Space on Cartesian Product is Vector Space | Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''$K$-vector space $K^n$'''.
Then $\struct {K^n, +, \times}_K$ is a $K$-vector space. | This is a special case of a direct product of vector spaces where each of the $G_k$ is the $K$-vector space $K$.
{{qed}} | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''[[Definition:Vector Space on Cartesian Product|$K$-vector space $K^n$]]'''.
Then $\struct {K^n, +, \times}_K$ is a [[Definition:Vector Space|$K$-vector space]]. | This is a special case of a [[Definition:Direct Product of Vector Spaces|direct product of vector spaces]] where each of the $G_k$ is the [[Definition:Vector Space|$K$-vector space]] $K$.
{{qed}} | Vector Space on Cartesian Product is Vector Space/Proof 2 | https://proofwiki.org/wiki/Vector_Space_on_Cartesian_Product_is_Vector_Space | https://proofwiki.org/wiki/Vector_Space_on_Cartesian_Product_is_Vector_Space/Proof_2 | [
"Examples of Vector Spaces",
"Vector Space on Cartesian Product"
] | [
"Definition:Division Ring",
"Definition:Vector Space on Cartesian Product",
"Definition:Vector Space"
] | [
"Definition:Direct Product of Vector Spaces",
"Definition:Vector Space"
] |
proofwiki-15797 | Ring of Polynomial Forms over Field is Vector Space | Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomials over $F$.
Then $F \sqbrk X$ is a vector space over $F$. | Let the operation $\times': F \to F \sqbrk X$ be defined as follows.
Let $x \in F$.
Let $\mathbf y \in F \sqbrk X$ be defined as:
:$\mathbf y = \ds \sum_{k \mathop = 0}^n y_k X^k$
where $n = \map \deg {\mathbf y}$ denotes the degree of $\mathbf y$
Thus:
:$x \times' \mathbf y := \ds x \sum_{k \mathop = 0}^n y_k X^k = \s... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] over $F$.
Then $F \sqbrk X$ is a [[Definition:Vector Space|vector space over $F$]]. | Let the [[Definition:Binary Operation|operation]] $\times': F \to F \sqbrk X$ be defined as follows.
Let $x \in F$.
Let $\mathbf y \in F \sqbrk X$ be defined as:
:$\mathbf y = \ds \sum_{k \mathop = 0}^n y_k X^k$
where $n = \map \deg {\mathbf y}$ denotes the [[Definition:Degree of Polynomial|degree]] of $\mathbf y$
... | Ring of Polynomial Forms over Field is Vector Space | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_over_Field_is_Vector_Space | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_over_Field_is_Vector_Space | [
"Examples of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Multiplicative Identity",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Vector Space"
] | [
"Definition:Operation/Binary Operation",
"Definition:Degree of Polynomial",
"Definition:Multiplication of Polynomials",
"Definition:Multiplication/Multiplier",
"Definition:Polynomial",
"Definition:Degree of Polynomial",
"Definition:Vector Space",
"Definition:Polynomial Addition",
"Definition:Integra... |
proofwiki-15798 | No Non-Trivial Norm on Rational Numbers is Complete | No non-trivial norm on the set of the rational numbers is complete. | By P-adic Norm not Complete on Rational Numbers, no $p$-adic norm $\norm{\,\cdot\,}_p$ on the set of the rational numbers, for any prime $p$, is complete.
By Rational Number Space is not Complete Metric Space, the absolute value $\size{\,\cdot\,}$ on the set of the rational numbers is not complete.
By Norm is Complete ... | No [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Set|set]] of the [[Definition:Rational Number|rational numbers]] is [[Definition:Complete Normed Division Ring|complete]]. | By [[P-adic Norm not Complete on Rational Numbers]], no [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on the [[Definition:Set|set]] of the [[Definition:Rational Number|rational numbers]], for any [[Definition:Prime Number|prime]] $p$, is [[Definition:Complete Normed Division Ring|complete]].
By [[Ratio... | No Non-Trivial Norm on Rational Numbers is Complete | https://proofwiki.org/wiki/No_Non-Trivial_Norm_on_Rational_Numbers_is_Complete | https://proofwiki.org/wiki/No_Non-Trivial_Norm_on_Rational_Numbers_is_Complete | [
"Normed Division Rings",
"Complete Metric Spaces"
] | [
"Definition:Trivial Norm/Division Ring/Nontrivial",
"Definition:Norm/Division Ring",
"Definition:Set",
"Definition:Rational Number",
"Definition:Complete Normed Division Ring"
] | [
"P-adic Norm not Complete on Rational Numbers",
"Definition:P-adic Norm",
"Definition:Set",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Complete Normed Division Ring",
"Rational Number Space is not Complete Metric Space",
"Definition:Absolute Value",
"Definition:Set",
"Def... |
proofwiki-15799 | Norm is Complete Iff Equivalent Norm is Complete | Let $R$ be a division ring.
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.
Then:
:$\struct {R,\norm {\,\cdot\,}_1}$ is complete {{iff}} $\struct {R,\norm {\,\cdot\,}_2}$ is complete. | By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.
By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ converges in $\nor... | Let $R$ be a [[Definition:Division Ring|division ring]].
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent norms]] on $R$.
Then:
:$\struct {R,\norm {\,\cdot\,}_1}$ is [[Definition:Complete Normed Division Ring|complete]] {{iff}} $\struct {R,\norm {\,\cdot\,}... | By [[Definition:Equivalent Division Ring Norms by Cauchy Sequence|Cauchy Sequence Equivalence]], for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Cauchy Sequence ... | Norm is Complete Iff Equivalent Norm is Complete | https://proofwiki.org/wiki/Norm_is_Complete_Iff_Equivalent_Norm_is_Complete | https://proofwiki.org/wiki/Norm_is_Complete_Iff_Equivalent_Norm_is_Complete | [
"Normed Division Rings",
"Complete Metric Spaces"
] | [
"Definition:Division Ring",
"Definition:Equivalent Division Ring Norms",
"Definition:Complete Normed Division Ring",
"Definition:Complete Normed Division Ring"
] | [
"Definition:Equivalent Division Ring Norms/Cauchy Sequence Equivalent",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Equivalent Division Ring Norms/Convergently Equivalent",
"Definition:Convergent Sequence/Normed Division Ring",
"Definit... |
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