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-900 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | {{begin-eqn}}
{{eqn | q = \exists m, n \in \Z
| l = m a + n b
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| o = \divides
| r = 1
| c = Set of Integer Combinations equals Set of Multiples of GCD
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r = 1... | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | {{begin-eqn}}
{{eqn | q = \exists m, n \in \Z
| l = m a + n b
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| o = \divides
| r = 1
| c = [[Set of Integer Combinations equals Set of Multiples of GCD]]
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r... | Integer Combination of Coprime Integers/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers/Necessary_Condition/Proof_1 | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Set of Integer Combinations equals Set of Multiples of GCD"
] |
proofwiki-901 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.
Let $d$ be a divisor of both $a$ and $b$.
Then:
:$d \divides m a + n b$
and so:
:$d \divides 1$
Thus:
:$d = \pm 1$
and so:
:$\gcd \set {a, b} = 1$
Thus, by definition, $a$ and $b$ are coprime.
{{qed}} | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.
Let $d$ be a [[Definition:Divisor of Integer|divisor]] of both $a$ and $b$.
Then:
:$d \divides m a + n b$
and so:
:$d \divides 1$
Thus:
:$d = \pm 1$
and so:
:$\gcd \set {a, b} = 1$
Thus, by definition, $a$ and $b$ are [[Definition:Coprime Intege... | Integer Combination of Coprime Integers/Necessary Condition/Proof 2 | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers/Necessary_Condition/Proof_2 | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] |
proofwiki-902 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | {{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r = 1
| c = {{Defof|Coprime Integers}}
}}
{{eqn | ll= \leadsto
| q = \exists m, n \in \Z
| l = m a + n b
| r = 1
| c = Bézout's Identity
}}
{{end-eqn}}
{{qed}} | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | {{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r = 1
| c = {{Defof|Coprime Integers}}
}}
{{eqn | ll= \leadsto
| q = \exists m, n \in \Z
| l = m a + n b
| r = 1
| c = [[Bézout's Identity]]
}}
{{end-eqn}}
{{qed}} | Integer Combination of Coprime Integers/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers/Sufficient_Condition/Proof_1 | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Bézout's Identity"
] |
proofwiki-903 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {a m + b n: m, n \in \Z}$
$S$ contains at least one strictly positive integer, because for example $a^2 + b^2 \in S$.
By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive.
L... | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {a m + b n: m, n \in \Z}$
$S$ contains at least one [[Definition:Strictly Positive Integer|strictly positive integer]], because for example $a^2 + b^2 \in S$.
By [[Set of Integers Bounded Below has Smallest Element]], let $d$ be the [[De... | Integer Combination of Coprime Integers/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers/Sufficient_Condition/Proof_2 | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Strictly Positive/Integer",
"Set of Integers Bounded Below has Smallest Element",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Division Theorem",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Definition:Strictly Positive/Integer",
"Def... |
proofwiki-904 | Integer Combination of Coprime Integers | Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
{{iff}}:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime. | Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {\lambda a + \mu b: \lambda, \mu \in \Z}$
$S$ contains at least one strictly positive integer, because for example:
:$a \in S$ (setting $\lambda = 1$ and $\mu = 0$)
:$b \in S$ (setting $\lambda = 0$ and $\mu = 1$)
By Set of Integers Bounded ... | Let $a, b \in \Z$ be [[Definition:Integer|integers]], not both [[Definition:Zero (Number)|zero]].
Then:
:$a$ and $b$ are [[Definition:Coprime Integers|coprime]]
{{iff}}:
:there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in... | Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {\lambda a + \mu b: \lambda, \mu \in \Z}$
$S$ contains at least one [[Definition:Strictly Positive Integer|strictly positive integer]], because for example:
:$a \in S$ (setting $\lambda = 1$ and $\mu = 0$)
:$b \in S$ (setting $\lambda = 0... | Integer Combination of Coprime Integers/Sufficient Condition/Proof 3 | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers | https://proofwiki.org/wiki/Integer_Combination_of_Coprime_Integers/Sufficient_Condition/Proof_3 | [
"Integer Combination of Coprime Integers",
"Coprime Integers",
"Integer Combinations"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Coprime/Integers",
"Definition:Integer Combination",
"Definition:Integer Combination",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Strictly Positive/Integer",
"Set of Integers Bounded Below has Smallest Element",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Division Theorem",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Definition:Strictly Positive/Integer",
"Def... |
proofwiki-905 | Divisor of Sum of Coprime Integers | Let $a, b, c \in \Z_{>0}$ such that:
:$a \perp b$ and $c \divides \paren {a + b}$.
where:
:$a \perp b$ denotes $a$ and $b$ are coprime
:$c \divides \paren {a + b}$ denotes that $c$ is a divisor of $a + b$.
Then $a \perp c$ and $b \perp c$.
That is, a divisor of the sum of two coprime integers is coprime to both. | Let $d \in \Z_{>0}: d \divides c \land d \divides a$.
Then:
{{begin-eqn}}
{{eqn | l = d
| o = \divides
| r = \paren {a + b}
| c = as $c \divides \paren {a + b}$
}}
{{eqn | ll= \leadsto
| l = d
| o = \divides
| r = \paren {a + b - a}
| c =
}}
{{eqn | ll= \leadsto
| l = d
... | Let $a, b, c \in \Z_{>0}$ such that:
:$a \perp b$ and $c \divides \paren {a + b}$.
where:
:$a \perp b$ denotes $a$ and $b$ are [[Definition:Coprime Integers|coprime]]
:$c \divides \paren {a + b}$ denotes that $c$ is a [[Definition:Divisor of Integer|divisor]] of $a + b$.
Then $a \perp c$ and $b \perp c$.
That is, a... | Let $d \in \Z_{>0}: d \divides c \land d \divides a$.
Then:
{{begin-eqn}}
{{eqn | l = d
| o = \divides
| r = \paren {a + b}
| c = as $c \divides \paren {a + b}$
}}
{{eqn | ll= \leadsto
| l = d
| o = \divides
| r = \paren {a + b - a}
| c =
}}
{{eqn | ll= \leadsto
| l = ... | Divisor of Sum of Coprime Integers | https://proofwiki.org/wiki/Divisor_of_Sum_of_Coprime_Integers | https://proofwiki.org/wiki/Divisor_of_Sum_of_Coprime_Integers | [
"Number Theory",
"Divisors",
"Coprime Integers"
] | [
"Definition:Coprime/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Coprime/Integers"
] | [
"Definition:Coprime/Integers"
] |
proofwiki-906 | Integers Divided by GCD are Coprime | Let $a, b \in \Z$ be integers which are not both zero.
Let $d$ be a common divisor of $a$ and $b$, that is:
:$\dfrac a d, \dfrac b d \in \Z$
Then:
:$\gcd \set {a, b} = d$
{{iff}}:
:$\gcd \set {\dfrac a d, \dfrac b d} = 1$
that is:
:$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$
where:
:$\gcd$ denotes... | Let $d = \gcd \set {a, b}$.
By definition of divisor:
:$d \divides a \iff \exists s \in \Z: a = d s$
:$d \divides b \iff \exists t \in \Z: b = d t$
So:
{{begin-eqn}}
{{eqn | q = \exists m, n \in \Z
| l = d
| r = m a + n b
| c = Bézout's Identity
}}
{{eqn | ll= \leadstoandfrom
| l = d
| r =... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] which are not both [[Definition:Zero (Number)|zero]].
Let $d$ be a [[Definition:Common Divisor|common divisor]] of $a$ and $b$, that is:
:$\dfrac a d, \dfrac b d \in \Z$
Then:
:$\gcd \set {a, b} = d$
{{iff}}:
:$\gcd \set {\dfrac a d, \dfrac b d} = 1$
that is:
:$\d... | Let $d = \gcd \set {a, b}$.
By definition of [[Definition:Divisor of Integer|divisor]]:
:$d \divides a \iff \exists s \in \Z: a = d s$
:$d \divides b \iff \exists t \in \Z: b = d t$
So:
{{begin-eqn}}
{{eqn | q = \exists m, n \in \Z
| l = d
| r = m a + n b
| c = [[Bézout's Identity]]
}}
{{eqn | ll= ... | Integers Divided by GCD are Coprime/Proof 1 | https://proofwiki.org/wiki/Integers_Divided_by_GCD_are_Coprime | https://proofwiki.org/wiki/Integers_Divided_by_GCD_are_Coprime/Proof_1 | [
"Greatest Common Divisor",
"Coprime Integers",
"Integers Divided by GCD are Coprime"
] | [
"Definition:Integer",
"Definition:Zero (Number)",
"Definition:Common Divisor",
"Definition:Greatest Common Divisor/Integers",
"Definition:Coprime/Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Bézout's Identity",
"Bézout's Identity"
] |
proofwiki-907 | Product of Coprime Factors | Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime.
Let both $a$ and $b$ be divisors of $c$.
Then $a b$ is also a divisor of $c$.
That is:
:$a \perp b \land a \divides c \land b \divides c \implies a b \divides c$ | By definition of divisor:
:$a \divides c \implies \exists r \in \Z: c = a r$
:$b \divides c \implies \exists s \in \Z: c = b s$
So:
{{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists m, n \in \Z
| l = m a + n b
| r = 1
| c = Integer Com... | Let $a, b, c \in \Z$ such that $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
Let both $a$ and $b$ be [[Definition:Divisor of Integer|divisors]] of $c$.
Then $a b$ is also a [[Definition:Divisor of Integer|divisor]] of $c$.
That is:
:$a \perp b \land a \divides c \land b \divides c \implies a b \divides c... | By definition of [[Definition:Divisor of Integer|divisor]]:
:$a \divides c \implies \exists r \in \Z: c = a r$
:$b \divides c \implies \exists s \in \Z: c = b s$
So:
{{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists m, n \in \Z
| l = m a + n b
... | Product of Coprime Factors | https://proofwiki.org/wiki/Product_of_Coprime_Factors | https://proofwiki.org/wiki/Product_of_Coprime_Factors | [
"Coprime Integers"
] | [
"Definition:Coprime/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Divisor (Algebra)/Integer",
"Integer Combination of Coprime Integers"
] |
proofwiki-908 | Existence of Lowest Common Multiple | Let $a, b \in \Z: a b \ne 0$.
The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | We prove its existence thus:
$a b \ne 0 \implies \size {a b} \ne 0$
Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.
So the lowest common multiple definitely exists, and we can say that:
:$0 < \lcm \set {a, b} \le \size {a b}$
Now we prove it is the lowest.
That is:
:$a \divides n \land b \divides n ... | Let $a, b \in \Z: a b \ne 0$.
The [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | We prove its existence thus:
$a b \ne 0 \implies \size {a b} \ne 0$
Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.
So the [[Definition:Lowest Common Multiple|lowest common multiple]] definitely exists, and we can say that:
:$0 < \lcm \set {a, b} \le \size {a b}$
Now we prove it is the lowest.... | Existence of Lowest Common Multiple/Proof 1 | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple/Proof_1 | [
"Lowest Common Multiple",
"Existence of Lowest Common Multiple"
] | [
"Definition:Lowest Common Multiple/Integers"
] | [
"Definition:Lowest Common Multiple",
"Definition:Positive/Integer",
"Definition:Common Multiple",
"Definition:Divisor (Algebra)/Integer",
"Definition:Common Multiple"
] |
proofwiki-909 | Existence of Lowest Common Multiple | Let $a, b \in \Z: a b \ne 0$.
The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | Either $a$ and $b$ are coprime or they are not.
Let:
:$a \perp b$
where $a \perp b$ denotes that $a$ and $b$ are coprime.
Let $a b = c$.
Then:
:$a \divides c, b \divides c$
where $a \divides c$ denotes that $a$ is a divisor of $c$.
Suppose both $a \divides d, b \divides d$ for some $d \in \N_{> 0}: d < c$.
Then:
:$\exi... | Let $a, b \in \Z: a b \ne 0$.
The [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | Either $a$ and $b$ are [[Definition:Coprime Integers|coprime]] or they are not.
Let:
:$a \perp b$
where $a \perp b$ denotes that $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
Let $a b = c$.
Then:
:$a \divides c, b \divides c$
where $a \divides c$ denotes that $a$ is a [[Definition:Divisor of Integer|divis... | Existence of Lowest Common Multiple/Proof 2 | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple/Proof_2 | [
"Lowest Common Multiple",
"Existence of Lowest Common Multiple"
] | [
"Definition:Lowest Common Multiple/Integers"
] | [
"Definition:Coprime/Integers",
"Definition:Coprime/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Ratio",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Intege... |
proofwiki-910 | Existence of Lowest Common Multiple | Let $a, b \in \Z: a b \ne 0$.
The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | Note that as Integer Divides Zero, both $a$ and $b$ are divisors of zero.
Thus by definition $0$ is a common multiple of $a$ and $b$.
Non-trivial common multiples of $a$ and $b$ exist.
Indeed, $a b$ and $-\paren {a b}$ are common multiples of $a$ and $b$.
Either $a b$ or $-\paren {a b}$ is strictly positive.
Let $S$ de... | Let $a, b \in \Z: a b \ne 0$.
The [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists. | Note that as [[Integer Divides Zero]], both $a$ and $b$ are [[Definition:Divisor of Integer|divisors]] of [[Definition:Zero (Number)|zero]].
Thus by definition $0$ is a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
Non-trivial [[Definition:Common Multiple|common multiples]] of $a$ and $b$ exist.
Ind... | Existence of Lowest Common Multiple/Proof 3 | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple/Proof_3 | [
"Lowest Common Multiple",
"Existence of Lowest Common Multiple"
] | [
"Definition:Lowest Common Multiple/Integers"
] | [
"Integer Divisor Results/Integer Divides Zero",
"Definition:Divisor (Algebra)/Integer",
"Definition:Zero (Number)",
"Definition:Common Multiple",
"Definition:Common Multiple",
"Definition:Common Multiple",
"Definition:Strictly Positive/Integer",
"Definition:Set",
"Definition:Strictly Positive/Intege... |
proofwiki-911 | Product of GCD and LCM | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$. | It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.
{{begin-eqn}}
{{eqn | l = d
| r = \gcd \set {a, b}
| c =
}}
{{eqn | ll= \leadsto
| l = d
| o = \divides
| r = a b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists n \in \Z_{... | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.
{{begin-eqn}}
{{eqn | l = d
| r = \gcd \set {a, b}
| c =
}}
{{eqn | ll= \leadsto
| l = d
| o = \divides
| r = a b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists n \in \Z_... | Product of GCD and LCM/Proof 1 | https://proofwiki.org/wiki/Product_of_GCD_and_LCM | https://proofwiki.org/wiki/Product_of_GCD_and_LCM/Proof_1 | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Product of GCD and LCM"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"Bézout's Identity"
] |
proofwiki-912 | Product of GCD and LCM | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$. | Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are coprime.
The existence of $m$ and $n$ are proved by Integers Divided by GCD are Coprime.
It follows that:
:$a = g m \divides g m n$
and:
:$b = g n \divides g m n$
So $g m n$ is a common multiple of $a$ and $b$.
Hence there exists an integer ... | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are [[Definition:Coprime Integers|coprime]].
The existence of $m$ and $n$ are proved by [[Integers Divided by GCD are Coprime]].
It follows that:
:$a = g m \divides g m n$
and:
:$b = g n \divides g m n$
So $g m n$ is a [[Definition:Common Mult... | Product of GCD and LCM/Proof 2 | https://proofwiki.org/wiki/Product_of_GCD_and_LCM | https://proofwiki.org/wiki/Product_of_GCD_and_LCM/Proof_2 | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Product of GCD and LCM"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Coprime/Integers",
"Integers Divided by GCD are Coprime",
"Definition:Common Multiple",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] |
proofwiki-913 | Product of GCD and LCM | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$. | Let $d := \gcd \set {a, b}$.
Then by definition of the GCD, there exist $j_1, j_2 \in \Z$ such that $a = d j_1$ and $b = d j_2$.
Because $d$ divides both $a$ and $b$, it must divide their product:
:$\exists l \in \Z$ such that $a b = d l$
Then we have:
{{begin-eqn}}
{{eqn | l = d l
| m = \paren {d j_1} b
... | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | Let $d := \gcd \set {a, b}$.
Then by definition of the [[Definition:Greatest Common Divisor of Integers|GCD]], there exist $j_1, j_2 \in \Z$ such that $a = d j_1$ and $b = d j_2$.
Because $d$ [[Definition:Divisor of Integer|divides]] both $a$ and $b$, it must [[Definition:Divisor of Integer|divide]] their [[Definiti... | Product of GCD and LCM/Proof 3 | https://proofwiki.org/wiki/Product_of_GCD_and_LCM | https://proofwiki.org/wiki/Product_of_GCD_and_LCM/Proof_3 | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Product of GCD and LCM"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Greatest Common Divisor/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Multiplication/Integers",
"Definition:Common Multiple",
"Definition:Common Multiple",
"Bézout's Identity",
"Definition:Lowest Common Multiple/Integers"
] |
proofwiki-914 | Product of GCD and LCM | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$. | From Fundamental Theorem of Arithmetic, let:
{{begin-eqn}}
{{eqn | l = m
| r = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}
}}
{{eqn | l = n
| r = {p_1}^{l_1} {p_2}^{l_2} \dotsm {p_r}r^{l_r}
| c =
}}
{{end-eqn}}
From LCM from Prime Decomposition:
:$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\m... | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | From [[Fundamental Theorem of Arithmetic]], let:
{{begin-eqn}}
{{eqn | l = m
| r = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}
}}
{{eqn | l = n
| r = {p_1}^{l_1} {p_2}^{l_2} \dotsm {p_r}r^{l_r}
| c =
}}
{{end-eqn}}
From [[LCM from Prime Decomposition]]:
:$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1... | Product of GCD and LCM/Proof 4 | https://proofwiki.org/wiki/Product_of_GCD_and_LCM | https://proofwiki.org/wiki/Product_of_GCD_and_LCM/Proof_4 | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Product of GCD and LCM"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"Fundamental Theorem of Arithmetic",
"LCM from Prime Decomposition",
"GCD from Prime Decomposition",
"Sum of Maximum and Minimum"
] |
proofwiki-915 | Product of GCD and LCM | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$. | Let $d := \gcd \set {a, b}$.
Then by definition of the GCD, there exist $r, s\in \Z$ such that $a = d r$ and $b = d s$.
Let $m = \dfrac {a b} d$.
Then:
:$m = a s = r b$
which makes $m$ a common multiple of $a$ and $b$.
Let $c \in \Z_{>0}$ be a common multiple of $a$ and $b$.
Let us say that:
:$c = a u = b v$
From Bézo... | :$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$. | Let $d := \gcd \set {a, b}$.
Then by definition of the [[Definition:Greatest Common Divisor of Integers|GCD]], there exist $r, s\in \Z$ such that $a = d r$ and $b = d s$.
Let $m = \dfrac {a b} d$.
Then:
:$m = a s = r b$
which makes $m$ a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
Let $c \in \... | Product of GCD and LCM/Proof 5 | https://proofwiki.org/wiki/Product_of_GCD_and_LCM | https://proofwiki.org/wiki/Product_of_GCD_and_LCM/Proof_5 | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Product of GCD and LCM"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Greatest Common Divisor/Integers",
"Definition:Common Multiple",
"Definition:Common Multiple",
"Bézout's Identity",
"Definition:Divisor (Algebra)/Integer",
"Absolute Value of Integer is not less than Divisors",
"Definition:Lowest Common Multiple/Integers"
] |
proofwiki-916 | Congruent to Zero iff Modulo is Divisor | Let $a, z \in \R$.
Then $a$ is congruent to $0$ modulo $z$ {{iff}} $a$ is an integer multiple of $z$.
:$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$
If $z \in \Z$, then further:
:$z \divides a \iff a \equiv 0 \pmod z$ | {{begin-eqn}}
{{eqn | q = \exists k \in \Z
| l = a
| r = k z
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists k \in \Z
| l = a
| r = 0 + k z
| c =
}}
{{end-eqn}}
Thus by definition of congruence, $a \equiv 0 \pmod z$ and the result is proved.
If $z$ is an integer, then by d... | Let $a, z \in \R$.
Then $a$ is [[Definition:Congruence (Number Theory)|congruent]] to $0$ modulo $z$ {{iff}} $a$ is an [[Definition:Integer Multiple|integer multiple]] of $z$.
:$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$
If $z \in \Z$, then further:
:$z \divides a \iff a \equiv 0 \pmod z$ | {{begin-eqn}}
{{eqn | q = \exists k \in \Z
| l = a
| r = k z
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists k \in \Z
| l = a
| r = 0 + k z
| c =
}}
{{end-eqn}}
Thus by definition of [[Definition:Congruence (Number Theory)|congruence]], $a \equiv 0 \pmod z$ and the resul... | Congruent to Zero iff Modulo is Divisor | https://proofwiki.org/wiki/Congruent_to_Zero_iff_Modulo_is_Divisor | https://proofwiki.org/wiki/Congruent_to_Zero_iff_Modulo_is_Divisor | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Integral Multiple/Real Numbers"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Category:Modulo Arithmetic"
] |
proofwiki-917 | Integer is Congruent Modulo Divisor to Remainder | Let $a \in \Z$.
Let $a$ have a remainder $r$ on division by $m$.
Then:
:$a \equiv r \pmod m$
where the notation denotes that $a$ and $r$ are congruent modulo $m$. | Let $a$ have a remainder $r$ on division by $m$.
Then:
:$\exists q \in \Z: a = q m + r$
Hence by definition of congruence modulo $m$:
:$a \equiv r \pmod m$
{{qed}} | Let $a \in \Z$.
Let $a$ have a [[Definition:Remainder|remainder]] $r$ on [[Definition:Integer Division|division]] by $m$.
Then:
:$a \equiv r \pmod m$
where the notation denotes that $a$ and $r$ are [[Definition:Congruence Modulo Integer|congruent modulo $m$]]. | Let $a$ have a [[Definition:Remainder|remainder]] $r$ on [[Definition:Integer Division|division]] by $m$.
Then:
:$\exists q \in \Z: a = q m + r$
Hence by definition of [[Definition:Congruence Modulo Integer|congruence modulo $m$]]:
:$a \equiv r \pmod m$
{{qed}} | Integer is Congruent Modulo Divisor to Remainder | https://proofwiki.org/wiki/Integer_is_Congruent_Modulo_Divisor_to_Remainder | https://proofwiki.org/wiki/Integer_is_Congruent_Modulo_Divisor_to_Remainder | [
"Modulo Arithmetic"
] | [
"Definition:Remainder",
"Definition:Integer Division",
"Definition:Congruence (Number Theory)/Integers"
] | [
"Definition:Remainder",
"Definition:Integer Division",
"Definition:Congruence (Number Theory)/Integers"
] |
proofwiki-918 | Integer is Congruent to Integer less than Modulus | Let $m \in \Z$.
Then each integer is congruent (modulo $m$) to precisely one of the integers $0, 1, \ldots, m - 1$. | === Existence ===
Let $a \in \Z$.
Then from the Division Theorem: $\exists r \in \set {0, 1, \ldots, m - 1}: a \equiv r \pmod m$.
{{qed|lemma}} | Let $m \in \Z$.
Then each [[Definition:Integer|integer]] is [[Definition:Congruence Modulo Integer|congruent (modulo $m$)]] to precisely one of the [[Definition:Integer|integers]] $0, 1, \ldots, m - 1$. | === Existence ===
Let $a \in \Z$.
Then from the [[Division Theorem]]: $\exists r \in \set {0, 1, \ldots, m - 1}: a \equiv r \pmod m$.
{{qed|lemma}} | Integer is Congruent to Integer less than Modulus | https://proofwiki.org/wiki/Integer_is_Congruent_to_Integer_less_than_Modulus | https://proofwiki.org/wiki/Integer_is_Congruent_to_Integer_less_than_Modulus | [
"Modulo Arithmetic",
"Complete Residue Systems"
] | [
"Definition:Integer",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Integer"
] | [
"Division Theorem",
"Division Theorem"
] |
proofwiki-919 | Modulo Addition is Well-Defined | Let $m \in \Z$ be an integer.
Let $\Z_m$ be the set of integers modulo $m$.
Let $\eqclass a m$ denote the equivalence class on $\Z_m$, for some $a \in \Z$.
The modulo addition operation on $\Z_m$, defined by the rule:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
is a well-defined operation.
That is:
:If $a \eq... | We need to show that if:
:$\eqclass {x'} m = \eqclass x m$
:$\eqclass {y'} m = \eqclass y m$
then:
:$\eqclass {x' + y'} m = \eqclass {x + y} m$
Since:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
it follows from the definition of set of integers modulo $m$ that:
:$x \equiv x' \pmod m$
and:
:... | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]].
Let $\eqclass a m$ denote the [[Definition:Equivalence Class|equivalence class]] on $\Z_m$, for some $a \in \Z$.
The [[Definition:Modulo Addition|modulo addition]] operation on $\Z_m$,... | We need to show that if:
:$\eqclass {x'} m = \eqclass x m$
:$\eqclass {y'} m = \eqclass y m$
then:
:$\eqclass {x' + y'} m = \eqclass {x + y} m$
Since:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
it follows from the definition of [[Definition:Integers Modulo m|set of integers modulo $m$]... | Modulo Addition is Well-Defined/Proof 1 | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined/Proof_1 | [
"Modulo Addition is Well-Defined",
"Modulo Addition"
] | [
"Definition:Integer",
"Definition:Integers Modulo m",
"Definition:Equivalence Class",
"Definition:Modulo Addition",
"Definition:Well-Defined/Operation"
] | [
"Definition:Integers Modulo m",
"Definition:Integer",
"Definition:Integers Modulo m"
] |
proofwiki-920 | Modulo Addition is Well-Defined | Let $m \in \Z$ be an integer.
Let $\Z_m$ be the set of integers modulo $m$.
Let $\eqclass a m$ denote the equivalence class on $\Z_m$, for some $a \in \Z$.
The modulo addition operation on $\Z_m$, defined by the rule:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
is a well-defined operation.
That is:
:If $a \eq... | The equivalence class $\eqclass a m$ is defined as:
:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
That is, the set of all integers which differ from $a$ by an integer multiple of $m$.
Thus the notation for addition of two set of integers modulo $m$ is not usually:
:$\eqclass a m +_m \eqclass b m$
What is mor... | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]].
Let $\eqclass a m$ denote the [[Definition:Equivalence Class|equivalence class]] on $\Z_m$, for some $a \in \Z$.
The [[Definition:Modulo Addition|modulo addition]] operation on $\Z_m$,... | The [[Definition:Equivalence Class|equivalence class]] $\eqclass a m$ is defined as:
:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
That is, the [[Definition:Set|set]] of all [[Definition:Integer|integers]] which differ from $a$ by an [[Definition:Integer Multiple|integer multiple]] of $m$.
Thus the notation... | Modulo Addition is Well-Defined/Proof 2 | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined/Proof_2 | [
"Modulo Addition is Well-Defined",
"Modulo Addition"
] | [
"Definition:Integer",
"Definition:Integers Modulo m",
"Definition:Equivalence Class",
"Definition:Modulo Addition",
"Definition:Well-Defined/Operation"
] | [
"Definition:Equivalence Class",
"Definition:Set",
"Definition:Integer",
"Definition:Integral Multiple/Real Numbers",
"Definition:Modulo Addition",
"Definition:Integers Modulo m"
] |
proofwiki-921 | Modulo Multiplication is Well-Defined | The multiplication modulo $m$ operation on $\Z_m$, the set of integers modulo $m$, defined by the rule:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
is a well-defined operation.
That is:
:If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a x \equiv b y \pmod m$. | We need to show that if:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
then:
:$\eqclass {x' y'} m = \eqclass {x y} m$
We have that:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
It follows from the definition of residue class modulo $m$ that:
:$x \equiv x' \pmod m$
... | The [[Definition:Modulo Multiplication|multiplication modulo $m$]] operation on $\Z_m$, the set of [[Definition:Integers Modulo m|integers modulo $m$]], defined by the rule:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
is a [[Definition:Well-Defined Operation|well-defined operation]].
That is:
:If $a \e... | We need to show that if:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
then:
:$\eqclass {x' y'} m = \eqclass {x y} m$
We have that:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
It follows from the definition of [[Definition:Residue Class|residue class modulo $m... | Modulo Multiplication is Well-Defined/Proof 1 | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined/Proof_1 | [
"Modulo Multiplication is Well-Defined",
"Modulo Multiplication",
"Examples of Well-Defined Mappings"
] | [
"Definition:Modulo Multiplication",
"Definition:Integers Modulo m",
"Definition:Well-Defined/Operation"
] | [
"Definition:Residue Class",
"Definition:Residue Class"
] |
proofwiki-922 | Modulo Multiplication is Well-Defined | The multiplication modulo $m$ operation on $\Z_m$, the set of integers modulo $m$, defined by the rule:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
is a well-defined operation.
That is:
:If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a x \equiv b y \pmod m$. | The equivalence class $\eqclass a m$ is defined as:
:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
that is, the set of all integers which differ from $a$ by an integer multiple of $m$.
Thus the notation for multiplication of two residue classes modulo $z$ is not usually $\eqclass a m \times_m \eqclass b m$.
W... | The [[Definition:Modulo Multiplication|multiplication modulo $m$]] operation on $\Z_m$, the set of [[Definition:Integers Modulo m|integers modulo $m$]], defined by the rule:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
is a [[Definition:Well-Defined Operation|well-defined operation]].
That is:
:If $a \e... | The [[Definition:Equivalence Class|equivalence class]] $\eqclass a m$ is defined as:
:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
that is, the [[Definition:Set|set]] of all [[Definition:Integer|integers]] which differ from $a$ by an [[Definition:Integer Multiple|integer multiple]] of $m$.
Thus the notation... | Modulo Multiplication is Well-Defined/Proof 2 | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined/Proof_2 | [
"Modulo Multiplication is Well-Defined",
"Modulo Multiplication",
"Examples of Well-Defined Mappings"
] | [
"Definition:Modulo Multiplication",
"Definition:Integers Modulo m",
"Definition:Well-Defined/Operation"
] | [
"Definition:Equivalence Class",
"Definition:Set",
"Definition:Integer",
"Definition:Integral Multiple/Real Numbers",
"Definition:Modulo Multiplication",
"Definition:Residue Class",
"Integer Multiplication Distributes over Addition"
] |
proofwiki-923 | Congruence of Product | Let $a, b, z \in \R$.
Let $a$ be congruent to $b$ modulo $z$, that is:
:$a \equiv b \pmod z$
Then:
:$\forall m \in \Z: m a \equiv m b \pmod z$ | Let $m \in \Z$ and $a \equiv b \pmod z$.
Suppose $m = 0$.
Then the {{RHS}} of the assertion degenerates to $0 \equiv 0 \pmod z$ which is trivially true.
Otherwise, from Congruence by Product of Moduli, we have:
:$a \equiv b \iff m a \equiv m b \pmod z$
As $m \in \Z$, it follows that $m z$ is an integer multiple of $z$.... | Let $a, b, z \in \R$.
Let $a$ be [[Definition:Congruence (Number Theory)|congruent to $b$ modulo $z$]], that is:
:$a \equiv b \pmod z$
Then:
:$\forall m \in \Z: m a \equiv m b \pmod z$ | Let $m \in \Z$ and $a \equiv b \pmod z$.
Suppose $m = 0$.
Then the {{RHS}} of the assertion degenerates to $0 \equiv 0 \pmod z$ which is trivially true.
Otherwise, from [[Congruence by Product of Moduli]], we have:
:$a \equiv b \iff m a \equiv m b \pmod z$
As $m \in \Z$, it follows that $m z$ is an [[Definition:In... | Congruence of Product | https://proofwiki.org/wiki/Congruence_of_Product | https://proofwiki.org/wiki/Congruence_of_Product | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)"
] | [
"Congruence by Product of Moduli",
"Definition:Integral Multiple/Real Numbers",
"Congruence by Divisor of Modulus"
] |
proofwiki-924 | Congruence of Powers | Let $a, b \in \R$ and $m \in \Z$.
Let $a$ be congruent to $b$ modulo $m$, that is:
:$a \equiv b \pmod m$
Then:
:$\forall n \in \Z_{\ge 0}: a^n \equiv b^n \pmod m$ | Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a \equiv b \implies a^k \equiv b^k \pmod m$
$\map P 0$ is trivially true, as $a^0 = b^0 = 1$.
$\map P 1$ is true, as this just says:
:$a \equiv b \pmod m$ | Let $a, b \in \R$ and $m \in \Z$.
Let $a$ be [[Definition:Congruence Modulo Integer|congruent to $b$ modulo $m$]], that is:
:$a \equiv b \pmod m$
Then:
:$\forall n \in \Z_{\ge 0}: a^n \equiv b^n \pmod m$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$a \equiv b \implies a^k \equiv b^k \pmod m$
$\map P 0$ is trivially true, as $a^0 = b^0 = 1$.
$\map P 1$ is true, as this just says:
:$a \equiv b \pmod m$ | Congruence of Powers | https://proofwiki.org/wiki/Congruence_of_Powers | https://proofwiki.org/wiki/Congruence_of_Powers | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)/Integers"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-925 | Modulo Addition is Associative | Addition modulo $m$ is associative:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m = \eqclass x m +_m \paren {\eqclass y m +_m \eqclass z m}$
where $\Z_m$ is the set of integers modulo $m$.
That is:
:$\forall x, y, z \in \Z: \paren {x + y} + z \equiv... | {{begin-eqn}}
{{eqn | l = \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m
| r = \eqclass {x + y} m +_m \eqclass z m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass {\paren {x + y} + z} m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass {x + \paren {y + z} } m
| c = Associa... | [[Definition:Modulo Addition|Addition modulo $m$]] is [[Definition:Associative Operation|associative]]:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m = \eqclass x m +_m \paren {\eqclass y m +_m \eqclass z m}$
where $\Z_m$ is the [[Definition:Intege... | {{begin-eqn}}
{{eqn | l = \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m
| r = \eqclass {x + y} m +_m \eqclass z m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass {\paren {x + y} + z} m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass {x + \paren {y + z} } m
| c = [[Assoc... | Modulo Addition is Associative | https://proofwiki.org/wiki/Modulo_Addition_is_Associative | https://proofwiki.org/wiki/Modulo_Addition_is_Associative | [
"Modulo Addition"
] | [
"Definition:Modulo Addition",
"Definition:Associative Operation",
"Definition:Integers Modulo m"
] | [
"Associative Law of Addition"
] |
proofwiki-926 | Modulo Addition is Commutative | Modulo addition is commutative:
:$\forall x, y, z \in \Z: x + y \pmod m = y + x \pmod m$ | From the definition of modulo addition, this is also written:
:$\forall m \in \Z: \forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m = \eqclass y m +_m \eqclass x m$
By Modulo Addition is Well-Defined, $+_m$ is well-defined.
Hence:
{{begin-eqn}}
{{eqn | l = \eqclass x m +_m \eqclass y m
| ... | [[Definition:Modulo Addition|Modulo addition]] is [[Definition:Commutative Operation|commutative]]:
:$\forall x, y, z \in \Z: x + y \pmod m = y + x \pmod m$ | From the definition of [[Definition:Modulo Addition|modulo addition]], this is also written:
:$\forall m \in \Z: \forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m = \eqclass y m +_m \eqclass x m$
By [[Modulo Addition is Well-Defined]], $+_m$ is [[Definition:Well-Defined|well-defined]].
Hence... | Modulo Addition is Commutative | https://proofwiki.org/wiki/Modulo_Addition_is_Commutative | https://proofwiki.org/wiki/Modulo_Addition_is_Commutative | [
"Modulo Addition"
] | [
"Definition:Modulo Addition",
"Definition:Commutative/Operation"
] | [
"Definition:Modulo Addition",
"Modulo Addition is Well-Defined",
"Definition:Well-Defined",
"Commutative Law of Addition"
] |
proofwiki-927 | Modulo Addition has Identity | Let $m \in \Z$ be an integer.
Then addition modulo $m$ has an identity:
:$\forall \eqclass x m \in \Z_m: \eqclass x m +_m \eqclass 0 m = \eqclass x m = \eqclass 0 m +_m \eqclass x m$
That is:
:$\forall a \in \Z: a + 0 \equiv a \equiv 0 + a \pmod m$ | {{begin-eqn}}
{{eqn | l = \eqclass x m +_m \eqclass 0 m
| r = \eqclass {x + 0} m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass x m
| c =
}}
{{eqn | r = \eqclass {0 + x} m
| c =
}}
{{eqn | r = \eqclass 0 m +_m \eqclass x m
| c = {{Defof|Modulo Addition}}
}}
{{end-eqn}}
Thus $\eq... | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Then [[Definition:Modulo Addition|addition modulo $m$]] has an [[Definition:Identity Element|identity]]:
:$\forall \eqclass x m \in \Z_m: \eqclass x m +_m \eqclass 0 m = \eqclass x m = \eqclass 0 m +_m \eqclass x m$
That is:
:$\forall a \in \Z: a + 0 \equiv a \eq... | {{begin-eqn}}
{{eqn | l = \eqclass x m +_m \eqclass 0 m
| r = \eqclass {x + 0} m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass x m
| c =
}}
{{eqn | r = \eqclass {0 + x} m
| c =
}}
{{eqn | r = \eqclass 0 m +_m \eqclass x m
| c = {{Defof|Modulo Addition}}
}}
{{end-eqn}}
Thus $\... | Modulo Addition has Identity | https://proofwiki.org/wiki/Modulo_Addition_has_Identity | https://proofwiki.org/wiki/Modulo_Addition_has_Identity | [
"Modulo Addition"
] | [
"Definition:Integer",
"Definition:Modulo Addition",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Modulo Addition"
] |
proofwiki-928 | Modulo Addition has Inverses | Let $m \in \Z$ be an integer.
Then addition modulo $m$ has inverses:
For each element $\eqclass x m \in \Z_m$, there exists the element $\eqclass {-x} m \in \Z_m$ with the property:
:$\eqclass x m +_m \eqclass {-x} m = \eqclass 0 m = \eqclass {-x} m +_m \eqclass x m$
where $\Z_m$ is the set of integers modulo $m$.
That... | Let $x \in \Z$ be arbitrary.
As $x \in \Z$ implies $-x \in \Z$, we have:
:$\eqclass {-x} m \in \Z_m$
Therefore:
{{begin-eqn}}
{{eqn | l = \eqclass x m +_m \eqclass {-x} m
| r = \eqclass {x + \paren {-x} } m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass 0 m
| c =
}}
{{eqn | r = \eqclass {\pa... | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Then [[Definition:Modulo Addition|addition modulo $m$]] has [[Definition:Inverse Element|inverses]]:
For each [[Definition:Element|element]] $\eqclass x m \in \Z_m$, there exists the element $\eqclass {-x} m \in \Z_m$ with the property:
:$\eqclass x m +_m \eqclas... | Let $x \in \Z$ be arbitrary.
As $x \in \Z$ implies $-x \in \Z$, we have:
:$\eqclass {-x} m \in \Z_m$
Therefore:
{{begin-eqn}}
{{eqn | l = \eqclass x m +_m \eqclass {-x} m
| r = \eqclass {x + \paren {-x} } m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \eqclass 0 m
| c =
}}
{{eqn | r = \eqclass {\... | Modulo Addition has Inverses | https://proofwiki.org/wiki/Modulo_Addition_has_Inverses | https://proofwiki.org/wiki/Modulo_Addition_has_Inverses | [
"Modulo Addition"
] | [
"Definition:Integer",
"Definition:Modulo Addition",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Definition:Integers Modulo m"
] | [] |
proofwiki-929 | Modulo Multiplication is Closed | Multiplication modulo $m$ is closed on the set of integers modulo $m$:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m \in \Z_m$. | From the definition of multiplication modulo $m$, we have:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
By the Division Theorem:
:$x y = q m + r$ where $0 \le r < m$
Therefore for all $0 \le r < m$:
:$\eqclass {x y} m = \eqclass r m$
Therefore from the definition of integers modulo $m$:
:$\eqclass {x y} m \... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Closed Algebraic Structure|closed]] on the set of [[Definition:Integers Modulo m|integers modulo $m$]]:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m \in \Z_m$. | From the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]], we have:
:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
By the [[Division Theorem]]:
:$x y = q m + r$ where $0 \le r < m$
Therefore for all $0 \le r < m$:
:$\eqclass {x y} m = \eqclass r m$
Therefore from the definition ... | Modulo Multiplication is Closed | https://proofwiki.org/wiki/Modulo_Multiplication_is_Closed | https://proofwiki.org/wiki/Modulo_Multiplication_is_Closed | [
"Modulo Multiplication"
] | [
"Definition:Modulo Multiplication",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Integers Modulo m"
] | [
"Definition:Modulo Multiplication",
"Division Theorem",
"Definition:Integers Modulo m"
] |
proofwiki-930 | Modulo Multiplication is Associative | Multiplication modulo $m$ is associative:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$
That is:
:$\forall x, y, z \in \Z_m: \paren {x \cdot_m y} \cdot_m z = x \cdot_m \p... | {{begin-eqn}}
{{eqn | l = \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m
| r = \eqclass {x y} m \times_m \eqclass z m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {\paren {x y} z} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {x \paren {y z} } m
... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Associative Operation|associative]]:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$
That ... | {{begin-eqn}}
{{eqn | l = \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m
| r = \eqclass {x y} m \times_m \eqclass z m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {\paren {x y} z} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {x \paren {y z} } m
... | Modulo Multiplication is Associative/Proof 1 | https://proofwiki.org/wiki/Modulo_Multiplication_is_Associative | https://proofwiki.org/wiki/Modulo_Multiplication_is_Associative/Proof_1 | [
"Modulo Multiplication",
"Modulo Multiplication is Associative"
] | [
"Definition:Modulo Multiplication",
"Definition:Associative Operation"
] | [
"Integer Multiplication is Associative"
] |
proofwiki-931 | Modulo Multiplication is Associative | Multiplication modulo $m$ is associative:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$
That is:
:$\forall x, y, z \in \Z_m: \paren {x \cdot_m y} \cdot_m z = x \cdot_m \p... | Let $j$ be the largest integer such that:
:$j m \le x y$
Let $p$ be the largest integer such that:
:$p m \le y z$
By definition of multiplication modulo $m$:
:$x \cdot_m y = x y - j m$
:$y \cdot_m z = y z - p m$
Let $k$ be the largest integer such that:
:$k m \le \paren {x y - j m} z$
Let $q$ be the largest integer suc... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Associative Operation|associative]]:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$
That ... | Let $j$ be the largest [[Definition:Integer|integer]] such that:
:$j m \le x y$
Let $p$ be the largest [[Definition:Integer|integer]] such that:
:$p m \le y z$
By definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]]:
:$x \cdot_m y = x y - j m$
:$y \cdot_m z = y z - p m$
Let $k$ be the larges... | Modulo Multiplication is Associative/Proof 2 | https://proofwiki.org/wiki/Modulo_Multiplication_is_Associative | https://proofwiki.org/wiki/Modulo_Multiplication_is_Associative/Proof_2 | [
"Modulo Multiplication",
"Modulo Multiplication is Associative"
] | [
"Definition:Modulo Multiplication",
"Definition:Associative Operation"
] | [
"Definition:Integer",
"Definition:Integer",
"Definition:Modulo Multiplication",
"Definition:Integer",
"Definition:Integer",
"Definition:Integer",
"Definition:Integer",
"Definition:Integer",
"Integer Multiplication is Associative"
] |
proofwiki-932 | Modulo Multiplication is Commutative | Multiplication modulo $m$ is commutative:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m = \eqclass y m \times_m \eqclass x m$ | {{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \eqclass y m
| r = \eqclass {x y} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {y x} m
| c = Integer Multiplication is Commutative
}}
{{eqn | r = \eqclass y m \times_m \eqclass x m
| c = {{Defof|Modulo Multiplication}}
}}
{{end-e... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Commutative Operation|commutative]]:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m = \eqclass y m \times_m \eqclass x m$ | {{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \eqclass y m
| r = \eqclass {x y} m
| c = {{Defof|Modulo Multiplication}}
}}
{{eqn | r = \eqclass {y x} m
| c = [[Integer Multiplication is Commutative]]
}}
{{eqn | r = \eqclass y m \times_m \eqclass x m
| c = {{Defof|Modulo Multiplication}}
}}
{{e... | Modulo Multiplication is Commutative | https://proofwiki.org/wiki/Modulo_Multiplication_is_Commutative | https://proofwiki.org/wiki/Modulo_Multiplication_is_Commutative | [
"Modulo Multiplication"
] | [
"Definition:Modulo Multiplication",
"Definition:Commutative/Operation"
] | [
"Integer Multiplication is Commutative"
] |
proofwiki-933 | Modulo Multiplication has Identity | Multiplication modulo $m$ has an identity:
:$\forall \eqclass x m \in \Z_m: \eqclass x m \times_m \eqclass 1 m = \eqclass x m = \eqclass 1 m \times_m \eqclass x m$ | Follows directly from the definition of multiplication modulo $m$:
{{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \eqclass 1 m
| r = \eqclass {x \times 1} m
| c =
}}
{{eqn | r = \eqclass x m
| c =
}}
{{eqn | r = \eqclass {1 \times x} m
| c =
}}
{{eqn | r = \eqclass 1 m \times_m \eqclass x m
... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] has an [[Definition:Identity Element|identity]]:
:$\forall \eqclass x m \in \Z_m: \eqclass x m \times_m \eqclass 1 m = \eqclass x m = \eqclass 1 m \times_m \eqclass x m$ | Follows directly from the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]]:
{{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \eqclass 1 m
| r = \eqclass {x \times 1} m
| c =
}}
{{eqn | r = \eqclass x m
| c =
}}
{{eqn | r = \eqclass {1 \times x} m
| c =
}}
{{eqn | r... | Modulo Multiplication has Identity | https://proofwiki.org/wiki/Modulo_Multiplication_has_Identity | https://proofwiki.org/wiki/Modulo_Multiplication_has_Identity | [
"Modulo Multiplication"
] | [
"Definition:Modulo Multiplication",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Modulo Multiplication",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Modulo Multiplication"
] |
proofwiki-934 | Modulo Multiplication Distributes over Modulo Addition | Multiplication modulo $m$ is distributive over addition modulo $m$:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m$:
:: $\eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m} = \paren {\eqclass x m \times_m \eqclass y m} +_m \paren {\eqclass x m \times_m \eqclass z m}$
:: $\paren {\eqclass x m +_m \... | Follows directly from the definition of multiplication modulo $m$ and addition modulo $m$:
{{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m}
| r = \eqclass x m \times_m \eqclass {y + z} m
| c =
}}
{{eqn | r = \eqclass {x \paren {y + z} } m
| c =
}}
{{eqn | r = \e... | [[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Modulo Addition|addition modulo $m$]]:
:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m$:
:: $\eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m} = \paren {\eqclass... | Follows directly from the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]] and [[Definition:Modulo Addition|addition modulo $m$]]:
{{begin-eqn}}
{{eqn | l = \eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m}
| r = \eqclass x m \times_m \eqclass {y + z} m
| c =
}}
{{eq... | Modulo Multiplication Distributes over Modulo Addition | https://proofwiki.org/wiki/Modulo_Multiplication_Distributes_over_Modulo_Addition | https://proofwiki.org/wiki/Modulo_Multiplication_Distributes_over_Modulo_Addition | [
"Modulo Multiplication",
"Modulo Addition",
"Examples of Distributive Operations"
] | [
"Definition:Modulo Multiplication",
"Definition:Distributive Operation",
"Definition:Modulo Addition",
"Definition:Integers Modulo m"
] | [
"Definition:Modulo Multiplication",
"Definition:Modulo Addition"
] |
proofwiki-935 | Intersection of Congruence Classes | Let $\RR_m$ denote congruence modulo $m$ on the set of integers $\Z$.
Then:
:$\RR_m \cap \RR_n = \RR_{\lcm \set {m, n} }$
where $\lcm \set {m, n}$ is the lowest common multiple of $m$ and $n$.
In the language of modulo arithmetic, this is equivalent to:
:$a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod... | Let $\tuple {a, b} \in \RR_m \cap \RR_n$.
That is, let $\tuple {a, b} \in \RR_m$ and $\tuple {a, b} \in \RR_n$.
That means, by definition of congruence:
:$a \equiv b \pmod m$
:$a \equiv b \pmod n$
Thus by definition of congruence:
:$\exists r, s \in \Z: a - b = r m, a - b = s n$
Let $d = \gcd \set {m, n}$ so that $m = ... | Let $\RR_m$ denote [[Definition:Congruence Modulo Integer|congruence modulo $m$]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$.
Then:
:$\RR_m \cap \RR_n = \RR_{\lcm \set {m, n} }$
where $\lcm \set {m, n}$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $m$... | Let $\tuple {a, b} \in \RR_m \cap \RR_n$.
That is, let $\tuple {a, b} \in \RR_m$ and $\tuple {a, b} \in \RR_n$.
That means, by definition of [[Definition:Congruence Modulo Integer|congruence]]:
:$a \equiv b \pmod m$
:$a \equiv b \pmod n$
Thus by definition of [[Definition:Congruence (Number Theory)/Integers/Integer... | Intersection of Congruence Classes | https://proofwiki.org/wiki/Intersection_of_Congruence_Classes | https://proofwiki.org/wiki/Intersection_of_Congruence_Classes | [
"Modulo Arithmetic",
"Set Intersection",
"Lowest Common Multiple",
"Intersection of Congruence Classes"
] | [
"Definition:Congruence (Number Theory)/Integers",
"Definition:Set",
"Definition:Integer",
"Definition:Lowest Common Multiple/Integers",
"Definition:Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)/Integers",
"Definition:Congruence (Number Theory)/Integers/Integer Multiple",
"Euclid's Lemma",
"Product of GCD and LCM"
] |
proofwiki-936 | Mappings Between Residue Classes | Let $\eqclass a m$ be the residue class of $a$ (modulo $m$).
Let $\phi: \Z_m \to \Z_n$ be a mapping given by:
:$\map \phi {\eqclass x m} = \eqclass x n$
Then $\phi$ is well defined {{iff}} $m$ is a divisor of $n$. | For $\phi$ to be well defined, we require that:
:$\forall x, y \in \Z_m: \eqclass x m = \eqclass y m \implies \map \phi {\eqclass x m} = \map \phi {\eqclass y m}$
Now:
:$\eqclass x m = \eqclass y m \implies x - y \divides m$
For $\map \phi {\eqclass x m} = \map \phi {\eqclass y m}$ we require that:
:$\eqclass x n = \eq... | Let $\eqclass a m$ be the [[Definition:Residue Class|residue class of $a$ (modulo $m$)]].
Let $\phi: \Z_m \to \Z_n$ be a [[Definition:Mapping|mapping]] given by:
:$\map \phi {\eqclass x m} = \eqclass x n$
Then $\phi$ is [[Definition:Well-Defined Mapping|well defined]] {{iff}} $m$ is a [[Definition:Divisor of Integer... | For $\phi$ to be [[Definition:Well-Defined Mapping|well defined]], we require that:
:$\forall x, y \in \Z_m: \eqclass x m = \eqclass y m \implies \map \phi {\eqclass x m} = \map \phi {\eqclass y m}$
Now:
:$\eqclass x m = \eqclass y m \implies x - y \divides m$
For $\map \phi {\eqclass x m} = \map \phi {\eqclass y m... | Mappings Between Residue Classes | https://proofwiki.org/wiki/Mappings_Between_Residue_Classes | https://proofwiki.org/wiki/Mappings_Between_Residue_Classes | [
"Residue Classes"
] | [
"Definition:Residue Class",
"Definition:Mapping",
"Definition:Well-Defined/Mapping",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Well-Defined/Mapping",
"Definition:Well-Defined/Mapping"
] |
proofwiki-937 | GCD from Congruence Modulo m | Let $a, b \in \Z, m \in \N$.
Let $a$ be congruent to $b$ modulo $m$.
Then the GCD of $a$ and $m$ is equal to the GCD of $b$ and $m$.
That is:
:$a \equiv b \pmod m \implies \gcd \set {a, m} = \gcd \set {b, m}$ | We have:
:$a \equiv b \pmod m \implies \exists k \in \Z: a = b + k m$
Thus:
:$a = b + k m$
and the result follows directly from GCD with Remainder.
{{qed}} | Let $a, b \in \Z, m \in \N$.
Let $a$ be [[Definition:Congruence Modulo Integer|congruent]] to $b$ modulo $m$.
Then the [[Definition:Greatest Common Divisor of Integers|GCD]] of $a$ and $m$ is equal to the [[Definition:Greatest Common Divisor of Integers|GCD]] of $b$ and $m$.
That is:
:$a \equiv b \pmod m \implies ... | We have:
:$a \equiv b \pmod m \implies \exists k \in \Z: a = b + k m$
Thus:
:$a = b + k m$
and the result follows directly from [[GCD with Remainder]].
{{qed}} | GCD from Congruence Modulo m | https://proofwiki.org/wiki/GCD_from_Congruence_Modulo_m | https://proofwiki.org/wiki/GCD_from_Congruence_Modulo_m | [
"Modulo Arithmetic",
"Greatest Common Divisor"
] | [
"Definition:Congruence (Number Theory)/Integers",
"Definition:Greatest Common Divisor/Integers",
"Definition:Greatest Common Divisor/Integers"
] | [
"GCD with Remainder"
] |
proofwiki-938 | Prime Number has 4 Integral Divisors | Let $p$ be an integer.
Then $p$ is a prime number {{iff}} $p$ has exactly four integral divisors: $1, -1, p, -p$. | === Necessary Condition ===
Let $p$ be a prime number from the definition that $p$ has exactly $2$ divisors which are positive integers.
From One Divides all Integers and Integer Divides Itself those positive integers are $1$ and $p$.
Also, we have $-1 \divides p$ and $-p \divides p$ from One Divides all Integers and I... | Let $p$ be an [[Definition:Integer|integer]].
Then $p$ is a [[Definition:Prime Number|prime number]] {{iff}} $p$ has exactly four [[Definition:Integer|integral]] [[Definition:Divisor of Integer|divisors]]: $1, -1, p, -p$. | === Necessary Condition ===
Let $p$ be a [[Definition:Prime Number/Definition 1|prime number]] from the definition that $p$ has exactly $2$ [[Definition:Divisor of Integer|divisors]] which are [[Definition:Positive Integer|positive integers]].
From [[One Divides all Integers]] and [[Integer Divides Itself]] those [[D... | Prime Number has 4 Integral Divisors | https://proofwiki.org/wiki/Prime_Number_has_4_Integral_Divisors | https://proofwiki.org/wiki/Prime_Number_has_4_Integral_Divisors | [
"Prime Numbers"
] | [
"Definition:Integer",
"Definition:Prime Number",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Prime Number/Definition 1",
"Definition:Divisor (Algebra)/Integer",
"Definition:Positive/Integer",
"Integer Divisor Results/One Divides all Integers",
"Integer Divisor Results/Integer Divides Itself",
"Definition:Positive/Integer",
"Integer Divisor Results/One Divides all Integers",
"Integ... |
proofwiki-939 | Prime not Divisor implies Coprime | Let $p, a \in \Z$.
If $p$ is a prime number then:
:$p \nmid a \implies p \perp a$
where:
:$p \nmid a$ denotes that $p$ does not divide $a$
:$p \perp a$ denotes that $p$ and $a$ are coprime.
It follows directly that if $p$ and $q$ are primes, then:
:$p \divides q \implies p = q$
:$p \ne q \implies p \perp q$ | Let $p \in \Bbb P, p \nmid a$.
We need to show that $\gcd \set {a, p} = 1$.
Let $\gcd \set {a, p} = d$.
As $d \divides p$, we must have $d = 1$ or $d = p$ by GCD with Prime.
But if $d = p$, then $p \divides a$ by definition of greatest common divisor.
So $d \ne p$ and therefore $d = 1$.
{{qed}} | Let $p, a \in \Z$.
If $p$ is a [[Definition:Prime Number|prime number]] then:
:$p \nmid a \implies p \perp a$
where:
:$p \nmid a$ denotes that $p$ does not [[Definition:Divisor of Integer|divide]] $a$
:$p \perp a$ denotes that $p$ and $a$ are [[Definition:Coprime Integers|coprime]].
It follows directly that if $p$ a... | Let $p \in \Bbb P, p \nmid a$.
We need to show that $\gcd \set {a, p} = 1$.
Let $\gcd \set {a, p} = d$.
As $d \divides p$, we must have $d = 1$ or $d = p$ by [[GCD with Prime]].
But if $d = p$, then $p \divides a$ by definition of [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
So $d \n... | Prime not Divisor implies Coprime/Proof 1 | https://proofwiki.org/wiki/Prime_not_Divisor_implies_Coprime | https://proofwiki.org/wiki/Prime_not_Divisor_implies_Coprime/Proof_1 | [
"Prime Numbers",
"Coprime Integers",
"Prime not Divisor implies Coprime"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Prime Number"
] | [
"GCD with Prime",
"Definition:Greatest Common Divisor/Integers"
] |
proofwiki-940 | Prime not Divisor implies Coprime | Let $p, a \in \Z$.
If $p$ is a prime number then:
:$p \nmid a \implies p \perp a$
where:
:$p \nmid a$ denotes that $p$ does not divide $a$
:$p \perp a$ denotes that $p$ and $a$ are coprime.
It follows directly that if $p$ and $q$ are primes, then:
:$p \divides q \implies p = q$
:$p \ne q \implies p \perp q$ | Let $p$ be a prime number.
Let $a \in \Z$ be such that $p$ is not a divisor of $a$.
{{AimForCont}} $p$ and $a$ are not coprime.
Then:
:$\exists c \in \Z_{>1}: c \divides p, c \divides a$
where $\divides$ denotes divisibility.
But then by definition of prime:
:$c = p$
Thus:
:$p \divides a$
The result follows by Proof by... | Let $p, a \in \Z$.
If $p$ is a [[Definition:Prime Number|prime number]] then:
:$p \nmid a \implies p \perp a$
where:
:$p \nmid a$ denotes that $p$ does not [[Definition:Divisor of Integer|divide]] $a$
:$p \perp a$ denotes that $p$ and $a$ are [[Definition:Coprime Integers|coprime]].
It follows directly that if $p$ a... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $a \in \Z$ be such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $a$.
{{AimForCont}} $p$ and $a$ are not [[Definition:Coprime Integers|coprime]].
Then:
:$\exists c \in \Z_{>1}: c \divides p, c \divides a$
where $\divides$ denotes [[Definitio... | Prime not Divisor implies Coprime/Proof 2 | https://proofwiki.org/wiki/Prime_not_Divisor_implies_Coprime | https://proofwiki.org/wiki/Prime_not_Divisor_implies_Coprime/Proof_2 | [
"Prime Numbers",
"Coprime Integers",
"Prime not Divisor implies Coprime"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Proof by Contradiction"
] |
proofwiki-941 | Composite Number has Two Divisors Less Than It | Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$.
Then:
:$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$
That is, a non-prime number greater than $1$ can be expressed as the product of two positive integers strictly greater than $1$ and less than $n$.
Note that these two numbers are not necessarily distinct. | Since $n \notin \mathbb P$, it has a positive factor $a$ such that $a \ne 1$ and $a \ne n$.
Hence $\exists b \in \Z: n = a b$.
Thus by definition of factor:
:$a \divides n$
where $\divides$ denotes divisibility.
From Divisor Relation on Positive Integers is Partial Ordering:
:$a \le n$
As $a \ne n$, it follows that $a ... | Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$.
Then:
:$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$
That is, a non-[[Definition:Prime Number|prime]] number greater than $1$ can be expressed as the product of two [[Definition:Positive Integer|positive integers]] strictly greater than $1$ and less than $... | Since $n \notin \mathbb P$, it has a [[Definition:Positive Integer|positive]] [[Definition:Divisor of Integer|factor]] $a$ such that $a \ne 1$ and $a \ne n$.
Hence $\exists b \in \Z: n = a b$.
Thus by definition of [[Definition:Divisor of Integer|factor]]:
:$a \divides n$
where $\divides$ denotes [[Definition:Diviso... | Composite Number has Two Divisors Less Than It | https://proofwiki.org/wiki/Composite_Number_has_Two_Divisors_Less_Than_It | https://proofwiki.org/wiki/Composite_Number_has_Two_Divisors_Less_Than_It | [
"Divisors"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Distinct/Plural"
] | [
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Divisor Relation on Positive Integers is Partial Ordering",
"Integer Divisor Results/One Divides all Integers",
"Divisor Relation on Positive Integers i... |
proofwiki-942 | Condition for Divisibility of Powers of Prime | Let $p$ be a prime.
Let $k, l \in \Z_{>0}$.
Then:
:$p^k \divides p^l \iff k \le l$ | === Necessary Condition ===
Let $k \le l$.
Then:
:$l - k \ge 0$
Thus $p^k, p^{l - k} \in \Z$ such that $p^l = p^k p^{l - k}$.
Thus:
:$p^k \divides p^l$
{{qed|lemma}} | Let $p$ be a [[Definition:Prime Number|prime]].
Let $k, l \in \Z_{>0}$.
Then:
:$p^k \divides p^l \iff k \le l$ | === Necessary Condition ===
Let $k \le l$.
Then:
:$l - k \ge 0$
Thus $p^k, p^{l - k} \in \Z$ such that $p^l = p^k p^{l - k}$.
Thus:
:$p^k \divides p^l$
{{qed|lemma}} | Condition for Divisibility of Powers of Prime | https://proofwiki.org/wiki/Condition_for_Divisibility_of_Powers_of_Prime | https://proofwiki.org/wiki/Condition_for_Divisibility_of_Powers_of_Prime | [
"Prime Numbers"
] | [
"Definition:Prime Number"
] | [] |
proofwiki-943 | Exponents of Primes in Prime Decomposition are Less iff Divisor | Let $a, b \in \Z_{>0}$.
Then $a \divides b$ {{iff}}:
:$(1): \quad$ every prime $p_i$ in the prime decomposition of $a$ appears in the prime decomposition of $b$
and:
:$(2): \quad$ the exponent of each $p_i$ in $a$ is less than or equal to its exponent in $b$. | Let $a, b \in \Z_{>0}$.
Let their prime decomposition be:
:$a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$
:$b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$ | Let $a, b \in \Z_{>0}$.
Then $a \divides b$ {{iff}}:
:$(1): \quad$ every [[Definition:Prime Number|prime]] $p_i$ in the [[Definition:Prime Decomposition|prime decomposition]] of $a$ appears in the [[Definition:Prime Decomposition|prime decomposition]] of $b$
and:
:$(2): \quad$ the [[Definition:Exponent|exponent]] of ... | Let $a, b \in \Z_{>0}$.
Let their [[Definition:Prime Decomposition|prime decomposition]] be:
:$a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$
:$b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$ | Exponents of Primes in Prime Decomposition are Less iff Divisor | https://proofwiki.org/wiki/Exponents_of_Primes_in_Prime_Decomposition_are_Less_iff_Divisor | https://proofwiki.org/wiki/Exponents_of_Primes_in_Prime_Decomposition_are_Less_iff_Divisor | [
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Prime Decomposition",
"Definition:Power (Algebra)/Exponent",
"Definition:Power (Algebra)/Exponent"
] | [
"Definition:Prime Decomposition",
"Definition:Prime Decomposition",
"Definition:Prime Decomposition",
"Definition:Prime Decomposition",
"Definition:Prime Decomposition",
"Definition:Prime Decomposition"
] |
proofwiki-944 | Not Coprime means Common Prime Factor | Let $a, b \in \Z$.
If $d \divides a$ and $d \divides b$ such that $d > 1$, then $a$ and $b$ have a common divisor which is prime. | As $d > 1$, it has a prime decomposition.
Thus there exists a prime $p$ such that $p \divides d$.
From Divisor Relation on Positive Integers is Partial Ordering, we have $p \divides d, d \divides a \implies p \divides a$, and similarly for $b$.
The result follows.
{{qed}}
Category:Coprime Integers
Category:Prime Number... | Let $a, b \in \Z$.
If $d \divides a$ and $d \divides b$ such that $d > 1$, then $a$ and $b$ have a [[Definition:Common Divisor of Integers|common divisor]] which is [[Definition:Prime Number|prime]]. | As $d > 1$, it has a [[Definition:Prime Decomposition|prime decomposition]].
Thus there exists a prime $p$ such that $p \divides d$.
From [[Divisor Relation on Positive Integers is Partial Ordering]], we have $p \divides d, d \divides a \implies p \divides a$, and similarly for $b$.
The result follows.
{{qed}}
[[Ca... | Not Coprime means Common Prime Factor | https://proofwiki.org/wiki/Not_Coprime_means_Common_Prime_Factor | https://proofwiki.org/wiki/Not_Coprime_means_Common_Prime_Factor | [
"Coprime Integers",
"Prime Numbers"
] | [
"Definition:Common Divisor/Integers",
"Definition:Prime Number"
] | [
"Definition:Prime Decomposition",
"Divisor Relation on Positive Integers is Partial Ordering",
"Category:Coprime Integers",
"Category:Prime Numbers"
] |
proofwiki-945 | Set of Divisors of Integer | Let $n \in \Z_{>1}$.
Let $n$ be expressed in its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
where $p_1 < p_2 < \dotsb < p_r$ are distinct primes and $k_1, k_2, \ldots, k_r$ are positive integers.
The set of divisors of $n$ is:
:$\set {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i \le k_i, i = 1,... | Each integer in the given set is a divisor of $n$ because:
:$(1): \quad \forall i: k_i - h_i \ge 0$
:$(2): \quad n = \paren {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r} } p_1^{k_1 - h_1} p_2^{k_2 - h_2} \ldots p_r^{k_r - h_r}$
from Exponents of Primes in Prime Decomposition are Less iff Divisor.
By the Fundamental Theorem of ... | Let $n \in \Z_{>1}$.
Let $n$ be expressed in its [[Definition:Prime Decomposition|prime decomposition]]:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
where $p_1 < p_2 < \dotsb < p_r$ are distinct [[Definition:Prime Number|primes]] and $k_1, k_2, \ldots, k_r$ are [[Definition:Positive Integer|positive integers]].
The ... | Each [[Definition:Integer|integer]] in the given set is a [[Definition:Divisor of Integer|divisor]] of $n$ because:
:$(1): \quad \forall i: k_i - h_i \ge 0$
:$(2): \quad n = \paren {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r} } p_1^{k_1 - h_1} p_2^{k_2 - h_2} \ldots p_r^{k_r - h_r}$
from [[Exponents of Primes in Prime Decom... | Set of Divisors of Integer | https://proofwiki.org/wiki/Set_of_Divisors_of_Integer | https://proofwiki.org/wiki/Set_of_Divisors_of_Integer | [
"Number Theory"
] | [
"Definition:Prime Decomposition",
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Set",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Exponents of Primes in Prime Decomposition are Less iff Divisor",
"Fundamental Theorem of Arithmetic",
"Definition:Integer",
"Definition:Integer",
"Definition:Set",
"Definition:Divisor (Algebra)/Integer",
"Divisor Relation on Positive In... |
proofwiki-946 | Sum Less Minimum is Maximum | For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \min \set {a, b} = \max \set {a, b}$ | From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \min \set {a, b} = \max \set {a, b}$ follows by subtracting $\min \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those ... | For all [[Definition:Number|numbers]] $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \min \set {a, b} = \max \set {a, b}$ | From [[Sum of Maximum and Minimum]] we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \min \set {a, b} = \max \set {a, b}$ follows by subtracting $\min \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as [[Definition:Subtraction|subtraction]]... | Sum Less Minimum is Maximum | https://proofwiki.org/wiki/Sum_Less_Minimum_is_Maximum | https://proofwiki.org/wiki/Sum_Less_Minimum_is_Maximum | [
"Algebra"
] | [
"Definition:Number"
] | [
"Sum of Maximum and Minimum",
"Definition:Subtraction",
"Category:Algebra"
] |
proofwiki-947 | Sum Less Maximum is Minimum | For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \max \set {a, b} = \min \set {a, b}$. | From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \max \set {a, b} = \min \set {a, b}$ follows by subtracting $\max \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those ... | For all [[Definition:Number|numbers]] $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \max \set {a, b} = \min \set {a, b}$. | From [[Sum of Maximum and Minimum]] we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \max \set {a, b} = \min \set {a, b}$ follows by subtracting $\max \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as [[Definition:Subtraction|subtraction]] ... | Sum Less Maximum is Minimum | https://proofwiki.org/wiki/Sum_Less_Maximum_is_Minimum | https://proofwiki.org/wiki/Sum_Less_Maximum_is_Minimum | [
"Algebra"
] | [
"Definition:Number"
] | [
"Sum of Maximum and Minimum",
"Definition:Subtraction",
"Category:Algebra"
] |
proofwiki-948 | GCD and LCM from Prime Decomposition | Let $m, n \in \Z$.
Let:
:$m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
:$n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}$
:$p_i \divides m \lor p_i \divides n, 1 \le i \le r$.
That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $m$ or $n$.
Note that if one of the primes $p_i$ does... | The proof of these results can be found in:
:GCD from Prime Decomposition
:LCM from Prime Decomposition
{{Qed}}
Category:Greatest Common Divisor
Category:Lowest Common Multiple
Category:Prime Numbers
iur6iagonkv1hmri44ycxc6lbtsc4ux | Let $m, n \in \Z$.
Let:
:$m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
:$n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}$
:$p_i \divides m \lor p_i \divides n, 1 \le i \le r$.
That is, the [[Definition:Prime Number|primes]] given in these [[Definition:Prime Decomposition|prime decompositions]] may be [[Definition:Divisor of I... | The proof of these results can be found in:
:[[GCD from Prime Decomposition]]
:[[LCM from Prime Decomposition]]
{{Qed}}
[[Category:Greatest Common Divisor]]
[[Category:Lowest Common Multiple]]
[[Category:Prime Numbers]]
iur6iagonkv1hmri44ycxc6lbtsc4ux | GCD and LCM from Prime Decomposition | https://proofwiki.org/wiki/GCD_and_LCM_from_Prime_Decomposition | https://proofwiki.org/wiki/GCD_and_LCM_from_Prime_Decomposition | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Definition:Prime Decomposition"
] | [
"GCD from Prime Decomposition",
"LCM from Prime Decomposition",
"Category:Greatest Common Divisor",
"Category:Lowest Common Multiple",
"Category:Prime Numbers"
] |
proofwiki-949 | GCD and LCM Distribute Over Each Other | Let $a, b, c \in \Z$.
Then:
: $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$
: $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$
That is, greatest common divisor and lowest common multiple are distributive over each other. | === LCM Distributive over GCD ===
Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.
Let $x = \lcm \set {a, \gcd \set {b, c} }$.
Then from GCD and LCM from Prime Decomposition, the exponent of $p_s$ in $x$ is $\max \set {s_a, \min \set {s_b, s... | Let $a, b, c \in \Z$.
Then:
: $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$
: $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$
That is, [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] and [[Definition:Lowest Common M... | === LCM Distributive over GCD ===
Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.
Let $x = \lcm \set {a, \gcd \set {b, c} }$.
Then from [[GCD and LCM from Prime Decomposition]], the exponent of $p_s$ in $x$ is $\max \set {s_a, \min \set ... | GCD and LCM Distribute Over Each Other | https://proofwiki.org/wiki/GCD_and_LCM_Distribute_Over_Each_Other | https://proofwiki.org/wiki/GCD_and_LCM_Distribute_Over_Each_Other | [
"Greatest Common Divisor",
"Lowest Common Multiple",
"Examples of Distributive Operations"
] | [
"Definition:Greatest Common Divisor/Integers",
"Definition:Lowest Common Multiple/Integers",
"Definition:Distributive Operation"
] | [
"GCD and LCM from Prime Decomposition",
"Max and Min Operations are Distributive over Each Other",
"Definition:Lowest Common Multiple/Integers",
"Definition:Distributive Operation",
"Definition:Greatest Common Divisor/Integers",
"GCD and LCM from Prime Decomposition",
"Max and Min Operations are Distrib... |
proofwiki-950 | Divisor Count Function from Prime Decomposition | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_0} n$ be the divisor count function of $n$.
Then:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ | We have:
:$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$
For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all.
By the Fundamental Theorem of Arithmetic and hence the uniquenes... | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_0} n$ be the [[Definition:Divisor Count Function|divisor count function]] of $n$.
Then:
:$\ds \map {\sigma_0} n = ... | We have:
:$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$
For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all.
By the [[Fundamental Theorem of Arithmetic]] and hence the un... | Divisor Count Function from Prime Decomposition/Proof 1 | https://proofwiki.org/wiki/Divisor_Count_Function_from_Prime_Decomposition | https://proofwiki.org/wiki/Divisor_Count_Function_from_Prime_Decomposition/Proof_1 | [
"Divisor Count Function",
"Divisor Count Function from Prime Decomposition"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Definition:Divisor Count Function"
] | [
"Fundamental Theorem of Arithmetic",
"Definition:Prime Decomposition",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-951 | Divisor Count Function from Prime Decomposition | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_0} n$ be the divisor count function of $n$.
Then:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ | From Divisor Count Function of Power of Prime we have:
:$\forall j \in \closedint 1 r: \map {\sigma_0} {p_j^{k_j} } = k_j + 1$
The result follows immediately from Divisor Count Function is Multiplicative.
{{qed}} | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_0} n$ be the [[Definition:Divisor Count Function|divisor count function]] of $n$.
Then:
:$\ds \map {\sigma_0} n = ... | From [[Divisor Count Function of Power of Prime]] we have:
:$\forall j \in \closedint 1 r: \map {\sigma_0} {p_j^{k_j} } = k_j + 1$
The result follows immediately from [[Divisor Count Function is Multiplicative]].
{{qed}} | Divisor Count Function from Prime Decomposition/Proof 2 | https://proofwiki.org/wiki/Divisor_Count_Function_from_Prime_Decomposition | https://proofwiki.org/wiki/Divisor_Count_Function_from_Prime_Decomposition/Proof_2 | [
"Divisor Count Function",
"Divisor Count Function from Prime Decomposition"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Definition:Divisor Count Function"
] | [
"Divisor Count Function of Power of Prime",
"Divisor Count Function is Multiplicative"
] |
proofwiki-952 | N less than M to the N | :$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$ | {{begin-eqn}}
{{eqn | l = n
| r = \underbrace {1 + 1 + \cdots + 1}_{\text {$n$ times} }
| c =
}}
{{eqn | o = <
| r = 1 + m + m^2 + \cdots + m^{n - 1}
| c = as $m > 1$
}}
{{eqn | r = \frac {m^n - 1} {m - 1}
| c = Sum of Geometric Sequence
}}
{{eqn | o = \le
| r = m^n - 1
| c = ... | :$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$ | {{begin-eqn}}
{{eqn | l = n
| r = \underbrace {1 + 1 + \cdots + 1}_{\text {$n$ times} }
| c =
}}
{{eqn | o = <
| r = 1 + m + m^2 + \cdots + m^{n - 1}
| c = as $m > 1$
}}
{{eqn | r = \frac {m^n - 1} {m - 1}
| c = [[Sum of Geometric Sequence]]
}}
{{eqn | o = \le
| r = m^n - 1
| ... | N less than M to the N | https://proofwiki.org/wiki/N_less_than_M_to_the_N | https://proofwiki.org/wiki/N_less_than_M_to_the_N | [
"Algebra"
] | [] | [
"Sum of Geometric Sequence"
] |
proofwiki-953 | Basis Representation Theorem | Let $b \in \Z: b > 1$.
For every $n \in \Z_{> 0}$, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k b^k$
:$(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_b$
:$(3): \quad r_t \ne 0$
This unique sequence is called ... | Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$.
We need to show that $\map {s_b} n = 1$ always.
Now, it is possible that some of the $r_k = 0$ in a particular representation.
So we may exclude these terms, and it will not affect the representation.
So, suppose:
:$n = r_k b^k + r_{k - 1} b^... | Let $b \in \Z: b > 1$.
For every $n \in \Z_{> 0}$, there exists [[Definition:Exactly One|one and only one]] [[Definition:Sequence|sequence]] $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k b^k$
:$(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_b$
:... | Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$.
We need to show that $\map {s_b} n = 1$ always.
Now, it is possible that some of the $r_k = 0$ in a particular [[Definition:Basis Representation|representation]].
So we may exclude these terms, and it will not affect the [[Definition:Basi... | Basis Representation Theorem | https://proofwiki.org/wiki/Basis_Representation_Theorem | https://proofwiki.org/wiki/Basis_Representation_Theorem | [
"Basis Representations",
"Number Bases",
"Named Theorems"
] | [
"Definition:Unique",
"Definition:Sequence",
"Definition:Basis Representation"
] | [
"Definition:Basis Representation",
"Definition:Basis Representation",
"Sum of Geometric Sequence",
"Definition:Basis Representation",
"Definition:Basis Representation",
"Definition:Basis Representation",
"Definition:Basis Representation",
"Definition:Basis Representation",
"N less than M to the N",
... |
proofwiki-954 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | By the Distributive Property:
:$a + a r + a r^2 + \cdots + a r^{n - 1} = a \paren {1 + r + r^2 + \cdots + r^{n - 1} }$
The result follows from Sum of Geometric Sequence.
{{qed}} | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | By the [[Distributive Property]]:
:$a + a r + a r^2 + \cdots + a r^{n - 1} = a \paren {1 + r + r^2 + \cdots + r^{n - 1} }$
The result follows from [[Sum of Geometric Sequence]].
{{qed}} | Sum of Geometric Sequence/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Corollary_1/Proof_1 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [
"Distributive Laws/Arithmetic",
"Sum of Geometric Sequence"
] |
proofwiki-955 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | {{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le j \mathop < n} a r^j
| r = a + \sum_{1 \mathop \le j \mathop < n} a r^j
| c =
}}
{{eqn | r = a + r \sum_{1 \mathop \le j \mathop < n} a r^{j-1}
| c = Distributive Property
}}
{{eqn | r = a + r \sum_{0 \mathop \le j \mathop < n - 1} a r^j
| c = Excha... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | {{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le j \mathop < n} a r^j
| r = a + \sum_{1 \mathop \le j \mathop < n} a r^j
| c =
}}
{{eqn | r = a + r \sum_{1 \mathop \le j \mathop < n} a r^{j-1}
| c = [[Distributive Property]]
}}
{{eqn | r = a + r \sum_{0 \mathop \le j \mathop < n - 1} a r^j
| c = [... | Sum of Geometric Sequence/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Corollary_1/Proof_2 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [
"Distributive Laws/Arithmetic",
"Exchange of Order of Summation"
] |
proofwiki-956 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \dfrac {x^1 - 1} {x - 1}
| r = 1
| c =
}}
{{eqn | r = 2^0
| c =
}}
{{eqn | r = \sum_{j \math... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \dfrac {x^1 - 1} {x - 1}
| r = 1
| c =
}}
{{eqn | r = 2^0
| c ... | Sum of Geometric Sequence/Proof 1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Proof_1 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Principle of Mathematical Induction"
] |
proofwiki-957 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x - 1} S_n
| r = x S_n - S_n
| c =
}}
{{eqn | r = x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j
|... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x - 1} S_n
| r = x S_n - S_n
| c =
}}
{{eqn | r = x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j
... | Sum of Geometric Sequence/Proof 2 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Proof_2 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [] |
proofwiki-958 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Set $a = x$ and $b = 1$:
:$\ds x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = ... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | From [[Difference of Two Powers]]:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Set $a = x$ and $b = 1$:
:$\ds x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x ... | Sum of Geometric Sequence/Proof 3 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Proof_3 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [
"Difference of Two Powers"
] |
proofwiki-959 | Sum of Geometric Sequence | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | === Lemma ===
{{:Sum of Geometric Sequence/Proof 4/Lemma}}{{qed|lemma}}
Then by the lemma:
{{begin-eqn}}
{{eqn | l = \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i
| r = 1 - x^n
}}
{{eqn | ll= \leadsto
| l = \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}
| r = \frac {x^n - 1} {x - 1}
}}
{... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ | === [[Sum of Geometric Sequence/Proof 4/Lemma|Lemma]] ===
{{:Sum of Geometric Sequence/Proof 4/Lemma}}{{qed|lemma}}
Then by the [[Sum of Geometric Sequence/Proof 4/Lemma|lemma]]:
{{begin-eqn}}
{{eqn | l = \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i
| r = 1 - x^n
}}
{{eqn | ll= \leadsto
| l = \sum_{i ... | Sum of Geometric Sequence/Proof 4 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Proof_4 | [
"Sums of Sequences",
"Geometric Sequences",
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field"
] | [
"Sum of Geometric Sequence/Proof 4/Lemma",
"Sum of Geometric Sequence/Proof 4/Lemma"
] |
proofwiki-960 | Congruence of Sum of Digits to Base Less 1 | Let $x \in \Z$, and $b \in \N, b > 1$.
Let $x$ be written in base $b$:
:$x = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Then:
:$\ds \map {s_b} x = \sum_{j \mathop = 0}^m r_j \equiv x \pmod {b - 1}$
where $\map {s_b} x$ is the digit sum of $x$ in base $b$ notation.
That is, the digit sum of any integer $x$ in base $b$... | Let $x \in \Z, x > 0$, and $b \in \N, b > 1$.
Then from the Basis Representation Theorem, $x$ can be expressed uniquely as:
:$\ds x = \sum_{j \mathop = 0}^m r_j b^j, r_0, r_1, \ldots, r_m \in \set {0, 1, \ldots, b - 1}$
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition $\ds \sum_{j \mathop ... | Let $x \in \Z$, and $b \in \N, b > 1$.
Let $x$ be [[Basis Representation Theorem|written in base $b$]]:
:$x = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Then:
:$\ds \map {s_b} x = \sum_{j \mathop = 0}^m r_j \equiv x \pmod {b - 1}$
where $\map {s_b} x$ is the [[Definition:Digit Sum|digit sum]] of $x$ [[Basis Repres... | Let $x \in \Z, x > 0$, and $b \in \N, b > 1$.
Then from the [[Basis Representation Theorem]], $x$ can be expressed uniquely as:
:$\ds x = \sum_{j \mathop = 0}^m r_j b^j, r_0, r_1, \ldots, r_m \in \set {0, 1, \ldots, b - 1}$
Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\m... | Congruence of Sum of Digits to Base Less 1 | https://proofwiki.org/wiki/Congruence_of_Sum_of_Digits_to_Base_Less_1 | https://proofwiki.org/wiki/Congruence_of_Sum_of_Digits_to_Base_Less_1 | [
"Number Theory"
] | [
"Basis Representation Theorem",
"Definition:Digit Sum",
"Basis Representation Theorem",
"Definition:Digit Sum",
"Definition:Integer",
"Basis Representation Theorem",
"Definition:Congruence (Number Theory)/Integers"
] | [
"Basis Representation Theorem",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-961 | Euler Phi Function of Integer | Let $n \in \Z_{>0}$, that is, a (strictly) positive integer.
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function.
Then for any $n \in \Z_{>0}$, we have:
:$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$
where $p_1, p_2, \ldots, p_r$ are the distinc... | If $n = 1$ the result holds by inspection.
Let $n \ge 2$.
We express $n$ in its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 < p_2 < \cdots < p_r$
as it is always possible to do.
By definition, all primes are coprime to each other.
Hence from Euler Phi Function is Multiplicative:
:$\map \phi n =... | Let $n \in \Z_{>0}$, that is, a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]].
Then for any $n \in \Z_{>0}$, we have:
:$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cd... | If $n = 1$ the result holds by inspection.
Let $n \ge 2$.
We express $n$ in its [[Definition:Prime Decomposition|prime decomposition]]:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 < p_2 < \cdots < p_r$
as it is always possible to do.
By definition, all primes are [[Definition:Coprime Integers|coprime]] to eac... | Euler Phi Function of Integer | https://proofwiki.org/wiki/Euler_Phi_Function_of_Integer | https://proofwiki.org/wiki/Euler_Phi_Function_of_Integer | [
"Euler Phi Function"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Euler Phi Function",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Prime Decomposition",
"Definition:Coprime/Integers",
"Euler Phi Function is Multiplicative",
"Euler Phi Function of Prime Power"
] |
proofwiki-962 | Euler Phi Function is Multiplicative | The Euler $\phi$ function is a multiplicative function:
:$m \perp n \implies \map \phi {m n} = \map \phi m \map \phi n$
where $m, n \in \Z_{>0}$. | From Euler Phi Function of Product, we have:
:$\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac {\map \gcd {m, n} } {\map \phi {\map \gcd {m, n} } } }$
From hypothesis, we have:
:$\map \gcd {m, n} = 1$
From Euler Phi Function of 1, we have $\map \phi 1 = 1$, giving the result.
{{qed}} | The [[Definition:Euler Phi Function|Euler $\phi$ function]] is a [[Definition:Multiplicative Arithmetic Function|multiplicative function]]:
:$m \perp n \implies \map \phi {m n} = \map \phi m \map \phi n$
where $m, n \in \Z_{>0}$. | From [[Euler Phi Function of Product]], we have:
:$\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac {\map \gcd {m, n} } {\map \phi {\map \gcd {m, n} } } }$
From hypothesis, we have:
:$\map \gcd {m, n} = 1$
From [[Euler Phi Function of 1]], we have $\map \phi 1 = 1$, giving the result.
{{qed}} | Euler Phi Function is Multiplicative/Proof 2 | https://proofwiki.org/wiki/Euler_Phi_Function_is_Multiplicative | https://proofwiki.org/wiki/Euler_Phi_Function_is_Multiplicative/Proof_2 | [
"Euler Phi Function is Multiplicative",
"Euler Phi Function",
"Multiplicative Functions"
] | [
"Definition:Euler Phi Function",
"Definition:Multiplicative Arithmetic Function"
] | [
"Euler Phi Function of Product",
"Euler Phi Function of 1"
] |
proofwiki-963 | Euler Phi Function of Prime Power | Let $p^n$ be a prime power for some prime number $p > 1$.
Then:
:$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function. | From Euler Phi Function of Prime:
:$\map \phi p = p - 1$
From Prime not Divisor implies Coprime:
:$k \perp p^n \iff p \nmid k$
There are $p^{n - 1}$ integers $k$ such that $1 \le k \le p^n$ which are divisible by $p$:
:$k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$
Therefore:
:$\map \phi {p^n} = p^n - p^{n -... | Let $p^n$ be a [[Definition:Prime Power|prime power]] for some [[Definition:Prime Number|prime number]] $p > 1$.
Then:
:$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]]. | From [[Euler Phi Function of Prime]]:
:$\map \phi p = p - 1$
From [[Prime not Divisor implies Coprime]]:
:$k \perp p^n \iff p \nmid k$
There are $p^{n - 1}$ [[Definition:Integer|integers]] $k$ such that $1 \le k \le p^n$ which are [[Definition:Divisor of Integer|divisible]] by $p$:
:$k \in \set {p, 2 p, 3 p, \ldots... | Euler Phi Function of Prime Power | https://proofwiki.org/wiki/Euler_Phi_Function_of_Prime_Power | https://proofwiki.org/wiki/Euler_Phi_Function_of_Prime_Power | [
"Euler Phi Function",
"Prime Numbers"
] | [
"Definition:Prime Power",
"Definition:Prime Number",
"Definition:Euler Phi Function"
] | [
"Euler Phi Function of Prime",
"Prime not Divisor implies Coprime",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-964 | Sum of Euler Phi Function over Divisors | Let $n \in \Z_{>0}$ be a strictly positive integer.
Then $\ds \sum_{d \mathop \divides n} \map \phi d = n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$
:$\map \phi d$ is the Euler $\phi$ function, the number of integers less than $d$ that are prime to $d$.
That is, the total... | Let us define:
:$S_d = \set {m \in \Z: 1 \le m \le n, \gcd \set {m, n} = d}$.
That is, $S_d$ is all the numbers less than or equal to $n$ whose GCD with $n$ is $d$.
Now from Integers Divided by GCD are Coprime we have:
:$\gcd \set {m, n} = d \iff \dfrac m d, \dfrac n d \in \Z: \dfrac m d \perp \dfrac n d$
So the number... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then $\ds \sum_{d \mathop \divides n} \map \phi d = n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the [[Definition:Sum Over Divisors|sum over all of the divisors]] of $n$
:$\map \phi d$ is the [[Definition:Euler Phi Fu... | Let us define:
:$S_d = \set {m \in \Z: 1 \le m \le n, \gcd \set {m, n} = d}$.
That is, $S_d$ is all the numbers less than or equal to $n$ whose [[Definition:Greatest Common Divisor of Integers|GCD]] with $n$ is $d$.
Now from [[Integers Divided by GCD are Coprime]] we have:
:$\gcd \set {m, n} = d \iff \dfrac m d, \dfr... | Sum of Euler Phi Function over Divisors | https://proofwiki.org/wiki/Sum_of_Euler_Phi_Function_over_Divisors | https://proofwiki.org/wiki/Sum_of_Euler_Phi_Function_over_Divisors | [
"Euler Phi Function"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Sum Over Divisors",
"Definition:Euler Phi Function",
"Definition:Integer",
"Definition:Coprime/Integers",
"Definition:Euler Phi Function",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Greatest Common Divisor/Integers",
"Integers Divided by GCD are Coprime",
"Definition:Positive/Integer",
"Definition:Coprime/Integers",
"Definition:Euler Phi Function",
"Definition:Pairwise Disjoint/Family",
"Sum Over Divisors Equals Sum Over Quotients"
] |
proofwiki-965 | Möbius Function is Multiplicative | The Möbius function $\mu$ is a multiplicative function:
:$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$
where $m, n \in \Z_{>0}$. | First note that we have $\map \mu 1 = 1$, which agrees with Value of Multiplicative Function at One.
Let $m, n \in \Z_{>0}$ such that $m \perp n$.
First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$.
Then either $m$ or $n$ has a factor $p^2$ where $p$ is prime.
Thus it will follow that $m n$ will also have ... | The [[Definition:Möbius Function|Möbius function]] $\mu$ is a [[Definition:Multiplicative Arithmetic Function|multiplicative function]]:
:$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$
where $m, n \in \Z_{>0}$. | First note that we have $\map \mu 1 = 1$, which agrees with [[Value of Multiplicative Function at One]].
Let $m, n \in \Z_{>0}$ such that $m \perp n$.
First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$.
Then either $m$ or $n$ has a factor $p^2$ where $p$ is [[Definition:Prime Number|prime]].
Thus it ... | Möbius Function is Multiplicative | https://proofwiki.org/wiki/Möbius_Function_is_Multiplicative | https://proofwiki.org/wiki/Möbius_Function_is_Multiplicative | [
"Möbius Function",
"Multiplicative Functions"
] | [
"Definition:Möbius Function",
"Definition:Multiplicative Arithmetic Function"
] | [
"Value of Multiplicative Function at One",
"Definition:Prime Number",
"Definition:Prime Number"
] |
proofwiki-966 | Euler Phi Function in terms of Möbius Function | Let $n \in \Z_{>0}$ be a strictly positive integer.
Then:
:$\ds \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$
:$\map \phi n$ is the Euler $\phi$ function, the number of integers less than $n$ that are prime to $n... | Sum of Möbius Function over Divisors says:
{{:Sum of Möbius Function over Divisors}}{{qed|lemma}}
Let $\map 1 k = 1$ be the constant mapping.
Then $\phi$ is defined as:
:$\ds \map \phi n = \sum_{\substack {k \mathop \perp n \\ 1 \mathop \le k \mathop \le n}} \map 1 k$
We have that $\floor {\dfrac 1 {\gcd \set {n, k} } ... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then:
:$\ds \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the [[Definition:Sum Over Divisors|sum over all of the divisors]] of $n$
:$\map \phi n$ is the [... | [[Sum of Möbius Function over Divisors]] says:
{{:Sum of Möbius Function over Divisors}}{{qed|lemma}}
Let $\map 1 k = 1$ be the [[Definition:Constant Mapping|constant mapping]].
Then $\phi$ is defined as:
:$\ds \map \phi n = \sum_{\substack {k \mathop \perp n \\ 1 \mathop \le k \mathop \le n}} \map 1 k$
We have tha... | Euler Phi Function in terms of Möbius Function | https://proofwiki.org/wiki/Euler_Phi_Function_in_terms_of_Möbius_Function | https://proofwiki.org/wiki/Euler_Phi_Function_in_terms_of_Möbius_Function | [
"Euler Phi Function",
"Möbius Function"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Sum Over Divisors",
"Definition:Euler Phi Function",
"Definition:Integer",
"Definition:Coprime/Integers",
"Definition:Möbius Function",
"Definition:Dirichlet Convolution",
"Definition:Identity Mapping"
] | [
"Sum of Möbius Function over Divisors",
"Definition:Constant Mapping",
"Definition:Summation",
"Sum of Möbius Function over Divisors",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-967 | Binomial Coefficient of Prime | Let $p$ be a prime number.
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a binomial coefficient. | Because:
:$\dbinom p k = \dfrac {p \paren {p - 1} \paren {p - 2} \cdots \paren {p - k + 1} } {k!}$
is an integer, we have that:
:$k! \divides p \paren {p - 1} \paren {p - 2} \cdots \paren {p - k + 1}$
But because $k < p$ it follows that:
:$k! \mathrel \perp p$
that is, that:
:$\gcd \set {k!, p} = 1$
So by Euclid's Lemm... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a [[Definition:Binomial Coefficient|binomial coefficient]]. | Because:
:$\dbinom p k = \dfrac {p \paren {p - 1} \paren {p - 2} \cdots \paren {p - k + 1} } {k!}$
is an [[Definition:Integer|integer]], we have that:
:$k! \divides p \paren {p - 1} \paren {p - 2} \cdots \paren {p - k + 1}$
But because $k < p$ it follows that:
:$k! \mathrel \perp p$
that is, that:
:$\gcd \set {k!, p... | Binomial Coefficient of Prime/Proof 1 | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime/Proof_1 | [
"Binomial Coefficient of Prime",
"Prime Numbers",
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient"
] | [
"Definition:Integer",
"Euclid's Lemma"
] |
proofwiki-968 | Binomial Coefficient of Prime | Let $p$ be a prime number.
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a binomial coefficient. | Lucas' Theorem gives:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
So, substituting $p$ for $n$:
:$\dbinom p k \equiv \dbinom {\floor {p / p} } {\floor {k / p} } \dbinom {p \bmod p} {k \bmod p} \pmod p$
But $p \bmod p = 0$ by definition.
Hence, if $0 < k < p$... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a [[Definition:Binomial Coefficient|binomial coefficient]]. | [[Lucas' Theorem]] gives:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
So, substituting $p$ for $n$:
:$\dbinom p k \equiv \dbinom {\floor {p / p} } {\floor {k / p} } \dbinom {p \bmod p} {k \bmod p} \pmod p$
But $p \bmod p = 0$ by [[Definition:Modulo Operat... | Binomial Coefficient of Prime/Proof 2 | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime/Proof_2 | [
"Binomial Coefficient of Prime",
"Prime Numbers",
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient"
] | [
"Lucas' Theorem",
"Definition:Modulo Operation",
"Definition:Binomial Coefficient"
] |
proofwiki-969 | Binomial Coefficient of Prime | Let $p$ be a prime number.
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a binomial coefficient. | By the definition of binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom p k
| r = \frac {p!} {k! \paren {n - k}!}
}}
{{eqn | ll= \leadstoandfrom
| l = p!
| r = k! \paren {p - k}! \binom p k
}}
{{end-eqn}}
Now, $p$ divides the {{LHS}} by Divisors of Factorial.
So $p$ must also divide the {{RHS}}.
B... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
where $\dbinom p k$ is defined as a [[Definition:Binomial Coefficient|binomial coefficient]]. | By the definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
{{begin-eqn}}
{{eqn | l = \binom p k
| r = \frac {p!} {k! \paren {n - k}!}
}}
{{eqn | ll= \leadstoandfrom
| l = p!
| r = k! \paren {p - k}! \binom p k
}}
{{end-eqn}}
Now, $p$ [[Definition:Divisor of Integer|divides]] the ... | Binomial Coefficient of Prime/Proof 3 | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime/Proof_3 | [
"Binomial Coefficient of Prime",
"Prime Numbers",
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient"
] | [
"Definition:Binomial Coefficient",
"Definition:Divisor (Algebra)/Integer",
"Divisors of Factorial",
"Definition:Divisor (Algebra)/Integer",
"Definition:By Hypothesis",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Euclid's Lemma"
] |
proofwiki-970 | Prime Power of Sum Modulo Prime | Let $p$ be a prime number.
Then:
:$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$ | Prime Power of Sum Modulo Prime | https://proofwiki.org/wiki/Prime_Power_of_Sum_Modulo_Prime | https://proofwiki.org/wiki/Prime_Power_of_Sum_Modulo_Prime | [
"Number Theory",
"Combinatorics",
"Prime Numbers",
"Proofs by Induction",
"Modulo Arithmetic"
] | [
"Definition:Prime Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-971 | Binomial Coefficient involving Power of Prime | : $\dbinom {p^n k} {p^n} \equiv k \pmod p$
where $\dbinom {p^n k} {p^n}$ is a binomial coefficient. | From Prime Power of Sum Modulo Prime we have:
:$(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$
We can write this:
:$\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$
By $(1)$ and Congruence of Powers, we therefore have:
:$\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k... | : $\dbinom {p^n k} {p^n} \equiv k \pmod p$
where $\dbinom {p^n k} {p^n}$ is a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Prime Power of Sum Modulo Prime]] we have:
:$(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$
We can write this:
:$\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$
By $(1)$ and [[Congruence of Powers]], we therefore have:
:$\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} +... | Binomial Coefficient involving Power of Prime/Proof 1 | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Power_of_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Power_of_Prime/Proof_1 | [
"Prime Numbers",
"Binomial Coefficients",
"Binomial Coefficient involving Power of Prime"
] | [
"Definition:Binomial Coefficient"
] | [
"Prime Power of Sum Modulo Prime",
"Congruence of Powers",
"Definition:Binomial Coefficient",
"Binomial Theorem"
] |
proofwiki-972 | Binomial Coefficient involving Power of Prime | : $\dbinom {p^n k} {p^n} \equiv k \pmod p$
where $\dbinom {p^n k} {p^n}$ is a binomial coefficient. | Lucas' Theorem states that for $n, k, p \in \Z$ and $p$ be a prime number, such that:
:$n = a_r p^r + \cdots + a_1 p + a_0$
:$k = b_r p^r + \cdots + b_1 p + b_0$
then:
:$\ds \binom n k \equiv \prod_{j \mathop = 0}^r \binom {a_j}{b_j} \pmod p$
Therefore:
{{begin-eqn}}
{{eqn | l = \binom {p^n k} {p^n}
| o = \equiv
... | : $\dbinom {p^n k} {p^n} \equiv k \pmod p$
where $\dbinom {p^n k} {p^n}$ is a [[Definition:Binomial Coefficient|binomial coefficient]]. | [[Lucas' Theorem]] states that for $n, k, p \in \Z$ and $p$ be a [[Definition:Prime Number|prime number]], such that:
:$n = a_r p^r + \cdots + a_1 p + a_0$
:$k = b_r p^r + \cdots + b_1 p + b_0$
then:
:$\ds \binom n k \equiv \prod_{j \mathop = 0}^r \binom {a_j}{b_j} \pmod p$
Therefore:
{{begin-eqn}}
{{eqn | l = \binom... | Binomial Coefficient involving Power of Prime/Proof 2 | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Power_of_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Power_of_Prime/Proof_2 | [
"Prime Numbers",
"Binomial Coefficients",
"Binomial Coefficient involving Power of Prime"
] | [
"Definition:Binomial Coefficient"
] | [
"Lucas' Theorem",
"Definition:Prime Number",
"Lucas' Theorem",
"Binomial Coefficient with One",
"Binomial Coefficient with Zero"
] |
proofwiki-973 | Cassini's Identity | :$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$ | === Basis for the Induction ===
We see that:
: $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \paren {-1}^1$
so the proposition holds for $n = 1$.
We also see that:
: $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \paren {-1}^2$
so the proposition holds for $n = 2$.
=== Induction Hypothesis ===
Suppose the proposition is true for $n = ... | :$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$ | === Basis for the Induction ===
We see that:
: $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \paren {-1}^1$
so the proposition holds for $n = 1$.
We also see that:
: $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \paren {-1}^2$
so the proposition holds for $n = 2$.
=== Induction Hypothesis ===
Suppose the proposition is true for $... | Cassini's Identity/Proof 1 | https://proofwiki.org/wiki/Cassini's_Identity | https://proofwiki.org/wiki/Cassini's_Identity/Proof_1 | [
"Fibonacci Numbers",
"Cassini's Identity"
] | [] | [
"Principle of Mathematical Induction"
] |
proofwiki-974 | Cassini's Identity | :$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$ | First we use this {{Lemma|Cassini's Identity}}:
=== Lemma ===
{{:Cassini's Identity/Lemma}}{{qed|lemma}}
Then the determinant of both sides is taken.
The {{LHS}} follows directly from Determinant of Order 2:
:<nowiki>$\begin{bmatrix}
F_{n + 1} & F_n \\
F_n & F_{n - 1}
\end{bmatrix} = F_{n + 1} F_{n - 1} - F... | :$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$ | First we use this {{Lemma|Cassini's Identity}}:
=== [[Cassini's Identity/Lemma|Lemma]] ===
{{:Cassini's Identity/Lemma}}{{qed|lemma}}
Then the [[Definition:Determinant of Matrix|determinant]] of both sides is taken.
The {{LHS}} follows directly from [[Determinant of Order 2]]:
:<nowiki>$\begin{bmatrix}
F_{n + 1} & ... | Cassini's Identity/Proof 2 | https://proofwiki.org/wiki/Cassini's_Identity | https://proofwiki.org/wiki/Cassini's_Identity/Proof_2 | [
"Fibonacci Numbers",
"Cassini's Identity"
] | [] | [
"Cassini's Identity/Lemma",
"Definition:Determinant/Matrix",
"Determinant/Examples/Order 2",
"Determinant of Matrix Product",
"Principle of Mathematical Induction"
] |
proofwiki-975 | Rational Numbers are Countably Infinite | The set $\Q$ of rational numbers is countably infinite. | The rational numbers are arranged thus:
:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3, \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3... | The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite Set|countably infinite]]. | The [[Definition:Rational Number|rational numbers]] are arranged thus:
:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3, \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfra... | Rational Numbers are Countably Infinite/Proof 1 | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite/Proof_1 | [
"Rational Numbers are Countably Infinite",
"Rational Numbers",
"Countably Infinite Sets"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Countably Infinite/Set"
] | [
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Bijection",
"Definition:Rational Number"
] |
proofwiki-976 | Rational Numbers are Countably Infinite | The set $\Q$ of rational numbers is countably infinite. | Let us define the mapping $\phi: \Q \to \Z \times \N$ as follows:
:$\forall \dfrac p q \in \Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$
where $\dfrac p q$ is in canonical form.
Then $\phi$ is clearly injective.
From Cartesian Product of Countable Sets is Countable, we have that $\Z \times \N$ is countably ... | The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite Set|countably infinite]]. | Let us define the mapping $\phi: \Q \to \Z \times \N$ as follows:
:$\forall \dfrac p q \in \Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$
where $\dfrac p q$ is [[Definition:Canonical Form of Rational Number|in canonical form]].
Then $\phi$ is clearly [[Definition:Injection|injective]].
From [[Cartesian Pro... | Rational Numbers are Countably Infinite/Proof 2 | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite/Proof_2 | [
"Rational Numbers are Countably Infinite",
"Rational Numbers",
"Countably Infinite Sets"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Countably Infinite/Set"
] | [
"Definition:Rational Number/Canonical Form",
"Definition:Injection",
"Cartesian Product of Countable Sets is Countable",
"Definition:Countable Set",
"Domain of Injection to Countable Set is Countable"
] |
proofwiki-977 | Rational Numbers are Countably Infinite | The set $\Q$ of rational numbers is countably infinite. | For each $n \in \N$, define $S_n$ to be the set:
:$S_n := \set {\dfrac m n: m \in \Z}$
By Integers are Countably Infinite, each $S_n$ is countably infinite.
Because each rational number can be written down with a positive denominator, it follows that:
:$\forall q \in \Q: \exists n \in \N: q \in S_n$
which is to say:
:$... | The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite Set|countably infinite]]. | For each $n \in \N$, define $S_n$ to be the [[Definition:Set|set]]:
:$S_n := \set {\dfrac m n: m \in \Z}$
By [[Integers are Countably Infinite]], each $S_n$ is [[Definition:Countably Infinite Set|countably infinite]].
Because each [[Definition:Rational Number|rational number]] can be written down with a [[Definition... | Rational Numbers are Countably Infinite/Proof 3 | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite/Proof_3 | [
"Rational Numbers are Countably Infinite",
"Rational Numbers",
"Countably Infinite Sets"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Countably Infinite/Set"
] | [
"Definition:Set",
"Integers are Countably Infinite",
"Definition:Countably Infinite/Set",
"Definition:Rational Number",
"Definition:Positive/Integer",
"Definition:Fraction/Denominator",
"Countable Union of Countable Sets is Countable",
"Definition:Countable Set",
"Definition:Infinite Set",
"Defini... |
proofwiki-978 | Rational Numbers are Countably Infinite | The set $\Q$ of rational numbers is countably infinite. | Let $Q_\pm = \set {q \in \Q: \pm q > 0}$.
For every $q \in Q_+$, there exists at least one pair $\tuple {m, n} \in \N \times \N$ such that $q = \dfrac m n$.
Therefore, we can find an injection $i: Q_+ \to \N \times \N$.
By Cartesian Product of Natural Numbers with Itself is Countable, $\N \times \N$ is countable.
Hence... | The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite Set|countably infinite]]. | Let $Q_\pm = \set {q \in \Q: \pm q > 0}$.
For every $q \in Q_+$, there exists at least one pair $\tuple {m, n} \in \N \times \N$ such that $q = \dfrac m n$.
Therefore, we can find an [[Definition:Injection|injection]] $i: Q_+ \to \N \times \N$.
By [[Cartesian Product of Natural Numbers with Itself is Countable]], $... | Rational Numbers are Countably Infinite/Proof 4 | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite | https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite/Proof_4 | [
"Rational Numbers are Countably Infinite",
"Rational Numbers",
"Countably Infinite Sets"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Countably Infinite/Set"
] | [
"Definition:Injection",
"Cartesian Product of Natural Numbers with Itself is Countable",
"Definition:Countable Set",
"Definition:Countable Set",
"Domain of Injection to Countable Set is Countable",
"Definition:Bijection",
"Definition:Countable Set",
"Definition:Countable Set"
] |
proofwiki-979 | Real Numbers form Totally Ordered Field | The set of real numbers $\R$ forms a totally ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$. | From Real Numbers form Field, we have that $\struct {\R, +, \times}$ forms a field.
From Ordering Properties of Real Numbers we have that $\struct {\R, +, \times, \le}$ is a totally ordered field.
{{qed}} | The [[Definition:Real Number|set of real numbers]] $\R$ forms a [[Definition:Totally Ordered Field|totally ordered field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$. | From [[Real Numbers form Field]], we have that $\struct {\R, +, \times}$ forms a [[Definition:Field (Abstract Algebra)|field]].
From [[Ordering Properties of Real Numbers]] we have that $\struct {\R, +, \times, \le}$ is a [[Definition:Totally Ordered Field|totally ordered field]].
{{qed}} | Real Numbers form Totally Ordered Field | https://proofwiki.org/wiki/Real_Numbers_form_Totally_Ordered_Field | https://proofwiki.org/wiki/Real_Numbers_form_Totally_Ordered_Field | [
"Examples of Fields",
"Ordered Fields",
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Totally Ordered Field",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers"
] | [
"Real Numbers form Field",
"Definition:Field (Abstract Algebra)",
"Ordering Properties of Real Numbers",
"Definition:Totally Ordered Field"
] |
proofwiki-980 | Rational Numbers form Subfield of Real Numbers | The (ordered) field $\struct {\Q, +, \times, \le}$ of rational numbers forms a subfield of the field of real numbers $\struct {\R, +, \times, \le}$.
That is, the field of real numbers $\struct {\R, +, \times, \le}$ is an extension of the rational numbers $\struct {\Q, +, \times, \le}$. | Recall that Rational Numbers form Totally Ordered Field.
Then from Rational Numbers form Subset of Real Numbers:
:$\Q \subseteq \R$
Hence the result by definition of subfield.
{{qed}} | The [[Definition:Ordered Field|(ordered) field]] $\struct {\Q, +, \times, \le}$ of [[Definition:Rational Number|rational numbers]] forms a [[Definition:Subfield|subfield]] of the [[Definition:Field of Real Numbers|field of real numbers]] $\struct {\R, +, \times, \le}$.
That is, the [[Definition:Field of Real Numbers|... | Recall that [[Rational Numbers form Totally Ordered Field]].
Then from [[Rational Numbers form Subset of Real Numbers]]:
:$\Q \subseteq \R$
Hence the result by definition of [[Definition:Subfield|subfield]].
{{qed}} | Rational Numbers form Subfield of Real Numbers | https://proofwiki.org/wiki/Rational_Numbers_form_Subfield_of_Real_Numbers | https://proofwiki.org/wiki/Rational_Numbers_form_Subfield_of_Real_Numbers | [
"Rational Numbers",
"Real Numbers",
"Examples of Subfields"
] | [
"Definition:Ordered Field",
"Definition:Rational Number",
"Definition:Subfield",
"Definition:Field of Real Numbers",
"Definition:Field of Real Numbers",
"Definition:Extension of Operation",
"Definition:Rational Number"
] | [
"Rational Numbers form Totally Ordered Field",
"Rational Numbers form Subset of Real Numbers",
"Definition:Subfield"
] |
proofwiki-981 | Order is Preserved on Positive Reals by Squaring | :$x < y \iff x^2 < y^2$ | === Necessary Condition ===
Assume $x < y$.
Then:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = x \times x < x \times y
| c = Real Number Ordering is Compatible with Multiplication
}}
{{eqn | l = x < y
| o = \implies
| r = x \times y < y \times y
| c = Real Number Ordering is ... | :$x < y \iff x^2 < y^2$ | === Necessary Condition ===
Assume $x < y$.
Then:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = x \times x < x \times y
| c = [[Real Number Ordering is Compatible with Multiplication]]
}}
{{eqn | l = x < y
| o = \implies
| r = x \times y < y \times y
| c = [[Real Number Orde... | Order is Preserved on Positive Reals by Squaring/Proof 1 | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring/Proof_1 | [
"Real Analysis",
"Inequalities",
"Order is Preserved on Positive Reals by Squaring"
] | [] | [
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Multiplication",
"Transitive Law",
"Real Number Ordering is Compatible with Addition",
"Difference of Two Squares"
] |
proofwiki-982 | Order is Preserved on Positive Reals by Squaring | :$x < y \iff x^2 < y^2$ | From Real Numbers form Totally Ordered Field, the real numbers form an ordered field.
The result follows from Order of Squares in Ordered Field.
{{qed}} | :$x < y \iff x^2 < y^2$ | From [[Real Numbers form Totally Ordered Field]], the [[Definition:Real Number|real numbers]] form an [[Definition:Ordered Field|ordered field]].
The result follows from [[Order of Squares in Ordered Field]].
{{qed}} | Order is Preserved on Positive Reals by Squaring/Proof 2 | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring/Proof_2 | [
"Real Analysis",
"Inequalities",
"Order is Preserved on Positive Reals by Squaring"
] | [] | [
"Real Numbers form Totally Ordered Field",
"Definition:Real Number",
"Definition:Ordered Field",
"Order of Squares in Ordered Field"
] |
proofwiki-983 | Order is Preserved on Positive Reals by Squaring | :$x < y \iff x^2 < y^2$ | From Real Numbers form Totally Ordered Field, the real numbers form a totally ordered field.
By definition, a totally ordered field is a totally ordered ring without proper zero divisors.
The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.
{{qed}} | :$x < y \iff x^2 < y^2$ | From [[Real Numbers form Totally Ordered Field]], the [[Definition:Real Number|real numbers]] form a [[Definition:Totally Ordered Field|totally ordered field]].
By definition, a [[Definition:Totally Ordered Field|totally ordered field]] is a [[Definition:Totally Ordered Ring|totally ordered ring]] without [[Definition... | Order is Preserved on Positive Reals by Squaring/Proof 3 | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring/Proof_3 | [
"Real Analysis",
"Inequalities",
"Order is Preserved on Positive Reals by Squaring"
] | [] | [
"Real Numbers form Totally Ordered Field",
"Definition:Real Number",
"Definition:Totally Ordered Field",
"Definition:Totally Ordered Field",
"Definition:Totally Ordered Ring",
"Definition:Proper Zero Divisor",
"Order of Squares in Totally Ordered Ring without Proper Zero Divisors"
] |
proofwiki-984 | Order is Preserved on Positive Reals by Squaring | :$x < y \iff x^2 < y^2$ | === Necessary Condition ===
Let $x < y$.
Then:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = x \times x < x \times y
| c = Real Number Ordering is Compatible with Multiplication
}}
{{eqn | l = x < y
| o = \implies
| r = x \times y < y \times y
| c = Real Number Ordering is Com... | :$x < y \iff x^2 < y^2$ | === Necessary Condition ===
Let $x < y$.
Then:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = x \times x < x \times y
| c = [[Real Number Ordering is Compatible with Multiplication]]
}}
{{eqn | l = x < y
| o = \implies
| r = x \times y < y \times y
| c = [[Real Number Orderin... | Order is Preserved on Positive Reals by Squaring/Proof 4 | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring | https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring/Proof_4 | [
"Real Analysis",
"Inequalities",
"Order is Preserved on Positive Reals by Squaring"
] | [] | [
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Multiplication",
"Transitive Law",
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Multiplication",
"Transitive Law",
"Definition:Contradiction",
"Proof... |
proofwiki-985 | Real Plus Epsilon | Let $a, b \in \R$, such that:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
where $\R_{>0}$ is the set of strictly positive real numbers.
That is:
:$\epsilon > 0$
Then:
:$a \le b$ | {{AimForCont}} $a > b$.
Then:
:$a - b > 0$
By hypothesis, we have:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
Let $\epsilon = a - b$.
Then:
:$a < b + \paren {a - b} \implies a < a$
The result follows by Proof by Contradiction.
{{qed}} | Let $a, b \in \R$, such that:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
where $\R_{>0}$ is the set of [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]].
That is:
:$\epsilon > 0$
Then:
:$a \le b$ | {{AimForCont}} $a > b$.
Then:
:$a - b > 0$
[[Definition:By Hypothesis|By hypothesis]], we have:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
Let $\epsilon = a - b$.
Then:
:$a < b + \paren {a - b} \implies a < a$
The result follows by [[Proof by Contradiction]].
{{qed}} | Real Plus Epsilon | https://proofwiki.org/wiki/Real_Plus_Epsilon | https://proofwiki.org/wiki/Real_Plus_Epsilon | [
"Real Analysis",
"Inequalities"
] | [
"Definition:Strictly Positive",
"Definition:Real Number"
] | [
"Definition:By Hypothesis",
"Proof by Contradiction"
] |
proofwiki-986 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | Let $n = a^2 - b^2$ for some $a, b \in \Z$.
By Square Modulo 4, both $a$ and $b$ are of the form $4 k$ or $4 k + 1$ for some integer $k$.
There are $4$ cases:
;$a \equiv b \equiv 0 \pmod 4$:
Then:
:$a^2 - b^2 \equiv 0 \pmod 4$
and so $n$ is in the form $4 k$.
;$a \equiv 0 \pmod 4$, $b \equiv 1 \pmod 4$:
Then:
:$a^2 - b... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | Let $n = a^2 - b^2$ for some $a, b \in \Z$.
By [[Square Modulo 4]], both $a$ and $b$ are of the form $4 k$ or $4 k + 1$ for some [[Definition:Integer|integer]] $k$.
There are $4$ cases:
;$a \equiv b \equiv 0 \pmod 4$:
Then:
:$a^2 - b^2 \equiv 0 \pmod 4$
and so $n$ is in the form $4 k$.
;$a \equiv 0 \pmod 4$, $... | Difference of Two Squares cannot equal 2 modulo 4/Proof 1 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares_cannot_equal_2_modulo_4/Proof_1 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"Square Modulo 4",
"Definition:Integer"
] |
proofwiki-987 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | Let $n_0 = c^2 - d^2$ for some $c, d \in \Z$.
Then:
:$n_0 = \paren {c + d} \paren {c - d}$
and so:
:$\paren {c + d} - \paren {c - d} = 2 d$
Therefore $n$ must be expressible as a product of two integers whose difference is even.
Now consider the integer $n \in \Z$ that satisfies $n \equiv 2 \pmod 4$.
$n$ is an even num... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | Let $n_0 = c^2 - d^2$ for some $c, d \in \Z$.
Then:
:$n_0 = \paren {c + d} \paren {c - d}$
and so:
:$\paren {c + d} - \paren {c - d} = 2 d$
Therefore $n$ must be expressible as a [[Definition:Multiplication|product]] of two [[Definition:Integer|integers]] whose [[Definition:Difference (Subtraction)|difference]] is [... | Difference of Two Squares cannot equal 2 modulo 4/Proof 2 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares_cannot_equal_2_modulo_4/Proof_2 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"Definition:Multiplication",
"Definition:Integer",
"Definition:Subtraction/Difference",
"Definition:Even Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Even Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Odd Integer",
... |
proofwiki-988 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | {{begin-eqn}}
{{eqn | l = \paren {x + y} \circ \paren {x + \paren {-y} }
| r = x \circ x + y \circ x + x \circ \paren {-y} + y \circ \paren {-y}
| c = Distributivity of $\circ$ over $+$ in a ring
}}
{{eqn | r = x \circ x + x \circ y + x \circ \paren {-y} + y \circ \paren {-y}
| c = $R$ is a commutativ... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | {{begin-eqn}}
{{eqn | l = \paren {x + y} \circ \paren {x + \paren {-y} }
| r = x \circ x + y \circ x + x \circ \paren {-y} + y \circ \paren {-y}
| c = [[Definition:Distributive Operation|Distributivity]] of $\circ$ over $+$ in a [[Definition:Ring (Abstract Algebra)|ring]]
}}
{{eqn | r = x \circ x + x \circ ... | Difference of Two Squares/Algebraic Proof 1 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares/Algebraic_Proof_1 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"Definition:Distributive Operation",
"Definition:Ring (Abstract Algebra)",
"Definition:Commutative Ring",
"Definition:Ring (Abstract Algebra)"
] |
proofwiki-989 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | This is a special case of Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \cdots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
The result follows by setting $n = 2$.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | This is a special case of [[Difference of Two Powers]]:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \cdots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
The result follows by setting $n = 2$.
{{qed}} | Difference of Two Squares/Algebraic Proof 2 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares/Algebraic_Proof_2 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"Difference of Two Powers"
] |
proofwiki-990 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | {{:Euclid:Proposition/II/5}}
:400px
Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.
Then the rectangle contained by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.
(That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.)
This is proved as follo... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | {{:Euclid:Proposition/II/5}}
:[[File:Euclid-II-5.png|400px]]
Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.
Then the [[Definition:Containment of Rectangle|rectangle contained]] by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.
(That is, let $x = AC, y = CD$. Then ... | Difference of Two Squares/Geometric Proof 1 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares/Geometric_Proof_1 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"File:Euclid-II-5.png",
"Definition:Quadrilateral/Rectangle/Containment",
"Construction of Square on Given Straight Line",
"Construction of Parallel Line",
"Construction of Parallel Line",
"Construction of Parallel Line",
"Complements of Parallelograms are Equal",
"Parallelograms with Equal Base and S... |
proofwiki-991 | Difference of Two Squares | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren ... | :360px
Let $\Box ABCD$ be a square of side length $x$.
Let $\Box DEFG$ be a square of side length $y$ where $y < x$
Let $EF$ be produced to $H$.
The area of $\Box ABCD$ is seen to be equal to the sum of:
:the area of the rectangle $AEHB$
:the area of the rectangle $FGCH$
:the area of the square $DEFG$
From Area of Squa... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard [[Definition:Number|sets of numbers]... | :[[File:Difference-of-Two-Squares.png|360px]]
Let $\Box ABCD$ be a [[Definition:Square (Geometry)|square]] of [[Definition:Side of Polygon|side]] [[Definition:Length of Line|length]] $x$.
Let $\Box DEFG$ be a [[Definition:Square (Geometry)|square]] of [[Definition:Side of Polygon|side]] [[Definition:Length of Line|le... | Difference of Two Squares/Geometric Proof 2 | https://proofwiki.org/wiki/Difference_of_Two_Squares | https://proofwiki.org/wiki/Difference_of_Two_Squares/Geometric_Proof_2 | [
"Difference of Two Squares",
"Commutative Rings",
"Square Function",
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Number"
] | [
"File:Difference-of-Two-Squares.png",
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Linear Measure/Length",
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Linear Measure/Length",
"Definition:Production",
"Definition:Area",
"Definition:Area",
... |
proofwiki-992 | Mediant is Between | Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $r, s \in \R$ be such that:
:$r < s$
and:
:$r = \dfrac a b, s = \dfrac c d$
where $a, b, c, d$ are real numbers such that $b > 0, d > 0$.
Because $b, d > 0$, it follows from Real Number Ordering is Compatible with Multiplication that:
:$b d > 0$
Thus:
{{begin-eqn}}
{{eqn | l = \frac a b
| o = <
| r = \f... | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $r, s \in \R$ be such that:
:$r < s$
and:
:$r = \dfrac a b, s = \dfrac c d$
where $a, b, c, d$ are [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Because $b, d > 0$, it follows from [[Real Number Ordering is Compatible with Multiplication]] that:
:$b d > 0$
Thus:
{{begin-eqn}}
{{eqn | l = \fr... | Mediant is Between | https://proofwiki.org/wiki/Mediant_is_Between | https://proofwiki.org/wiki/Mediant_is_Between | [
"Mediant is Between",
"Real Analysis",
"Mediants"
] | [
"Definition:Real Number"
] | [
"Definition:Real Number",
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Addition",
"Reciprocal of Strictly Positive Real Number is Strictly Positive",
"Real Number Ordering is Compatible with Mul... |
proofwiki-993 | Mediant is Between | Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $p, q, r, s \in \R$ such that $q > 0, s > 0$.
Then from Mediant is Between:
:$\dfrac p q < \dfrac {p + r} {q + s} < \dfrac r s$
In order to present this in the form required by the Squeeze Theorem for Functions, we weaken the ordering:
:$(1): \quad \dfrac p q \le \dfrac {p + r} {q + s} \le \dfrac r s$
Let $f$, $g$ ... | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $p, q, r, s \in \R$ such that $q > 0, s > 0$.
Then from [[Mediant is Between]]:
:$\dfrac p q < \dfrac {p + r} {q + s} < \dfrac r s$
In order to present this in the form required by the [[Squeeze Theorem for Functions]], we [[Definition:Weak Ordering|weaken]] the [[Definition:Ordering|ordering]]:
:$(1): \quad \d... | Mediant is Between/Corollary 2/Proof 1 | https://proofwiki.org/wiki/Mediant_is_Between | https://proofwiki.org/wiki/Mediant_is_Between/Corollary_2/Proof_1 | [
"Mediant is Between",
"Real Analysis",
"Mediants"
] | [
"Definition:Real Number"
] | [
"Mediant is Between",
"Squeeze Theorem/Functions",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Constant Mapping",
"Definition:Real Function",
"Squeeze Theorem/Functions"
] |
proofwiki-994 | Mediant is Between | Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | {{begin-eqn}}
{{eqn | l = \dfrac {a + c} {b + d}
| r = \dfrac a b \times \dfrac {b \paren {a + c} } {a \paren {b + d} }
| c = multiplying by $\dfrac a b \times \dfrac b a$
}}
{{eqn | r = \dfrac a b \times \dfrac {1 + c / a} {1 + d / b}
| c = dividing top and bottom by $a b$
}}
{{eqn | l = \dfrac {a + ... | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | {{begin-eqn}}
{{eqn | l = \dfrac {a + c} {b + d}
| r = \dfrac a b \times \dfrac {b \paren {a + c} } {a \paren {b + d} }
| c = multiplying by $\dfrac a b \times \dfrac b a$
}}
{{eqn | r = \dfrac a b \times \dfrac {1 + c / a} {1 + d / b}
| c = dividing [[Definition:Numerator|top]] and [[Definition:Denom... | Mediant is Between/Corollary 2/Proof 2 | https://proofwiki.org/wiki/Mediant_is_Between | https://proofwiki.org/wiki/Mediant_is_Between/Corollary_2/Proof_2 | [
"Mediant is Between",
"Real Analysis",
"Mediants"
] | [
"Definition:Real Number"
] | [
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator"
] |
proofwiki-995 | Mediant is Between | Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let:
:$t := \dfrac a b = \dfrac c d$
Then:
:$a = t c$
and:
:$c = t d$
Hence:
:$\dfrac {a + b}{c + d} = \dfrac {t \paren {c + d} }{c + d} = t$
{{qed}} | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let:
:$t := \dfrac a b = \dfrac c d$
Then:
:$a = t c$
and:
:$c = t d$
Hence:
:$\dfrac {a + b}{c + d} = \dfrac {t \paren {c + d} }{c + d} = t$
{{qed}} | Mediant is Between/Corollary 2/Proof 3 | https://proofwiki.org/wiki/Mediant_is_Between | https://proofwiki.org/wiki/Mediant_is_Between/Corollary_2/Proof_3 | [
"Mediant is Between",
"Real Analysis",
"Mediants"
] | [
"Definition:Real Number"
] | [] |
proofwiki-996 | Mediant is Between | Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $\epsilon \in \R_{>0}$.
Then:
:$\dfrac a b < \dfrac {c + \epsilon} d$
By Mediant is Between:
:$\dfrac a b < \dfrac {a + c + \epsilon} {b + d} < \dfrac {c + \epsilon} d$
By Inequality of Sequences Preserved in Limit, letting $\epsilon \to 0$:
:$\dfrac a b \le \dfrac {a + c} {b + d} \le \dfrac c d$
Since {{hypothesis... | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$ | Let $\epsilon \in \R_{>0}$.
Then:
:$\dfrac a b < \dfrac {c + \epsilon} d$
By [[Mediant is Between]]:
:$\dfrac a b < \dfrac {a + c + \epsilon} {b + d} < \dfrac {c + \epsilon} d$
By [[Inequality of Sequences Preserved in Limit]], letting $\epsilon \to 0$:
:$\dfrac a b \le \dfrac {a + c} {b + d} \le \dfrac c d$
Since ... | Mediant is Between/Corollary 2/Proof 4 | https://proofwiki.org/wiki/Mediant_is_Between | https://proofwiki.org/wiki/Mediant_is_Between/Corollary_2/Proof_4 | [
"Mediant is Between",
"Real Analysis",
"Mediants"
] | [
"Definition:Real Number"
] | [
"Mediant is Between",
"Inequality Rule for Real Sequences"
] |
proofwiki-997 | Real Numbers Between Epsilons | Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$.
Then $a = b$. | From Real Plus Epsilon:
:$b < a + \epsilon \implies b \le a$
From Real Number Ordering is Compatible with Addition:
:$a - \epsilon < b \implies a < b + \epsilon$
Then from Real Plus Epsilon:
:$a < b + \epsilon \implies a \le b$
The result follows.
{{qed}} | Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$.
Then $a = b$. | From [[Real Plus Epsilon]]:
:$b < a + \epsilon \implies b \le a$
From [[Real Number Ordering is Compatible with Addition]]:
:$a - \epsilon < b \implies a < b + \epsilon$
Then from [[Real Plus Epsilon]]:
:$a < b + \epsilon \implies a \le b$
The result follows.
{{qed}} | Real Numbers Between Epsilons | https://proofwiki.org/wiki/Real_Numbers_Between_Epsilons | https://proofwiki.org/wiki/Real_Numbers_Between_Epsilons | [
"Real Analysis"
] | [] | [
"Real Plus Epsilon",
"Real Number Ordering is Compatible with Addition",
"Real Plus Epsilon"
] |
proofwiki-998 | Existence of Integer Below Given Real Number | Let $x$ be a real number.
Then there exists an integer less than $x$:
:$\forall x \in \R: \exists n \in \Z: n < x$ | {{WLOG}}, assume that $x < 0$.
From the Axiom of Archimedes:
:$\exists m \in \N: m > -x$
By Real Numbers form Totally Ordered Field, we have that $\R$ is an ordered field.
Therefore by property $(3)$ of Properties of Ordered Field, $\Z \owns -m < x$.
{{qed}}
Category:Analysis
pyvgsf2jdou7chjgadituqrihwohxo3 | Let $x$ be a [[Definition:Real Number|real number]].
Then there exists an [[Definition:Integer|integer]] less than $x$:
:$\forall x \in \R: \exists n \in \Z: n < x$ | {{WLOG}}, assume that $x < 0$.
From the [[Axiom of Archimedes]]:
:$\exists m \in \N: m > -x$
By [[Real Numbers form Totally Ordered Field]], we have that $\R$ is an [[Definition:Ordered Field|ordered field]].
Therefore by property $(3)$ of [[Properties of Ordered Field]], $\Z \owns -m < x$.
{{qed}}
[[Category:Analy... | Existence of Integer Below Given Real Number | https://proofwiki.org/wiki/Existence_of_Integer_Below_Given_Real_Number | https://proofwiki.org/wiki/Existence_of_Integer_Below_Given_Real_Number | [
"Analysis"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Axiom of Archimedes",
"Real Numbers form Totally Ordered Field",
"Definition:Ordered Field",
"Properties of Ordered Field",
"Category:Analysis"
] |
proofwiki-999 | Axiom of Archimedes | Let $x$ be a real number.
Then there exists a natural number greater than $x$.
:$\forall x \in \R: \exists n \in \N: n > x$
That is, the set of natural numbers is unbounded above. | Let $x \in \R$.
Let $S$ be the set of all natural numbers less than or equal to $x$:
:$S = \set {a \in \N: a \le x}$
It is possible that $S = \O$.
Suppose $0 \le x$.
Then by definition, $0 \in S$.
But $S = \O$, so this is a contradiction.
From the Trichotomy Law for Real Numbers it follows that $0 > x$.
Thus we have th... | Let $x$ be a [[Definition:Real Number|real number]].
Then there exists a [[Definition:Natural Numbers|natural number]] greater than $x$.
:$\forall x \in \R: \exists n \in \N: n > x$
That is, the [[Definition:Natural Numbers|set of natural numbers]] is [[Definition:Bounded Above Set|unbounded above]]. | Let $x \in \R$.
Let $S$ be the set of all [[Definition:Natural Numbers|natural numbers]] less than or equal to $x$:
:$S = \set {a \in \N: a \le x}$
It is possible that $S = \O$.
Suppose $0 \le x$.
Then by definition, $0 \in S$.
But $S = \O$, so this is a [[Proof by Contradiction|contradiction]].
From the [[Trich... | Axiom of Archimedes | https://proofwiki.org/wiki/Axiom_of_Archimedes | https://proofwiki.org/wiki/Axiom_of_Archimedes | [
"Axiom of Archimedes",
"Real Analysis",
"Number Theory"
] | [
"Definition:Real Number",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Bounded Above Set"
] | [
"Definition:Natural Numbers",
"Proof by Contradiction",
"Trichotomy Law for Real Numbers",
"Definition:Bounded Above Set",
"Continuum Property",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set"
] |
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