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-8200 | Bernoulli's Inequality | Let $x \in \R$ be a real number such that $x > -1$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\paren {1 + x}^n \ge 1 + n x$ | Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\paren {1 + x}^n \ge 1 + nx$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\paren {1 + x}^0 \ge 1$
so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P ... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > -1$.
Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\paren {1 + x}^n \ge 1 + n x$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {1 + x}^n \ge 1 + nx$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\paren {1 + x}^0 \ge 1$
so $\map P 0$ holds.
This is our [[Definition:Basis ... | Bernoulli's Inequality/Proof 1 | https://proofwiki.org/wiki/Bernoulli's_Inequality | https://proofwiki.org/wiki/Bernoulli's_Inequality/Proof_1 | [
"Bernoulli's Inequality",
"Inequalities",
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Positive/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Bernoulli's Inequality/Proof 1",
"Principle of Mathematical Induction"
] |
proofwiki-8201 | Bernoulli's Inequality | Let $x \in \R$ be a real number such that $x > -1$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\paren {1 + x}^n \ge 1 + n x$ | Let $y = 1 + x$.
Then $y \ge 0$, and:
:$\paren {1 + x}^n = 1 + \paren {y^n - 1}$
If $y \ge 1$, then by Sum of Geometric Sequence:
:$\ds y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$
If $y < 1$, then by Sum of Geometric Sequence:
:$\ds y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > -1$.
Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\paren {1 + x}^n \ge 1 + n x$ | Let $y = 1 + x$.
Then $y \ge 0$, and:
:$\paren {1 + x}^n = 1 + \paren {y^n - 1}$
If $y \ge 1$, then by [[Sum of Geometric Sequence]]:
:$\ds y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$
If $y < 1$, then by [[Sum of Geometric Sequence]]:
:$\ds y^n - 1 = -\paren {1 - y} \sum_{k ... | Bernoulli's Inequality/Proof 2 | https://proofwiki.org/wiki/Bernoulli's_Inequality | https://proofwiki.org/wiki/Bernoulli's_Inequality/Proof_2 | [
"Bernoulli's Inequality",
"Inequalities",
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Positive/Integer"
] | [
"Sum of Geometric Sequence",
"Sum of Geometric Sequence"
] |
proofwiki-8202 | Categories of Elements of Ring | Let $\left({R, +, \circ}\right)$ be a ring.
The elements of $R$ are partitioned into three classes:
:$(1): \quad$ the zero
:$(2): \quad$ the units
:$(3): \quad$ the proper elements. | By definition, a proper element is a non-zero element which has no product inverse.
Also by definition, a unit is an element which does have a product inverse.
Because $0 \circ x = 0$ there can be no $x \in R$ such that $0 \times x = 1$, and so $0$ is not a unit.
Hence the result.
{{qed}}
Category:Ring Theory
76cxzcd0t... | Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
The [[Definition:Element|elements]] of $R$ are [[Definition:Partition (Set Theory)|partitioned]] into three classes:
:$(1): \quad$ the [[Definition:Ring Zero|zero]]
:$(2): \quad$ the [[Definition:Unit of Ring|units]]
:$(3): \quad$ the [... | By definition, a [[Definition:Proper Element of Ring|proper element]] is a non-[[Definition:Ring Zero|zero]] element which has no [[Definition:Ring Product Inverse|product inverse]].
Also by definition, a [[Definition:Unit of Ring|unit]] is an element which does have a [[Definition:Ring Product Inverse|product inverse... | Categories of Elements of Ring | https://proofwiki.org/wiki/Categories_of_Elements_of_Ring | https://proofwiki.org/wiki/Categories_of_Elements_of_Ring | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Element",
"Definition:Set Partition",
"Definition:Ring Zero",
"Definition:Unit of Ring",
"Definition:Proper Element of Ring"
] | [
"Definition:Proper Element of Ring",
"Definition:Ring Zero",
"Definition:Product Inverse",
"Definition:Unit of Ring",
"Definition:Product Inverse",
"Definition:Unit of Ring",
"Category:Ring Theory"
] |
proofwiki-8203 | Multiple of Ring Product | Let $\struct {R, +, \circ}$ be a ring.
Let $x, y \in \struct {R, +, \circ}$.
Then:
:$\forall n \in \Z_{> 0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$
where $n \cdot x$ denotes the $n$th multiple of $x$. | By definition:
:$\ds n \cdot x := \sum_{j \mathop = 1}^n x$
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n \cdot x} \circ y
| r = \paren {\sum_{j \mathop = 1}^n x} \circ y
| c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | n = 1
| r = \sum_{j \mathop = 1}^n \paren {x \circ y}
| c ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $x, y \in \struct {R, +, \circ}$.
Then:
:$\forall n \in \Z_{> 0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$
where $n \cdot x$ denotes the [[Definition:Integral Multiple/Rings and Fields|$n$th... | By definition:
:$\ds n \cdot x := \sum_{j \mathop = 1}^n x$
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n \cdot x} \circ y
| r = \paren {\sum_{j \mathop = 1}^n x} \circ y
| c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | n = 1
| r = \sum_{j \mathop = 1}^n \paren {x \circ y}
| ... | Multiple of Ring Product | https://proofwiki.org/wiki/Multiple_of_Ring_Product | https://proofwiki.org/wiki/Multiple_of_Ring_Product | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Integral Multiple/Rings and Fields"
] | [
"General Distributivity Theorem",
"General Distributivity Theorem",
"Category:Ring Theory"
] |
proofwiki-8204 | Ordering is Preserved on Integers by Addition | The usual ordering on the integers is preserved by the operation of addition:
:$\forall a, b, c, d, \in \Z: a \le b, c \le d \implies a + c \le b + d$ | Recall that Integers form Ordered Integral Domain.
Then from Relation Induced by Strict Positivity Property is Compatible with Addition:
:$\forall x, y, z \in \Z: x \le y \implies \paren {x + z} \le \paren {y + z}$
:$\forall x, y, z \in \Z: x \le y \implies \paren {z + x} \le \paren {z + y}$
So:
{{begin-eqn}}
{{eqn | l... | The [[Definition:Usual Ordering|usual ordering]] on the [[Definition:Integer|integers]] is preserved by the operation of [[Definition:Integer Addition|addition]]:
:$\forall a, b, c, d, \in \Z: a \le b, c \le d \implies a + c \le b + d$ | Recall that [[Integers form Ordered Integral Domain]].
Then from [[Relation Induced by Strict Positivity Property is Compatible with Addition]]:
:$\forall x, y, z \in \Z: x \le y \implies \paren {x + z} \le \paren {y + z}$
:$\forall x, y, z \in \Z: x \le y \implies \paren {z + x} \le \paren {z + y}$
So:
{{begin-eq... | Ordering is Preserved on Integers by Addition | https://proofwiki.org/wiki/Ordering_is_Preserved_on_Integers_by_Addition | https://proofwiki.org/wiki/Ordering_is_Preserved_on_Integers_by_Addition | [
"Integer Addition",
"Orderings on Integers"
] | [
"Definition:Usual Ordering",
"Definition:Integer",
"Definition:Addition/Integers"
] | [
"Integers form Ordered Integral Domain",
"Relation Induced by Strict Positivity Property is Compatible with Addition",
"Relation Induced by Strict Positivity Property is Compatible with Addition",
"Relation Induced by Strict Positivity Property is Compatible with Addition"
] |
proofwiki-8205 | Power Set of Finite Set is Finite | Let $S$ be a finite set.
Then the power set of $S$ is likewise finite. | Let $S$ be a finite set.
Then by definition:
:$\exists n \in \N: \card S = n$
where $\card S$ denotes the cardinality of $S$.
From Cardinality of Power Set of Finite Set:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$.
As $n \in \N$ it follows that $2^n \in \N$ and so $\powerset S$ is als... | Let $S$ be a [[Definition:Finite Set|finite set]].
Then the [[Definition:Power Set|power set]] of $S$ is likewise [[Definition:Finite Set|finite]]. | Let $S$ be a [[Definition:Finite Set|finite set]].
Then by definition:
:$\exists n \in \N: \card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
From [[Cardinality of Power Set of Finite Set]]:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the [[Definition:Power Set|power ... | Power Set of Finite Set is Finite | https://proofwiki.org/wiki/Power_Set_of_Finite_Set_is_Finite | https://proofwiki.org/wiki/Power_Set_of_Finite_Set_is_Finite | [
"Power Set"
] | [
"Definition:Finite Set",
"Definition:Power Set",
"Definition:Finite Set"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Cardinality of Power Set of Finite Set",
"Definition:Power Set",
"Definition:Finite Set",
"Category:Power Set"
] |
proofwiki-8206 | Common Divisor Divides Difference | Let $c$ be a common divisor of two integers $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a - b}$ | Let $c \divides a \land c \divides b$.
From Common Divisor Divides Integer Combination:
:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
Putting $p = 1$ and $q = -1$:
:$c \divides \paren {a - b}$
{{qed}} | Let $c$ be a [[Definition:Common Divisor of Integers|common divisor]] of two [[Definition:Integer|integers]] $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a - b}$ | Let $c \divides a \land c \divides b$.
From [[Common Divisor Divides Integer Combination]]:
:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
Putting $p = 1$ and $q = -1$:
:$c \divides \paren {a - b}$
{{qed}} | Common Divisor Divides Difference/Proof 1 | https://proofwiki.org/wiki/Common_Divisor_Divides_Difference | https://proofwiki.org/wiki/Common_Divisor_Divides_Difference/Proof_1 | [
"Divisors",
"Common Divisor Divides Difference"
] | [
"Definition:Common Divisor/Integers",
"Definition:Integer"
] | [
"Common Divisor Divides Integer Combination"
] |
proofwiki-8207 | Common Divisor Divides Difference | Let $c$ be a common divisor of two integers $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a - b}$ | {{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | ll= \leadsto
| q = \exists x \in \Z
| l = a
| r = x c
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{eqn | ll= \leadsto
| q = \exists y \in \Z
| l = b
| r = y ... | Let $c$ be a [[Definition:Common Divisor of Integers|common divisor]] of two [[Definition:Integer|integers]] $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a - b}$ | {{begin-eqn}}
{{eqn | l = c
| o = \divides
| r = a
}}
{{eqn | ll= \leadsto
| q = \exists x \in \Z
| l = a
| r = x c
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = c
| o = \divides
| r = b
}}
{{eqn | ll= \leadsto
| q = \exists y \in \Z
| l = b
| r = y ... | Common Divisor Divides Difference/Proof 2 | https://proofwiki.org/wiki/Common_Divisor_Divides_Difference | https://proofwiki.org/wiki/Common_Divisor_Divides_Difference/Proof_2 | [
"Divisors",
"Common Divisor Divides Difference"
] | [
"Definition:Common Divisor/Integers",
"Definition:Integer"
] | [
"Integer Multiplication Distributes over Addition"
] |
proofwiki-8208 | Divisor Divides Multiple | Let $a, b$ be integers.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$\forall c \in \Z: a \divides b c$ | Let $a \divides b$.
From Integer Divides Zero:
:$a \divides 0$
Thus $a$ is a common divisor of $b$ and $0$.
From Common Divisor Divides Integer Combination:
:$\forall p, q \in \Z: a \divides \paren {p \cdot b + q \cdot 0}$
Putting $p = c$ and $q = 1$ (for example):
:$a \divides \paren {c b + 0}$
Hence the result.
{{qed... | Let $a, b$ be [[Definition:Integer|integers]].
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\forall c \in \Z: a \divides b c$ | Let $a \divides b$.
From [[Integer Divides Zero]]:
:$a \divides 0$
Thus $a$ is a [[Definition:Common Divisor of Integers|common divisor]] of $b$ and $0$.
From [[Common Divisor Divides Integer Combination]]:
:$\forall p, q \in \Z: a \divides \paren {p \cdot b + q \cdot 0}$
Putting $p = c$ and $q = 1$ (for example):... | Divisor Divides Multiple/Proof 1 | https://proofwiki.org/wiki/Divisor_Divides_Multiple | https://proofwiki.org/wiki/Divisor_Divides_Multiple/Proof_1 | [
"Divisors",
"Divisor Divides Multiple"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Divisor Results/Integer Divides Zero",
"Definition:Common Divisor/Integers",
"Common Divisor Divides Integer Combination"
] |
proofwiki-8209 | Divisor Divides Multiple | Let $a, b$ be integers.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$\forall c \in \Z: a \divides b c$ | {{begin-eqn}}
{{eqn | l = a
| o = \divides
| r = b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists x \in \Z
| l = b
| r = x a
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
| l = b c
| r = x c a
| c =
}}
{{eqn | ll= \leadsto
| q = \exists z \in... | Let $a, b$ be [[Definition:Integer|integers]].
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\forall c \in \Z: a \divides b c$ | {{begin-eqn}}
{{eqn | l = a
| o = \divides
| r = b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists x \in \Z
| l = b
| r = x a
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
| l = b c
| r = x c a
| c =
}}
{{eqn | ll= \leadsto
| q = \exists z \in... | Divisor Divides Multiple/Proof 2 | https://proofwiki.org/wiki/Divisor_Divides_Multiple | https://proofwiki.org/wiki/Divisor_Divides_Multiple/Proof_2 | [
"Divisors",
"Divisor Divides Multiple"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [] |
proofwiki-8210 | Divisor Divides Multiple | Let $a, b$ be integers.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$\forall c \in \Z: a \divides b c$ | We have that Integers form Integral Domain.
The result then follows from Multiple of Divisor in Integral Domain Divides Multiple.
{{qed}} | Let $a, b$ be [[Definition:Integer|integers]].
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\forall c \in \Z: a \divides b c$ | We have that [[Integers form Integral Domain]].
The result then follows from [[Multiple of Divisor in Integral Domain Divides Multiple]].
{{qed}} | Multiple of Divisor Divides Multiple/Proof 1 | https://proofwiki.org/wiki/Divisor_Divides_Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple/Proof_1 | [
"Divisors",
"Divisor Divides Multiple"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integers form Integral Domain",
"Multiple of Divisor in Integral Domain Divides Multiple"
] |
proofwiki-8211 | Divisor Divides Multiple | Let $a, b$ be integers.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$\forall c \in \Z: a \divides b c$ | By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.
Then:
:$\paren {a d} c = b c$
that is:
:$\paren {a c} d = b c$
which follows because Integer Multiplication is Commutative and Integer Multiplication is Associative.
Hence the result.
{{qed}} | Let $a, b$ be [[Definition:Integer|integers]].
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\forall c \in \Z: a \divides b c$ | By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.
Then:
:$\paren {a d} c = b c$
that is:
:$\paren {a c} d = b c$
which follows because [[Integer Multiplication is Commutative]] and [[Integer Multiplication is Associative]].
Hence the result.
{{qed}} | Multiple of Divisor Divides Multiple/Proof 2 | https://proofwiki.org/wiki/Divisor_Divides_Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple/Proof_2 | [
"Divisors",
"Divisor Divides Multiple"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Multiplication is Commutative",
"Integer Multiplication is Associative"
] |
proofwiki-8212 | Non-Zero Integer has Unique Positive Integer Associate | Let $a \in \Z$ be an integer such that $a \ne 0$.
Then $a$ has a unique associate $b \in \Z_{>0}$. | Let $a \in \Z_{\ne 0}$.
By Integer Divides its Absolute Value:
:$a \divides \size a$ and $\size a \divides a$
Hence $\size a$ is an associate of $a$.
Now we prove its uniqueness.
Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$.
Let $a \sim b$ and $a \sim c$ where $\sim$ denotes the relation of associatehood.
By... | Let $a \in \Z$ be an [[Definition:Integer|integer]] such that $a \ne 0$.
Then $a$ has a [[Definition:Unique|unique]] [[Definition:Associate of Integer|associate]] $b \in \Z_{>0}$. | Let $a \in \Z_{\ne 0}$.
By [[Integer Divides its Absolute Value]]:
:$a \divides \size a$ and $\size a \divides a$
Hence $\size a$ is an [[Definition:Associate of Integer|associate]] of $a$.
Now we prove its [[Definition:Unique|uniqueness]].
Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$.
Let $a \sim b$ a... | Non-Zero Integer has Unique Positive Integer Associate | https://proofwiki.org/wiki/Non-Zero_Integer_has_Unique_Positive_Integer_Associate | https://proofwiki.org/wiki/Non-Zero_Integer_has_Unique_Positive_Integer_Associate | [
"Integers"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Associate/Integers"
] | [
"Integer Divisor Results/Integer Divides its Absolute Value",
"Definition:Associate/Integers",
"Definition:Unique",
"Definition:Associate/Integers",
"Definition:Associate/Integers",
"Divisor Relation is Antisymmetric/Corollary/Proof 2"
] |
proofwiki-8213 | One is not Prime | The integer $1$ (one) is not a prime number. | By definition, a prime number is a positive integer which has exactly $2$ divisors which are themselves positive integers.
From Divisors of One, the only divisors of $1$ are $1$ and $-1$.
So the only divisor of $1$ which is a positive integer is $1$.
As $1$ has only one such divisor, it is not classified as a prime num... | The [[Definition:Integer|integer]] [[1|$1$ (one)]] is not a [[Definition:Prime Number|prime number]]. | By definition, a [[Definition:Prime Number/Definition 1|prime number]] is a [[Definition:Positive Integer|positive integer]] which has exactly $2$ [[Definition:Divisor of Integer|divisors]] which are themselves [[Definition:Positive Integer|positive integers]].
From [[Divisors of One]], the only [[Definition:Divisor o... | One is not Prime/Proof 1 | https://proofwiki.org/wiki/One_is_not_Prime | https://proofwiki.org/wiki/One_is_not_Prime/Proof_1 | [
"One is not Prime",
"Prime Numbers",
"1"
] | [
"Definition:Integer",
"1",
"Definition:Prime Number"
] | [
"Definition:Prime Number/Definition 1",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Positive/Integer",
"Divisors of One",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra... |
proofwiki-8214 | One is not Prime | The integer $1$ (one) is not a prime number. | From Divisor Sum of Prime Number, the sum $\map {\sigma_1} p$ of all the positive integer divisors of a prime number $p$ is $p + 1$.
But from Divisor Sum of 1, $\map {\sigma_1} 1 = 1$.
If $1$ were to be classified as prime, then $\map {\sigma_1} 1$ would be an exception to the rule that $\map {\sigma_1} p = p + 1$.
{{q... | The [[Definition:Integer|integer]] [[1|$1$ (one)]] is not a [[Definition:Prime Number|prime number]]. | From [[Divisor Sum of Prime Number]], the [[Definition:Integer Addition|sum]] $\map {\sigma_1} p$ of all the [[Definition:Positive Integer|positive integer]] [[Definition:Divisor of Integer|divisors]] of a [[Definition:Prime Number|prime number]] $p$ is $p + 1$.
But from [[Divisor Sum of 1]], $\map {\sigma_1} 1 = 1$.
... | One is not Prime/Proof 2 | https://proofwiki.org/wiki/One_is_not_Prime | https://proofwiki.org/wiki/One_is_not_Prime/Proof_2 | [
"One is not Prime",
"Prime Numbers",
"1"
] | [
"Definition:Integer",
"1",
"Definition:Prime Number"
] | [
"Divisor Sum of Prime Number",
"Definition:Addition/Integers",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Divisor Sum of 1",
"Definition:Prime Number"
] |
proofwiki-8215 | Prefix of String is Substring | Let $S$ be a string.
Let $T$ be a prefix of $S$.
Then $T$ is a substring of $S$. | By definition of substring, there exists a string $T'$ such that:
:$S = TT'$
Hence $S$ is the concatenation of the null string, $T$, and $T'$.
Thus by definition of substring, $T$ is a substring of $S$.
{{qed}}
Category:Prefixes
Category:Substrings
1ffntq581uimopz9dy0fh6decupnmhm | Let $S$ be a [[Definition:String|string]].
Let $T$ be a [[Definition:Prefix|prefix]] of $S$.
Then $T$ is a [[Definition:Substring|substring]] of $S$. | By definition of [[Definition:Substring|substring]], there exists a [[Definition:String|string]] $T'$ such that:
:$S = TT'$
Hence $S$ is the [[Definition:Concatenation (Formal Systems)|concatenation]] of the [[Definition:Null String|null string]], $T$, and $T'$.
Thus by definition of [[Definition:Substring|substrin... | Prefix of String is Substring | https://proofwiki.org/wiki/Prefix_of_String_is_Substring | https://proofwiki.org/wiki/Prefix_of_String_is_Substring | [
"Prefixes",
"Substrings"
] | [
"Definition:String",
"Definition:Prefix",
"Definition:Substring"
] | [
"Definition:Substring",
"Definition:String",
"Definition:Concatenation (Formal Systems)",
"Definition:Null String",
"Definition:Substring",
"Definition:Substring",
"Category:Prefixes",
"Category:Substrings"
] |
proofwiki-8216 | Ring Zero is Idempotent | Let $\struct {R, +, \circ}$ be a ring whose ring zero is $0_R$.
Then $0_R$ is an idempotent element of $R$ under the ring product $\circ$:
:$0_R \circ 0_R = 0_R$ | By Ring Product with Zero (applied to $0_R$):
:$0_R \circ 0_R = 0_R$
which was to be proven.
{{qed}}
Category:Ring Theory
aazu4uwonejgx8f8if9b80vsklxwxl1 | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|ring zero]] is $0_R$.
Then $0_R$ is an [[Definition:Idempotent Element|idempotent element]] of $R$ under the [[Definition:Ring Product|ring product]] $\circ$:
:$0_R \circ 0_R = 0_R$ | By [[Ring Product with Zero]] (applied to $0_R$):
:$0_R \circ 0_R = 0_R$
which was to be proven.
{{qed}}
[[Category:Ring Theory]]
aazu4uwonejgx8f8if9b80vsklxwxl1 | Ring Zero is Idempotent | https://proofwiki.org/wiki/Ring_Zero_is_Idempotent | https://proofwiki.org/wiki/Ring_Zero_is_Idempotent | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Idempotence/Element",
"Definition:Ring (Abstract Algebra)/Product"
] | [
"Ring Product with Zero",
"Category:Ring Theory"
] |
proofwiki-8217 | Prime iff Coprime to all Smaller Positive Integers | Let $p$ be a prime number.
Then:
:$\forall x \in \Z, 0 < x < p: x \perp p$
That is, $p$ is relatively prime to all smaller (strictly) positive integers. | From Prime not Divisor implies Coprime, if $p$ does not divide an integer $x$, it is relatively prime to $x$.
From Absolute Value of Integer is not less than Divisors: Corollary, $p$ does not divide an integer smaller than $p$.
It follows that $p$ is relatively prime to all smaller (strictly) positive integers.
The spe... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\forall x \in \Z, 0 < x < p: x \perp p$
That is, $p$ is [[Definition:Coprime Integers|relatively prime]] to all smaller [[Definition:Strictly Positive Integer|(strictly) positive integers]]. | From [[Prime not Divisor implies Coprime]], if $p$ does not [[Definition:Divisor of Integer|divide]] an [[Definition:Integer|integer]] $x$, it is [[Definition:Coprime Integers|relatively prime]] to $x$.
From [[Absolute Value of Integer is not less than Divisors/Corollary|Absolute Value of Integer is not less than Divi... | Prime iff Coprime to all Smaller Positive Integers | https://proofwiki.org/wiki/Prime_iff_Coprime_to_all_Smaller_Positive_Integers | https://proofwiki.org/wiki/Prime_iff_Coprime_to_all_Smaller_Positive_Integers | [
"Prime Numbers",
"Coprime Integers"
] | [
"Definition:Prime Number",
"Definition:Coprime/Integers",
"Definition:Strictly Positive/Integer"
] | [
"Prime not Divisor implies Coprime",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Coprime/Integers",
"Absolute Value of Integer is not less than Divisors/Corollary",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Coprime/Integers",
"Definition:... |
proofwiki-8218 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | This result can be split into two parts: | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | This result can be split into two parts: | Division Theorem/Positive Divisor/Positive Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [] |
proofwiki-8219 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [
"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-8220 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | 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 ... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [
"Basis Representation Theorem",
"Definition:Unique",
"Definition:Number Base"
] |
proofwiki-8221 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [
"Definition:Integer",
"Definition:Closed Interval/Integer Interval",
"Definition:Strictly Increasing/Sequence",
"Definition:Positive/Integer",
"Strictly Increasing Sequence induces Partition"
] |
proofwiki-8222 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | 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 ... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [
"Proof by Contradiction",
"Definition:Unique"
] |
proofwiki-8223 | Division Theorem/Positive Divisor/Positive Dividend | <onlyinclude>
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the ''... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor/Positive_Dividend | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique"
] | [
"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-8224 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | This result can be split into two parts: | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | This result can be split into two parts: | Division Theorem/Positive Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [] |
proofwiki-8225 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"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-8226 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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 ... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Definition:Floor Function"
] |
proofwiki-8227 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"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-8228 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"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-8229 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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 ... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Basis Representation Theorem",
"Definition:Unique",
"Definition:Number Base"
] |
proofwiki-8230 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Definition:Integer",
"Definition:Closed Interval/Integer Interval",
"Definition:Strictly Increasing/Sequence",
"Definition:Positive/Integer",
"Strictly Increasing Sequence induces Partition"
] |
proofwiki-8231 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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 ... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Proof by Contradiction",
"Definition:Unique"
] |
proofwiki-8232 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"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-8233 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
{{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1}} | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Division Theorem/Positive Divisor/Existence"
] |
proofwiki-8234 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"Division Theorem/Positive Divisor/Existence",
"Definition:Unique"
] |
proofwiki-8235 | Division Theorem/Positive Divisor | <onlyinclude>
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equation:
:$a$ is the '''dividend'''
:$b$ is the '''divisor'''
:$q$ is the... | 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... | <onlyinclude>
For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
<onlyinclude>
In the above equa... | 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/Positive_Divisor | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Division/Dividend",
"Definition:Division/Divisor",
"Definition:Quotient (Integer Division)",
"Definition:Remainder",
"Definition:Remainder"
] | [
"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-8236 | Division Theorem/Positive Divisor/Positive Dividend/Existence | For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $a \ge 0$ and $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Positive_Dividend/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"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-8237 | Division Theorem/Positive Divisor/Positive Dividend/Existence | For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $a \ge 0$ and $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Positive_Dividend/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer"
] | [
"Basis Representation Theorem",
"Definition:Unique",
"Definition:Number Base"
] |
proofwiki-8238 | Division Theorem/Positive Divisor/Positive Dividend/Existence | For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $a \ge 0$ and $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Positive_Dividend/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer"
] | [
"Definition:Integer",
"Definition:Closed Interval/Integer Interval",
"Definition:Strictly Increasing/Sequence",
"Definition:Positive/Integer",
"Strictly Increasing Sequence induces Partition"
] |
proofwiki-8239 | Division Theorem/Positive Divisor/Positive Dividend/Uniqueness | For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $a \ge 0$ and $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Positive_Dividend/Uniqueness | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"Proof by Contradiction",
"Definition:Unique"
] |
proofwiki-8240 | Division Theorem/Positive Divisor/Positive Dividend/Uniqueness | For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $a \ge 0$ and $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Positive_Dividend/Uniqueness | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"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-8241 | Division Theorem/Positive Divisor/Existence/Proof 2 | For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Existence/Proof_2 | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer"
] | [
"Definition:Floor Function"
] |
proofwiki-8242 | Division Theorem/Positive Divisor/Existence | For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"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-8243 | Division Theorem/Positive Divisor/Existence | For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer"
] | [
"Definition:Floor Function"
] |
proofwiki-8244 | Division Theorem/Positive Divisor/Existence | For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Existence | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"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-8245 | Division Theorem/Positive Divisor/Uniqueness/Proof 2 | For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Uniqueness/Proof_2 | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"Division Theorem/Positive Divisor/Existence",
"Definition:Unique"
] |
proofwiki-8246 | Division Theorem/Positive Divisor/Uniqueness | For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Uniqueness | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_1 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"Division Theorem/Positive Divisor/Existence"
] |
proofwiki-8247 | Division Theorem/Positive Divisor/Uniqueness | For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Uniqueness | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_2 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"Division Theorem/Positive Divisor/Existence",
"Definition:Unique"
] |
proofwiki-8248 | Division Theorem/Positive Divisor/Uniqueness | For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Uniqueness | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"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-8249 | Division Theorem/Positive Divisor/Existence/Proof 3 | For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Existence/Proof_3 | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Existence/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"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-8250 | Division Theorem/Positive Divisor/Uniqueness/Proof 3 | For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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 $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ | 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/Positive_Divisor/Uniqueness/Proof_3 | https://proofwiki.org/wiki/Division_Theorem/Positive_Divisor/Uniqueness/Proof_3 | [
"Division Theorem"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Unique"
] | [
"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-8251 | Integer Coprime to all Factors is Coprime to Whole | Let $a, b \in \Z$.
Let $\ds b = \prod_{j \mathop = 1}^r b_j$
Let $a$ be coprime to each of $b_1, \ldots, b_r$.
Then $a$ is coprime to $b$. | From Integer Combination of Coprime Integers:
:$\forall j \in \set {1, 2, \ldots, r}: a x_j + b_j y_j = 1$
for some $x_j, y_j \in \Z$.
Thus:
:$\ds \prod_{j \mathop = 1}^r b_j y_j = \prod_{j \mathop = 1}^r \paren {1 - a x_j}$
But $\ds \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ is of the form $1 - a z$.
Thus:
{{begin-eq... | Let $a, b \in \Z$.
Let $\ds b = \prod_{j \mathop = 1}^r b_j$
Let $a$ be [[Definition:Coprime Integers|coprime]] to each of $b_1, \ldots, b_r$.
Then $a$ is [[Definition:Coprime Integers|coprime]] to $b$. | From [[Integer Combination of Coprime Integers]]:
:$\forall j \in \set {1, 2, \ldots, r}: a x_j + b_j y_j = 1$
for some $x_j, y_j \in \Z$.
Thus:
:$\ds \prod_{j \mathop = 1}^r b_j y_j = \prod_{j \mathop = 1}^r \paren {1 - a x_j}$
But $\ds \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ is of the form $1 - a z$.
Thus:
{{b... | Integer Coprime to all Factors is Coprime to Whole | https://proofwiki.org/wiki/Integer_Coprime_to_all_Factors_is_Coprime_to_Whole | https://proofwiki.org/wiki/Integer_Coprime_to_all_Factors_is_Coprime_to_Whole | [
"Integer Coprime to Factors is Coprime to Whole"
] | [
"Definition:Coprime/Integers",
"Definition:Coprime/Integers"
] | [
"Integer Combination of Coprime Integers",
"Integer Combination of Coprime Integers",
"Definition:Coprime/Integers"
] |
proofwiki-8252 | All Factors Divide Integer then Whole Divides Integer | Let $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ be a finite subset of the integers.
Let $S$ be pairwise coprime.
Let:
:$\forall j \in \set {1, 2, \ldots, r}: a_r \divides b$
where $\divides$ denotes divisibility.
Then:
:$\ds \prod_{j \mathop = 1}^r a_j \divides b$ | Proof by induction:
In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is pairwise coprime.
For all $r \in \N_{> 1}$, let $\map P r$ be the proposition:
:$\ds \prod_{j \mathop = 1}^r a_j \divides b$ | Let $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of the [[Definition:Integer|integers]].
Let $S$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]].
Let:
:$\forall j \in \set {1, 2, \ldots, r}: a_r \divides b$
where $\divides$ denotes [[D... | Proof by [[Principle of Mathematical Induction|induction]]:
In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is [[Definition:Pairwise Coprime Integers|pairwise coprime]].
For all $r \in \N_{> 1}$, let $\map P r$ be the [[Definition:Proposition|proposition]]:
:$\ds \p... | All Factors Divide Integer then Whole Divides Integer | https://proofwiki.org/wiki/All_Factors_Divide_Integer_then_Whole_Divides_Integer | https://proofwiki.org/wiki/All_Factors_Divide_Integer_then_Whole_Divides_Integer | [
"Coprime Integers"
] | [
"Definition:Finite Set",
"Definition:Subset",
"Definition:Integer",
"Definition:Pairwise Coprime/Integers",
"Definition:Divisor (Algebra)/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Pairwise Coprime/Integers",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-8253 | Piecewise Continuous Function with One-Sided Limits is Bounded | Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be a piecewise continuous function with one-sided limits.
Then $f$ is a bounded piecewise continuous function. | Let $f$ be a piecewise continuous function with one-sided limits.
By definition, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
For every $i \in \s... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ be a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]].
Then $f$ is a [[Definition:Bounded Piecewise ... | Let $f$ be a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]].
By definition, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that $f$ is [[D... | Piecewise Continuous Function with One-Sided Limits is Bounded | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Bounded | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Bounded | [
"Piecewise Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Piecewise Continuous Function/Bounded"
] | [
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Subdivision of Interval/Finite",
"Definition:Continuous Real Function/Open Interval",
"Definition:Real Function",
"Definition:Domain (Set Theory)/Mapping",
"Definition:One-Sided Limit of Real Function",
"Definition:Piecewise Contin... |
proofwiki-8254 | Upper Bounds for Prime Numbers/Result 1 | :$\forall n \in \N: \map p n \le 2^{2^{n - 1} }$ | Proof by strong induction:
Let us write $p_n = \map p n$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map p n \le 2^{2^{n - 1} }$ | :$\forall n \in \N: \map p n \le 2^{2^{n - 1} }$ | Proof by [[Second Principle of Mathematical Induction|strong induction]]:
Let us write $p_n = \map p n$.
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map p n \le 2^{2^{n - 1} }$ | Upper Bounds for Prime Numbers/Result 1 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_1 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_1 | [
"Upper Bounds for Prime Numbers"
] | [] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Second Principle of Mathematical Induction"
] |
proofwiki-8255 | Upper Bounds for Prime Numbers/Result 2 | :$\forall n \in \N: \map p n \le \paren {p \paren {n - 1} }^{n - 1} + 1$ | Let us write $p_n = \map p n$.
Let us take $N = p_1 p_2 \cdots p_n + 1$.
By the same argument as in Euclid's Theorem, we have that either $N$ is prime, or it is not.
If $N$ is prime, then either $N = p_{n + 1}$ or not, in which case $N > p_{n + 1}$.
In the second case, $N$ has a prime factor not in $\set {p_1, p_2, \ld... | :$\forall n \in \N: \map p n \le \paren {p \paren {n - 1} }^{n - 1} + 1$ | Let us write $p_n = \map p n$.
Let us take $N = p_1 p_2 \cdots p_n + 1$.
By the same argument as in [[Euclid's Theorem]], we have that either $N$ is [[Definition:Prime Number|prime]], or it is not.
If $N$ is [[Definition:Prime Number|prime]], then either $N = p_{n + 1}$ or not, in which case $N > p_{n + 1}$.
In the... | Upper Bounds for Prime Numbers/Result 2 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_2 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_2 | [
"Upper Bounds for Prime Numbers"
] | [] | [
"Euclid's Theorem",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Factor",
"Definition:Prime Factor",
"Definition:Prime Number"
] |
proofwiki-8256 | Upper Bounds for Prime Numbers/Result 3 | : $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$ | Let us write $p_n = p \left({n}\right)$.
From Bertrand's Conjecture, for each $n \ge 2$ there exists a prime $p$ such that $n < p < 2 n$.
For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
: $p_n < 2^n$
$P(1)$ is the statement:
:$p_1 = 2 = 2^1$
As this does not fulfil the criterion:
:$p \left({n}\rig... | : $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$ | Let us write $p_n = p \left({n}\right)$.
From [[Bertrand's Conjecture]], for each $n \ge 2$ there exists a [[Definition:Prime Number|prime]] $p$ such that $n < p < 2 n$.
For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
: $p_n < 2^n$
$P(1)$ is the statement:
:$p_1 = ... | Upper Bounds for Prime Numbers/Result 3 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_3 | https://proofwiki.org/wiki/Upper_Bounds_for_Prime_Numbers/Result_3 | [
"Upper Bounds for Prime Numbers"
] | [] | [
"Bertrand-Chebyshev Theorem",
"Definition:Prime Number",
"Definition:Proposition",
"Bertrand-Chebyshev Theorem"
] |
proofwiki-8257 | Product of Integers of form 4n + 1 | Let $m, n \in \Z$ such that both $m$ and $n$ are of the form $4 k + 1$ where $k \in \Z$.
Then $m n$ is also of the form $4 k + 1$. | Let $m = 4 k_1 + 1, n = 4 k_2 + 1$.
Then:
{{begin-eqn}}
{{eqn | l = m n
| r = \paren {4 k_1 + 1} \paren {4 k_2 + 1}
| c =
}}
{{eqn | r = 4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2 + 1
| c = Integer Multiplication Distributes over Addition
}}
{{eqn | r = \paren {4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2} + 1
| c... | Let $m, n \in \Z$ such that both $m$ and $n$ are of the form $4 k + 1$ where $k \in \Z$.
Then $m n$ is also of the form $4 k + 1$. | Let $m = 4 k_1 + 1, n = 4 k_2 + 1$.
Then:
{{begin-eqn}}
{{eqn | l = m n
| r = \paren {4 k_1 + 1} \paren {4 k_2 + 1}
| c =
}}
{{eqn | r = 4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2 + 1
| c = [[Integer Multiplication Distributes over Addition]]
}}
{{eqn | r = \paren {4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2} + 1
... | Product of Integers of form 4n + 1 | https://proofwiki.org/wiki/Product_of_Integers_of_form_4n_+_1 | https://proofwiki.org/wiki/Product_of_Integers_of_form_4n_+_1 | [
"Arithmetic",
"Number Theory"
] | [] | [
"Integer Multiplication Distributes over Addition",
"Integer Addition is Associative",
"Integer Multiplication Distributes over Addition",
"Integer Multiplication is Commutative"
] |
proofwiki-8258 | Infinite Number of Primes of form 4n - 1 | There are infinitely many prime numbers of the form $4 n - 1$. | {{AimForCont}} the contrary.
That is, suppose there is a finite number of prime numbers of the form $4 n - 1$.
Let there be $k$ of them: $p_1, p_2, \ldots, p_k$.
Let $S = \set {p_1, p_2, \ldots, p_k}$.
Let $N$ be constructed as:
:$\ds N = 4 \prod_{i \mathop = 1}^k p_i - 1$
If $N$ is a prime number, then it is of the fo... | There are [[Definition:Infinite Set|infinitely many]] [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$. | {{AimForCont}} the contrary.
That is, suppose there is a [[Definition:Finite Set|finite number]] of [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$.
Let there be $k$ of them: $p_1, p_2, \ldots, p_k$.
Let $S = \set {p_1, p_2, \ldots, p_k}$.
Let $N$ be constructed as:
:$\ds N = 4 \prod_{i \mathop = 1}... | Infinite Number of Primes of form 4n - 1 | https://proofwiki.org/wiki/Infinite_Number_of_Primes_of_form_4n_-_1 | https://proofwiki.org/wiki/Infinite_Number_of_Primes_of_form_4n_-_1 | [
"Arithmetic",
"Number Theory",
"Prime Numbers"
] | [
"Definition:Infinite Set",
"Definition:Prime Number"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Composite Number",
"Definition:Prime Factor",
"Product of Integers of form 4n + 1",
"Definition:Prime Factor",
"Euclidean Algorithm",
"Definition:Divisor (Algebra)/Integer",
"Def... |
proofwiki-8259 | Integer Coprime to Modulus iff Linear Congruence to 1 exists | Let $a, m \in \Z$.
The linear congruence:
:$a x \equiv 1 \pmod m$
has a solution $x$ {{iff}} $a$ and $m$ are coprime. | {{begin-eqn}}
{{eqn | l = a x
| o = \equiv
| r = 1
| rr= \pmod m
| c =
}}
{{eqn | ll=\leadstoandfrom
| q = \exists y \in \Z
| l = a x + m y
| r = 1
| c = {{Defof|Congruence Modulo Integer}}
}}
{{end-eqn}}
From Integer Combination of Coprime Integers:
:$a \perp m \iff \ex... | Let $a, m \in \Z$.
The [[Definition:Linear Congruence|linear congruence]]:
:$a x \equiv 1 \pmod m$
has a solution $x$ {{iff}} $a$ and $m$ are [[Definition:Coprime Integers|coprime]]. | {{begin-eqn}}
{{eqn | l = a x
| o = \equiv
| r = 1
| rr= \pmod m
| c =
}}
{{eqn | ll=\leadstoandfrom
| q = \exists y \in \Z
| l = a x + m y
| r = 1
| c = {{Defof|Congruence Modulo Integer}}
}}
{{end-eqn}}
From [[Integer Combination of Coprime Integers]]:
:$a \perp m \i... | Integer Coprime to Modulus iff Linear Congruence to 1 exists | https://proofwiki.org/wiki/Integer_Coprime_to_Modulus_iff_Linear_Congruence_to_1_exists | https://proofwiki.org/wiki/Integer_Coprime_to_Modulus_iff_Linear_Congruence_to_1_exists | [
"Modulo Arithmetic"
] | [
"Definition:Linear Congruence",
"Definition:Coprime/Integers"
] | [
"Integer Combination of Coprime Integers",
"Definition:Coprime/Integers"
] |
proofwiki-8260 | Integer Coprime to Modulus iff Linear Congruence to 1 exists/Corollary | Let $p$ be a prime number.
The linear congruence:
:$a x \equiv 1 \pmod p$
has a solution $x$ {{iff}} $a \not \equiv 0 \pmod p$. | By definition of congruence:
:$a \not \equiv 0 \pmod p \iff p \nmid a$
where $p \nmid a$ denotes that $p$ is not a divisor of $a$.
From Prime not Divisor implies Coprime:
:$p \nmid a \iff p \perp a$
where $p \perp a$ denotes that $p$ and $a$ are coprime.
The result follows from Integer Coprime to Modulus iff Linear Con... | Let $p$ be a [[Definition:Prime Number|prime number]].
The [[Definition:Linear Congruence|linear congruence]]:
:$a x \equiv 1 \pmod p$
has a solution $x$ {{iff}} $a \not \equiv 0 \pmod p$. | By definition of [[Definition:Congruence (Number Theory)|congruence]]:
:$a \not \equiv 0 \pmod p \iff p \nmid a$
where $p \nmid a$ denotes that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $a$.
From [[Prime not Divisor implies Coprime]]:
:$p \nmid a \iff p \perp a$
where $p \perp a$ denotes that $p$ and ... | Integer Coprime to Modulus iff Linear Congruence to 1 exists/Corollary | https://proofwiki.org/wiki/Integer_Coprime_to_Modulus_iff_Linear_Congruence_to_1_exists/Corollary | https://proofwiki.org/wiki/Integer_Coprime_to_Modulus_iff_Linear_Congruence_to_1_exists/Corollary | [
"Modulo Arithmetic"
] | [
"Definition:Prime Number",
"Definition:Linear Congruence"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Divisor (Algebra)/Integer",
"Prime not Divisor implies Coprime",
"Definition:Coprime/Integers",
"Integer Coprime to Modulus iff Linear Congruence to 1 exists"
] |
proofwiki-8261 | Solution of Linear Congruence/Existence | $a x \equiv b \pmod n$ has at least one solution {{iff}}:
:$\gcd \set {a, n} \divides b$
that is, {{iff}} $\gcd \set {a, n}$ is a divisor of $b$. | Consider the linear congruence $a x \equiv b \pmod n$.
Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$.
Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of congruence.
Thus $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$.
On the other hand, if $x = x_0, y = y_0$ is a so... | $a x \equiv b \pmod n$ has at least one [[Definition:Solution to Polynomial Congruence|solution]] {{iff}}:
:$\gcd \set {a, n} \divides b$
that is, {{iff}} $\gcd \set {a, n}$ is a [[Definition:Divisor of Integer|divisor]] of $b$. | Consider the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$.
Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$.
Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of [[Definition:Congruence (Number Theory)|congruence]].
Thus $x = x_0, y = y_0$ is a solution to the [[Definition:Li... | Solution of Linear Congruence/Existence | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Existence | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Existence | [
"Solution of Linear Congruence"
] | [
"Definition:Polynomial Congruence/Solution",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Linear Congruence",
"Definition:Congruence (Number Theory)",
"Definition:Diophantine Equation/Linear Diophantine Equation",
"Definition:Diophantine Equation/Linear Diophantine Equation",
"Definition:Linear Congruence",
"Definition:Diophantine Equation/Linear Diophantine Equation",
"Solution ... |
proofwiki-8262 | Solution of Linear Congruence/Number of Solutions | Let $\gcd \set {a, n} = d$.
Then $a x \equiv b \pmod n$ has $d$ solutions which are given by the unique solution modulo $\dfrac n d$ of the congruence:
:$\dfrac a d x \equiv \dfrac b d \paren {\bmod \dfrac n d}$ | From Solution of Linear Congruence: Existence:
:the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$
is the same problem as:
:the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$.
From Integers Divided by GCD are Coprime:
:$\gcd \set {\dfra... | Let $\gcd \set {a, n} = d$.
Then $a x \equiv b \pmod n$ has $d$ solutions which are given by the [[Definition:Unique|unique]] solution modulo $\dfrac n d$ of the [[Definition:Linear Congruence|congruence]]:
:$\dfrac a d x \equiv \dfrac b d \paren {\bmod \dfrac n d}$ | From [[Solution of Linear Congruence/Existence|Solution of Linear Congruence: Existence]]:
:the problem of finding all integers satisfying the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$
is the same problem as:
:the problem of finding all the $x$ values in the [[Definition:Linear Diophant... | Solution of Linear Congruence/Number of Solutions | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Number_of_Solutions | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Number_of_Solutions | [
"Solution of Linear Congruence"
] | [
"Definition:Unique",
"Definition:Linear Congruence"
] | [
"Solution of Linear Congruence/Existence",
"Definition:Linear Congruence",
"Definition:Diophantine Equation/Linear Diophantine Equation",
"Integers Divided by GCD are Coprime",
"Division Theorem"
] |
proofwiki-8263 | Solution of Linear Congruence/Unique iff Coprime to Modulus | $a x \equiv b \pmod n$ has a unique solution {{iff}} $\gcd \set {a, n} = 1$. | === Sufficient Condition ===
{{ProofWanted}} | $a x \equiv b \pmod n$ has a [[Definition:Unique|unique]] solution {{iff}} $\gcd \set {a, n} = 1$. | === Sufficient Condition ===
{{ProofWanted}} | Solution of Linear Congruence/Unique iff Coprime to Modulus | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Unique_iff_Coprime_to_Modulus | https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Unique_iff_Coprime_to_Modulus | [
"Solution of Linear Congruence"
] | [
"Definition:Unique"
] | [] |
proofwiki-8264 | Ring Direct Product of Modulo Integers is Isomorphic to Ring Modulo Product iff Coprime | Let $m, n \in \Z_{>1}$.
Let $\struct {\Z_m, +_m, \times_m}$ and $\struct {\Z_n, +_n, \times_n}$ be the rings of integers modulo $m$ and $n$ respectively.
Let $\struct {\Z_m \times \Z_n}$ be the direct product of $\Z_m$ and $\Z_n$.
Let $\struct {\Z_{m n}, +_{m n}, \times_{m n} }$ be the ring of integers modulo $mn$.
The... | {{ProofWanted|First a homomorphism is established between $\struct {\Z_m \times \Z_n}$ and $\struct {\Z_m \times \Z_n}$. Then it is demonstrated that $\phi: \struct {\Z_m \times \Z_n} \to \Z_{m n}$ is a bijection. Some of this has already been done, I'm fairly sure.}} | Let $m, n \in \Z_{>1}$.
Let $\struct {\Z_m, +_m, \times_m}$ and $\struct {\Z_n, +_n, \times_n}$ be the [[Definition:Ring of Integers Modulo m|rings of integers modulo $m$ and $n$]] respectively.
Let $\struct {\Z_m \times \Z_n}$ be the [[Definition:Ring Direct Product|direct product]] of $\Z_m$ and $\Z_n$.
Let $\stru... | {{ProofWanted|First a homomorphism is established between $\struct {\Z_m \times \Z_n}$ and $\struct {\Z_m \times \Z_n}$. Then it is demonstrated that $\phi: \struct {\Z_m \times \Z_n} \to \Z_{m n}$ is a bijection. Some of this has already been done, I'm fairly sure.}} | Ring Direct Product of Modulo Integers is Isomorphic to Ring Modulo Product iff Coprime | https://proofwiki.org/wiki/Ring_Direct_Product_of_Modulo_Integers_is_Isomorphic_to_Ring_Modulo_Product_iff_Coprime | https://proofwiki.org/wiki/Ring_Direct_Product_of_Modulo_Integers_is_Isomorphic_to_Ring_Modulo_Product_iff_Coprime | [
"Commutative Algebra",
"Modulo Arithmetic"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Ring Direct Product",
"Definition:Ring of Integers Modulo m",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Coprime/Integers"
] | [] |
proofwiki-8265 | Trivial Norm on Division Ring is Norm | Let $\struct {R, +, \circ}$ be a division ring, and denote its ring zero by $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$ \norm x = \begin{cases} 0 & \text { if } x = 0_R\\ 1 & \text { otherwise} \end{cases}$
defines a norm on $R$. | Proving each of the norm axioms one by one: | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]], and denote its [[Definition:Ring Zero|ring zero]] by $0_R$.
Then the [[Definition:Trivial Norm on Division Ring|trivial norm]] $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$ \norm x = \begin{cases} 0 & \text { if } x = 0_R\\... | Proving each of the [[Definition:Norm on Division Ring|norm axioms]] one by one: | Trivial Norm on Division Ring is Norm | https://proofwiki.org/wiki/Trivial_Norm_on_Division_Ring_is_Norm | https://proofwiki.org/wiki/Trivial_Norm_on_Division_Ring_is_Norm | [
"Division Rings",
"Trivial Norms"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Trivial Norm/Division Ring",
"Definition:Norm/Division Ring"
] | [
"Definition:Norm/Division Ring"
] |
proofwiki-8266 | Trivial Norm on Division Ring is Non-Archimedean | Let $\struct {R, +, \circ}$ be a division ring whose ring zero is $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$\norm x = \begin{cases}
0 & : x = 0_R \\
1 & : \text{ otherwise}
\end{cases}$
is non-archimedean:
:$\norm {x + y} \le \max \set {\norm x, \norm y}$ | Let $x, y = 0_R$.
Then:
:$\norm x, \norm y = 0$
Therefore:
:$\max \set {\norm x, \norm y} = 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \norm {x + y}
| r = \norm {0_R + 0_R}
| c =
}}
{{eqn | r = \norm {0_R}
| c =
}}
{{eqn | r = 0
| c =
}}
{{eqn | r = \max \set {\norm x, \norm y}
| c =
}}
{{e... | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|ring zero]] is $0_R$.
Then the [[Definition:Trivial Norm on Division Ring|trivial norm]] $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$\norm x = \begin{cases}
0 & : x = 0_R \\
1 & : \text{ othe... | Let $x, y = 0_R$.
Then:
:$\norm x, \norm y = 0$
Therefore:
:$\max \set {\norm x, \norm y} = 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \norm {x + y}
| r = \norm {0_R + 0_R}
| c =
}}
{{eqn | r = \norm {0_R}
| c =
}}
{{eqn | r = 0
| c =
}}
{{eqn | r = \max \set {\norm x, \norm y}
| c =
}}... | Trivial Norm on Division Ring is Non-Archimedean | https://proofwiki.org/wiki/Trivial_Norm_on_Division_Ring_is_Non-Archimedean | https://proofwiki.org/wiki/Trivial_Norm_on_Division_Ring_is_Non-Archimedean | [
"Trivial Norms"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Trivial Norm/Division Ring",
"Definition:Non-Archimedean/Norm (Division Ring)"
] | [
"Category:Trivial Norms"
] |
proofwiki-8267 | Ring of Integers Modulo m is Ring | For all $m \in \N: m \ge 2$, the ring of integers modulo $m$:
:$\struct {\Z_m, +_m, \times_m}$
is a commutative ring with unity $\eqclass 1 m$.
The zero of $\struct {\Z_m, +_m, \times_m}$ is $\eqclass 0 m$. | First we check the ring axioms:
:{{Ring-axiom|A}}
::The Integers Modulo $m$ under Addition form Abelian Group.
::From Modulo Addition has Identity, $\eqclass 0 m$ is the identity of the additive group $\struct {\Z_m, +_m}$.
From Integers Modulo m under Multiplication form Commutative Monoid:
:{{Ring-axiom|M0}}:
::$\str... | For all $m \in \N: m \ge 2$, the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]]:
:$\struct {\Z_m, +_m, \times_m}$
is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\eqclass 1 m$]].
The [[Definition:Ring Zero|zero]] of $\struct {\Z_m, +_m, \times_m}$ is $\eqclass 0 m$. | First we check the [[Axiom:Ring Axioms|ring axioms]]:
:{{Ring-axiom|A}}
::The [[Integers Modulo m under Addition form Abelian Group |Integers Modulo $m$ under Addition form Abelian Group]].
::From [[Modulo Addition has Identity]], $\eqclass 0 m$ is the [[Definition:Identity Element|identity]] of the [[Definition:Addit... | Ring of Integers Modulo m is Ring | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_m_is_Ring | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_m_is_Ring | [
"Ring of Integers Modulo m",
"Commutative Algebra"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero"
] | [
"Axiom:Ring Axioms",
"Integers Modulo m under Addition form Abelian Group ",
"Modulo Addition has Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Additive Group of Ring",
"Integers Modulo m under Multiplication form Commutative Monoid",
"Definition:Closure (Abstract A... |
proofwiki-8268 | Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece | Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits:
{{:Definition:Piecewise Continuous Function with One-Sided Limits}}
Then:
:for all $i \in \set {1, 2, \ldots, n}$, $f$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$. | We have that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
Since $f$ is piecewise continuous with one-sided limits, the one-sided limits:
:$\ds \lim_{x \mathop \to {x_{i−1} }^+} \map f x$
and:
:$\ds \lim_{x \mathop \to {x_i}^-} \map f x$
exist for every $i \in \set {1, 2, ... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]]:
{{:Definition:Piecewise Continuous Function with One-Sided... | We have that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
Since $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]], the [[Definition:One-Sided Limit... | Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Uniformly_Continuous_on_Each_Piece | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Uniformly_Continuous_on_Each_Piece | [
"Piecewise Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Uniform Continuity/Real Function"
] | [
"Definition:Continuous Real Function/Open Interval",
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:One-Sided Limit of Real Function",
"Extendability Theorem for Function Continuous on Open Interval",
"Definition:Continuous Real Function/Closed Interval",
"Continuous Function on ... |
proofwiki-8269 | Principle of Definition by Structural Induction | Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar with unique parsability.
A definition $\map D \phi$ (in the metalanguage of $\LL$) for all well-formed formulas $\phi$ of $\LL$ is uniquely specified by:
:$(1): \quad$ A definition $\map D a$ for each letter $a$ of $\LL$
:$(2): \quad... | We apply the Principle of Structural Induction on the following statement $\map P \phi$:
:The definition is specified for $\phi$
The given hypotheses verify the conditions for the Principle of Structural Induction.
It follows that $\map D \phi$ is specified for each WFF $\phi$.
Moreover, by virtue of the unique parsabi... | Let $\LL$ be a [[Definition:Formal Language|formal language]].
Let the [[Definition:Formal Grammar|formal grammar]] of $\LL$ be a [[Definition:Bottom-Up Grammar|bottom-up grammar]] with [[Definition:Unique Parsability|unique parsability]].
A [[Definition:Definition|definition]] $\map D \phi$ (in the [[Definition:Met... | We apply the [[Principle of Structural Induction]] on the following [[Definition:Statement|statement]] $\map P \phi$:
:The definition is specified for $\phi$
The given hypotheses verify the conditions for the [[Principle of Structural Induction]].
It follows that $\map D \phi$ is specified for each [[Definition:Wel... | Principle of Definition by Structural Induction | https://proofwiki.org/wiki/Principle_of_Definition_by_Structural_Induction | https://proofwiki.org/wiki/Principle_of_Definition_by_Structural_Induction | [
"Formal Languages"
] | [
"Definition:Formal Language",
"Definition:Formal Grammar",
"Definition:Formal Grammar/Bottom-Up",
"Definition:Unique Parsability",
"Definition:Definition",
"Definition:Metalanguage",
"Definition:Well-Formed Formula",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Rule of Formation",
"De... | [
"Principle of Structural Induction",
"Definition:Statement",
"Principle of Structural Induction",
"Definition:Well-Formed Formula",
"Definition:Unique Parsability",
"Definition:Well-Formed Formula",
"Definition:Unique",
"Category:Formal Languages"
] |
proofwiki-8270 | Principle of Structural Induction | Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar.
Let $\map P \phi$ be a statement (in the metalanguage of $\LL$) about well-formed formulas $\phi$ of $\LL$.
Then $P$ is true for all WFFs of $\LL$ {{iff}} both:
:$\map P a$ is true for all letters $a$ of $\LL$,
and, for each rule of... | Let $\phi$ be a WFF of $\LL$.
Then $\phi$ is the result of applying finitely many rules of formation of $\LL$.
If we can show that the result of each rule of formation satisfies $P$, we will have finished.
Suppose now that for a rule of formation $\mathbf R$, all preceding rules have produced WFFs satisfying $P$.
By th... | Let $\LL$ be a [[Definition:Formal Language|formal language]].
Let the [[Definition:Formal Grammar|formal grammar]] of $\LL$ be a [[Definition:Bottom-Up Grammar|bottom-up grammar]].
Let $\map P \phi$ be a [[Definition:Statement|statement]] (in the [[Definition:Metalanguage|metalanguage]] of $\LL$) about [[Definition:... | Let $\phi$ be a [[Definition:Well-Formed Formula|WFF]] of $\LL$.
Then $\phi$ is the result of applying finitely many [[Definition:Rule of Formation|rules of formation]] of $\LL$.
If we can show that the result of each [[Definition:Rule of Formation|rule of formation]] satisfies $P$, we will have finished.
Suppose n... | Principle of Structural Induction | https://proofwiki.org/wiki/Principle_of_Structural_Induction | https://proofwiki.org/wiki/Principle_of_Structural_Induction | [
"Formal Languages"
] | [
"Definition:Formal Language",
"Definition:Formal Grammar",
"Definition:Formal Grammar/Bottom-Up",
"Definition:Statement",
"Definition:Metalanguage",
"Definition:Well-Formed Formula",
"Definition:Well-Formed Formula",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Rule of Formation",
"De... | [
"Definition:Well-Formed Formula",
"Definition:Rule of Formation",
"Definition:Rule of Formation",
"Definition:Rule of Formation",
"Definition:Well-Formed Formula",
"Definition:Formal Grammar/Bottom-Up",
"Definition:Formal Grammar",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Well-Forme... |
proofwiki-8271 | Integers under Multiplication form Monoid | The set of integers under multiplication $\struct {\Z, \times}$ is a monoid. | From Integers under Multiplication form Semigroup, $\struct {\Z, \times}$ is a semigroup.
From Integer Multiplication Identity is One, $\struct {\Z, \times}$ has an identity element, which is $1$.
All the criteria for being a monoid are thus seen to be fulfilled.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Monoid|monoid]]. | From [[Integers under Multiplication form Semigroup]], $\struct {\Z, \times}$ is a [[Definition:Semigroup|semigroup]].
From [[Integer Multiplication Identity is One]], $\struct {\Z, \times}$ has an [[Definition:Identity Element|identity element]], which is $1$.
All the criteria for being a [[Definition:Monoid|monoid]... | Integers under Multiplication form Monoid | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Monoid | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Monoid | [
"Integer Multiplication",
"Examples of Monoids"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Monoid"
] | [
"Integers under Multiplication form Semigroup",
"Definition:Semigroup",
"Integer Multiplication Identity is One",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid"
] |
proofwiki-8272 | Integers under Multiplication do not form Group | The set of integers under multiplication $\struct {\Z, \times}$ does not form a group. | In order to be classified as a group, the algebraic structure $\struct {\Z, \times}$ needs to fulfil the group axioms.
From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ forms a monoid.
Therefore {{Group-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|2}} are satisfied.
However, from Invertible Integ... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ does not form a [[Definition:Group|group]]. | In order to be classified as a [[Definition:Group|group]], the [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\Z, \times}$ needs to fulfil the [[Axiom:Group Axioms|group axioms]].
From [[Integers under Multiplication form Monoid]], $\struct {\Z, \times}$ forms a [[Definition:Monoid... | Integers under Multiplication do not form Group | https://proofwiki.org/wiki/Integers_under_Multiplication_do_not_form_Group | https://proofwiki.org/wiki/Integers_under_Multiplication_do_not_form_Group | [
"Integer Multiplication",
"Examples of Groups"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Algebraic Structure/One Operation",
"Axiom:Group Axioms",
"Integers under Multiplication form Monoid",
"Definition:Monoid",
"Invertible Integers under Multiplication",
"Definition:Integer",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Multiplication/In... |
proofwiki-8273 | Integers under Multiplication form Semigroup | The set of integers under multiplication $\struct {\Z, \times}$ is a semigroup. | === {{Semigroup-axiom|0|nolink}} ===
Integer Multiplication is Closed, fulfilling {{Semigroup-axiom|0}}.
{{qed|lemma}} | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Semigroup|semigroup]]. | === {{Semigroup-axiom|0|nolink}} ===
[[Integer Multiplication is Closed]], fulfilling {{Semigroup-axiom|0}}.
{{qed|lemma}} | Integers under Multiplication form Semigroup | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Semigroup | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Semigroup | [
"Integer Multiplication",
"Examples of Semigroups"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Semigroup"
] | [
"Integer Multiplication is Closed"
] |
proofwiki-8274 | Bottom-Up Form of Top-Down Grammar defines same Formal Language | Let $\LL$ be a formal language.
Let $\TT$ be a top-down grammar for $\LL$.
Let $\BB$ be the bottom-up form of $\TT$.
Then $\BB$ is also a formal grammar for $\LL$. | Let $\phi$ be a $\BB$-WFF.
If $\phi$ is a letter, then it is {{afortiori}} a $\TT$-WFF.
For, it may be formed by replacing the starting metasymbol of $\TT$ by $\phi$.
Suppose that $\phi$ is formed from WFFs $\phi_1, \ldots, \phi_n$ by the rule of formation $\mathbf R_\BB$ of $\BB$.
Suppose also that each of $\phi_1, \l... | Let $\LL$ be a [[Definition:Formal Language|formal language]].
Let $\TT$ be a [[Definition:Top-Down Grammar|top-down grammar]] for $\LL$.
Let $\BB$ be the [[Definition:Bottom-Up Form of Top-Down Grammar|bottom-up form]] of $\TT$.
Then $\BB$ is also a [[Definition:Formal Grammar|formal grammar]] for $\LL$. | Let $\phi$ be a $\BB$-[[Definition:Well-Formed Formula|WFF]].
If $\phi$ is a [[Definition:Letter of Formal Language|letter]], then it is {{afortiori}} a $\TT$-[[Definition:Well-Formed Formula|WFF]].
For, it may be formed by replacing the starting [[Definition:Metasymbol|metasymbol]] of $\TT$ by $\phi$.
Suppose tha... | Bottom-Up Form of Top-Down Grammar defines same Formal Language | https://proofwiki.org/wiki/Bottom-Up_Form_of_Top-Down_Grammar_defines_same_Formal_Language | https://proofwiki.org/wiki/Bottom-Up_Form_of_Top-Down_Grammar_defines_same_Formal_Language | [
"Formal Languages"
] | [
"Definition:Formal Language",
"Definition:Formal Grammar/Top-Down",
"Definition:Bottom-Up Form of Top-Down Grammar",
"Definition:Formal Grammar"
] | [
"Definition:Well-Formed Formula",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Well-Formed Formula",
"Definition:Metalanguage/Metasymbol",
"Definition:Well-Formed Formula",
"Definition:Rule of Formation",
"Definition:Well-Formed Formula",
"Definition:Rule of Formation",
"Definition:Coll... |
proofwiki-8275 | Equivalence of Definitions of Characteristic of Ring | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
{{TFAE|def = Characteristic of Ring}} | === Definition 1 is equivalent to Definition 3 ===
By definition of order, the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the identity of the additive group $\struct {R, +}$ of $\struct {R, +, \circ}$.
Hence if the order of $1_R$ is finite, the definitions coincide.
If the order of $... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
{{TFAE|def = Characteristic of Ring}} | === Definition 1 is equivalent to Definition 3 ===
By definition of [[Definition:Order of Group Element|order]], the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the [[Definition:Identity of Group|identity]] of the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$ ... | Equivalence of Definitions of Characteristic of Ring | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Characteristic_of_Ring | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Characteristic_of_Ring | [
"Characteristics of Rings"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Order of Group Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Additive Group of Ring"
] |
proofwiki-8276 | Canonical Form of Rational Number is Unique | The canonical form of a rational number is unique. | Let $r \in \Q$ be a rational number.
Let $\dfrac p q$ and $\dfrac {p'} {q'}$ be two canonical forms of $r$.
{{WLOG}}, assume $q \le q'$.
By Equality of Rational Numbers:
:$p q' = p' q$
Therefore, $q'$ divides $p' q$.
As $\dfrac {p'} {q'}$ is a canonical form of $r$, by definition $q'$ is coprime to $p'$.
By Euclid's Le... | The [[Definition:Canonical Form of Rational Number|canonical form]] of a [[Definition:Rational Number|rational number]] is [[Definition:Unique|unique]]. | Let $r \in \Q$ be a [[Definition:Rational Number|rational number]].
Let $\dfrac p q$ and $\dfrac {p'} {q'}$ be two [[Definition:Canonical Form of Rational Number|canonical forms]] of $r$.
{{WLOG}}, assume $q \le q'$.
By [[Equality of Rational Numbers]]:
:$p q' = p' q$
Therefore, $q'$ [[Definition:Divisor of Intege... | Canonical Form of Rational Number is Unique | https://proofwiki.org/wiki/Canonical_Form_of_Rational_Number_is_Unique | https://proofwiki.org/wiki/Canonical_Form_of_Rational_Number_is_Unique | [
"Canonical Form of Rational Number"
] | [
"Definition:Rational Number/Canonical Form",
"Definition:Rational Number",
"Definition:Unique"
] | [
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form",
"Equality of Rational Numbers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Rational Number/Canonical Form",
"Definition:Coprime/Integers",
"Euclid's Lemma",
"Definition:Divisor (Algebra)/Integer",
"Definition:Positiv... |
proofwiki-8277 | Existence of Canonical Form of Rational Number | Let $r \in \Q$.
Then:
:$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$
That is, every rational number can be expressed in its canonical form. | We have that the set of rational numbers is the field of quotients of the set of integers.
From Divided by Positive Element of Field of Quotients:
:$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$
Now if $s \perp t$, our task is complete.
Otherwise, let:
:$\gcd \set {s, t} = d$
where $\gcd \set {s, t}$ denotes the gre... | Let $r \in \Q$.
Then:
:$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$
That is, every [[Definition:Rational Number|rational number]] can be expressed in its [[Definition:Canonical Form of Rational Number|canonical form]]. | We have that the [[Definition:Rational Number|set of rational numbers]] is the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Set|set]] of [[Definition:Integer|integers]].
From [[Divided by Positive Element of Field of Quotients]]:
:$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$
Now if $... | Existence of Canonical Form of Rational Number | https://proofwiki.org/wiki/Existence_of_Canonical_Form_of_Rational_Number | https://proofwiki.org/wiki/Existence_of_Canonical_Form_of_Rational_Number | [
"Canonical Form of Rational Number"
] | [
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form"
] | [
"Definition:Rational Number",
"Definition:Field of Quotients",
"Definition:Set",
"Definition:Integer",
"Divided by Positive Element of Field of Quotients",
"Definition:Greatest Common Divisor/Integers",
"Integers Divided by GCD are Coprime"
] |
proofwiki-8278 | Existence of Dyadic Rational between two Rationals | Let $a$ and $b$ be rational numbers such that $a < b$.
Then there exist integers $m$ and $r$ such that:
:$a < \dfrac m {2^r} < b$
That is, there exists a dyadic rational between any pair of rational numbers. | As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the Axiom of Archimedes:
:$\exists r \in \N: r > \dfrac 1 {b - a}$
Notice that $2^r > r$.
Thus we also have:
:$2^r > \dfrac 1 {b - a}$
Let $M := \set {x \in \Z: x > a 2^r}$.
By Set of Integers Bounded Below has Smallest Eleme... | Let $a$ and $b$ be [[Definition:Rational Number|rational numbers]] such that $a < b$.
Then there exist [[Definition:Integer|integers]] $m$ and $r$ such that:
:$a < \dfrac m {2^r} < b$
That is, there exists a [[Definition:Dyadic Rational|dyadic rational]] between any pair of [[Definition:Rational Number|rational num... | As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the [[Axiom of Archimedes]]:
:$\exists r \in \N: r > \dfrac 1 {b - a}$
Notice that $2^r > r$.
Thus we also have:
:$2^r > \dfrac 1 {b - a}$
Let $M := \set {x \in \Z: x > a 2^r}$.
By [[Set of Integers Bounded Below has Sm... | Existence of Dyadic Rational between two Rationals | https://proofwiki.org/wiki/Existence_of_Dyadic_Rational_between_two_Rationals | https://proofwiki.org/wiki/Existence_of_Dyadic_Rational_between_two_Rationals | [
"Rational Numbers"
] | [
"Definition:Rational Number",
"Definition:Integer",
"Definition:Dyadic Rational",
"Definition:Rational Number"
] | [
"Axiom of Archimedes",
"Set of Integers Bounded Below has Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Ordering of Reciprocals"
] |
proofwiki-8279 | Field with 4 Elements has only Order 2 Elements | Let $\struct {\GF, +, \times}$ be a field which has exactly $4$ elements.
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the zero of $\GF$. | By definition of field, both the algebraic structures $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are (abelian) groups, where $\GF^* := \GF \setminus \set 0$.
By definition:
:$\struct {\GF, +}$ is of order $4$
:$\struct {\GF^*, \times}$ is of order $3$.
From Classification of Groups of Order up to 15, there are on... | Let $\struct {\GF, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] which has exactly $4$ [[Definition:Element|elements]].
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the [[Definition:Field Zero|zero]] of $\GF$. | By definition of [[Definition:Field (Abstract Algebra)|field]], both the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are [[Definition:Abelian Group|(abelian) groups]], where $\GF^* := \GF \setminus \set 0$.
By definition:
:$\struct {\GF, +... | Field with 4 Elements has only Order 2 Elements/Proof 1 | https://proofwiki.org/wiki/Field_with_4_Elements_has_only_Order_2_Elements | https://proofwiki.org/wiki/Field_with_4_Elements_has_only_Order_2_Elements/Proof_1 | [
"Galois Fields",
"Field with 4 Elements has only Order 2 Elements"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Element",
"Definition:Field Zero"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Structure/One Operation",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Order of Structure",
"Classification of Groups of Order up to 15",
"Definition:Order of Group Element",
"Definition:Field (Abstract Algebra)",... |
proofwiki-8280 | Field with 4 Elements has only Order 2 Elements | Let $\struct {\GF, +, \times}$ be a field which has exactly $4$ elements.
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the zero of $\GF$. | Let $\struct {\GF, +, \times}$ be a field of order $4$ whose zero is $0_\GF$.
By definition, $\GF$ is a Galois field.
The additive group $\struct {\GF, +}$ of $\GF$ can be one of two:
:$(1): \quad$ the cyclic group of order $4$, generated by the identity of $\struct {\GF, +}$ which is $0_\GF$
or:
:$(2): \quad$ the Klei... | Let $\struct {\GF, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] which has exactly $4$ [[Definition:Element|elements]].
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the [[Definition:Field Zero|zero]] of $\GF$. | Let $\struct {\GF, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Order of Structure|order $4$]] whose [[Definition:Field Zero|zero]] is $0_\GF$.
By definition, $\GF$ is a [[Definition:Galois Field|Galois field]].
The [[Definition:Additive Group of Field|additive group]] $\struct {\GF,... | Field with 4 Elements has only Order 2 Elements/Proof 2 | https://proofwiki.org/wiki/Field_with_4_Elements_has_only_Order_2_Elements | https://proofwiki.org/wiki/Field_with_4_Elements_has_only_Order_2_Elements/Proof_2 | [
"Galois Fields",
"Field with 4 Elements has only Order 2 Elements"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Element",
"Definition:Field Zero"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Order of Structure",
"Definition:Field Zero",
"Definition:Galois Field",
"Definition:Additive Group of Field",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Identity (Abstract Algebra)/T... |
proofwiki-8281 | Count of Truth Functions | There are $2^{\paren {2^k} }$ distinct truth functions on $k$ variables. | Let $f: \mathbb B^k \to \mathbb B$ be a truth function.
The domain of $f$ has $2^k$ elements, from Cardinality of Cartesian Product of Finite Sets.
The result follows from Cardinality of Set of All Mappings.
{{qed}} | There are $2^{\paren {2^k} }$ [[Definition:Distinct Elements|distinct]] [[Definition:Truth Function|truth functions]] on $k$ variables. | Let $f: \mathbb B^k \to \mathbb B$ be a [[Definition:Truth Function|truth function]].
The [[Definition:Domain of Mapping|domain]] of $f$ has $2^k$ elements, from [[Cardinality of Cartesian Product of Finite Sets]].
The result follows from [[Cardinality of Set of All Mappings]].
{{qed}} | Count of Truth Functions | https://proofwiki.org/wiki/Count_of_Truth_Functions | https://proofwiki.org/wiki/Count_of_Truth_Functions | [
"Combinatorics",
"Truth Functions"
] | [
"Definition:Distinct/Plural",
"Definition:Truth Function"
] | [
"Definition:Truth Function",
"Definition:Domain (Set Theory)/Mapping",
"Cardinality of Cartesian Product of Finite Sets",
"Cardinality of Set of All Mappings"
] |
proofwiki-8282 | Unary Truth Functions | There are $4$ distinct unary truth functions:
:$(1): \quad$ The constant function $\map f p = \F$
:$(2): \quad$ The constant function $\map f p = \T$
:$(3): \quad$ The identity function $\map f p = p$
:$(4): \quad$ The logical not function $\map f p = \neg p$ | From Count of Truth Functions there are $2^{\paren {2^1} } = 4$ distinct truth functions on $1$ variable.
These can be depicted in a truth table as follows:
:<nowiki>$\begin{array}{|c|cccc|} \hline
p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
\hline
\T & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\hline
\end{array}... | There are $4$ distinct [[Definition:Unary|unary]] [[Definition:Truth Function|truth functions]]:
:$(1): \quad$ The [[Definition:Constant Mapping|constant function]] $\map f p = \F$
:$(2): \quad$ The [[Definition:Constant Mapping|constant function]] $\map f p = \T$
:$(3): \quad$ The [[Definition:Identity Mapping|identi... | From [[Count of Truth Functions]] there are $2^{\paren {2^1} } = 4$ distinct [[Definition:Truth Function|truth functions]] on $1$ variable.
These can be depicted in a [[Definition:Truth Table|truth table]] as follows:
:<nowiki>$\begin{array}{|c|cccc|} \hline
p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
\hline
\T & \... | Unary Truth Functions | https://proofwiki.org/wiki/Unary_Truth_Functions | https://proofwiki.org/wiki/Unary_Truth_Functions | [
"Truth Functions"
] | [
"Definition:Operation/Unary Operation",
"Definition:Truth Function",
"Definition:Constant Mapping",
"Definition:Constant Mapping",
"Definition:Identity Mapping",
"Definition:Logical Not"
] | [
"Count of Truth Functions",
"Definition:Truth Function",
"Definition:Truth Table",
"Definition:Constant Mapping",
"Definition:Identity Mapping",
"Definition:Logical Not",
"Definition:Constant Mapping",
"Definition:Operation/Unary Operation",
"Definition:Truth Function"
] |
proofwiki-8283 | Binary Truth Functions | There are $16$ distinct binary truth functions:
* Two constant operations:
** $\map {f_\F} {p, q} = \F$
** $\map {f_\T} {p, q} = \T$
* Two projections:
** $\map {\pr_1} {p, q} = p$
** $\map {\pr_2} {p, q} = q$
* Two negated projections:
** $\map {\overline {\pr_1} } {p, q} = \neg p$
** $\map {\overline {\pr_2} } {p, q}... | From Count of Truth Functions there are $2^{\paren {2^2} } = 16$ distinct truth functions on $2$ variables.
These can be depicted in a truth table as follows:
$\begin{array}{|r|cccc|} \hline
p & \T & \T & \F & \F \\
q & \T & \F & \T & \F \\
\hline
\map {f_\T... | There are $16$ [[Definition:Distinct Elements|distinct]] [[Definition:Binary Operation|binary]] [[Definition:Truth Function|truth functions]]:
* Two [[Definition:Constant Operation|constant operations]]:
** $\map {f_\F} {p, q} = \F$
** $\map {f_\T} {p, q} = \T$
* Two [[Definition:Projection (Mapping Theory)|projectio... | From [[Count of Truth Functions]] there are $2^{\paren {2^2} } = 16$ distinct [[Definition:Truth Function|truth functions]] on $2$ variables.
These can be depicted in a [[Definition:Truth Table|truth table]] as follows:
$\begin{array}{|r|cccc|} \hline
p & \T & \T & \F & \F \\
q ... | Binary Truth Functions | https://proofwiki.org/wiki/Binary_Truth_Functions | https://proofwiki.org/wiki/Binary_Truth_Functions | [
"Truth Functions"
] | [
"Definition:Distinct/Plural",
"Definition:Operation/Binary Operation",
"Definition:Truth Function",
"Definition:Constant Operation",
"Definition:Projection (Mapping Theory)",
"Definition:Logical Not",
"Definition:Projection (Mapping Theory)",
"Definition:Conjunction",
"Definition:Disjunction",
"De... | [
"Count of Truth Functions",
"Definition:Truth Function",
"Definition:Truth Table"
] |
proofwiki-8284 | Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition | Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational addition. | Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a + b
| r = \frac {p n_1} {d_1} + \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 d_2 + p n_2 d_1} {d_1 d... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $A_p$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] which, when expressed in [[Definition:Canonical Form of Rational Number|canonical form]] has a [[Definition:Numerator|numerator]] which is [[Definition:Divisor of Intege... | Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a + b
| r = \frac {p n_1} {d_1} + \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 d_2 + p n_2 d_1} {d... | Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition | https://proofwiki.org/wiki/Set_of_Rational_Numbers_whose_Numerator_Divisible_by_p_is_Closed_under_Addition | https://proofwiki.org/wiki/Set_of_Rational_Numbers_whose_Numerator_Divisible_by_p_is_Closed_under_Addition | [
"Prime Numbers",
"Rational Addition",
"Algebraic Closure"
] | [
"Definition:Prime Number",
"Definition:Set",
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form",
"Definition:Fraction/Numerator",
"Definition:Divisor (Algebra)/Integer",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Rational Numbers"
] | [
"Euclid's Lemma for Prime Divisors",
"Prime not Divisor implies Coprime",
"Definition:Coprime/Integers",
"Definition:Rational Number/Canonical Form",
"Definition:Divisor (Algebra)/Integer",
"Definition:Fraction/Numerator"
] |
proofwiki-8285 | Inverse Hyperbolic Sine Logarithmic Formulation | {{finish|The true story is more complicated than this. The inverse hyperbolic sine is a multifunction. So is $\ln$ in the complex plane. And $\sqrt {z^2 + 1}$ has two values, pos and neg, so you need to justify which one is taken. The best context to put this page is probably into Definition:Inverse Hyperbolic Sine/Com... | {{begin-eqn}}
{{eqn | l = z
| r = \sinh \arsinh z
}}
{{eqn | ll= \leadstoandfrom
| l = z
| r = \frac {e^{\arsinh z} - e^{-\arsinh z} } 2
| c = {{Defof|Inverse Hyperbolic Sine}}
}}
{{eqn | ll= \leadstoandfrom
| l = 2 z e^{\arsinh z}
| r = e^{2 \arsinh z} - 1
| c = Multiplication... | {{finish|The true story is more complicated than this. The [[Definition:Inverse Hyperbolic Sine|inverse hyperbolic sine]] is a multifunction. So is $\ln$ in the complex plane. And $\sqrt {z^2 + 1}$ has two values, pos and neg, so you need to justify which one is taken. The best context to put this page is probably into... | {{begin-eqn}}
{{eqn | l = z
| r = \sinh \arsinh z
}}
{{eqn | ll= \leadstoandfrom
| l = z
| r = \frac {e^{\arsinh z} - e^{-\arsinh z} } 2
| c = {{Defof|Inverse Hyperbolic Sine}}
}}
{{eqn | ll= \leadstoandfrom
| l = 2 z e^{\arsinh z}
| r = e^{2 \arsinh z} - 1
| c = Multiplication... | Inverse Hyperbolic Sine Logarithmic Formulation | https://proofwiki.org/wiki/Inverse_Hyperbolic_Sine_Logarithmic_Formulation | https://proofwiki.org/wiki/Inverse_Hyperbolic_Sine_Logarithmic_Formulation | [
"Inverse Hyperbolic Sine"
] | [
"Definition:Inverse Hyperbolic Sine",
"Definition:Inverse Hyperbolic Sine/Complex/Principal Branch",
"Definition:Complex Number",
"Definition:Inverse Hyperbolic Sine"
] | [
"Solution to Quadratic Equation",
"Category:Inverse Hyperbolic Sine"
] |
proofwiki-8286 | Value of Cauchy Determinant | Let $D_n$ be a Cauchy determinant of order $n$:
:<nowiki>$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdot... | Take the version of the Cauchy matrix defined such that $a_{i j} = \dfrac 1 {x_i + y_j}$.
Subtract column $1$ from each of columns $2$ to $n$.
Thus:
{{begin-eqn}}
{{eqn | l = a_{ij}
| o = \gets
| r = \frac 1 {x_i + y_j} - \frac 1 {x_i + y_1}
| c =
}}
{{eqn | r = \frac {\paren {x_i + y_1} - \paren {x_... | Let $D_n$ be a [[Definition:Cauchy Determinant|Cauchy determinant of order $n$]]:
:<nowiki>$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + ... | Take the version of the [[Definition:Cauchy Matrix|Cauchy matrix]] defined such that $a_{i j} = \dfrac 1 {x_i + y_j}$.
Subtract [[Definition:Column of Matrix|column]] $1$ from each of [[Definition:Column of Matrix|column]]s $2$ to $n$.
Thus:
{{begin-eqn}}
{{eqn | l = a_{ij}
| o = \gets
| r = \frac 1 {x_i... | Value of Cauchy Determinant/Proof 1 | https://proofwiki.org/wiki/Value_of_Cauchy_Determinant | https://proofwiki.org/wiki/Value_of_Cauchy_Determinant/Proof_1 | [
"Value of Cauchy Determinant",
"Cauchy Matrices",
"Determinants"
] | [
"Definition:Cauchy Determinant",
"Definition:Determinant/Matrix"
] | [
"Definition:Cauchy Matrix",
"Definition:Matrix/Column",
"Definition:Matrix/Column",
"Multiple of Row Added to Row of Determinant",
"Definition:Determinant/Matrix",
"Definition:Matrix/Row",
"Definition:Matrix/Column",
"Determinant with Row Multiplied by Constant",
"Definition:Matrix/Row",
"Definiti... |
proofwiki-8287 | Value of Cauchy Determinant | Let $D_n$ be a Cauchy determinant of order $n$:
:<nowiki>$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdot... | Let:
{{begin-eqn}}
{{eqn | l = C
| r = <nowiki>\begin {bmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & ... | Let $D_n$ be a [[Definition:Cauchy Determinant|Cauchy determinant of order $n$]]:
:<nowiki>$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + ... | Let:
{{begin-eqn}}
{{eqn | l = C
| r = <nowiki>\begin {bmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots &... | Value of Cauchy Determinant/Proof 2 | https://proofwiki.org/wiki/Value_of_Cauchy_Determinant | https://proofwiki.org/wiki/Value_of_Cauchy_Determinant/Proof_2 | [
"Value of Cauchy Determinant",
"Cauchy Matrices",
"Determinants"
] | [
"Definition:Cauchy Determinant",
"Definition:Determinant/Matrix"
] | [
"Vandermonde Matrix Identity for Cauchy Matrix",
"Definition:Polynomial/Complex Numbers",
"Vandermonde Matrix Identity for Cauchy Matrix",
"Determinant of Matrix Product",
"Effect of Elementary Row Operations on Determinant",
"Matrix Product with Adjugate Matrix",
"Determinant of Matrix Product",
"Val... |
proofwiki-8288 | Integers under Addition form Abelian Group | The set of integers under addition $\struct {\Z, +}$ forms an abelian group. | From the definition of the integers, the algebraic structure $\struct {\Z, +}$ is an isomorphic copy of the inverse completion of $\struct {\N, +}$.
From Natural Numbers under Addition form Commutative Semigroup, $\struct {\N, +}$ is a commutative semigroup.
From Natural Number Addition is Cancellable all elements of $... | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Addition|addition]] $\struct {\Z, +}$ forms an [[Definition:Abelian Group|abelian group]]. | From the definition of the [[Definition:Integer/Formal Definition|integers]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, +}$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] copy of the [[Definition:Inverse Completion|inverse completion]] of $\struct {\N, +}$.
From [[Natural... | Integers under Addition form Abelian Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Abelian_Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Abelian_Group | [
"Additive Group of Integers",
"Integer Addition",
"Examples of Abelian Groups"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Abelian Group"
] | [
"Definition:Integer/Formal Definition",
"Definition:Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Inverse Completion",
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:Commutative Semigroup",
"Natural Number Addition is Cancellable",
"Definition... |
proofwiki-8289 | Piecewise Continuous Function with One-Sided Limits is Darboux Integrable | Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$. | We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.
From Piecewise Continuous Function with One-Sided Limits is Bounded, $f$ is a bounded piecewise continuous function.
The result follows from Bounded Piecewise Continuous Function is Darboux Integrable.
{{qed}} | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
Then $f$ is [[Definition:Darboux Int... | We are given that $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
From [[Piecewise Continuous Function with One-Sided Limits is Bounded]], $f$ is a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous f... | Piecewise Continuous Function with One-Sided Limits is Darboux Integrable/Proof 1 | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Darboux_Integrable | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Darboux_Integrable/Proof_1 | [
"Piecewise Continuous Function with One-Sided Limits is Darboux Integrable",
"Piecewise Continuous Functions",
"Darboux Integrable Functions",
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Darboux Integrable Function"
] | [
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Piecewise Continuous Function with One-Sided Limits is Bounded",
"Definition:Piecewise Continuous Function/Bounded",
"Bounded Piecewise Continuous Function is Darboux Integrable"
] |
proofwiki-8290 | Piecewise Continuous Function with One-Sided Limits is Darboux Integrable | Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$. | We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.
Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
Then $f$ is [[Definition:Darboux Int... | We are given that $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, s... | Piecewise Continuous Function with One-Sided Limits is Darboux Integrable/Proof 2 | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Darboux_Integrable | https://proofwiki.org/wiki/Piecewise_Continuous_Function_with_One-Sided_Limits_is_Darboux_Integrable/Proof_2 | [
"Piecewise Continuous Function with One-Sided Limits is Darboux Integrable",
"Piecewise Continuous Functions",
"Darboux Integrable Functions",
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Darboux Integrable Function"
] | [
"Definition:Piecewise Continuous Function/One-Sided Limits",
"Definition:Subdivision of Interval/Finite",
"Definition:Continuous Real Function/Open Interval",
"Definition:Real Interval/Open",
"Definition:Subdivision of Interval/Finite",
"Principle of Mathematical Induction",
"Definition:Real Interval/Op... |
proofwiki-8291 | Infinite Sequence Property of Well-Founded Relation/Forward Implication | Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation.
Then there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$ | Let $\RR$ be a well-founded relation.
{{AimForCont}} there exists an infinite sequence $\sequence {a_n}$ in $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Let $T = \set {a_0, a_1, a_2, \ldots}$.
Let $a_k \in T$ be a minimal element of $T$.
That is:
:$\fo... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
Let $\RR$ be a [[Definition:Well-Founded Relation|well-founded relation]].
Then there exists no [[Definition:Infinite Sequence|infinite sequence]] $\sequence {a_n}$ of [[Definition:Element|elements]] of $S$ such that:
:$\forall n \... | Let $\RR$ be a [[Definition:Well-Founded Relation|well-founded relation]].
{{AimForCont}} there exists an [[Definition:Sequence|infinite sequence]] $\sequence {a_n}$ in $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Let $T = \set {a_0, a_1, a_2, \ldots... | Infinite Sequence Property of Well-Founded Relation/Forward Implication | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Forward_Implication | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Forward_Implication | [
"Infinite Sequence Property of Well-Founded Relation"
] | [
"Definition:Relational Structure",
"Definition:Well-Founded Relation",
"Definition:Sequence/Infinite Sequence",
"Definition:Element"
] | [
"Definition:Well-Founded Relation",
"Definition:Sequence",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Proof by Contradiction",
"Definition:Sequence"
] |
proofwiki-8292 | Extendability Theorem for Function Continuous on Open Interval | Let $f$ be a continuous real function that is defined on an open interval $\openint a b$.
Let $g$ be a real function that satisfies:
:$g$ is defined on $\closedint a b$
:$g$ is continuous on $\closedint a b$
:$g = f$ on $\openint a b$.
Then $g$ exists {{iff}} $\ds \lim_{x \mathop \to a^+} \map f x$ and $\ds \lim_{x \ma... | === Necessary Condition ===
Assume that $g$ exists.
We need to prove that the limits $\ds \lim_{x \mathop \to a^+} \map f x$ and $\ds \lim_{x \mathop \to b^-} \map f x$ exist.
$g$ is continuous at the end points $a$ and $b$ of its domain as $g$ is continuous on $\closedint a b$.
$g$ is right-continuous at $a$ and left-... | Let $f$ be a [[Definition:Continuous Real Function on Open Interval|continuous real function]] that is defined on an [[Definition:Open Real Interval|open interval]] $\openint a b$.
Let $g$ be a [[Definition:Real Function|real function]] that satisfies:
:$g$ is defined on $\closedint a b$
:$g$ is [[Definition:Continu... | === Necessary Condition ===
Assume that $g$ exists.
We need to prove that the limits $\ds \lim_{x \mathop \to a^+} \map f x$ and $\ds \lim_{x \mathop \to b^-} \map f x$ exist.
$g$ is [[Definition:Continuous Real Function at Point|continuous]] at the [[Definition:Endpoint of Real Interval|end points]] $a$ and $b$ of... | Extendability Theorem for Function Continuous on Open Interval | https://proofwiki.org/wiki/Extendability_Theorem_for_Function_Continuous_on_Open_Interval | https://proofwiki.org/wiki/Extendability_Theorem_for_Function_Continuous_on_Open_Interval | [
"Continuous Real Functions"
] | [
"Definition:Continuous Real Function/Open Interval",
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Continuous Real Function/Closed Interval"
] | [
"Definition:Continuous Real Function/Point",
"Definition:Real Interval/Endpoints",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Continuous Real Function/Closed Interval",
"Definition:Continuous Real Function/Right-Continuous",
"Definition:Continuous Real Function/Left-Continuous",
"Definition:C... |
proofwiki-8293 | Integrability Theorem for Functions Continuous on Open Intervals | Let $f$ be a real function defined on a closed interval $\closedint a b$ such that $a < b$.
Let $f$ be continuous on $\openint a b$.
Let the one-sided limits $\ds \lim_{x \mathop \to a^+} \map f x$ and $\ds \lim_{x \mathop \to b^-} \map f x$ exist.
Then $f$ is Darboux integrable on $\closedint a b$. | We start by showing that $f$ is bounded.
A real function $g$ defined on $\closedint a b$ exists by Extendability Theorem for Function Continuous on Open Interval such that:
:$g$ equals $f$ on $\openint a b$
:$g$ is continuous on $\closedint a b$
Therefore:
:$g$ is bounded on $\closedint a b$ by Continuous Function on C... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ such that $a < b$.
Let $f$ be [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$.
Let the [[Definition:One-Sided Limit of Real Function|one-sided... | We start by showing that $f$ is [[Definition:Bounded Real-Valued Function|bounded]].
A [[Definition:Real Function|real function]] $g$ defined on $\closedint a b$ exists by [[Extendability Theorem for Function Continuous on Open Interval]] such that:
:$g$ equals $f$ on $\openint a b$
:$g$ is [[Definition:Continuous R... | Integrability Theorem for Functions Continuous on Open Intervals | https://proofwiki.org/wiki/Integrability_Theorem_for_Functions_Continuous_on_Open_Intervals | https://proofwiki.org/wiki/Integrability_Theorem_for_Functions_Continuous_on_Open_Intervals | [
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Continuous Real Function/Open Interval",
"Definition:One-Sided Limit of Real Function",
"Definition:Darboux Integrable Function"
] | [
"Definition:Bounded Mapping/Real-Valued",
"Definition:Real Function",
"Extendability Theorem for Function Continuous on Open Interval",
"Definition:Continuous Real Function/Closed Interval",
"Definition:Bounded Mapping/Real-Valued",
"Continuous Function on Compact Subspace of Euclidean Space is Bounded",
... |
proofwiki-8294 | Set of Rational Numbers whose Numerator Divisible by p is Closed under Multiplication | Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational multiplication. | Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a \times b
| r = \frac {p n_1} {d_1} \times \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 p n_2} {d_1 d... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $A_p$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] which, when expressed in [[Definition:Canonical Form of Rational Number|canonical form]] has a [[Definition:Numerator|numerator]] which is [[Definition:Divisor of Intege... | Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a \times b
| r = \frac {p n_1} {d_1} \times \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 p n_2} {d... | Set of Rational Numbers whose Numerator Divisible by p is Closed under Multiplication | https://proofwiki.org/wiki/Set_of_Rational_Numbers_whose_Numerator_Divisible_by_p_is_Closed_under_Multiplication | https://proofwiki.org/wiki/Set_of_Rational_Numbers_whose_Numerator_Divisible_by_p_is_Closed_under_Multiplication | [
"Prime Numbers",
"Rational Multiplication",
"Algebraic Closure"
] | [
"Definition:Prime Number",
"Definition:Set",
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form",
"Definition:Fraction/Numerator",
"Definition:Divisor (Algebra)/Integer",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Rational Numbers"
] | [
"Euclid's Lemma for Prime Divisors",
"Prime not Divisor implies Coprime",
"Definition:Coprime/Integers",
"Definition:Rational Number/Canonical Form",
"Definition:Divisor (Algebra)/Integer",
"Definition:Fraction/Numerator"
] |
proofwiki-8295 | Integer Less One divides Power Less One | Let $q, n \in \Z_{>0}$.
Then:
:$\paren {q - 1} \divides \paren {q^n - 1}$
where $\divides$ denotes divisibility. | From Sum of Geometric Sequence:
:$\ds \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n - 1} q^k$
That is:
:$q^n - 1 = r \paren {q - 1}$
where $r = 1 + q + q^2 + \cdots + q^{n - 1}$.
As Integer Addition is Closed and Integer Multiplication is Closed, it follows that $r \in \Z$.
Hence the result by definition of divisor... | Let $q, n \in \Z_{>0}$.
Then:
:$\paren {q - 1} \divides \paren {q^n - 1}$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. | From [[Sum of Geometric Sequence]]:
:$\ds \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n - 1} q^k$
That is:
:$q^n - 1 = r \paren {q - 1}$
where $r = 1 + q + q^2 + \cdots + q^{n - 1}$.
As [[Integer Addition is Closed]] and [[Integer Multiplication is Closed]], it follows that $r \in \Z$.
Hence the result by defin... | Integer Less One divides Power Less One | https://proofwiki.org/wiki/Integer_Less_One_divides_Power_Less_One | https://proofwiki.org/wiki/Integer_Less_One_divides_Power_Less_One | [
"Number Theory"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Sum of Geometric Sequence",
"Integer Addition is Closed",
"Integer Multiplication is Closed",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-8296 | Number Plus One divides Power Plus One iff Odd | Let $q, n \in \Z_{>0}$.
Then:
:$\paren {q + 1} \divides \paren {q^n + 1}$
{{iff}} $n$ is odd.
In the above, $\divides$ denotes divisibility. | Let $n$ be odd.
Then from Sum of Odd Positive Powers:
:$\ds q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$
Let $n$ be even.
Consider the equation:
:$q^n + 1 = 0$
We have:
:$\paren {-1}^n + 1 = 2$
So $-1$ is not a root of $q^n + 1 = 0$.
By Polynomial Factor Theorem, $q + 1$ is not a divisor of ... | Let $q, n \in \Z_{>0}$.
Then:
:$\paren {q + 1} \divides \paren {q^n + 1}$
{{iff}} $n$ is [[Definition:Odd Integer|odd]].
In the above, $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. | Let $n$ be [[Definition:Odd Integer|odd]].
Then from [[Sum of Odd Positive Powers]]:
:$\ds q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$
Let $n$ be [[Definition:Even Integer|even]].
Consider the equation:
:$q^n + 1 = 0$
We have:
:$\paren {-1}^n + 1 = 2$
So $-1$ is not a root of $q^n + 1... | Number Plus One divides Power Plus One iff Odd | https://proofwiki.org/wiki/Number_Plus_One_divides_Power_Plus_One_iff_Odd | https://proofwiki.org/wiki/Number_Plus_One_divides_Power_Plus_One_iff_Odd | [
"Number Theory"
] | [
"Definition:Odd Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Odd Integer",
"Sum of Odd Positive Powers",
"Definition:Even Integer",
"Polynomial Factor Theorem",
"Definition:Divisor (Algebra)/Integer",
"Definition:Even Integer"
] |
proofwiki-8297 | Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function | Let $f$ be a real-valued function.
Then:
:$f$ is bounded on the union of a finite number of sets within the domain of $f$
{{iff}}:
:$f$ is bounded on each of the sets. | Let $S_i$, $i \in \set {1, \ldots, n}$, $n \in \N_{>0}$, denote sets within the domain of $f$.
Define $S = \ds \bigcup_{i \mathop = 1}^n S_i$. | Let $f$ be a [[Definition:Real-Valued Function|real-valued function]].
Then:
:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the [[Definition:Set Union|union]] of a [[Definition:Finite Set|finite]] number of [[Definition:Set|sets]] within the [[Definition:domain|domain]] of $f$
{{iff}}:
:$f$ is [[Defin... | Let $S_i$, $i \in \set {1, \ldots, n}$, $n \in \N_{>0}$, denote sets within the [[Definition:Domain of Mapping|domain]] of $f$.
Define $S = \ds \bigcup_{i \mathop = 1}^n S_i$. | Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function | https://proofwiki.org/wiki/Mapping_is_Bounded_on_Union_iff_Bounded_on_Each_Component/Real-Valued_Function | https://proofwiki.org/wiki/Mapping_is_Bounded_on_Union_iff_Bounded_on_Each_Component/Real-Valued_Function | [
"Bounded Real-Valued Functions",
"Real-Valued Functions"
] | [
"Definition:Real-Valued Function",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Set Union",
"Definition:Finite Set",
"Definition:Set",
"Definition:domain",
"Definition:Bounded Mapping/Real-Valued"
] | [
"Definition:Domain (Set Theory)/Mapping"
] |
proofwiki-8298 | Parity of Integer equals Parity of Positive Power | Let $p \in \Z$ be an integer.
Let $n \in \Z_{>0}$ be a strictly positive integer.
Then $p$ is even {{iff}} $p^n$ is even.
That is, the parity of an integer equals the parity of all its (strictly) positive powers. | Proof by induction:
For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
:For all $p \in \Z$, $p$ is even {{iff}} $p^n$ is even.
First it is worth confirming that $P \left({0}\right)$ does not hold:
:$\forall p \in \Z: p^0 = 1$
which is not even whatever the parity of $p$.
$P \left({1}\right)$ is true... | Let $p \in \Z$ be an [[Definition:Integer|integer]].
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then $p$ is [[Definition:Even Integer|even]] {{iff}} $p^n$ is [[Definition:Even Integer|even]].
That is, the [[Definition:Parity|parity]] of an [[Definition:Integer|inte... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:For all $p \in \Z$, $p$ is [[Definition:Even Integer|even]] {{iff}} $p^n$ is [[Definition:Even Integer|even]].
First it is worth confirming that $P \left({0}\... | Parity of Integer equals Parity of Positive Power | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_Positive_Power | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_Positive_Power | [
"Integers"
] | [
"Definition:Integer",
"Definition:Strictly Positive/Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Parity",
"Definition:Integer",
"Definition:Parity",
"Definition:Strictly Positive/Integer",
"Definition:Power (Algebra)"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Parity",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Tautology",
"Definition:Even Integer",
"Definition:Even Integer... |
proofwiki-8299 | Two divides Power Plus One iff Odd | Let $q, n \in \Z_{>0}$.
Then:
:$2 \divides \paren {q^n + 1}$
{{iff}} $q$ is odd.
In the above, $\divides$ denotes divisibility. | By Parity of Integer equals Parity of Positive Power, $q^n$ is even {{iff}} $q$ is even.
Thus it follows that $q^n + 1$ is even {{iff}} $q$ is odd.
The result follows by definition of even integer.
{{qed}} | Let $q, n \in \Z_{>0}$.
Then:
:$2 \divides \paren {q^n + 1}$
{{iff}} $q$ is [[Definition:Odd Integer|odd]].
In the above, $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. | By [[Parity of Integer equals Parity of Positive Power]], $q^n$ is [[Definition:Even Integer|even]] {{iff}} $q$ is [[Definition:Even Integer|even]].
Thus it follows that $q^n + 1$ is [[Definition:Even Integer|even]] {{iff}} $q$ is [[Definition:Odd Integer|odd]].
The result follows by definition of [[Definition:Even I... | Two divides Power Plus One iff Odd | https://proofwiki.org/wiki/Two_divides_Power_Plus_One_iff_Odd | https://proofwiki.org/wiki/Two_divides_Power_Plus_One_iff_Odd | [
"Number Theory"
] | [
"Definition:Odd Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Parity of Integer equals Parity of Positive Power",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.