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-800 | Infinite Set is Equivalent to Proper Subset | A set is infinite {{iff}} it is equivalent to one of its proper subsets. | Let $X$ be a set which has a proper subset $Y$ such that:
:$\card X = \card Y$
where $\card X$ denotes the cardinality of $X$.
Then:
:$\exists \alpha \in \complement_X \paren Y$
and
:$Y \subsetneqq Y \cup \set \alpha \subseteq X$
The inclusion mappings:
:$i_Y: Y \to X: \forall y \in Y: i \paren y = y$
:$i_{Y \cup \set ... | A [[Definition:Set|set]] is [[Definition:Infinite Set|infinite]] {{iff}} it is [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]]. | Let $X$ be a [[Definition:Set|set]] which has a [[Definition:Proper Subset|proper subset]] $Y$ such that:
:$\card X = \card Y$
where $\card X$ denotes the [[Definition:Cardinality|cardinality]] of $X$.
Then:
:$\exists \alpha \in \complement_X \paren Y$
and
:$Y \subsetneqq Y \cup \set \alpha \subseteq X$
The [[Definit... | Infinite Set is Equivalent to Proper Subset/Proof 3 | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset/Proof_3 | [
"Infinite Set is Equivalent to Proper Subset",
"Set Equivalence",
"Infinite Sets",
"Proper Subsets"
] | [
"Definition:Set",
"Definition:Infinite Set",
"Definition:Set Equivalence",
"Definition:Proper Subset"
] | [
"Definition:Set",
"Definition:Proper Subset",
"Definition:Cardinality",
"Definition:Inclusion Mapping",
"Definition:Infinite Set",
"Definition:Infinite Set",
"Definition:Object",
"Definition:Bijection",
"Definition:Restriction/Mapping",
"Injection to Image is Bijection",
"Definition:Proper Subse... |
proofwiki-801 | Cartesian Product of Countable Sets is Countable | The cartesian product of two countable sets is countable. | Let $S_1, S_2, \ldots, S_k$ be countable sets. Equivalently, there exists an injection $f_i: S_i \to \N$ for each $i \in \{ 1, \dots, k \}$.
Therefore, the mapping $g: S_1 \times S_2 \times \cdots \times S_k \to \N^k$ defined by (for $x_i \in S_i$):
:$\map g {x_1, x_2, \dotsc, x_k} = \tuple {\map {f_1} {x_1}, \map {f_2... | The [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]]. | Let $S_1, S_2, \ldots, S_k$ be [[Definition:Countable|countable sets]]. Equivalently, there exists an [[Definition:Injection|injection]] $f_i: S_i \to \N$ for each $i \in \{ 1, \dots, k \}$.
Therefore, the [[Definition:Mapping|mapping]] $g: S_1 \times S_2 \times \cdots \times S_k \to \N^k$ defined by (for $x_i \in S_i... | Cartesian Product of Countable Sets is Countable/Corollary/Proof 1 | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable/Corollary/Proof_1 | [
"Set Theory",
"Cartesian Product",
"Countable Sets",
"Cartesian Product of Countable Sets is Countable"
] | [
"Definition:Cartesian Product",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection",
"Definition:Prime Number",
"Fundamental Theorem of Arithmetic",
"Definition:Injection",
"Composite of Injections is Injection",
"Definition:Injection"
] |
proofwiki-802 | Cartesian Product of Countable Sets is Countable | The cartesian product of two countable sets is countable. | Proof by induction:
=== Basis for the Induction ===
When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable.
So shown for basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$
where $f_k$ is a... | The [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]]. | Proof by [[Principle of Mathematical Induction|induction]]:
=== Basis for the Induction ===
When $k = 2$, the case is the same as [[Cartesian Product of Countable Sets is Countable]].
So shown for [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
=== Induction Hypothesis ===
... | Cartesian Product of Countable Sets is Countable/Corollary/Proof 2 | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable/Corollary/Proof_2 | [
"Set Theory",
"Cartesian Product",
"Countable Sets",
"Cartesian Product of Countable Sets is Countable"
] | [
"Definition:Cartesian Product",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Principle of Mathematical Induction",
"Cartesian Product of Countable Sets is Countable",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Definition:Injection",
"Definition:Injection",
"Principle of Mathematical Induction",
"Cartesian Product of Countable Sets is Counta... |
proofwiki-803 | Cartesian Product of Countable Sets is Countable | The cartesian product of two countable sets is countable. | Let $S, T$ be countable sets.
From the definition of countable, there exists a injection from $S$ to $\N$, and similarly one from $T$ to $\N$.
Hence there exists an injection $g$ from $S \times T$ to $\N^2$.
Now let us investigate the cardinality of $\N^2$.
From the Fundamental Theorem of Arithmetic, every natural numb... | The [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]]. | Let $S, T$ be [[Definition:Countable Set|countable sets]].
From the definition of [[Definition:Countable Set|countable]], there exists a [[Definition:Injection|injection]] from $S$ to $\N$, and similarly one from $T$ to $\N$.
Hence there exists an [[Definition:Injection|injection]] $g$ from $S \times T$ to $\N^2$.
N... | Cartesian Product of Countable Sets is Countable/Formal Proof 1 | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable/Formal_Proof_1 | [
"Set Theory",
"Cartesian Product",
"Countable Sets",
"Cartesian Product of Countable Sets is Countable"
] | [
"Definition:Cartesian Product",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Countable Set",
"Definition:Injection",
"Definition:Injection",
"Definition:Cardinality",
"Fundamental Theorem of Arithmetic",
"Definition:Natural Numbers",
"Definition:Prime Decomposition",
"Definition:Function",
"Definition:Injection",
"Definition:Counta... |
proofwiki-804 | Cartesian Product of Countable Sets is Countable | The cartesian product of two countable sets is countable. | Let $S = \set {s_0, s_1, s_2, \dotsc}$ and $T = \set {t_0, t_1, t_2, \dotsc}$ be countable sets.
If both $S$ and $T$ are finite, the result follows immediately.
Suppose either of $S$ or $T$ (or both) is countably infinite.
We can write the elements of $S \times T$ in the form of an infinite table:
$\quad \begin{array} ... | The [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]]. | Let $S = \set {s_0, s_1, s_2, \dotsc}$ and $T = \set {t_0, t_1, t_2, \dotsc}$ be [[Definition:Countable Set|countable sets]].
If both $S$ and $T$ are [[Definition:Finite Set|finite]], the result follows immediately.
Suppose either of $S$ or $T$ (or both) is [[Definition:Countably Infinite Set|countably infinite]].
... | Cartesian Product of Countable Sets is Countable/Informal Proof | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable/Informal_Proof | [
"Set Theory",
"Cartesian Product",
"Countable Sets",
"Cartesian Product of Countable Sets is Countable"
] | [
"Definition:Cartesian Product",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Countably Infinite/Set",
"Definition:Element",
"Definition:Infinite Set",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Bijection"
] |
proofwiki-805 | Composition of Sequence with Mapping | Let $\sequence {a_j}_{j \mathop \in B}$ be a sequence.
Let $\sigma: A \to B$ be a mapping, where $A \subseteq \N$.
Then $\sequence {a_j} \circ \sigma$ is a sequence whose value at each $k \in A$ is $a_{\map \sigma k}$.
Thus $\sequence {a_j} \circ \sigma$ is denoted $\sequence {a_{\map \sigma k} }_{k \mathop \in A}$. | By definition, a sequence is a mapping whose domain is a subset of $\N$.
Let the range of $\sequence {a_j}_{j \mathop \in B}$ be $S$.
Thus $\sequence {a_j}_{j \mathop \in B}$ can be expressed using the mapping $f: B \to S$ as:
:$\forall j \in B: \map f j = a_j$
Let $k \in A$.
Then $\map \sigma k \in B$.
By definition o... | Let $\sequence {a_j}_{j \mathop \in B}$ be a [[Definition:Sequence|sequence]].
Let $\sigma: A \to B$ be a [[Definition:Mapping|mapping]], where $A \subseteq \N$.
Then $\sequence {a_j} \circ \sigma$ is a [[Definition:Sequence|sequence]] whose value at each $k \in A$ is $a_{\map \sigma k}$.
Thus $\sequence {a_j} \cir... | By definition, a [[Definition:Sequence|sequence]] is a [[Definition:Mapping|mapping]] whose [[Definition:Domain of Mapping|domain]] is a [[Definition:Subset|subset]] of $\N$.
Let the [[Definition:Range of Sequence|range]] of $\sequence {a_j}_{j \mathop \in B}$ be $S$.
Thus $\sequence {a_j}_{j \mathop \in B}$ can be e... | Composition of Sequence with Mapping | https://proofwiki.org/wiki/Composition_of_Sequence_with_Mapping | https://proofwiki.org/wiki/Composition_of_Sequence_with_Mapping | [
"Sequences"
] | [
"Definition:Sequence",
"Definition:Mapping",
"Definition:Sequence"
] | [
"Definition:Sequence",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Subset",
"Definition:Range of Sequence",
"Definition:Mapping",
"Definition:Composition of Mappings",
"Category:Sequences"
] |
proofwiki-806 | General Operation from Binary Operation | Let $\struct {S, \oplus}$ be a magma.
Then there a unique sequence $\sequence {\oplus_k}_{k \mathop \ge 1}$ such that:
:$(1): \quad \forall n \in \N_{>0}: \oplus_n$ is an $n$-ary operation on $S$ such that:
:$(2): \quad \forall \tuple {a_1, \ldots, a_k} \in S^k: \map {\oplus_k} {a_1, \ldots, a_k} = \begin{cases}
a : & ... | Let $\Bbb S = \leftset {\odot:}$ for some $n \in \N_{>0}$, $\odot$ is an $n$-ary operation on $\rightset S$.
Let $s: \Bbb S \to \Bbb S$ be the mapping defined as follows.
Let $\odot$ be any $n$-ary operation defined on $\Bbb S$.
Then $\map s \odot$ is the $\paren {n + 1}$-ary operation defined by:
:$\forall \tuple {a_1... | Let $\struct {S, \oplus}$ be a [[Definition:Magma|magma]].
Then there a unique [[Definition:Sequence|sequence]] $\sequence {\oplus_k}_{k \mathop \ge 1}$ such that:
:$(1): \quad \forall n \in \N_{>0}: \oplus_n$ is an [[Definition:Operation on Set|$n$-ary operation]] on $S$ such that:
:$(2): \quad \forall \tuple {a_1... | Let $\Bbb S = \leftset {\odot:}$ for some $n \in \N_{>0}$, $\odot$ is an [[Definition:Operation on Set|$n$-ary operation]] on $\rightset S$.
Let $s: \Bbb S \to \Bbb S$ be the [[Definition:Mapping|mapping]] defined as follows.
Let $\odot$ be any [[Definition:Operation on Set|$n$-ary operation]] defined on $\Bbb S$.
... | General Operation from Binary Operation | https://proofwiki.org/wiki/General_Operation_from_Binary_Operation | https://proofwiki.org/wiki/General_Operation_from_Binary_Operation | [
"Abstract Algebra"
] | [
"Definition:Magma",
"Definition:Sequence",
"Definition:Operation/Operation on Set",
"Definition:Operation/Binary Operation",
"Definition:Term of Sequence",
"Definition:Sequence",
"Definition:N-Ary Operation Induced by Binary Operation"
] | [
"Definition:Operation/Operation on Set",
"Definition:Mapping",
"Definition:Operation/Operation on Set",
"Principle of Recursive Definition",
"Definition:Unique",
"Definition:Sequence",
"Definition:Operation/Unary Operation",
"Definition:Set",
"Definition:Operation/Operation on Set",
"Definition:Or... |
proofwiki-807 | Strictly Increasing Sequence induces Partition | Let $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ be a strictly increasing finite sequence of natural numbers.
Let:
:$\forall k \in \closedint 1 n: A_k := \closedint {r_{k - 1} + 1} {r_k}$
Then:
:$\set {A_k: k \in \closedint 1 n}$ is a partition of $\closedint {r_0 + 1} {r_n}$. | First we show that the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint.
Let $j \in \closedint 1 n$.
We have that:
:$\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing
and:
:$0 \le j - 1 < j \le n$
Thus by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$r_0 \le r_{j... | Let $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ be a [[Definition:Strictly Increasing Sequence|strictly increasing]] [[Definition:Finite Sequence|finite sequence]] of [[Definition:Natural Numbers|natural numbers]].
Let:
:$\forall k \in \closedint 1 n: A_k := \closedint {r_{k - 1} + 1} {r_k}$
Then:
:$\set {A_k... | First we show that the [[Definition:Element|elements]] of $\set {A_k: k \in \closedint 1 n}$ are [[Definition:Disjoint Sets|disjoint]].
Let $j \in \closedint 1 n$.
We have that:
:$\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is [[Definition:Strictly Increasing Sequence|strictly increasing]]
and:
:$0 \le j - 1 < j... | Strictly Increasing Sequence induces Partition | https://proofwiki.org/wiki/Strictly_Increasing_Sequence_induces_Partition | https://proofwiki.org/wiki/Strictly_Increasing_Sequence_induces_Partition | [
"Increasing Sequences"
] | [
"Definition:Strictly Increasing/Sequence",
"Definition:Finite Sequence",
"Definition:Natural Numbers",
"Definition:Set Partition"
] | [
"Definition:Element",
"Definition:Disjoint Sets",
"Definition:Strictly Increasing/Sequence",
"Sum with One is Immediate Successor in Naturally Ordered Semigroup",
"Definition:Strictly Increasing/Sequence",
"Definition:Disjoint Sets",
"Definition:Smallest Element",
"Sum with One is Immediate Successor ... |
proofwiki-808 | Fundamental Principle of Counting | Let $A$ be a finite set.
Let $\sequence {B_n}$ be a sequence of distinct finite subsets of $A$ which form a partition of $A$.
Let $p_k = \size {B_k}$ for each $k \in \closedint 1 n$.
Then:
:$\ds \size A = \sum_{k \mathop = 1}^n p_k$
That is, the sum of the numbers of elements in the subsets of a partition of a set is e... | Let $r_0 = 0$, and let:
:$\ds \forall k \in \closedint 1 n: r_k = \sum_{j \mathop = 1}^k {p_j}$
Then:
:$r_{k - 1} + p_k = r_k$
so:
:$r_{k - 1} < r_k$.
Thus by Isomorphism to Closed Interval, $\closedint {r_{k - 1} } {r_k}$ has $r_k - r_{k - 1} = p_k$ elements.
As a consequence, there exists a bijection $\sigma_k: B_k \... | Let $A$ be a [[Definition:Finite Set|finite set]].
Let $\sequence {B_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ which form a [[Definition:Set Partition|partition]] of $A$.
Let $p_k = \size {B_k}$ for each $k \in \closed... | Let $r_0 = 0$, and let:
:$\ds \forall k \in \closedint 1 n: r_k = \sum_{j \mathop = 1}^k {p_j}$
Then:
:$r_{k - 1} + p_k = r_k$
so:
:$r_{k - 1} < r_k$.
Thus by [[Isomorphism to Closed Interval]], $\closedint {r_{k - 1} } {r_k}$ has $r_k - r_{k - 1} = p_k$ [[Definition:Element|elements]].
As a consequence, there exis... | Fundamental Principle of Counting | https://proofwiki.org/wiki/Fundamental_Principle_of_Counting | https://proofwiki.org/wiki/Fundamental_Principle_of_Counting | [
"Fundamental Principle of Counting",
"Sequences",
"Fundamental Theorems",
"Set Theory",
"Combinatorics",
"Counting Arguments"
] | [
"Definition:Finite Set",
"Definition:Sequence of Distinct Terms",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Set Partition",
"Definition:Element",
"Definition:Subset",
"Definition:Set Partition",
"Definition:Set",
"Definition:Element",
"Definition:Set"
] | [
"Isomorphism to Closed Interval",
"Definition:Element",
"Definition:Bijection",
"Definition:Mapping",
"Strictly Increasing Sequence on Ordered Set",
"Definition:Strictly Increasing/Sequence",
"Definition:Natural Numbers",
"Strictly Increasing Sequence induces Partition",
"Definition:Bijection",
"I... |
proofwiki-809 | Odd Number Theorem | :$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
That is, the sum of the first $n$ odd numbers is the $n$th square number. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = n^2
| c = Odd Number Theorem
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
| r = n^2 + n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j}... | :$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
That is, the sum of the first $n$ [[Definition:Odd Integer|odd numbers]] is the $n$th [[Definition:Square Number|square number]]. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = n^2
| c = [[Odd Number Theorem]]
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
| r = n^2 + n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {... | Closed Form for Triangular Numbers/Proof using Odd Number Theorem | https://proofwiki.org/wiki/Odd_Number_Theorem | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_using_Odd_Number_Theorem | [
"Square Numbers",
"Proofs by Induction",
"Sums of Sequences",
"Named Theorems",
"Odd Number Theorem"
] | [
"Definition:Odd Integer",
"Definition:Square Number"
] | [
"Odd Number Theorem"
] |
proofwiki-810 | Odd Number Theorem | :$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
That is, the sum of the first $n$ odd numbers is the $n$th square number. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = 2 \sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^n 1
| c = Linear Combination of Indexed Summations
}}
{{eqn | r = 2 \paren {\frac {n \paren {n + 1} } 2} - n
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = n^2 + n - n
}}
... | :$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
That is, the sum of the first $n$ [[Definition:Odd Integer|odd numbers]] is the $n$th [[Definition:Square Number|square number]]. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = 2 \sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^n 1
| c = [[Linear Combination of Indexed Summations]]
}}
{{eqn | r = 2 \paren {\frac {n \paren {n + 1} } 2} - n
| c = [[Closed Form for Triangular Numbers]]
}}
{{eqn | r = n^2 + n... | Odd Number Theorem/Proof 2 | https://proofwiki.org/wiki/Odd_Number_Theorem | https://proofwiki.org/wiki/Odd_Number_Theorem/Proof_2 | [
"Square Numbers",
"Proofs by Induction",
"Sums of Sequences",
"Named Theorems",
"Odd Number Theorem"
] | [
"Definition:Odd Integer",
"Definition:Square Number"
] | [
"Linear Combination of Indexed Summations",
"Closed Form for Triangular Numbers"
] |
proofwiki-811 | General Associativity Theorem | If an operation is associative on $3$ entities, then it is associative on any number of them. | The cases where $n = 1$ and $n = 2$ are clear.
Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.
If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction.
Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \circ y} = a_1 \circ \paren {a_2 \circ a_3 \cir... | If an [[Definition:Operation|operation]] is [[Definition:Associative Operation|associative]] on $3$ entities, then it is [[Definition:Associative Operation|associative]] on any number of them. | The cases where $n = 1$ and $n = 2$ are clear.
Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.
If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by [[Principle of Mathematical Induction|induction]].
Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \cir... | General Associativity Theorem/Formulation 2/Proof 1 | https://proofwiki.org/wiki/General_Associativity_Theorem | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2/Proof_1 | [
"General Associativity Theorem",
"Associativity",
"Named Theorems"
] | [
"Definition:Operation",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-812 | General Associativity Theorem | If an operation is associative on $3$ entities, then it is associative on any number of them. | Proof by strong induction:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$.
$P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$.
$P \left({2}\right)$ is the cas... | If an [[Definition:Operation|operation]] is [[Definition:Associative Operation|associative]] on $3$ entities, then it is [[Definition:Associative Operation|associative]] on any number of them. | Proof by [[Second Principle of Mathematical Induction|strong induction]]:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:The [[General Associativity Theorem]] holds for all [[Definition:Composite (Abstract Algebra)|composites]] $a_1 \circ a_2 \circ \cdots \circ a_r$ ... | General Associativity Theorem/Formulation 2/Proof 2 | https://proofwiki.org/wiki/General_Associativity_Theorem | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2/Proof_2 | [
"General Associativity Theorem",
"Associativity",
"Named Theorems"
] | [
"Definition:Operation",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"General Associativity Theorem",
"Definition:Iterated Binary Operation",
"Definition:Associative Operation",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"General Associativity Theorem",
"Definit... |
proofwiki-813 | General Commutativity Theorem | Let $\struct {S, \circ}$ be a semigroup.
Let $\family {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $S$.
Suppose that:
:$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$
Then for every permutation $\sigma: \N_n \to \N_n$:
:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_... | The proof will proceed by the Principle of Mathematical Induction on $\N_{>0}$.
Let $T$ be the set of all $n \in \N_{>0}$ such that:
:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$
holds for all sequences $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of $n$ elements of $S$ whi... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\family {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Finite Sequence|sequence]] of [[Definition:Element|elements]] of $S$.
Suppose that:
:$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$
Then for every [[Definition:Pe... | The proof will proceed by the [[Principle of Mathematical Induction]] on $\N_{>0}$.
Let $T$ be the [[Definition:Set|set]] of all $n \in \N_{>0}$ such that:
:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$
holds for all [[Definition:Finite Sequence|sequences]] $\sequence {a_k}_{1 \... | General Commutativity Theorem | https://proofwiki.org/wiki/General_Commutativity_Theorem | https://proofwiki.org/wiki/General_Commutativity_Theorem | [
"Named Theorems",
"Commutativity"
] | [
"Definition:Semigroup",
"Definition:Finite Sequence",
"Definition:Element",
"Definition:Permutation",
"Definition:Initial Segment of Natural Numbers/One-Based"
] | [
"Principle of Mathematical Induction",
"Definition:Set",
"Definition:Finite Sequence",
"Definition:Element",
"Definition:Permutation",
"Definition:Finite Sequence",
"Definition:Element",
"Definition:Permutation",
"Definition:Finite Sequence",
"Definition:Element",
"Definition:Permutation",
"De... |
proofwiki-814 | General Distributivity Theorem | :$\ds \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$ | Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
We have that $\struct {R, \cir... | :$\ds \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le ... | General Distributivity Theorem | https://proofwiki.org/wiki/General_Distributivity_Theorem | https://proofwiki.org/wiki/General_Distributivity_Theorem | [
"Abstract Algebra",
"Summations",
"Named Theorems",
"Distributive Operations",
"General Distributivity Theorem"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Ringoid (Abstract Algebra)",
"Definition:Ringoid (Abstract Algebra)",
"Principle of Mathematical Induction"
] |
proofwiki-815 | Associativity on Indexing Set | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\family {x_\alpha}_{\alpha \mathop \in A}$ be a family of terms of $S$ indexed by a finite non-empty set $A$.
Let $\family {B_k}_{1 \mathop \le k \mathop \le n}$ be a family of distinct subsets of $A$ forming a partition of $A$.
Then:
:$\ds \prod_{k \mathop = 1}... | For each $k \in \N_{>0}$, let $\card {B_k} = p_k$.
Let:
:$r_0 = 0$
:$\ds \forall k \in \N_{>0}: r_k = \sum_{j \mathop = 1}^k {p_j}$
and:
:$p = r_n$
Then:
:$r_k - r_{k - 1} = p_k$
So, by Isomorphism to Closed Interval, both integer intervals $\closedint 1 {p_k}$ and $\closedint {r_{k - 1} + 1} {r_k}$ have $p_k$ elements... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\family {x_\alpha}_{\alpha \mathop \in A}$ be a [[Definition:Indexed Family|family of terms of $S$ indexed by]] a [[Definition:Finite Set|finite]] [[Definition:Non-Empty Set|non-empty set]] $A$.
Let $\family {B_k}_{1 \matho... | For each $k \in \N_{>0}$, let $\card {B_k} = p_k$.
Let:
:$r_0 = 0$
:$\ds \forall k \in \N_{>0}: r_k = \sum_{j \mathop = 1}^k {p_j}$
and:
:$p = r_n$
Then:
:$r_k - r_{k - 1} = p_k$
So, by [[Isomorphism to Closed Interval]], both [[Definition:Integer Interval|integer intervals]] $\closedint 1 {p_k}$ and $\closedint {r_... | Associativity on Indexing Set | https://proofwiki.org/wiki/Associativity_on_Indexing_Set | https://proofwiki.org/wiki/Associativity_on_Indexing_Set | [
"Indexed Families",
"Associativity"
] | [
"Definition:Commutative Semigroup",
"Definition:Indexing Set/Family",
"Definition:Finite Set",
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family of Distinct Elements",
"Definition:Subset",
"Definition:Set Partition"
] | [
"Isomorphism to Closed Interval",
"Definition:Closed Interval/Integer Interval",
"Definition:Element",
"Unique Isomorphism between Equivalent Finite Totally Ordered Sets",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Totally Ordered Set",
"Definition:Ordering",
"Definition:Bijection",
"S... |
proofwiki-816 | External Direct Product of Groups is Group | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups whose identity elements are $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the external direct product of $G_1$ and $G_2$.
Then $\struct {G_1 \times G_2, \circ}$ is a group whose identity element is $\tuple {e_1, e_2}$. | Taking the group axioms in turn: | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identity elements]] are $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|external direct product]] of $G_1$ and $G_2$.
Then $\struct... | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | External Direct Product of Groups is Group | https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group | https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group | [
"Group Direct Products",
"External Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-817 | External Direct Product of Projection with Canonical Injection | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {S_1 \times S_2, \circ}$ be the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
Let:
:$\pr_1: \struct {S_1 \times S_2, \circ} \to \struct {... | Let $\tuple {s_1, s_2} \in S_1 \times S_2$.
So, $s_1 \in S_1$ and $s_2 \in S_2$.
From the definition of the canonical injection, we have:
:$\map {\inj_1} {s_1} = \tuple {s_1, e_2}$
:$\map {\inj_2} {s_2} = \tuple {e_1, s_2}$
From the definition of the first projection:
:$\map {\pr_1} {s_1, e_2} = s_1$
and similarly from... | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identity elements]] $e_1$ and $e_2$ respectively.
Let $\struct {S_1 \times S_2, \circ}$ be the [[Definition:External Direct Product|external direct ... | Let $\tuple {s_1, s_2} \in S_1 \times S_2$.
So, $s_1 \in S_1$ and $s_2 \in S_2$.
From the definition of the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]], we have:
:$\map {\inj_1} {s_1} = \tuple {s_1, e_2}$
:$\map {\inj_2} {s_2} = \tuple {e_1, s_2}$
From the definition of the [[Definiti... | External Direct Product of Projection with Canonical Injection | https://proofwiki.org/wiki/External_Direct_Product_of_Projection_with_Canonical_Injection | https://proofwiki.org/wiki/External_Direct_Product_of_Projection_with_Canonical_Injection | [
"Canonical Injections",
"External Direct Products",
"Projections"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:External Direct Product",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Canonical Injection (Abstrac... | [
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Identity Mapping"
] |
proofwiki-818 | Inverse Completion of Natural Numbers | There exists an inverse completion of the natural numbers under addition. | The set of natural numbers under addition can be denoted $\left ({\N, +}\right)$.
From Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\left ({\N, +}\right)$ is a commutative monoid.
Therefore by definition of commutative monoid, $\left ({\N, +}\right)$ is a commutative semigroup.
From ... | There exists an [[Definition:Inverse Completion|inverse completion]] of the [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Addition|addition]]. | The set of [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Addition|addition]] can be denoted $\left ({\N, +}\right)$.
From [[Natural Numbers under Addition form Commutative Monoid]], the [[Definition:Algebraic Structure|algebraic structure]] $\left ({\N, +}\right)$ is a [[Definition:C... | Inverse Completion of Natural Numbers | https://proofwiki.org/wiki/Inverse_Completion_of_Natural_Numbers | https://proofwiki.org/wiki/Inverse_Completion_of_Natural_Numbers | [
"Inverse Completions",
"Natural Numbers"
] | [
"Definition:Inverse Completion",
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Natural Numbers under Addition form Commutative Monoid",
"Definition:Algebraic Structure",
"Definition:Commutative Monoid",
"Definition:Commutative Monoid",
"Definition:Commutative Semigroup",
"Natural Number Addition is Cancellable... |
proofwiki-819 | L'Hôpital's Rule | Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ {{WRT|Differentiation}} $x$.
Let:
:$\ds \lim_{x \mathop \to a^+} \map f x = \lim_{x \mathop \to a^+} \map g x = 0$
Then:
:$\ds \l... | Let $l = \ds \lim_{x \mathop \to a^+} \frac {\map {f'} x}{\map {g'} x}$.
Let $\epsilon \in \R_{>0}$.
By the definition of limit, we ought to find a $\delta \in \R_{>0}$ such that:
:$\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map f x} {\map g x} - l} < \epsilon$
Fix $\delta$ such that:
:$\forall x... | Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the [[Definition:Derivative|derivative]] of $... | Let $l = \ds \lim_{x \mathop \to a^+} \frac {\map {f'} x}{\map {g'} x}$.
Let $\epsilon \in \R_{>0}$.
By the definition of [[Definition:Limit of Real Function|limit]], we ought to find a $\delta \in \R_{>0}$ such that:
:$\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map f x} {\map g x} - l} < \eps... | L'Hôpital's Rule/Proof 1 | https://proofwiki.org/wiki/L'Hôpital's_Rule | https://proofwiki.org/wiki/L'Hôpital's_Rule/Proof_1 | [
"L'Hôpital's Rule",
"Differential Calculus",
"Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Real Interval/Open",
"Definition:Derivative"
] | [
"Definition:Limit of Real Function",
"Definition:Limit of Real Function",
"Definition:Continuous Real Function/Right-Continuous",
"Definition:Continuous Real Function/Half Open Interval",
"Definition:Continuous Real Function/Closed Interval",
"Definition:Differentiable Mapping/Real Function/Interval",
"... |
proofwiki-820 | L'Hôpital's Rule | Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ {{WRT|Differentiation}} $x$.
Let:
:$\ds \lim_{x \mathop \to a^+} \map f x = \lim_{x \mathop \to a^+} \map g x = 0$
Then:
:$\ds \l... | Take the Cauchy Mean Value Theorem with $b = x$:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$
Then if $\map f a = \map g a = 0$ we have:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map g x}$
... | Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the [[Definition:Derivative|derivative]] of $... | Take the [[Cauchy Mean Value Theorem]] with $b = x$:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$
Then if $\map f a = \map g a = 0$ we have:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map ... | L'Hôpital's Rule/Proof 2 | https://proofwiki.org/wiki/L'Hôpital's_Rule | https://proofwiki.org/wiki/L'Hôpital's_Rule/Proof_2 | [
"L'Hôpital's Rule",
"Differential Calculus",
"Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Real Interval/Open",
"Definition:Derivative"
] | [
"Cauchy Mean Value Theorem",
"Limit of Function in Interval",
"Limit of Composite Function"
] |
proofwiki-821 | Natural Numbers under Addition form Commutative Monoid | The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is a commutative monoid whose identity is zero. | Consider the natural numbers $\N$ defined as the naturally ordered semigroup.
From the definition of the naturally ordered semigroup, it follows that $\struct {\N, +}$ is a commutative semigroup.
From the definition of zero, $\struct {\N, +}$ has $0 \in \N$ as the identity, hence is a monoid.
{{qed}} | The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Addition|addition]] $+$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Ele... | Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
From the definition of the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]], it follows that $\struct {\N, +}$ is a [[Definition:Commutative Semi... | Natural Numbers under Addition form Commutative Monoid | https://proofwiki.org/wiki/Natural_Numbers_under_Addition_form_Commutative_Monoid | https://proofwiki.org/wiki/Natural_Numbers_under_Addition_form_Commutative_Monoid | [
"Natural Number Addition",
"Examples of Commutative Monoids"
] | [
"Definition:Algebraic Structure",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Zero (Number)"
] | [
"Definition:Natural Numbers",
"Definition:Naturally Ordered Semigroup",
"Definition:Naturally Ordered Semigroup",
"Definition:Commutative Semigroup",
"Definition:Zero (Number)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid"
] |
proofwiki-822 | Natural Numbers form Commutative Semiring | The semiring of natural numbers $\struct {\N, +, \times}$ forms a commutative semiring. | The algebraic structure $\struct {\N, +}$ is a commutative monoid from Natural Numbers under Addition form Commutative Monoid.
Then we have:
:$(1): \quad$ Natural Number Multiplication is Closed
:$(2): \quad$ Natural Number Multiplication is Associative
:$(3): \quad$ Natural Number Multiplication is Commutative
:$(4): ... | The [[Definition:Semiring of Natural Numbers|semiring of natural numbers]] $\struct {\N, +, \times}$ forms a [[Definition:Commutative Semiring|commutative semiring]]. | The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ is a [[Definition:Commutative Monoid|commutative monoid]] from [[Natural Numbers under Addition form Commutative Monoid]].
Then we have:
:$(1): \quad$ [[Natural Number Multiplication is Closed]]
:$(2): \quad$ [[Natural Number Multiplication ... | Natural Numbers form Commutative Semiring | https://proofwiki.org/wiki/Natural_Numbers_form_Commutative_Semiring | https://proofwiki.org/wiki/Natural_Numbers_form_Commutative_Semiring | [
"Natural Numbers",
"Commutative Semirings"
] | [
"Definition:Semiring of Natural Numbers",
"Definition:Commutative Semiring"
] | [
"Definition:Algebraic Structure",
"Definition:Commutative Monoid",
"Natural Numbers under Addition form Commutative Monoid",
"Natural Number Multiplication is Closed",
"Natural Number Multiplication is Associative",
"Natural Number Multiplication is Commutative",
"Natural Number Multiplication Distribut... |
proofwiki-823 | Integer Multiplication is Well-Defined | Integer multiplication is well-defined. | Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
:$\forall u \in \N_{>0}: \map \phi u = u'$
where $u' \in \Z$ is the (strictly) positive integer $\eqclass {b + u, b} {}$.
Let... | [[Definition:Integer Multiplication|Integer multiplication]] is [[Definition:Well-Defined Operation|well-defined]]. | Consider the [[Definition:Integer/Formal Definition|formal definition of the integers]]: $x = \eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
Consider the [[Definition:Mapping|mapping]] $\phi: ... | Integer Multiplication is Well-Defined/Proof 2 | https://proofwiki.org/wiki/Integer_Multiplication_is_Well-Defined | https://proofwiki.org/wiki/Integer_Multiplication_is_Well-Defined/Proof_2 | [
"Integer Multiplication is Well-Defined",
"Integer Multiplication",
"Examples of Well-Defined Mappings"
] | [
"Definition:Multiplication/Integers",
"Definition:Well-Defined/Operation"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Mapping",
"Definition:Strictly Positive/Integer"
] |
proofwiki-824 | Integer Multiplication is Closed | The set of integers is closed under multiplication:
:$\forall a, b \in \Z: a \times b \in \Z$ | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]]:
:$\forall a, b \in \Z: a \times b \in \Z$ | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Integer Multiplication is Closed | https://proofwiki.org/wiki/Integer_Multiplication_is_Closed | https://proofwiki.org/wiki/Integer_Multiplication_is_Closed | [
"Integer Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Integers"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Definition:Multiplication/Integers",
"Definition:Mult... |
proofwiki-825 | Integer Multiplication is Commutative | The operation of multiplication on the set of integers $\Z$ is commutative:
:$\forall x, y \in \Z: x \times y = y \times x$ | From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = x \times y
| r = \eqclass {a, b} {}\times \eqclass {c, d} {}
| c = {{Def... | The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Commutative Operation|commutative]]:
:$\forall x, y \in \Z: x \times y = y \times x$ | From the [[Definition:Integer/Formal Definition|formal definition of integers]], $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for s... | Integer Multiplication is Commutative | https://proofwiki.org/wiki/Integer_Multiplication_is_Commutative | https://proofwiki.org/wiki/Integer_Multiplication_is_Commutative | [
"Integer Multiplication",
"Examples of Commutative Operations",
"Commutative Law of Multiplication"
] | [
"Definition:Multiplication/Integers",
"Definition:Set",
"Definition:Integer",
"Definition:Commutative/Operation"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Natural Number Multiplication is Commutative",
"Natural Number Addition is Commutative"
] |
proofwiki-826 | Integer Multiplication is Associative | The operation of multiplication on the set of integers $\Z$ is associative:
:$\forall x, y, z \in \Z: x \times \paren {y \times z} = \paren {x \times y} \times z$ | From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$ for some $x, y, z \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = x \times \paren {y \times z}
| r = \eqclass {a, b... | The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Associative Operation|associative]]:
:$\forall x, y, z \in \Z: x \times \paren {y \times z} = \paren {x \times y} \times z$ | From the [[Definition:Integer/Formal Definition|formal definition of integers]], $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z =... | Integer Multiplication is Associative | https://proofwiki.org/wiki/Integer_Multiplication_is_Associative | https://proofwiki.org/wiki/Integer_Multiplication_is_Associative | [
"Integer Multiplication",
"Examples of Associative Operations",
"Associative Law of Multiplication"
] | [
"Definition:Multiplication/Integers",
"Definition:Set",
"Definition:Integer",
"Definition:Associative Operation"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Natural Number Multiplication Distributes over Addition",
"Natural Number Addition is Commutative",
"Natural Number Addition is Associative",
"Natural Number Multiplication... |
proofwiki-827 | Integer Multiplication Distributes over Addition | The operation of multiplication on the set of integers $\Z$ is distributive over addition:
:$\forall x, y, z \in \Z: x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
:$\forall x, y, z \in \Z: \paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$ | Let us define $\Z$ as in the formal definition of integers.
That is, $\Z$ is an inverse completion of $\N$.
From Natural Numbers form Commutative Semiring, we have that:
:All elements of $\N$ are cancellable for addition
:Addition and multiplication are commutative and associative on the natural numbers $\N$
:Natural n... | The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Distributive Operation|distributive]] over [[Definition:Integer Addition|addition]]:
:$\forall x, y, z \in \Z: x \times \paren {y + z} = \paren {x \times y} + \par... | Let us define $\Z$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\Z$ is an [[Definition:Inverse Completion|inverse completion]] of $\N$.
From [[Natural Numbers form Commutative Semiring]], we have that:
:All [[Definition:Element|elements]] of $\N$ are [[Definition:Cance... | Integer Multiplication Distributes over Addition | https://proofwiki.org/wiki/Integer_Multiplication_Distributes_over_Addition | https://proofwiki.org/wiki/Integer_Multiplication_Distributes_over_Addition | [
"Integer Addition",
"Integer Multiplication",
"Examples of Distributive Operations"
] | [
"Definition:Multiplication/Integers",
"Definition:Set",
"Definition:Integer",
"Definition:Distributive Operation",
"Definition:Addition/Integers"
] | [
"Definition:Integer/Formal Definition",
"Definition:Inverse Completion",
"Natural Numbers form Commutative Semiring",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Addition/Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"De... |
proofwiki-828 | Integer Multiplication Identity is One | The identity of integer multiplication is $1$:
:$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$ | Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by $\tuple {... | The [[Definition:Identity Element|identity]] of [[Definition:Integer Multiplication|integer multiplication]] is $1$:
:$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$ | Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Integer Multiplication Identity is One | https://proofwiki.org/wiki/Integer_Multiplication_Identity_is_One | https://proofwiki.org/wiki/Integer_Multiplication_Identity_is_One | [
"Integer Multiplication"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Multiplication/Integers"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Construction of Inverse Completion",
"Definition:Natu... |
proofwiki-829 | Integer Multiplication has Zero | The set of integers under multiplication $\struct {\Z, \times}$ has a zero element, which is $0$. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ has a [[Definition:Zero Element|zero element]], which is $0$. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Integer Multiplication has Zero | https://proofwiki.org/wiki/Integer_Multiplication_has_Zero | https://proofwiki.org/wiki/Integer_Multiplication_has_Zero | [
"Integer Multiplication"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Zero Element"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Construction of Inverse Completion/Identity of Quotient... |
proofwiki-830 | Ring of Integers has no Zero Divisors | The integers have no zero divisors:
:$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$ | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {... | The [[Definition:Integer|integers]] have no [[Definition:Zero Divisor of Ring|zero divisors]]:
:$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$ | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Ring of Integers has no Zero Divisors | https://proofwiki.org/wiki/Ring_of_Integers_has_no_Zero_Divisors | https://proofwiki.org/wiki/Ring_of_Integers_has_no_Zero_Divisors | [
"Integers"
] | [
"Definition:Integer",
"Definition:Zero Divisor/Ring"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Construction of Inverse Completion/Identity of Quotient... |
proofwiki-831 | Integers form Integral Domain | The set of integers $\Z$ form an integral domain under addition and multiplication. | First we note that the integers form a commutative ring with unity whose zero is $0$ and whose unity is $1$.
Next we see that the $\struct {\Z, +, \times}$ has no divisors of zero.
So, by definition, the algebraic structure $\struct {\Z, +, \times}$ is an integral domain whose zero is $0$ and whose unity is $1$.
{{Qed}... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ form an [[Definition:Integral Domain|integral domain]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]]. | First we note that the [[Integers form Commutative Ring with Unity|integers form a commutative ring with unity]] whose [[Integer Multiplication has Zero|zero is $0$]] and whose [[Integer Multiplication Identity is One|unity is $1$]].
Next we see that the [[Ring of Integers has no Zero Divisors|$\struct {\Z, +, \times}... | Integers form Integral Domain | https://proofwiki.org/wiki/Integers_form_Integral_Domain | https://proofwiki.org/wiki/Integers_form_Integral_Domain | [
"Integers",
"Examples of Integral Domains"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Integral Domain",
"Definition:Addition/Integers",
"Definition:Multiplication/Integers"
] | [
"Integers form Commutative Ring with Unity",
"Integer Multiplication has Zero",
"Integer Multiplication Identity is One",
"Ring of Integers has no Zero Divisors",
"Definition:Algebraic Structure",
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring"
] |
proofwiki-832 | Positive Integer is Well-Defined | "Positive" as applied to an integer is well-defined. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {... | "[[Definition:Positive Integer|Positive]]" as applied to an [[Definition:Integer|integer]] is [[Definition:Well-Defined Relation|well-defined]]. | Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natu... | Positive Integer is Well-Defined | https://proofwiki.org/wiki/Positive_Integer_is_Well-Defined | https://proofwiki.org/wiki/Positive_Integer_is_Well-Defined | [
"Integers"
] | [
"Definition:Positive/Integer",
"Definition:Integer",
"Definition:Well-Defined/Relation"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Congruence Relation",
"Definition:Congruence Relation",
"Definition:Integer/Formal Definition/Notation",
"Category:Integers"
] |
proofwiki-833 | Natural Numbers are Non-Negative Integers | Let $m \in \Z$. Then:
:$(1): \quad m \in \N \iff 0 \le m$
:$(2): \quad m \in \N_{> 0} \iff 0 < m$
:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
That is, the natural numbers are precisely those integers which are greater than or equal to zero. | Let $m \in \N$.
Then by definition of $0$:
:$0 \le m$
Conversely, let $m \in \Z: 0 \le m$.
Then:
:$\exists x, y \in \N: m = x - y$
Thus:
:$y \le m + y = x$
By Naturally Ordered Semigroup: NO 3:
:$\exists z \in \N: z + y = x = m + y$
From Naturally Ordered Semigroup: NO 2, $y$ is cancellable.
Hence:
$m = z \in \N_{> 0}$... | Let $m \in \Z$. Then:
:$(1): \quad m \in \N \iff 0 \le m$
:$(2): \quad m \in \N_{> 0} \iff 0 < m$
:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
That is, the [[Definition:Natural Numbers|natural numbers]] are precisely those [[Definition:Integer|integers]] which are greater than or equal to [[Definition:Zero (Algebr... | Let $m \in \N$.
Then by definition of $0$:
:$0 \le m$
Conversely, let $m \in \Z: 0 \le m$.
Then:
:$\exists x, y \in \N: m = x - y$
Thus:
:$y \le m + y = x$
By [[Definition:Naturally Ordered Semigroup|Naturally Ordered Semigroup: NO 3]]:
:$\exists z \in \N: z + y = x = m + y$
From [[Definition:Naturally Ordered Se... | Natural Numbers are Non-Negative Integers/Proof 1 | https://proofwiki.org/wiki/Natural_Numbers_are_Non-Negative_Integers | https://proofwiki.org/wiki/Natural_Numbers_are_Non-Negative_Integers/Proof_1 | [
"Integers",
"Natural Numbers",
"Natural Numbers are Non-Negative Integers"
] | [
"Definition:Natural Numbers",
"Definition:Integer",
"Definition:Zero (Number)"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Naturally Ordered Semigroup",
"Definition:Cancellable Element",
"Ordering of Inverses in Ordered Monoid"
] |
proofwiki-834 | Natural Numbers are Non-Negative Integers | Let $m \in \Z$. Then:
:$(1): \quad m \in \N \iff 0 \le m$
:$(2): \quad m \in \N_{> 0} \iff 0 < m$
:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
That is, the natural numbers are precisely those integers which are greater than or equal to zero. | Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Let $x \in \Z_{>0}$ be a (strictly) positive integer.
Thus by definition:
:$x > 0$
This is equivalent to the condition that $a > b$.
Hence:
:$x = \eqclass {b + u, b} {}$
for some $u \in ... | Let $m \in \Z$. Then:
:$(1): \quad m \in \N \iff 0 \le m$
:$(2): \quad m \in \N_{> 0} \iff 0 < m$
:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
That is, the [[Definition:Natural Numbers|natural numbers]] are precisely those [[Definition:Integer|integers]] which are greater than or equal to [[Definition:Zero (Algebr... | Consider the [[Definition:Integer/Formal Definition|formal definition of the integers]]: $x = \eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
Let $x \in \Z_{>0}$ be a [[Definition:Strictly Posit... | Natural Numbers are Non-Negative Integers/Proof 2 | https://proofwiki.org/wiki/Natural_Numbers_are_Non-Negative_Integers | https://proofwiki.org/wiki/Natural_Numbers_are_Non-Negative_Integers/Proof_2 | [
"Integers",
"Natural Numbers",
"Natural Numbers are Non-Negative Integers"
] | [
"Definition:Natural Numbers",
"Definition:Integer",
"Definition:Zero (Number)"
] | [
"Definition:Integer/Formal Definition",
"Definition:Equivalence Class",
"Definition:Ordered Pair",
"Definition:Natural Numbers",
"Definition:Strictly Positive/Integer",
"Definition:Mapping",
"Definition:Strictly Positive/Integer",
"Definition:Well-Defined/Mapping",
"Definition:Injection",
"Definit... |
proofwiki-835 | Subtraction on Integers is Extension of Natural Numbers | Integer subtraction is an extension of the definition of subtraction on the natural numbers. | Let $m, n \in \N: m \le n$.
From natural number subtraction, $\exists p \in \N: m + p = n$ such that $n - m = p$.
As $m, n, p \in \N$, it follows that $m, n, p \in \Z$ as well.
However, as $\Z$ is the inverse completion of $\N$, it follows that $-m \in \Z$ as well, so it makes sense to express the following:
{{begin-eq... | [[Definition:Integer Subtraction|Integer subtraction]] is an extension of the definition of [[Definition:Natural Number Subtraction|subtraction]] on the [[Definition:Natural Numbers|natural numbers]]. | Let $m, n \in \N: m \le n$.
From [[Definition:Natural Number Subtraction|natural number subtraction]], $\exists p \in \N: m + p = n$ such that $n - m = p$.
As $m, n, p \in \N$, it follows that $m, n, p \in \Z$ as well.
However, as $\Z$ is the [[Definition:Inverse Completion|inverse completion]] of $\N$, it follows t... | Subtraction on Integers is Extension of Natural Numbers | https://proofwiki.org/wiki/Subtraction_on_Integers_is_Extension_of_Natural_Numbers | https://proofwiki.org/wiki/Subtraction_on_Integers_is_Extension_of_Natural_Numbers | [
"Integers",
"Subtraction"
] | [
"Definition:Subtraction/Integers",
"Definition:Subtraction/Natural Numbers",
"Definition:Natural Numbers"
] | [
"Definition:Subtraction/Natural Numbers",
"Definition:Inverse Completion",
"Definition:Cancellable Element",
"Category:Integers",
"Category:Subtraction"
] |
proofwiki-836 | Non-Zero Integers are Cancellable for Multiplication | Every non-zero integer is cancellable for multiplication.
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.
By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication.
{{qed|le... | Every non-zero [[Definition:Integer|integer]] is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From [[Natural Numbers are Non-Negative Integers]], $x \in \N_{> 0}$.
By the [[Extension Theorem for Distributive Operations]] and [[Ordering on Natural Numbers is Compatible with Multiplication]], $x$ is [[Definition:Cancellabl... | Non-Zero Integers are Cancellable for Multiplication/Proof 1 | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication/Proof_1 | [
"Integers",
"Non-Zero Integers are Cancellable for Multiplication"
] | [
"Definition:Integer",
"Definition:Cancellable Element",
"Definition:Multiplication/Integers"
] | [
"Natural Numbers are Non-Negative Integers",
"Extension Theorem for Distributive Operations",
"Ordering on Natural Numbers is Compatible with Multiplication",
"Definition:Cancellable Element",
"Definition:Multiplication/Integers",
"Integers form Integral Domain",
"Definition:Ring (Abstract Algebra)",
... |
proofwiki-837 | Non-Zero Integers are Cancellable for Multiplication | Every non-zero integer is cancellable for multiplication.
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $y, z \in \Z: y \ne z$.
{{begin-eqn}}
{{eqn | l = y
| o = \ne
| r = z
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y - z
| o = \ne
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \paren {y - z}
| o = \ne
| r = 0
| c = Ring of Integers has no Zer... | Every non-zero [[Definition:Integer|integer]] is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $y, z \in \Z: y \ne z$.
{{begin-eqn}}
{{eqn | l = y
| o = \ne
| r = z
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y - z
| o = \ne
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \paren {y - z}
| o = \ne
| r = 0
| c = [[Ring of Integers has no ... | Non-Zero Integers are Cancellable for Multiplication/Proof 2 | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication/Proof_2 | [
"Integers",
"Non-Zero Integers are Cancellable for Multiplication"
] | [
"Definition:Integer",
"Definition:Cancellable Element",
"Definition:Multiplication/Integers"
] | [
"Ring of Integers has no Zero Divisors",
"Integer Multiplication Distributes over Addition/Corollary",
"Rule of Transposition"
] |
proofwiki-838 | Non-Zero Integers are Cancellable for Multiplication | Every non-zero integer is cancellable for multiplication.
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.
By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication.
{{qed|le... | Every non-zero [[Definition:Integer|integer]] is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$ | Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From [[Natural Numbers are Non-Negative Integers]], $x \in \N_{> 0}$.
By the [[Extension Theorem for Distributive Operations]] and [[Ordering on Natural Numbers is Compatible with Multiplication]], $x$ is [[Definition:Cancellabl... | Non-Zero Integers are Cancellable for Multiplication/Proof 3 | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication | https://proofwiki.org/wiki/Non-Zero_Integers_are_Cancellable_for_Multiplication/Proof_3 | [
"Integers",
"Non-Zero Integers are Cancellable for Multiplication"
] | [
"Definition:Integer",
"Definition:Cancellable Element",
"Definition:Multiplication/Integers"
] | [
"Natural Numbers are Non-Negative Integers",
"Extension Theorem for Distributive Operations",
"Ordering on Natural Numbers is Compatible with Multiplication",
"Definition:Cancellable Element",
"Integers form Integral Domain",
"Definition:Ring (Abstract Algebra)",
"Negative of Ring Negative",
"Product ... |
proofwiki-839 | Multiplicative Ordering on Integers | Let $x, y, z \in \Z$ such that $z > 0$.
Then:
:$x < y \iff z x < z y$
:$x \le y \iff z x \le z y$ | Let $z > 0$.
Let $M_z: \Z \to \Z$ be the mapping defined as:
:$\forall x \in \Z: \map {M_z} x = z x$
It is sufficient to show that $M_z$ is an order embedding from $\struct {\Z, +, \le}$ to itself.
By Monomorphism from Total Ordering, it is sufficient to show that:
:$x < y \implies z x < z y$
Let $x < y$.
Then:
:$0 < y... | Let $x, y, z \in \Z$ such that $z > 0$.
Then:
:$x < y \iff z x < z y$
:$x \le y \iff z x \le z y$ | Let $z > 0$.
Let $M_z: \Z \to \Z$ be the mapping defined as:
:$\forall x \in \Z: \map {M_z} x = z x$
It is sufficient to show that $M_z$ is an [[Definition:Order Embedding|order embedding]] from $\struct {\Z, +, \le}$ to itself.
By [[Monomorphism from Total Ordering]], it is sufficient to show that:
:$x < y \implie... | Multiplicative Ordering on Integers | https://proofwiki.org/wiki/Multiplicative_Ordering_on_Integers | https://proofwiki.org/wiki/Multiplicative_Ordering_on_Integers | [
"Integer Multiplication",
"Orderings on Integers"
] | [] | [
"Definition:Order Embedding",
"Monomorphism from Total Ordering",
"Natural Numbers are Non-Negative Integers",
"Ordering on Natural Numbers is Compatible with Multiplication"
] |
proofwiki-840 | Invertible Integers under Multiplication | The only invertible elements of $\Z$ for multiplication (that is, units of $\Z$) are $1$ and $-1$. | Let $x > 0$ and $x y > 0$.
{{AimForCont}} first that $y \le 0$.
Then from Multiplicative Ordering on Integers and Ring Product with Zero:
:$x y \le x \, 0 = 0$
From this contradiction we deduce that $y > 0$.
Let $x > 0$ and $x y = 1$.
Then:
:$y > 0$
and by Natural Numbers are Non-Negative Integers:
:$y \in \N$
Hence by... | The only [[Definition:Invertible Element|invertible elements]] of $\Z$ for [[Definition:Integer Multiplication|multiplication]] (that is, [[Definition:Unit of Ring|units of $\Z$]]) are $1$ and $-1$. | Let $x > 0$ and $x y > 0$.
{{AimForCont}} first that $y \le 0$.
Then from [[Multiplicative Ordering on Integers]] and [[Ring Product with Zero]]:
:$x y \le x \, 0 = 0$
From this [[Definition:Contradiction|contradiction]] we deduce that $y > 0$.
Let $x > 0$ and $x y = 1$.
Then:
:$y > 0$
and by [[Natural Numbers a... | Invertible Integers under Multiplication | https://proofwiki.org/wiki/Invertible_Integers_under_Multiplication | https://proofwiki.org/wiki/Invertible_Integers_under_Multiplication | [
"Integer Multiplication",
"Units of Rings"
] | [
"Definition:Invertible Element",
"Definition:Multiplication/Integers",
"Definition:Unit of Ring"
] | [
"Multiplicative Ordering on Integers",
"Ring Product with Zero",
"Definition:Contradiction",
"Natural Numbers are Non-Negative Integers",
"Invertible Elements under Natural Number Multiplication",
"Definition:Invertible Element",
"Definition:Multiplication/Natural Numbers",
"Natural Numbers are Non-Ne... |
proofwiki-841 | Totally Ordered Abelian Group Isomorphism | Let $\struct {\Z', +', \le'}$ be a totally ordered abelian group.
Let $0'$ be the identity of $\struct {\Z', +', \le'}$.
Let $\N' = \set {x \in \Z': x \ge' 0'}$.
Let $\Z'$ contain at least two elements.
Let $\N'$ be well-ordered for the ordering induced on $\N'$ by $\le'$.
Then the mapping $g: \Z \to \Z'$ defined by:
:... | First we establish that $g$ is a homomorphism.
Suppose $z \in \Z'$ such that $z \ne 0'$.
Then by Ordering of Inverses in Ordered Monoid, either $z >' 0'$ or $-z >' 0'$.
Thus either:
:$z \in \N' \setminus \set {0'}$
or:
:$-z \in \N' \setminus \set {0'}$
and thus $\N' \setminus \set {0'}$ is not empty.
Therefore $\N' \se... | Let $\struct {\Z', +', \le'}$ be a [[Definition:Totally Ordered Structure|totally ordered]] [[Definition:Abelian Group|abelian group]].
Let $0'$ be the [[Definition:Identity Element|identity]] of $\struct {\Z', +', \le'}$.
Let $\N' = \set {x \in \Z': x \ge' 0'}$.
Let $\Z'$ contain at least two elements.
Let $\N'$ ... | First we establish that $g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Suppose $z \in \Z'$ such that $z \ne 0'$.
Then by [[Ordering of Inverses in Ordered Monoid]], either $z >' 0'$ or $-z >' 0'$.
Thus either:
:$z \in \N' \setminus \set {0'}$
or:
:$-z \in \N' \setminus \set {0'}$
and thus $\N'... | Totally Ordered Abelian Group Isomorphism | https://proofwiki.org/wiki/Totally_Ordered_Abelian_Group_Isomorphism | https://proofwiki.org/wiki/Totally_Ordered_Abelian_Group_Isomorphism | [
"Isomorphisms (Abstract Algebra)",
"Abelian Groups"
] | [
"Definition:Totally Ordered Structure",
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Well-Ordered Set",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Smallest Element"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Ordering of Inverses in Ordered Monoid",
"Definition:Empty Set",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Semigroup",
"Definition:Naturally Ordered Semigroup",
"Naturally Ordered Semigroup is Unique",
"Definition:Restricti... |
proofwiki-842 | Integers under Addition form Totally Ordered Group | Then the ordered structure $\struct {\Z, +, \le}$ is a totally ordered group. | === $\struct {\Z, +, \le}$ is an Ordered Structure === | Then the [[Definition:Ordered Structure|ordered structure]] $\struct {\Z, +, \le}$ is a [[Definition:Totally Ordered Group|totally ordered group]]. | === $\struct {\Z, +, \le}$ is an Ordered Structure === | Integers under Addition form Totally Ordered Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Totally_Ordered_Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Totally_Ordered_Group | [
"Integer Addition",
"Additive Group of Integers"
] | [
"Definition:Ordered Structure",
"Definition:Totally Ordered Group"
] | [] |
proofwiki-843 | Integers form Totally Ordered Ring | The structure $\struct {\Z, +, \times, \le}$ is a totally ordered ring. | From Integers form Commutative Ring, $\struct {\Z, +, \times}$ is a commutative ring.
We need to show that $\le$ is a compatible ordering on $\Z$.
That is, that:
:$(1): \quad \le$ is compatible with $+$
:$(2): \quad \forall x, y \in \Z: 0 \le x, 0 \le y \implies 0 \le x \times y$
The first one follows from the fact tha... | The [[Definition:Algebraic Structure|structure]] $\struct {\Z, +, \times, \le}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]]. | From [[Integers form Commutative Ring]], $\struct {\Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
We need to show that $\le$ is a [[Definition:Ordering Compatible with Ring Structure|compatible ordering]] on $\Z$.
That is, that:
:$(1): \quad \le$ is [[Definition:Relation Compatible with Opera... | Integers form Totally Ordered Ring | https://proofwiki.org/wiki/Integers_form_Totally_Ordered_Ring | https://proofwiki.org/wiki/Integers_form_Totally_Ordered_Ring | [
"Integers",
"Totally Ordered Rings"
] | [
"Definition:Algebraic Structure",
"Definition:Totally Ordered Ring"
] | [
"Integers form Commutative Ring",
"Definition:Commutative Ring",
"Definition:Ordering Compatible with Ring Structure",
"Definition:Relation Compatible with Operation",
"Integers under Addition form Totally Ordered Group",
"Multiplicative Ordering on Integers",
"Definition:Relation Compatible with Operat... |
proofwiki-844 | Congruences on Rational Numbers | There are only two congruence relations on the field of rational numbers $\struct {\Q, +, \times}$:
:$(1): \quad$ The diagonal relation $\Delta_\Q$
:$(2): \quad$ The trivial relation $\Q \times \Q$. | From:
:Diagonal Relation is Universally Congruent and
:Trivial Relation is Universally Congruent
we know that both these relations are congruent with both addition and multiplication on $\Q$.
Now we need to show that these are the ''only'' such relations.
Let $\RR$ be a congruence on $\Q$, such that $\RR \ne \Delta_\Q$... | There are only two [[Definition:Congruence Relation|congruence relations]] on the [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Rational Number|rational numbers]] $\struct {\Q, +, \times}$:
:$(1): \quad$ The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_\Q$
:$(2): \quad$ The [[Definition:T... | From:
:[[Diagonal Relation is Universally Congruent]] and
:[[Trivial Relation is Universally Congruent]]
we know that both these [[Definition:Relation|relations]] are [[Definition:Congruence Relation|congruent]] with both [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]... | Congruences on Rational Numbers | https://proofwiki.org/wiki/Congruences_on_Rational_Numbers | https://proofwiki.org/wiki/Congruences_on_Rational_Numbers | [
"Congruence Relations",
"Rational Numbers"
] | [
"Definition:Congruence Relation",
"Definition:Field (Abstract Algebra)",
"Definition:Rational Number",
"Definition:Diagonal Relation",
"Definition:Trivial Relation"
] | [
"Diagonal Relation is Universally Congruent",
"Trivial Relation is Universally Congruent",
"Definition:Relation",
"Definition:Congruence Relation",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers",
"Definition:Relation",
"Definition:Congruence Relation",
"Definiti... |
proofwiki-845 | Number of Elements in Partition | Let $S$ be a set.
Let there be a partition on $S$ of $n$ subsets, each of which has $m$ elements.
Then:
:$\card S = n m$ | Let the partition of $S$ be $S_1, S_2, \ldots, S_n$.
Then:
:$\forall k \in \set {1, 2, \ldots, n}: \card {S_k} = m$
By definition of multiplication:
:$\ds \sum_{k \mathop = 1}^n \card {S_k} = n m$
and the result follows from the Fundamental Principle of Counting.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let there be a [[Definition:Set Partition|partition]] on $S$ of $n$ [[Definition:Subset|subsets]], each of which has $m$ [[Definition:Element|elements]].
Then:
:$\card S = n m$ | Let the [[Definition:Set Partition|partition]] of $S$ be $S_1, S_2, \ldots, S_n$.
Then:
:$\forall k \in \set {1, 2, \ldots, n}: \card {S_k} = m$
By definition of [[Definition:Integer Multiplication|multiplication]]:
:$\ds \sum_{k \mathop = 1}^n \card {S_k} = n m$
and the result follows from the [[Fundamental Princip... | Number of Elements in Partition | https://proofwiki.org/wiki/Number_of_Elements_in_Partition | https://proofwiki.org/wiki/Number_of_Elements_in_Partition | [
"Combinatorics"
] | [
"Definition:Set",
"Definition:Set Partition",
"Definition:Subset",
"Definition:Element"
] | [
"Definition:Set Partition",
"Definition:Multiplication/Integers",
"Fundamental Principle of Counting"
] |
proofwiki-846 | Cardinality of Complement | Let $T \subseteq S$ such that $\card S = n, \card T = m$.
Then:
:$\card {\relcomp S T} = \card {S \setminus T} = n - m$
where:
:$\relcomp S T$ denotes the complement of $T$ relative to $S$
:$S \setminus T$ denotes the difference between $S$ and $T$. | The result is obvious for $S = T$ or $T = \O$.
Otherwise, $\set {T, S \setminus T}$ is a partition of $S$.
Let $\card {S \setminus T} = p$.
Then by the Fundamental Principle of Counting:
:$m + p = n$
and the result follows.
{{qed}} | Let $T \subseteq S$ such that $\card S = n, \card T = m$.
Then:
:$\card {\relcomp S T} = \card {S \setminus T} = n - m$
where:
:$\relcomp S T$ denotes the [[Definition:Relative Complement|complement of $T$ relative to $S$]]
:$S \setminus T$ denotes the [[Definition:Set Difference|difference between $S$ and $T$]]. | The result is obvious for $S = T$ or $T = \O$.
Otherwise, $\set {T, S \setminus T}$ is a [[Definition:Set Partition|partition]] of $S$.
Let $\card {S \setminus T} = p$.
Then by the [[Fundamental Principle of Counting]]:
:$m + p = n$
and the result follows.
{{qed}} | Cardinality of Complement | https://proofwiki.org/wiki/Cardinality_of_Complement | https://proofwiki.org/wiki/Cardinality_of_Complement | [
"Combinatorics",
"Set Difference",
"Cardinality"
] | [
"Definition:Relative Complement",
"Definition:Set Difference"
] | [
"Definition:Set Partition",
"Fundamental Principle of Counting"
] |
proofwiki-847 | Cardinality of Cartesian Product of Finite Sets | Let $S \times T$ be the cartesian product of two finite sets $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes cardinality. | Let $\card S = n$ and $\card T = m$.
If either $n = 0$ or $m = 0$, then from Cartesian Product is Empty iff Factor is Empty:
:$S \times T = \O$
and the result holds, as $n m = 0 = \card \O$ from Cardinality of Empty Set.
So, we assume that $n > 0$ and $m > 0$.
For each $a \in S$, we define the mapping $g_a: T \to \set ... | Let $S \times T$ be the [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Finite Set|finite sets]] $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes [[Definition:Cardinality|cardinality]]. | Let $\card S = n$ and $\card T = m$.
If either $n = 0$ or $m = 0$, then from [[Cartesian Product is Empty iff Factor is Empty]]:
:$S \times T = \O$
and the result holds, as $n m = 0 = \card \O$ from [[Cardinality of Empty Set]].
So, we assume that $n > 0$ and $m > 0$.
For each $a \in S$, we define the [[Definition... | Cardinality of Cartesian Product of Finite Sets | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets | [
"Combinatorics",
"Cartesian Product",
"Cardinality",
"Cardinality of Cartesian Product"
] | [
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Cartesian Product is Empty iff Factor is Empty",
"Cardinality of Empty Set",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Set Partition",
"Definition:Set",
"Number of Elements in Partition"
] |
proofwiki-848 | Cardinality of Cartesian Product of Finite Sets | Let $S \times T$ be the cartesian product of two finite sets $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes cardinality. | {{begin-eqn}}
{{eqn | l = \card {S \times T}
| r = \card S \times \card T
| c = Cardinality of Cartesian Product of Finite Sets
}}
{{eqn | r = \card T \times \card S
| c = Integer Multiplication is Commutative
}}
{{eqn | r = \card {T \times S}
| c = Cardinality of Cartesian Product of Finite Set... | Let $S \times T$ be the [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Finite Set|finite sets]] $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes [[Definition:Cardinality|cardinality]]. | {{begin-eqn}}
{{eqn | l = \card {S \times T}
| r = \card S \times \card T
| c = [[Cardinality of Cartesian Product of Finite Sets]]
}}
{{eqn | r = \card T \times \card S
| c = [[Integer Multiplication is Commutative]]
}}
{{eqn | r = \card {T \times S}
| c = [[Cardinality of Cartesian Product of ... | Cardinality of Cartesian Product of Finite Sets/Corollary/Proof 1 | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets/Corollary/Proof_1 | [
"Combinatorics",
"Cartesian Product",
"Cardinality",
"Cardinality of Cartesian Product"
] | [
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Cardinality of Cartesian Product of Finite Sets",
"Integer Multiplication is Commutative",
"Cardinality of Cartesian Product of Finite Sets"
] |
proofwiki-849 | Cardinality of Cartesian Product of Finite Sets | Let $S \times T$ be the cartesian product of two finite sets $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes cardinality. | Let $f: S \times T \to T \times S$ be the mapping defined as:
:$\forall \tuple {s, t} \in S \times T: \map f {s, t} = \tuple {t, s}$
which is shown to be bijective as follows:
{{begin-eqn}}
{{eqn | l = \map f {s_1, t_1}
| r = \map f {s_2, t_2}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {t_1, s_1}
... | Let $S \times T$ be the [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Finite Set|finite sets]] $S$ and $T$.
Then:
:$\card {S \times T} = \card S \times \card T$
where $\card S$ denotes [[Definition:Cardinality|cardinality]]. | Let $f: S \times T \to T \times S$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall \tuple {s, t} \in S \times T: \map f {s, t} = \tuple {t, s}$
which is shown to be [[Definition:Bijection|bijective]] as follows:
{{begin-eqn}}
{{eqn | l = \map f {s_1, t_1}
| r = \map f {s_2, t_2}
| c =
}}
{{eq... | Cardinality of Cartesian Product of Finite Sets/Corollary/Proof 2 | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets | https://proofwiki.org/wiki/Cardinality_of_Cartesian_Product_of_Finite_Sets/Corollary/Proof_2 | [
"Combinatorics",
"Cartesian Product",
"Cardinality",
"Cardinality of Cartesian Product"
] | [
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Set Equivalence"
] |
proofwiki-850 | Cardinality of Power Set of Finite Set | Let $S$ be a set such that:
:$\card S = n$
where $\card S$ denotes the cardinality of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$. | Given an element $x$ of $S$, each subset of $S$ either includes $x$ or does not include $x$ (this follows directly from the definition of a set), which gives us two possibilities.
The same reasoning holds for any element of $S$.
One can intuitively see that this means that there are $\ds \underbrace {2 \times 2 \times ... | Let $S$ be a [[Definition:Set|set]] such that:
:$\card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | Given an element $x$ of $S$, each [[Definition:Subset|subset]] of $S$ either includes $x$ or does not include $x$ (this follows directly from the [[Definition:Set|definition of a set]]), which gives us two possibilities.
The same reasoning holds for any [[Definition:Element|element]] of $S$.
One can intuitively see t... | Cardinality of Power Set of Finite Set/Informal Proof | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set/Informal_Proof | [
"Cardinality of Power Set of Finite Set",
"Cardinality of Power Set",
"Power Set",
"Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Power Set"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Element",
"Definition:Element"
] |
proofwiki-851 | Cardinality of Power Set of Finite Set | Let $S$ be a set such that:
:$\card S = n$
where $\card S$ denotes the cardinality of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$. | Let $T = \set {0, 1}$.
For each $A \in \powerset S$, we consider the characteristic function $\chi_A: S \to T$ defined as:
:<nowiki>$\forall x \in S: \map {\chi_A} x = \begin{cases}
1 & : x \in A \\
0 & : x \notin A
\end{cases}$</nowiki>
Now consider the mapping $f: \powerset S \to T^S$:
:$\forall A \in \powerset S: \m... | Let $S$ be a [[Definition:Set|set]] such that:
:$\card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | Let $T = \set {0, 1}$.
For each $A \in \powerset S$, we consider the [[Definition:Characteristic Function of Set|characteristic function]] $\chi_A: S \to T$ defined as:
:<nowiki>$\forall x \in S: \map {\chi_A} x = \begin{cases}
1 & : x \in A \\
0 & : x \notin A
\end{cases}$</nowiki>
Now consider the [[Definition:Ma... | Cardinality of Power Set of Finite Set/Proof 1 | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set/Proof_1 | [
"Cardinality of Power Set of Finite Set",
"Cardinality of Power Set",
"Power Set",
"Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Power Set"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Mapping",
"Definition:Set of All Mappings",
"Definition:Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Identity Mapping",
"Definition:Preimage/Mapping/Subset",
"Defi... |
proofwiki-852 | Cardinality of Power Set of Finite Set | Let $S$ be a set such that:
:$\card S = n$
where $\card S$ denotes the cardinality of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$. | Enumerating the subsets of $S$ is equivalent to counting all of the ways of selecting $k$ out of the $n$ elements of $S$ with $k = 0, 1, \ldots, n$.
So, from Cardinality of Set of Subsets, the number we are looking for is:
:$\ds \card {\powerset S} = \sum_{k \mathop = 0}^n \binom n k$
But from the binomial theorem:
:$\... | Let $S$ be a [[Definition:Set|set]] such that:
:$\card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | Enumerating the [[Definition:Subset|subsets]] of $S$ is equivalent to counting all of the ways of selecting $k$ out of the $n$ elements of $S$ with $k = 0, 1, \ldots, n$.
So, from [[Cardinality of Set of Subsets]], the number we are looking for is:
:$\ds \card {\powerset S} = \sum_{k \mathop = 0}^n \binom n k$
But f... | Cardinality of Power Set of Finite Set/Proof 2 | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set/Proof_2 | [
"Cardinality of Power Set of Finite Set",
"Cardinality of Power Set",
"Power Set",
"Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Power Set"
] | [
"Definition:Subset",
"Cardinality of Set of Subsets",
"Binomial Theorem"
] |
proofwiki-853 | Cardinality of Power Set of Finite Set | Let $S$ be a set such that:
:$\card S = n$
where $\card S$ denotes the cardinality of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$. | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\card S = n \implies \card {\powerset S} = 2^n$
Do not confuse $\map P n$, which is a propositional function on $\N$, with $\powerset S$, the power set of $S$.
=== Basis for the Induction ===
From Cardinality of Empty Set:
: $S = \O \iff \card... | Let $S$ be a [[Definition:Set|set]] such that:
:$\card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$,
Then:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\card S = n \implies \card {\powerset S} = 2^n$
Do not confuse $\map P n$, which is a [[Definition:Propositional Function|propositional function]] on $\N$, with $\powerset S$... | Cardinality of Power Set of Finite Set/Proof 3 | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Power_Set_of_Finite_Set/Proof_3 | [
"Cardinality of Power Set of Finite Set",
"Cardinality of Power Set",
"Power Set",
"Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Power Set"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Propositional Function",
"Definition:Power Set",
"Cardinality of Empty Set",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Definition:Subset",
"Defin... |
proofwiki-854 | Cardinality of Set of Injections | Let $S$ and $T$ be finite sets.
The number of injections from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is:
:${}^m P_n = \begin {cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end {cases}$ | Let $m > n$.
By the Pigeonhole Principle, there can be no injection from $S$ to $T$ when $\card S > \card T$.
Once $\card T$ elements of $S$ have been used up, there is no element of $T$ left for the remaining elements of $S$ to be mapped to such that they all still map to different elements of $T$.
Let $m = 0$.
The on... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
The number of [[Definition:Injection|injections]] from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is:
:${}^m P_n = \begin {cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end {cases}$ | Let $m > n$.
By the [[Pigeonhole Principle]], there can be no [[Definition:Injection|injection]] from $S$ to $T$ when $\card S > \card T$.
Once $\card T$ [[Definition:Element|elements]] of $S$ have been used up, there is no [[Definition:Element|element]] of $T$ left for the remaining [[Definition:Element|elements]] o... | Cardinality of Set of Injections/Formal Proof | https://proofwiki.org/wiki/Cardinality_of_Set_of_Injections | https://proofwiki.org/wiki/Cardinality_of_Set_of_Injections/Formal_Proof | [
"Cardinality of Set of Injections",
"Injections",
"Combinatorics",
"Counting Arguments",
"Cardinality"
] | [
"Definition:Finite Set",
"Definition:Injection"
] | [
"Dirichlet's Box Principle/Corollary",
"Definition:Injection",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Injection",
"Definition:Injection",
"Cardinality of Set of All Mappings",
"Principle of Mathematical Induction",
"Cardinality of Set ... |
proofwiki-855 | Cardinality of Set of Bijections | Let $S$ and $T$ be finite sets with the same cardinality:
:$\size S = \size T = n$
Then there are $n!$ bijections from $S$ to $T$. | Follows directly from Cardinality of Set of Injections and Equivalence of Mappings between Finite Sets of Same Cardinality.
{{Qed}} | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]] with the same [[Definition:Cardinality|cardinality]]:
:$\size S = \size T = n$
Then there are $n!$ [[Definition:Bijection|bijections]] from $S$ to $T$. | Follows directly from [[Cardinality of Set of Injections]] and [[Equivalence of Mappings between Finite Sets of Same Cardinality]].
{{Qed}} | Cardinality of Set of Bijections | https://proofwiki.org/wiki/Cardinality_of_Set_of_Bijections | https://proofwiki.org/wiki/Cardinality_of_Set_of_Bijections | [
"Combinatorics",
"Bijections",
"Cardinality",
"Cardinality of Set of Bijections"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Bijection"
] | [
"Cardinality of Set of Injections",
"Equivalence of Mappings between Finite Sets of Same Cardinality"
] |
proofwiki-856 | Cardinality of Set of Subsets | Let $S$ be a set such that $\card S = n$.
Let $m \le n$.
Then the number of subsets $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | For each $X \subseteq \N_n$ and $Y \subseteq S$, let $\map B {X, Y}$ be the set of all bijections from $X$ onto $Y$.
Let $\Bbb S$ be the set of all subsets of $S$ with $m$ elements.
By Cardinality of Power Set of Finite Set and Cardinality of Subset of Finite Set, $\Bbb S$ is finite, so let $s = \card {\Bbb S}$.
Let $\... | Let $S$ be a [[Definition:Set|set]] such that $\card S = n$.
Let $m \le n$.
Then the number of [[Definition:Subset|subsets]] $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | For each $X \subseteq \N_n$ and $Y \subseteq S$, let $\map B {X, Y}$ be the [[Definition:Set|set]] of all [[Definition:Bijection|bijections]] from $X$ onto $Y$.
Let $\Bbb S$ be the [[Definition:Set of Sets|set]] of all [[Definition:Subset|subsets]] of $S$ with $m$ [[Definition:Element|elements]].
By [[Cardinality of ... | Cardinality of Set of Subsets/Proof 1 | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets/Proof_1 | [
"Combinations",
"Combinatorics",
"Cardinality of Set of Subsets"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Definition:Set",
"Definition:Bijection",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Element",
"Cardinality of Power Set of Finite Set",
"Cardinality of Subset of Finite Set",
"Definition:Finite Set",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Bijection",
"Cardinal... |
proofwiki-857 | Cardinality of Set of Subsets | Let $S$ be a set such that $\card S = n$.
Let $m \le n$.
Then the number of subsets $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | Let $\dbinom n m$ denote the number of subsets of $m$ elements of $S$.
From Number of Permutations, the number of $m$-permutations of $S$ is:
:${}^m P_n = \dfrac {n!} {\paren {n - m}!}$
Consider the way ${}^m P_n$ can be calculated.
First one makes the selection of which $m$ elements of $S$ are to be arranged.
This num... | Let $S$ be a [[Definition:Set|set]] such that $\card S = n$.
Let $m \le n$.
Then the number of [[Definition:Subset|subsets]] $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | Let $\dbinom n m$ denote the number of [[Definition:Subset|subsets]] of $m$ [[Definition:Element|elements]] of $S$.
From [[Number of Permutations]], the number of [[Definition:Permutation (Ordered Selection)|$m$-permutations of $S$]] is:
:${}^m P_n = \dfrac {n!} {\paren {n - m}!}$
Consider the way ${}^m P_n$ can be c... | Cardinality of Set of Subsets/Proof 2 | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets/Proof_2 | [
"Combinations",
"Combinatorics",
"Cardinality of Set of Subsets"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Definition:Subset",
"Definition:Element",
"Number of Permutations",
"Definition:Permutation/Ordered Selection",
"Number of Permutations",
"Product Rule for Counting",
"Number of Permutations",
"Product Rule for Counting"
] |
proofwiki-858 | Cardinality of Set of Subsets | Let $S$ be a set such that $\card S = n$.
Let $m \le n$.
Then the number of subsets $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
:$B_r := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defin... | Let $S$ be a [[Definition:Set|set]] such that $\card S = n$.
Let $m \le n$.
Then the number of [[Definition:Subset|subsets]] $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | Let $\N_n$ denote the [[Definition:Set|set]] $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the [[Definition:Symmetric Group on n Letters|symmetric group]] on $\N_n$.
Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the [[Definition:Set|set]] of all [[Definition:Subset|subsets]] of $\N_n$ of [[Definition:C... | Cardinality of Set of Subsets/Proof 3 | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets/Proof_3 | [
"Combinations",
"Combinatorics",
"Cardinality of Set of Subsets"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Definition:Set",
"Definition:Symmetric Group/n Letters",
"Definition:Set",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Group Action of Symmetric Group/Subset",
"Definition:Group Action",
"Definition:Stabilizer",
"Definiti... |
proofwiki-859 | Cardinality of Set of Subsets | Let $S$ be a set such that $\card S = n$.
Let $m \le n$.
Then the number of subsets $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \dbinom 1 m
| r = \dfrac {1!} {1! \paren {m - 1}!}
| c =
}}
{{eqn | r = \begin ... | Let $S$ be a [[Definition:Set|set]] such that $\card S = n$.
Let $m \le n$.
Then the number of [[Definition:Subset|subsets]] $T$ of $S$ such that $\card T = m$ is:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \dbinom 1 m
| ... | Cardinality of Set of Subsets/Proof 4 | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets | https://proofwiki.org/wiki/Cardinality_of_Set_of_Subsets/Proof_4 | [
"Combinations",
"Combinatorics",
"Cardinality of Set of Subsets"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element... |
proofwiki-860 | Set of Integers Bounded Below by Integer has Smallest Element | Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\Z, \le}$.
Then $S$ has a smallest element. | We have that $S$ is bounded below in $\Z$.
So:
: $\exists m \in \Z: \forall s \in S: m \le s$
Hence:
: $\forall s \in S: 0 \le s - m$
Thus:
:$T = \set {s - m: s \in S} \subseteq \N$
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
:$\paren {\forall s \in S: b_T \l... | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\le$ be the [[Definition:Ordering on Integers|ordering on the integers]].
Let $\O \subset S \subseteq \Z$ such that $S$ is [[Definition:Bounded Below Set|bounded below]] in $\struct {\Z, \le}$.
Then $S$ has a [[Definition:Smallest Element|smallest element... | We have that $S$ is [[Definition:Bounded Below Set|bounded below]] in $\Z$.
So:
: $\exists m \in \Z: \forall s \in S: m \le s$
Hence:
: $\forall s \in S: 0 \le s - m$
Thus:
:$T = \set {s - m: s \in S} \subseteq \N$
The [[Well-Ordering Principle]] gives that $T$ has a [[Definition:Smallest Element|smallest element]... | Set of Integers Bounded Below by Integer has Smallest Element | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Below_by_Integer_has_Smallest_Element | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Below_by_Integer_has_Smallest_Element | [
"Set of Integers Bounded Below has Smallest Element"
] | [
"Definition:Integer",
"Definition:Ordering on Integers",
"Definition:Bounded Below Set",
"Definition:Smallest Element"
] | [
"Definition:Bounded Below Set",
"Well-Ordering Principle",
"Definition:Smallest Element",
"Definition:Smallest Element"
] |
proofwiki-861 | Set of Integers Bounded Above by Integer has Greatest Element | Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$.
Then $S$ has a greatest element. | $S$ is bounded above, so $\exists M \in \Z: \forall s \in S: s \le M$.
Hence $\forall s \in S: 0 \le M - s$.
Thus the set $T = \set {M - s: s \in S} \subseteq \N$.
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
:$\paren {\forall s \in S: b_T \le M - s} \land \pa... | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\le$ be the [[Definition:Ordering on Integers|ordering on the integers]].
Let $\O \subset S \subseteq \Z$ such that $S$ is [[Definition:Bounded Above Set|bounded above]] in $\struct {\Z, \le}$.
Then $S$ has a [[Definition:Greatest Element|greatest element... | $S$ is [[Definition:Bounded Above Set|bounded above]], so $\exists M \in \Z: \forall s \in S: s \le M$.
Hence $\forall s \in S: 0 \le M - s$.
Thus the set $T = \set {M - s: s \in S} \subseteq \N$.
The [[Well-Ordering Principle]] gives that $T$ has a [[Definition:Smallest Element|smallest element]], which we can call... | Set of Integers Bounded Above by Integer has Greatest Element/Proof 1 | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Above_by_Integer_has_Greatest_Element | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Above_by_Integer_has_Greatest_Element/Proof_1 | [
"Number Theory",
"Integers",
"Set of Integers Bounded Above has Greatest Element"
] | [
"Definition:Integer",
"Definition:Ordering on Integers",
"Definition:Bounded Above Set",
"Definition:Greatest Element"
] | [
"Definition:Bounded Above Set",
"Well-Ordering Principle",
"Definition:Smallest Element",
"Definition:Greatest Element",
"Definition:Greatest Element"
] |
proofwiki-862 | Set of Integers Bounded Above by Integer has Greatest Element | Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$.
Then $S$ has a greatest element. | Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$.
Hence we can define the set $S' = \set {-s: s \in S}$.
$S'$ is bounded below by $-M$.
So from Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a smallest element, $-g_S$, say, where $\forall s \in S: -g_S \le -s$.
Therefore... | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\le$ be the [[Definition:Ordering on Integers|ordering on the integers]].
Let $\O \subset S \subseteq \Z$ such that $S$ is [[Definition:Bounded Above Set|bounded above]] in $\struct {\Z, \le}$.
Then $S$ has a [[Definition:Greatest Element|greatest element... | Since $S$ is [[Definition:Bounded Above Set|bounded above]], $\exists M \in \Z: \forall s \in S: s \le M$.
Hence we can define the set $S' = \set {-s: s \in S}$.
$S'$ is [[Definition:Bounded Below Set|bounded below]] by $-M$.
So from [[Set of Integers Bounded Below by Integer has Smallest Element]], $S'$ has a [[Def... | Set of Integers Bounded Above by Integer has Greatest Element/Proof 2 | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Above_by_Integer_has_Greatest_Element | https://proofwiki.org/wiki/Set_of_Integers_Bounded_Above_by_Integer_has_Greatest_Element/Proof_2 | [
"Number Theory",
"Integers",
"Set of Integers Bounded Above has Greatest Element"
] | [
"Definition:Integer",
"Definition:Ordering on Integers",
"Definition:Bounded Above Set",
"Definition:Greatest Element"
] | [
"Definition:Bounded Above Set",
"Definition:Bounded Below Set",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Smallest Element",
"Definition:Greatest Element"
] |
proofwiki-863 | Principle of Least Counterexample | Let $\map P n$ be a condition on $n \in \set {x \in \Z: x \ge m \in \Z}$.
Suppose that:
:$\neg \paren {\forall n \ge m: \map P n}$
(That is, not all $n \ge m$ satisfy $\map P n$.)
Then there exists a '''least counterexample''', that is a smallest integral value of $n$ for which $\neg \map P n$. | Let $S = \set {n \in \Z: n \ge m \in \Z: \neg \map P n}$.
That is, $S$ is the set of all elements of $\Z$ not less than $m$ for which the condition is false.
Since:
:$\neg \paren {\forall n \ge m: \map P n}$
it follows that:
:$S \ne \O$
Also, $S \subseteq \Z$ and is bounded below (by $m$).
Therefore $S$ has a smallest ... | Let $\map P n$ be a [[Definition:Propositional Function|condition]] on $n \in \set {x \in \Z: x \ge m \in \Z}$.
Suppose that:
:$\neg \paren {\forall n \ge m: \map P n}$
(That is, not all $n \ge m$ satisfy $\map P n$.)
Then there exists a '''least counterexample''', that is a [[Definition:Smallest Element|smallest]]... | Let $S = \set {n \in \Z: n \ge m \in \Z: \neg \map P n}$.
That is, $S$ is the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $\Z$ not less than $m$ for which the [[Definition:Propositional Function|condition]] is false.
Since:
:$\neg \paren {\forall n \ge m: \map P n}$
it follows that:
:$S \ne \O$
... | Principle of Least Counterexample | https://proofwiki.org/wiki/Principle_of_Least_Counterexample | https://proofwiki.org/wiki/Principle_of_Least_Counterexample | [
"Proof Techniques"
] | [
"Definition:Propositional Function",
"Definition:Smallest Element",
"Definition:Integer"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Propositional Function",
"Definition:Bounded Below Set",
"Set of Integers Bounded Below by Integer has Smallest Element"
] |
proofwiki-864 | Absolute Value is Bounded Below by Zero | Let $x \in \R$ be a real number.
Then the absolute value $\size x$ of $x$ is bounded below by $0$. | Let $x \ge 0$.
Then $\size x = x \ge 0$.
Let $x < 0$.
Then $\size x = -x > 0$.
The result follows.
{{qed}} | Let $x \in \R$ be a [[Definition:Real Number|real number]].
Then the [[Definition:Absolute Value|absolute value]] $\size x$ of $x$ is [[Definition:Bounded Below Mapping|bounded below]] by $0$. | Let $x \ge 0$.
Then $\size x = x \ge 0$.
Let $x < 0$.
Then $\size x = -x > 0$.
The result follows.
{{qed}} | Absolute Value is Bounded Below by Zero | https://proofwiki.org/wiki/Absolute_Value_is_Bounded_Below_by_Zero | https://proofwiki.org/wiki/Absolute_Value_is_Bounded_Below_by_Zero | [
"Absolute Value Function"
] | [
"Definition:Real Number",
"Definition:Absolute Value",
"Definition:Bounded Below Mapping"
] | [] |
proofwiki-865 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
So, if there is a counterexample to be found, it will have a degree.
{{AimForCont}} there exists at least one counterexample.
By a version of the Well-Ordering Principle, we can assign a number $m$ to the lowest degree poss... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
So, if there is a [[Definition:Counterexample|counterexample]] to be found, it will have a [[Definition:Degree of Polynomial over Field|degree]].
{{AimForCont}} there exists at least one [[Definition:Counterexample|count... | Division Theorem for Polynomial Forms over Field/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Counterexample",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"Well-Ordering Principle",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"Definition:Counterexample",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"De... |
proofwiki-866 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Suppose $\map \deg f < \map \deg d$.
Then we take $\map q X = 0$ and $\map r X = \map f X$ and the result holds.
Otherwise, $\map \deg f \ge \map \deg d$.
Let:
:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
We can subtract from $f$ a suitable multiple of $d... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Suppose $\map \deg f < \map \deg d$.
Then we take $\map q X = 0$ and $\map r X = \map f X$ and the result holds.
Otherwise, $\map \deg f \ge \map \deg d$.
Let:
:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
We can subtract from $f$ a suitable multiple... | Division Theorem for Polynomial Forms over Field/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Degree of Polynomial/Field",
"Definition:Degree of Polynomial/Field",
"Definition:Degree of Polynomial/Field",
"Second Principle of Mathematical Induction"
] |
proofwiki-867 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
=== Basis for the Induction ===
$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.
This is shown trivially to be true... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
=== Basis for the Induction ===
$\map P 0$ is the statement ... | Division Theorem for Polynomial Forms over Field/Proof 3 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_3 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Coefficient of Polynomial",
"Division Theorem for Polynomial Forms over Field/Proof 3",
"Degree of Sum of Polynomials",
"... |
proofwiki-868 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From Division Theorem: Positive Divisor: Positive Dividend: Existence:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
That is, the result holds for positive $a$.
{{qed|lemma}}
It remains to be shown that the statement holds for $a < 0$.
From Division Theorem: Positive Divisor: Pos... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From [[Division Theorem/Positive Divisor/Positive Dividend/Existence|Division Theorem: Positive Divisor: Positive Dividend: Existence]]:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
That is, the result holds for [[Definition:Positive Integer|positive]] $a$.
{{qed|lemma}}
It r... | Division Theorem/Positive Divisor/Existence/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor/Positive Dividend/Existence",
"Definition:Positive/Integer",
"Division Theorem/Positive Divisor/Positive Dividend",
"Definition:Absolute Value",
"Definition:Absolute Value",
"Definition:Unique",
"Definition:Unique"
] |
proofwiki-869 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let:
: $q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$
where $\floor {\, \cdot \,}$ denotes the floor function.
Thus $q \in \Z$ and $t \in \hointr 0 1$.
So:
:$\dfrac a b = q + t$
and so:
:$(1): \quad a = q b + r$
where $r = t d$.
Since $a, q, b \in \Z$, it follows from $(1)$ that:
:$r = a - q b$
and so ... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let:
: $q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$
where $\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]].
Thus $q \in \Z$ and $t \in \hointr 0 1$.
So:
:$\dfrac a b = q + t$
and so:
:$(1): \quad a = q b + r$
where $r = t d$.
Since $a, q, b \in \Z$, it follows from $... | Division Theorem/Positive Divisor/Existence/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Floor Function"
] |
proofwiki-870 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let there exist $q \in Z$ such that $a - b q = 0$.
Then $a = b q$ as required, with $r = 0$.
Otherwise, let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer:
:$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$
Setting $z = 0$ it is seen that $a \in S$, so $S \ne... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let there exist $q \in Z$ such that $a - b q = 0$.
Then $a = b q$ as required, with $r = 0$.
Otherwise, let $S$ be defined as the [[Definition:Set|set]] of all [[Definition:Positive Integer|positive integers]] of the form $a - z b$ where $z$ is an [[Definition:Integer|integer]]:
:$S = \set {x \in \Z_{\ge 0}: \exists... | Division Theorem/Positive Divisor/Existence/Proof 3 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_3 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Integer",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element"
] |
proofwiki-871 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.
Let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer:
:$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$
By setting $z = 0$ we have that $a \in S$.
Thus $S \ne \O$.
We have that $S$ is bounded below by... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.
Let $S$ be defined as the [[Definition:Set|set]] of all [[Definition:Positive Integer|positive integers]] of the form $a - z b$ where $z$ is an [[Definition:Integer|integer]]:
:$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$
By setting $z = 0$ ... | Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Integer",
"Definition:Bounded Below Set",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Smallest Element",
"Definition:Bounded Below Set",
"Proof by Contradiction",
"Definition:Smallest Element"
] |
proofwiki-872 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a = 0$.
It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.
{{qed|lemma}}
Let $a > 0$ and $b = 1$.
Then from the condition $0 \le r < b$ it follows that $r = 0$ and hence $q = a$.
{{qed|lemma}}
Let $a > 0$ and $b > 1$.
By the Basis Representation Theorem, $a$ has a unique ... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a = 0$.
It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.
{{qed|lemma}}
Let $a > 0$ and $b = 1$.
Then from the condition $0 \le r < b$ it follows that $r = 0$ and hence $q = a$.
{{qed|lemma}}
Let $a > 0$ and $b > 1$.
By the [[Basis Representation Theorem]], $a$ ha... | Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Basis Representation Theorem",
"Definition:Unique",
"Definition:Number Base"
] |
proofwiki-873 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.
When $a = 0$, the integers $q = r = 0$ satisfy the conditions of the theorem.
Let $a > 0$.
For each $k \in \closedint 0 {a + 1}$, let $r_k = b k$.
(Note that here, as elsewhere in this proof, $\closedint 0 {a + 1}$ denotes an integer interval.)
Then $\sequence... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.
When $a = 0$, the [[Definition:Integer|integers]] $q = r = 0$ satisfy the conditions of the theorem.
Let $a > 0$.
For each $k \in \closedint 0 {a + 1}$, let $r_k = b k$.
(Note that here, as elsewhere in this proof, $\closedint 0 {a + 1}$ denotes an [[Defi... | Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 3 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_3 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Integer",
"Definition:Closed Interval/Integer Interval",
"Definition:Strictly Increasing/Sequence",
"Definition:Positive/Integer",
"Strictly Increasing Sequence induces Partition"
] |
proofwiki-874 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$.
That is:
{{begin-eqn}}
{{eqn | l = a
| r = q_1 b + r_1, 0 \le r_1 < b
| c =
}}
{{eqn | l = a
| r = q_2 b + r_2, 0 \le r_2 < b
| c =
}}
{{end-eqn}}
This gives:
:$0 = b \paren {q_1 - q_2} + \paren ... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$.
That is:
{{begin-eqn}}
{{eqn | l = a
| r = q_1 b + r_1, 0 \le r_1 < b
| c =
}}
{{eqn | l = a
| r = q_2 b + r_2, 0 \le r_2 < b
| c =
}}
{{end-eqn}}
This gives:
:$0 = b \paren {q_1 - q_2} + \pa... | Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Proof by Contradiction",
"Definition:Unique"
] |
proofwiki-875 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by Division Theorem: Positive Divisor: Positive Dividend: Existence that such $q$ and $r$ exist.
Let $a = 0$.
It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.
{{qed|lemma}}
Let $a > 0$ and $b = 1$.
Then from the condition $0 \le r < b$ it follows that $r = 0$.
He... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by [[Division Theorem/Positive Divisor/Positive Dividend/Existence|Division Theorem: Positive Divisor: Positive Dividend: Existence]] that such $q$ and $r$ exist.
Let $a = 0$.
It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.
{{qed|lemma}}
Let $a > 0$ and $b ... | Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor/Positive Dividend/Existence",
"Basis Representation Theorem",
"Definition:Unique",
"Definition:Number Base",
"Definition:Number Base",
"Definition:Number Base",
"Basis Representation Theorem",
"Definition:Unique"
] |
proofwiki-876 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
{{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1}} | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
{{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1}} | Division Theorem/Positive Divisor/Uniqueness/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor/Existence"
] |
proofwiki-877 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$.
Thus:
:$\dfrac a b = q + \dfrac r b$
and:
:$0 \le \dfrac r b \le \dfrac {b - 1} b < 1$
So:
:$q = \floor {\dfrac a b}$
and so:
:$r = a - b \floor {\dfrac a b}$
Thus, given $a... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$.
Thus:
:$\dfrac a b = q + \dfrac r b$
and:
:$0 \le \dfrac r b \le \dfrac {b - 1} b < 1$
So:
:$q = \floor {\dfrac a b}$
and... | Division Theorem/Positive Divisor/Uniqueness/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor/Existence",
"Definition:Unique"
] |
proofwiki-878 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
Suppose that:
:$a = b q_1 + r_1 = b q_2 + r_2$
where both $0 \le r_1 < b$ and $0 \le r_2 < b$.
{{WLOG}}, suppose $r_1 \ge r_2$.
Then:
:$r_1 - r_2 = b \paren {q_2 - q_1}$
That is:
:$b \divides \paren {r_2 - r_1}$
where $\divides$ d... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
Suppose that:
:$a = b q_1 + r_1 = b q_2 + r_2$
where both $0 \le r_1 < b$ and $0 \le r_2 < b$.
{{WLOG}}, suppose $r_1 \ge r_2$.
Then:
:$r_1 - r_2 = b \paren {q_2 - q_1}$
That is:... | Division Theorem/Positive Divisor/Uniqueness/Proof 3 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_3 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor/Existence",
"Definition:Divisor (Algebra)/Integer",
"Absolute Value of Integer is not less than Divisors/Corollary",
"Integer Divisor Results/Integer Divides Zero"
] |
proofwiki-879 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From Division Theorem: Positive Divisor:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
That is, the result holds for positive $b$.
{{qed|lemma}}
It remains to show that the result also holds for negative values of $b$.
Let $b < 0$.
Consider:
:$\size b = -b > 0$
where $\size b$ denotes th... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | From [[Division Theorem/Positive Divisor|Division Theorem: Positive Divisor]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
That is, the result holds for [[Definition:Positive Integer|positive]] $b$.
{{qed|lemma}}
It remains to show that the result also holds for [[Definition:Negativ... | Division Theorem/Proof 1 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Proof_1 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem/Positive Divisor",
"Definition:Positive/Integer",
"Definition:Negative/Integer",
"Definition:Absolute Value",
"Division Theorem/Positive Divisor",
"Definition:Integer",
"Definition:Unique",
"Division Theorem/Positive Divisor/Uniqueness"
] |
proofwiki-880 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Consider the set of integer multiples $x \size b$ of $\size b$ less than or equal to $a$:
:$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$
We have that:
:$-\size a \size b \le -\size a \le a$
and so $M \ne \O$.
From Set of Integers Bounded Above by Integer has Greatest Element, $M$ has a greatest eleme... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | Consider the set of [[Definition:Divisor of Integer|integer multiples]] $x \size b$ of $\size b$ less than or equal to $a$:
:$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$
We have that:
:$-\size a \size b \le -\size a \le a$
and so $M \ne \O$.
From [[Set of Integers Bounded Above by Integer has Grea... | Division Theorem/Proof 2 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Proof_2 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Divisor (Algebra)/Integer",
"Set of Integers Bounded Above by Integer has Greatest Element",
"Definition:Greatest Element"
] |
proofwiki-881 | Division Theorem | For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | === Existence ===
Consider the arithmetic sequence:
:$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$
which extends in both directions.
Then by the Well-Ordering Principle, there must exist a smallest non-negative element, denoted by $r$.
So $r = a - q b$ for some $q \in \Z$.
$r$ must be in the int... | For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$ | === Existence ===
Consider the [[Definition:Arithmetic Sequence|arithmetic sequence]]:
:$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$
which extends in both directions.
Then by the [[Well-Ordering Principle]], there must exist a smallest [[Definition:Positive Integer|non-negative]] element, de... | Division Theorem/Proof 3 | https://proofwiki.org/wiki/Division_Theorem | https://proofwiki.org/wiki/Division_Theorem/Proof_3 | [
"Division Theorem",
"Divisors",
"Named Theorems"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Definition:Arithmetic Sequence",
"Well-Ordering Principle",
"Definition:Positive/Integer",
"Definition:Positive/Integer",
"Definition:Arithmetic Sequence",
"Definition:Unique"
] |
proofwiki-882 | Odd Integer 2n + 1 | Let $m$ be an odd integer.
Then there exists exactly one integer $n$ such that $2 n + 1 = m$. | Follows directly from the Division Theorem.
{{qed}}
Category:Odd Integers
mpi9q6vy7fvjpowo6jxu7u75i7r217j | Let $m$ be an [[Definition:Odd Integer|odd integer]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Integer|integer]] $n$ such that $2 n + 1 = m$. | Follows directly from the [[Division Theorem]].
{{qed}}
[[Category:Odd Integers]]
mpi9q6vy7fvjpowo6jxu7u75i7r217j | Odd Integer 2n + 1 | https://proofwiki.org/wiki/Odd_Integer_2n_+_1 | https://proofwiki.org/wiki/Odd_Integer_2n_+_1 | [
"Odd Integers"
] | [
"Definition:Odd Integer",
"Definition:Unique",
"Definition:Integer"
] | [
"Division Theorem",
"Category:Odd Integers"
] |
proofwiki-883 | Integer Divisor Results | Let $m, n \in \Z$ be integers.
Let $m \divides n$ denote that $m$ is a divisor of $n$.
The following results all hold: | From Integer Multiplication Identity is One:
:$\forall n \in \Z: 1 \cdot n = n = n \cdot 1$
thus demonstrating that $n$ is a divisor of itself.
{{qed}} | Let $m, n \in \Z$ be [[Definition:Integer|integers]].
Let $m \divides n$ denote that $m$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
The following results all hold: | From [[Integer Multiplication Identity is One]]:
:$\forall n \in \Z: 1 \cdot n = n = n \cdot 1$
thus demonstrating that $n$ is a [[Definition:Divisor of Integer|divisor]] of itself.
{{qed}} | Integer Divisor Results/Integer Divides Itself/Proof 1 | https://proofwiki.org/wiki/Integer_Divisor_Results | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself/Proof_1 | [
"Divisors",
"Integers",
"Integer Divisor Results"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Multiplication Identity is One",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-884 | Integer Divisor Results | Let $m, n \in \Z$ be integers.
Let $m \divides n$ denote that $m$ is a divisor of $n$.
The following results all hold: | As the set of integers form an integral domain, the concept '''divides''' is fully applicable to the integers.
Therefore this result follows directly from Element of Integral Domain is Divisor of Itself.
{{qed}} | Let $m, n \in \Z$ be [[Definition:Integer|integers]].
Let $m \divides n$ denote that $m$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
The following results all hold: | As the [[Integers form Integral Domain|set of integers form an integral domain]], the concept '''[[Definition:Divisor of Ring Element|divides]]''' is fully applicable to the [[Definition:Integer|integers]].
Therefore this result follows directly from [[Element of Integral Domain is Divisor of Itself]].
{{qed}} | Integer Divisor Results/Integer Divides Itself/Proof 2 | https://proofwiki.org/wiki/Integer_Divisor_Results | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself/Proof_2 | [
"Divisors",
"Integers",
"Integer Divisor Results"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integers form Integral Domain",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Integer",
"Element of Integral Domain is Divisor of Itself"
] |
proofwiki-885 | Zero Divides Zero | Let $n \in \Z$ be an integer.
Then:
:$0 \divides n \implies n = 0$
That is, zero is the only integer divisible by zero. | {{begin-eqn}}
{{eqn | l = 0
| o = \divides
| r = n
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q \in \Z
| l = n
| r = q \times 0
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
| l = n
| r = 0
| c = Integers have no zero divisors, as Integers form ... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Then:
:$0 \divides n \implies n = 0$
That is, [[Definition:Zero (Number)|zero]] is the only [[Definition:Integer|integer]] [[Definition:Divisor of Integer|divisible]] by [[Definition:Zero (Number)|zero]]. | {{begin-eqn}}
{{eqn | l = 0
| o = \divides
| r = n
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q \in \Z
| l = n
| r = q \times 0
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
| l = n
| r = 0
| c = Integers have no [[Definition:Zero Divisor of Rin... | Zero Divides Zero | https://proofwiki.org/wiki/Zero_Divides_Zero | https://proofwiki.org/wiki/Zero_Divides_Zero | [
"Divisors"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Zero (Number)"
] | [
"Definition:Zero Divisor/Ring",
"Integers form Integral Domain"
] |
proofwiki-886 | Absolute Value of Integer is not less than Divisors | A (non-zero) integer is greater than or equal to its divisors in magnitude:
:$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$ | Suppose $a \divides c$ for some $c \ne 0$.
From Negative of Absolute Value:
: $a \le \size a$
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \divides
| r = c
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q \in \Z
| l = c
| r = a q
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \lead... | A (non-[[Definition:Zero (Number)|zero]]) [[Definition:Integer|integer]] is greater than or equal to its [[Definition:Divisor of Integer|divisors]] in [[Definition:Absolute Value|magnitude]]:
:$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$ | Suppose $a \divides c$ for some $c \ne 0$.
From [[Negative of Absolute Value]]:
: $a \le \size a$
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \divides
| r = c
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q \in \Z
| l = c
| r = a q
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll=... | Absolute Value of Integer is not less than Divisors | https://proofwiki.org/wiki/Absolute_Value_of_Integer_is_not_less_than_Divisors | https://proofwiki.org/wiki/Absolute_Value_of_Integer_is_not_less_than_Divisors | [
"Number Theory",
"Divisors"
] | [
"Definition:Zero (Number)",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Absolute Value"
] | [
"Negative of Absolute Value"
] |
proofwiki-887 | Divisor Relation on Positive Integers is Partial Ordering | The divisor relation is a partial ordering of $\Z_{>0}$. | Checking in turn each of the criteria for an ordering: | The [[Definition:Divisor of Integer|divisor relation]] is a [[Definition:Partial Ordering|partial ordering]] of $\Z_{>0}$. | Checking in turn each of the criteria for an [[Definition:Ordering|ordering]]: | Divisor Relation on Positive Integers is Partial Ordering | https://proofwiki.org/wiki/Divisor_Relation_on_Positive_Integers_is_Partial_Ordering | https://proofwiki.org/wiki/Divisor_Relation_on_Positive_Integers_is_Partial_Ordering | [
"Divisors",
"Integers",
"Examples of Orderings"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Partial Ordering"
] | [
"Definition:Ordering",
"Definition:Ordering"
] |
proofwiki-888 | Common Divisor in Integral Domain Divides Linear Combination | Let $\struct {D, +, \times}$ be an integral domain.
Let $c$ be a common divisor of two elements $a$ and $b$ of $D$.
That is:
:$a, b, c \in D: c \divides a \land c \divides b$
Then:
:$\forall p, q \in D: c \divides \paren {p \times a + q \times b}$ | {{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | ll= \leadsto
| q = \exists x \in D
| l = a
| r = x \times c
| c = {{Defof|Divisor of Ring Element}}
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{eqn | ll= \leadsto
| q = \exists y \in D
| l = b
... | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
Let $c$ be a [[Definition:Common Divisor of Ring Elements|common divisor]] of two elements $a$ and $b$ of $D$.
That is:
:$a, b, c \in D: c \divides a \land c \divides b$
Then:
:$\forall p, q \in D: c \divides \paren {p \times a + q... | {{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | ll= \leadsto
| q = \exists x \in D
| l = a
| r = x \times c
| c = {{Defof|Divisor of Ring Element}}
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{eqn | ll= \leadsto
| q = \exists y \in D
| l = b
... | Common Divisor in Integral Domain Divides Linear Combination | https://proofwiki.org/wiki/Common_Divisor_in_Integral_Domain_Divides_Linear_Combination | https://proofwiki.org/wiki/Common_Divisor_in_Integral_Domain_Divides_Linear_Combination | [
"Divisibility"
] | [
"Definition:Integral Domain",
"Definition:Common Divisor/Integral Domain"
] | [
"Definition:Distributive Operation"
] |
proofwiki-889 | Existence of Greatest Common Divisor | Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.
Then the greatest common divisor of $a$ and $b$ exists. | === Proof of Existence ===
This is proved in Greatest Common Divisor is at least $1$.
{{qed|lemma}}
=== Proof of there being a Largest ===
{{WLOG}}, suppose $a \ne 0$.
First we note that from Absolute Value of Integer is not less than Divisors:
:$\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \siz... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ne 0$ or $b \ne 0$.
Then the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ exists. | === Proof of Existence ===
This is proved in [[Greatest Common Divisor is at least 1|Greatest Common Divisor is at least $1$]].
{{qed|lemma}}
=== Proof of there being a Largest ===
{{WLOG}}, suppose $a \ne 0$.
First we note that from [[Absolute Value of Integer is not less than Divisors]]:
:$\forall c \in \Z: \fo... | Existence of Greatest Common Divisor/Proof 1 | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor/Proof_1 | [
"Greatest Common Divisor",
"Existence of Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Greatest Common Divisor is at least 1",
"Absolute Value of Integer is not less than Divisors",
"Definition:Zero (Number)",
"Definition:Greatest Common Divisor/Integers",
"Definition:Common Divisor",
"Definition:Bounded Above Set",
"Set of Integers Bounded Above by Integer has Greatest Element",
"Defi... |
proofwiki-890 | Existence of Greatest Common Divisor | Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.
Then the greatest common divisor of $a$ and $b$ exists. | By definition of greatest common divisor, we aim to show that there exists $c \in \Z_{>0}$ such that:
{{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{end-eqn}}
and:
:$d \divides a, d \divides b \implies d \divides c$
Consider the set $S$:
:$S = \se... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ne 0$ or $b \ne 0$.
Then the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ exists. | By definition of [[Definition:Greatest Common Divisor of Integers|greatest common divisor]], we aim to show that there exists $c \in \Z_{>0}$ such that:
{{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{end-eqn}}
and:
:$d \divides a, d \divides b \i... | Existence of Greatest Common Divisor/Proof 2 | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor/Proof_2 | [
"Greatest Common Divisor",
"Existence of Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Greatest Common Divisor/Integers",
"Definition:Set",
"Definition:Empty Set",
"Well-Ordering Principle",
"Common Divisor Divides Integer Combination",
"Division Theorem",
"Definition:Mutatis Mutandis",
"Definition:Unique"
] |
proofwiki-891 | Existence of Greatest Common Divisor | Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.
Then the greatest common divisor of $a$ and $b$ exists. | From Integers form Integral Domain, we have that $\Z$ is an integral domain.
From Euclidean Domain is GCD Domain, $a$ and $b$ have a greatest common divisor $c$.
This proves existence.
From Ring of Integers is Principal Ideal Domain, we have that $\Z$ is a principal ideal domain.
Suppose $c$ and $c'$ are both greatest ... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ne 0$ or $b \ne 0$.
Then the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ exists. | From [[Integers form Integral Domain]], we have that $\Z$ is an [[Definition:Integral Domain|integral domain]].
From [[Euclidean Domain is GCD Domain]], $a$ and $b$ have a [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] $c$.
This proves existence.
From [[Ring of Integers is Principal Idea... | Existence of Greatest Common Divisor/Proof 3 | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor/Proof_3 | [
"Greatest Common Divisor",
"Existence of Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Integers form Integral Domain",
"Definition:Integral Domain",
"Euclidean Domain is GCD Domain",
"Definition:Greatest Common Divisor/Integers",
"Ring of Integers is Principal Ideal Domain",
"Definition:Principal Ideal Domain",
"Definition:Greatest Common Divisor/Integers",
"Greatest Common Divisors in... |
proofwiki-892 | Existence of Greatest Common Divisor | Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.
Then the greatest common divisor of $a$ and $b$ exists. | From the Euclidean Algorithm, we have calculated a sequence $\tuple {r_1, r_2, \ldots r_{n - 2}, r_{n - 1}, r_n}$ such that:
:$b > r_1 > r_2 > \dotsb > r_{n - 2} > r_{n - 1} > r_n = 0$
We have that:
:$r_{n - 1} \divides a$
and:
:$r_{n - 1} \divides b$
Working backwards from the final equation, we see that:
:$r_k \divid... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ne 0$ or $b \ne 0$.
Then the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ exists. | From the [[Euclidean Algorithm]], we have calculated a sequence $\tuple {r_1, r_2, \ldots r_{n - 2}, r_{n - 1}, r_n}$ such that:
:$b > r_1 > r_2 > \dotsb > r_{n - 2} > r_{n - 1} > r_n = 0$
We have that:
:$r_{n - 1} \divides a$
and:
:$r_{n - 1} \divides b$
Working backwards from the final equation, we see that:
:$r_k ... | Existence of Greatest Common Divisor/Proof 4 | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor | https://proofwiki.org/wiki/Existence_of_Greatest_Common_Divisor/Proof_4 | [
"Greatest Common Divisor",
"Existence of Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Euclidean Algorithm",
"Principle of Mathematical Induction",
"Euclidean Algorithm",
"Definition:Greatest Common Divisor/Integers",
"Definition:Greatest Common Divisor/Integers",
"Definition:Greatest Common Divisor/Integers",
"Definition:Unique"
] |
proofwiki-893 | Greatest Common Divisor is at least 1 | Let $a, b \in \Z$ be integers.
The greatest common divisor of $a$ and $b$ is at least $1$:
:$\forall a, b \in \Z_{\ne 0}: \gcd \set {a, b} \ge 1$ | From One Divides all Integers:
:$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$
and so:
:$1 \le \gcd \set {a, b}$
as required.
{{qed}}
Category:Greatest Common Divisor
2nrji97dz25wb80mslueng8xuvr7oza | Let $a, b \in \Z$ be [[Definition:Integer|integers]].
The [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ is at least $1$:
:$\forall a, b \in \Z_{\ne 0}: \gcd \set {a, b} \ge 1$ | From [[One Divides all Integers]]:
:$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$
and so:
:$1 \le \gcd \set {a, b}$
as required.
{{qed}}
[[Category:Greatest Common Divisor]]
2nrji97dz25wb80mslueng8xuvr7oza | Greatest Common Divisor is at least 1 | https://proofwiki.org/wiki/Greatest_Common_Divisor_is_at_least_1 | https://proofwiki.org/wiki/Greatest_Common_Divisor_is_at_least_1 | [
"Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Integer Divisor Results/One Divides all Integers",
"Category:Greatest Common Divisor"
] |
proofwiki-894 | GCD of Integer and Divisor | Let $a, b \in \Z_{>0}$ be strictly positive integers.
Then:
:$a \divides b \implies \gcd \set {a, b} = a$ | We have:
:$a \divides b$ {{hypothesis}}
:$a \divides a$ from Integer Divides Itself.
Thus $a$ is a common divisor of $a$ and $b$.
Then from Absolute Value of Integer is not less than Divisors:
:$\forall x \in \Z: x \divides a \implies x \le \size a$
As $a$ and $b$ are both positive, the result follows.
{{qed}} | Let $a, b \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|strictly positive integers]].
Then:
:$a \divides b \implies \gcd \set {a, b} = a$ | We have:
:$a \divides b$ {{hypothesis}}
:$a \divides a$ from [[Integer Divides Itself]].
Thus $a$ is a [[Definition:Common Divisor of Integers|common divisor]] of $a$ and $b$.
Then from [[Absolute Value of Integer is not less than Divisors]]:
:$\forall x \in \Z: x \divides a \implies x \le \size a$
As $a$ and $b$ ... | GCD of Integer and Divisor | https://proofwiki.org/wiki/GCD_of_Integer_and_Divisor | https://proofwiki.org/wiki/GCD_of_Integer_and_Divisor | [
"Greatest Common Divisor"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Integer Divisor Results/Integer Divides Itself",
"Definition:Common Divisor/Integers",
"Absolute Value of Integer is not less than Divisors",
"Definition:Positive/Integer"
] |
proofwiki-895 | GCD for Negative Integers | Let $a$ and $b$ be integers.
:$\gcd \set {a, b} = \gcd \set {\size a, b} = \gcd \set {a, \size b} = \gcd \set {\size a, \size b}$
where $\gcd$ denotes the greatest common divisor.
Alternatively, this can be put:
:$\gcd \set {a, b} = \gcd \set {-a, b} = \gcd \set {a, -b} = \gcd \set {-a, -b}$
which follows directly from... | Note that $\size a = \pm a$.
Suppose that:
:$u \divides a$
where $\divides$ denotes divisibility.
Then:
:$\exists q \in \Z: a = q u$
Then:
:$\size a = \pm q u = \paren {\pm q} u \implies u \divides \size a$
So every divisor of $a$ is a divisor of $\size a$.
Similarly, note that:
:$a = \pm \size a$
so every divisor of $... | Let $a$ and $b$ be [[Definition:Integer|integers]].
:$\gcd \set {a, b} = \gcd \set {\size a, b} = \gcd \set {a, \size b} = \gcd \set {\size a, \size b}$
where $\gcd$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
Alternatively, this can be put:
:$\gcd \set {a, b} = \gcd \se... | Note that $\size a = \pm a$.
Suppose that:
:$u \divides a$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\exists q \in \Z: a = q u$
Then:
:$\size a = \pm q u = \paren {\pm q} u \implies u \divides \size a$
So every [[Definition:Divisor of Integer|divisor]] of $a$ is a [[Definition... | GCD for Negative Integers | https://proofwiki.org/wiki/GCD_for_Negative_Integers | https://proofwiki.org/wiki/GCD_for_Negative_Integers | [
"Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Common Divisor/Integers"
] |
proofwiki-896 | GCD with Zero | Let $a \in \Z$ be an integer such that $a \ne 0$.
Then:
:$\gcd \set {a, 0} = \size a$
where $\gcd$ denotes greatest common divisor (GCD). | Follows from:
:Integer Divides Zero
and:
:GCD for Negative Integers.
{{qed}} | Let $a \in \Z$ be an [[Definition:Integer|integer]] such that $a \ne 0$.
Then:
:$\gcd \set {a, 0} = \size a$
where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor (GCD)]]. | Follows from:
:[[Integer Divides Zero]]
and:
:[[GCD for Negative Integers]].
{{qed}} | GCD with Zero | https://proofwiki.org/wiki/GCD_with_Zero | https://proofwiki.org/wiki/GCD_with_Zero | [
"Greatest Common Divisor"
] | [
"Definition:Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Integer Divisor Results/Integer Divides Zero",
"GCD for Negative Integers"
] |
proofwiki-897 | Set of Integer Combinations equals Set of Multiples of GCD | The set of all integer combinations of $a$ and $b$ is precisely the set of all integer multiples of the GCD of $a$ and $b$:
:$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$ | === Necessary Condition ===
Let $d = \gcd \set {a, b}$.
Then:
:$d \divides c \implies \exists m \in \Z: c = m d$
So:
{{begin-eqn}}
{{eqn | q = \exists p, q \in \Z
| l = d
| r = p a + q b
| c = Bézout's Identity
}}
{{eqn | ll= \leadsto
| l = c
| r = m d
| c =
}}
{{eqn | r = m p a + m... | The [[Definition:Set|set]] of all [[Definition:Integer Combination|integer combinations]] of $a$ and $b$ is precisely the set of all [[Definition:Divisor of Integer|integer multiples]] of the [[Definition:Greatest Common Divisor of Integers|GCD]] of $a$ and $b$:
:$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: ... | === Necessary Condition ===
Let $d = \gcd \set {a, b}$.
Then:
:$d \divides c \implies \exists m \in \Z: c = m d$
So:
{{begin-eqn}}
{{eqn | q = \exists p, q \in \Z
| l = d
| r = p a + q b
| c = [[Bézout's Identity]]
}}
{{eqn | ll= \leadsto
| l = c
| r = m d
| c =
}}
{{eqn | r = ... | Set of Integer Combinations equals Set of Multiples of GCD | https://proofwiki.org/wiki/Set_of_Integer_Combinations_equals_Set_of_Multiples_of_GCD | https://proofwiki.org/wiki/Set_of_Integer_Combinations_equals_Set_of_Multiples_of_GCD | [
"Integer Combinations",
"Greatest Common Divisor"
] | [
"Definition:Set",
"Definition:Integer Combination",
"Definition:Divisor (Algebra)/Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Bézout's Identity"
] |
proofwiki-898 | GCD with Remainder | Let $a, b \in \Z$.
Let $q, r \in \Z$ such that $a = q b + r$.
Then:
:$\gcd \set {a, b} = \gcd \set {b, r}$
where $\gcd \set {a, b}$ is the greatest common divisor of $a$ and $b$. | {{begin-eqn}}
{{eqn | o =
| r = \gcd \set {a, b} \divides a \land \gcd \set {a, b} \divides b
| c = {{Defof|Greatest Common Divisor of Integers}}
}}
{{eqn | ll= \leadsto
| o =
| r = \gcd \set {a, b} \divides \paren {a - q b}
| c = Common Divisor Divides Integer Combination
}}
{{eqn | ll=... | Let $a, b \in \Z$.
Let $q, r \in \Z$ such that $a = q b + r$.
Then:
:$\gcd \set {a, b} = \gcd \set {b, r}$
where $\gcd \set {a, b}$ is the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | {{begin-eqn}}
{{eqn | o =
| r = \gcd \set {a, b} \divides a \land \gcd \set {a, b} \divides b
| c = {{Defof|Greatest Common Divisor of Integers}}
}}
{{eqn | ll= \leadsto
| o =
| r = \gcd \set {a, b} \divides \paren {a - q b}
| c = [[Common Divisor Divides Integer Combination]]
}}
{{eqn |... | GCD with Remainder | https://proofwiki.org/wiki/GCD_with_Remainder | https://proofwiki.org/wiki/GCD_with_Remainder | [
"Greatest Common Divisor"
] | [
"Definition:Greatest Common Divisor/Integers"
] | [
"Common Divisor Divides Integer Combination",
"Common Divisor Divides Integer Combination"
] |
proofwiki-899 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | The proof can conveniently be divided into two parts: | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | The proof can conveniently be divided into two parts: | Integer Combination of Coprime Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [] |
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