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-2000 | Euler Phi Function of Product with Prime | Let $p$ be prime and $n \in \Z: n \ge 1$.
Then:
:$\map \phi {p n} = \begin{cases} \paren {p - 1} \map \phi n & : p \nmid n \\ p \map \phi n & : p \divides n \end{cases}$
where:
:$\map \phi n$ denotes the Euler $\phi$ function of $n$
:$\divides$ denotes divisibility.
Thus for all $n \ge 1$ and for any prime $p$, we have... | First suppose that $p \nmid n$.
Then by Prime not Divisor implies Coprime, $p \perp n$.
So by Euler Phi Function is Multiplicative, $\map \phi {p n} = \map \phi p \map \phi n$.
It follows from Euler Phi Function of Prime that $\map \phi {p n} = \paren {p - 1} \map \phi n$.
Now suppose that $p \divides n$.
Then $n = p^k... | Let $p$ be [[Definition:Prime Number|prime]] and $n \in \Z: n \ge 1$.
Then:
:$\map \phi {p n} = \begin{cases} \paren {p - 1} \map \phi n & : p \nmid n \\ p \map \phi n & : p \divides n \end{cases}$
where:
:$\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $n$
:$\divides$ denotes [[De... | First suppose that $p \nmid n$.
Then by [[Prime not Divisor implies Coprime]], $p \perp n$.
So by [[Euler Phi Function is Multiplicative]], $\map \phi {p n} = \map \phi p \map \phi n$.
It follows from [[Euler Phi Function of Prime]] that $\map \phi {p n} = \paren {p - 1} \map \phi n$.
Now suppose that $p \divides ... | Euler Phi Function of Product with Prime | https://proofwiki.org/wiki/Euler_Phi_Function_of_Product_with_Prime | https://proofwiki.org/wiki/Euler_Phi_Function_of_Product_with_Prime | [
"Euler Phi Function"
] | [
"Definition:Prime Number",
"Definition:Euler Phi Function",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer"
] | [
"Prime not Divisor implies Coprime",
"Euler Phi Function is Multiplicative",
"Euler Phi Function of Prime",
"Euler Phi Function is Multiplicative",
"Euler Phi Function of Prime Power",
"Euler Phi Function is Multiplicative",
"Euler Phi Function of Prime Power",
"Category:Euler Phi Function"
] |
proofwiki-2001 | Euler Phi Function is Even for Argument greater than 2 | Let $n \in \Z: n \ge 1$.
Let $\map \phi n$ be the Euler $\phi$ function of $n$.
Then $\map \phi n$ is even {{iff}} $n > 2$. | We have:
{{begin-eqn}}
{{eqn | l = \map \phi 1
| r = 1
| c = {{EulerPhiLink|1}}
}}
{{eqn | l = \map \phi 2
| r = 1
| c = {{EulerPhiLink|2}}
}}
{{end-eqn}}
Now let $n \ge 3$.
There are two possibilities: | Let $n \in \Z: n \ge 1$.
Let $\map \phi n$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $n$.
Then $\map \phi n$ is [[Definition:Even Integer|even]] {{iff}} $n > 2$. | We have:
{{begin-eqn}}
{{eqn | l = \map \phi 1
| r = 1
| c = {{EulerPhiLink|1}}
}}
{{eqn | l = \map \phi 2
| r = 1
| c = {{EulerPhiLink|2}}
}}
{{end-eqn}}
Now let $n \ge 3$.
There are two possibilities: | Euler Phi Function is Even for Argument greater than 2 | https://proofwiki.org/wiki/Euler_Phi_Function_is_Even_for_Argument_greater_than_2 | https://proofwiki.org/wiki/Euler_Phi_Function_is_Even_for_Argument_greater_than_2 | [
"Euler Phi Function"
] | [
"Definition:Euler Phi Function",
"Definition:Even Integer"
] | [] |
proofwiki-2002 | Cauchy-Goursat Theorem | Let $f: D \to \C$ be a holomorphic function, where $D \subseteq \C$ is a simply connected domain.
Let $C$ be a closed contour in $D$.
Then:
:$\ds \oint_C \map f z \rd z = 0$ | {{CircularStructure|See below. No need to resolve this.}}
This proof assumes that the derivative $f' : D \to \C$ is continuous.
This assumption is always true, as can be seen from Holomorphic Function is Continuously Differentiable.
However, the proof of that theorem depends on Cauchy's Integral Formula, which again de... | Let $f: D \to \C$ be a [[Definition:Holomorphic Function|holomorphic function]], where $D \subseteq \C$ is a [[Definition:Simply Connected Domain|simply connected domain]].
Let $C$ be a [[Definition:Closed Contour (Complex Plane)|closed contour]] in $D$.
Then:
:$\ds \oint_C \map f z \rd z = 0$ | {{CircularStructure|See below. No need to resolve this.}}
This proof assumes that the [[Definition:Derivative of Complex Function|derivative]] $f' : D \to \C$ is [[Definition:Continuous Complex Function|continuous]].
This assumption is always true, as can be seen from [[Holomorphic Function is Continuously Differentia... | Cauchy-Goursat Theorem/Proof 2 | https://proofwiki.org/wiki/Cauchy-Goursat_Theorem | https://proofwiki.org/wiki/Cauchy-Goursat_Theorem/Proof_2 | [
"Cauchy-Goursat Theorem",
"Complex Contour Integrals"
] | [
"Definition:Holomorphic Function",
"Definition:Connected Domain (Complex Analysis)/Simply Connected Domain",
"Definition:Contour/Closed/Complex Plane"
] | [
"Definition:Derivative/Complex Function",
"Definition:Continuous Complex Function",
"Holomorphic Function is Continuously Differentiable",
"Cauchy's Integral Formula",
"Cauchy-Goursat Theorem",
"Definition:Circular Proof",
"Definition:Continuously Differentiable",
"Cauchy-Riemann Equations",
"Defini... |
proofwiki-2003 | Divisor Sum of Integer | Let $n$ be an integer such that $n \ge 2$.
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Then:
:$\ds... | We have that the Divisor Sum Function is Multiplicative.
From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:
:$\map f n = \map f {p_1^{k_1} } \map f {p_2^{k_2} } \ldots \map f {p_r^{k_r} }$
From Divisor Sum of Power of Prime, we have:
:$\ds \map {\sigma_1} {p_i^{k_i} } = \frac {p... | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let $\map {\sigma_1} n$ be the [[Definition:Divisor Sum Function|divisor sum]] of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive [[Definition:Divisor of Integer|divisors]] of $n$.
Let the [[Definition:Prime Decomposition|prime decomp... | We have that the [[Divisor Sum Function is Multiplicative]].
From [[Value of Multiplicative Function is Product of Values of Prime Power Factors]], we have:
:$\map f n = \map f {p_1^{k_1} } \map f {p_2^{k_2} } \ldots \map f {p_r^{k_r} }$
From [[Divisor Sum of Power of Prime]], we have:
:$\ds \map {\sigma_1} {p_i^{k_i... | Divisor Sum of Integer | https://proofwiki.org/wiki/Divisor_Sum_of_Integer | https://proofwiki.org/wiki/Divisor_Sum_of_Integer | [
"Divisor Sum of Integer",
"Divisor Sum Function"
] | [
"Definition:Integer",
"Definition:Divisor Sum Function",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Decomposition"
] | [
"Divisor Sum Function is Multiplicative",
"Value of Multiplicative Function is Product of Values of Prime Power Factors",
"Divisor Sum of Power of Prime"
] |
proofwiki-2004 | Product of Divisor Sum and Euler Phi Functions | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
Let $\map \phi n$ be the Euler phi function of $n$.
Then:
:$\ds \map {\sigma_1} n \map \phi n = n^2 \prod_{1 \mathop \le i \mathop \le... | {{improve|Better to start with $\ds \map \phi n {{=}} n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ and then a lot of the gnarliness goes away}}
From Euler Phi Function of Integer:
:$\ds \map \phi n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}$
From Divisor Sum of Integer:
:$\ds \map ... | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
Let $\map {\sigma_1} n$ be the [[Definition:Divisor Sum Function|divisor sum function]] of $n$.
Let $\map \phi n$ be the [[Definition... | {{improve|Better to start with $\ds \map \phi n {{=}} n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ and then a lot of the gnarliness goes away}}
From [[Euler Phi Function of Integer]]:
:$\ds \map \phi n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}$
From [[Divisor Sum of Integer]]:
:... | Product of Divisor Sum and Euler Phi Functions | https://proofwiki.org/wiki/Product_of_Divisor_Sum_and_Euler_Phi_Functions | https://proofwiki.org/wiki/Product_of_Divisor_Sum_and_Euler_Phi_Functions | [
"Divisor Sum Function",
"Euler Phi Function"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Definition:Divisor Sum Function",
"Definition:Euler Phi Function"
] | [
"Euler Phi Function of Integer",
"Divisor Sum of Integer",
"Category:Divisor Sum Function",
"Category:Euler Phi Function"
] |
proofwiki-2005 | Primitive Root is Generator of Reduced Residue System | Let $a$ be a primitive root of $n$.
Then:
:$\set {a, a^2, a^3, \ldots, a^{\map \phi n} }$
where $\map \phi n$ is the Euler phi function of $n$, is the reduced residue system of $n$.
Thus the first $\map \phi n$ powers of $a$ "generates" the reduced residue system of $n$.
We say that $a$ is a '''generator''' of the redu... | Let $R = \set {a, a^2, a^3, \ldots, a^{\map \phi n} }$.
Each element of $R$ is coprime to $n$ as $a \perp n$.
Suppose $a^r \equiv a^s \pmod n$ where $1 \le r \le s \le \map \phi n$.
Then $a^{r - s} \equiv 1 \pmod n$.
From the definition of primitive root, the multiplicative order of $a$ modulo $n$ is $\map \phi n$.
So ... | Let $a$ be a [[Definition:Primitive Root (Number Theory)|primitive root of $n$]].
Then:
:$\set {a, a^2, a^3, \ldots, a^{\map \phi n} }$
where $\map \phi n$ is the [[Definition:Euler Phi Function|Euler phi function]] of $n$, is the [[Definition:Reduced Residue System|reduced residue system]] of $n$.
Thus the first $\... | Let $R = \set {a, a^2, a^3, \ldots, a^{\map \phi n} }$.
Each element of $R$ is [[Definition:Coprime Integers|coprime]] to $n$ as $a \perp n$.
Suppose $a^r \equiv a^s \pmod n$ where $1 \le r \le s \le \map \phi n$.
Then $a^{r - s} \equiv 1 \pmod n$.
From the definition of [[Definition:Primitive Root (Number Theory)|... | Primitive Root is Generator of Reduced Residue System | https://proofwiki.org/wiki/Primitive_Root_is_Generator_of_Reduced_Residue_System | https://proofwiki.org/wiki/Primitive_Root_is_Generator_of_Reduced_Residue_System | [
"Number Theory"
] | [
"Definition:Primitive Root (Number Theory)",
"Definition:Euler Phi Function",
"Definition:Reduced Residue System",
"Definition:Reduced Residue System",
"Definition:Reduced Residue System"
] | [
"Definition:Coprime/Integers",
"Definition:Primitive Root (Number Theory)",
"Definition:Multiplicative Order of Integer",
"Integer to Power of Multiple of Order",
"Definition:Congruence (Number Theory)",
"Definition:Congruence (Number Theory)",
"Definition:Reduced Residue System",
"Category:Number The... |
proofwiki-2006 | Order Modulo n of Power of Integer | Let $a$ have multiplicative order $c$ modulo $n$.
Then for any $k \ge 1$, $a^k$ has multiplicative order $\dfrac c {\gcd \set {c, k}}$ modulo $n$. | Let $a$ have multiplicative order $c$ modulo $n$.
Consider $a^k$ and let $d = \gcd \set {c, k}$.
Let $c = d c'$ and $k = d k'$ where $\gcd \set {c', k'} = 1$ from Integers Divided by GCD are Coprime.
We want to show that the multiplicative order $a^k$ modulo $n$ is $c'$.
Let the order $a^k$ modulo $n$ be $r$.
Then:
{{b... | Let $a$ have [[Definition:Multiplicative Order of Integer|multiplicative order $c$ modulo $n$]].
Then for any $k \ge 1$, $a^k$ has [[Definition:Multiplicative Order of Integer|multiplicative order]] $\dfrac c {\gcd \set {c, k}}$ modulo $n$. | Let $a$ have [[Definition:Multiplicative Order of Integer|multiplicative order $c$ modulo $n$]].
Consider $a^k$ and let $d = \gcd \set {c, k}$.
Let $c = d c'$ and $k = d k'$ where $\gcd \set {c', k'} = 1$ from [[Integers Divided by GCD are Coprime]].
We want to show that the [[Definition:Multiplicative Order of Inte... | Order Modulo n of Power of Integer | https://proofwiki.org/wiki/Order_Modulo_n_of_Power_of_Integer | https://proofwiki.org/wiki/Order_Modulo_n_of_Power_of_Integer | [
"Number Theory"
] | [
"Definition:Multiplicative Order of Integer",
"Definition:Multiplicative Order of Integer"
] | [
"Definition:Multiplicative Order of Integer",
"Integers Divided by GCD are Coprime",
"Definition:Multiplicative Order of Integer",
"Integer to Power of Multiple of Order",
"Euclid's Lemma",
"Divisor Relation on Positive Integers is Partial Ordering",
"Category:Number Theory"
] |
proofwiki-2007 | Divisor Count of Square-Free Integer is Power of 2 | Let $n$ be a square-free integer.
Let $\sigma_0: \Z \to \Z$ be the divisor count Function.
Then $\map {\sigma_0} n = 2^r$ for some $r \ge 1$.
The converse is not true in general.
That is, if $\map {\sigma_0} n = 2^r$ for some $r \ge 1$, it is not necessarily the case that $n$ is square-free. | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
Then from Divisor Count Function from Prime Decomposition we have that:
:$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$
Let $n$ be square-free.
Then by definition:
:$\forall ... | Let $n$ be a [[Definition:Square-Free Integer|square-free integer]].
Let $\sigma_0: \Z \to \Z$ be the [[Definition:Divisor Count Function|divisor count Function]].
Then $\map {\sigma_0} n = 2^r$ for some $r \ge 1$.
The [[Definition:Converse Statement|converse]] is not true in general.
That is, if $\map {\sigma_0} ... | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
Then from [[Divisor Count Function from Prime Decomposition]] we have that:
:$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {... | Divisor Count of Square-Free Integer is Power of 2 | https://proofwiki.org/wiki/Divisor_Count_of_Square-Free_Integer_is_Power_of_2 | https://proofwiki.org/wiki/Divisor_Count_of_Square-Free_Integer_is_Power_of_2 | [
"Divisor Count Function",
"Square-Free Integers"
] | [
"Definition:Square-Free Integer",
"Definition:Divisor Count Function",
"Definition:Converse Statement",
"Definition:Square-Free Integer"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Divisor Count Function from Prime Decomposition",
"Definition:Square-Free Integer",
"Definition:Square-Free Integer",
"Definition:Converse Statement",
"Proof by Counterexample",
"Definition:Prime Number",
"Definition:Square-Free Integer",
"C... |
proofwiki-2008 | Divisor Count Function is Odd Iff Argument is Square | Let $\sigma_0: \Z \to \Z$ denote the divisor count function.
Then $\map {\sigma_0} n$ is odd {{iff}} $n$ is square. | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
Then from Divisor Count Function from Prime Decomposition we have that:
:$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$ | Let $\sigma_0: \Z \to \Z$ denote the [[Definition:Divisor Count Function|divisor count function]].
Then $\map {\sigma_0} n$ is [[Definition:Odd Integer|odd]] {{iff}} $n$ is [[Definition:Square Number|square]]. | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
Then from [[Divisor Count Function from Prime Decomposition]] we have that:
:$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {... | Divisor Count Function is Odd Iff Argument is Square | https://proofwiki.org/wiki/Divisor_Count_Function_is_Odd_Iff_Argument_is_Square | https://proofwiki.org/wiki/Divisor_Count_Function_is_Odd_Iff_Argument_is_Square | [
"Divisor Count Function",
"Square Numbers"
] | [
"Definition:Divisor Count Function",
"Definition:Odd Integer",
"Definition:Square Number"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Divisor Count Function from Prime Decomposition"
] |
proofwiki-2009 | Divisor Sum is Odd iff Argument is Square or Twice Square | Let $\sigma_1: \Z \to \Z$ be the divisor sum function.
Then $\map {\sigma_1} n$ is odd {{iff}} $n$ is either square or twice a square. | Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
| r = \prod_{i \mathop = 1}^r \frac {p_i^{k_i + 1} - 1} {p_i - 1}
| c = Divisor Sum of Integer
}}
{{eqn | r = \prod_{i \mathop = 1}^... | Let $\sigma_1: \Z \to \Z$ be the [[Definition:Divisor Sum Function|divisor sum function]].
Then $\map {\sigma_1} n$ is [[Definition:Odd Integer|odd]] {{iff}} $n$ is either [[Definition:Square Number|square]] or twice a [[Definition:Square Number|square]]. | Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
| r = \prod_{i \mathop = 1}^r \frac {p_i^{k_i + 1} - 1} {p_i - 1}
| c =... | Divisor Sum is Odd iff Argument is Square or Twice Square | https://proofwiki.org/wiki/Divisor_Sum_is_Odd_iff_Argument_is_Square_or_Twice_Square | https://proofwiki.org/wiki/Divisor_Sum_is_Odd_iff_Argument_is_Square_or_Twice_Square | [
"Square Numbers",
"Divisor Sum Function",
"Odd Integers"
] | [
"Definition:Divisor Sum Function",
"Definition:Odd Integer",
"Definition:Square Number",
"Definition:Square Number"
] | [
"Definition:Integer",
"Definition:Prime Decomposition",
"Divisor Sum of Integer",
"Sum of Geometric Sequence",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Odd Integer",
... |
proofwiki-2010 | Number of Quadratic Residues of Prime | Let $p$ be an odd prime.
Then $p$ has $\dfrac {p - 1} 2$ quadratic residues and $\dfrac {p - 1} 2$ quadratic non-residues.
The quadratic residues are congruent modulo $p$ to the integers $1^2, 2^2, \ldots, \paren {\dfrac {p - 1} 2}$. | The quadratic residues of $p$ are the integers which result from the evaluation of the squares:
:$1^2, 2^2, \ldots, \paren {p - 1}^2$ modulo $p$
But:
:$r^2 = \paren {-r}^2$
and so these $p - 1$ integers fall into congruent pairs modulo $p$, namely:
{{begin-eqn}}
{{eqn | l = 1^2
| o = \equiv
| r = \paren {p ... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Then $p$ has $\dfrac {p - 1} 2$ [[Definition:Quadratic Residue|quadratic residues]] and $\dfrac {p - 1} 2$ [[Definition:Quadratic Non-Residue|quadratic non-residues]].
The [[Definition:Quadratic Residue|quadratic residues]] are [[Definition:Congruence Modulo Integer|c... | The [[Definition:Quadratic Residue|quadratic residues]] of $p$ are the [[Definition:Integer|integers]] which result from the evaluation of the [[Definition:Square Number|squares]]:
:$1^2, 2^2, \ldots, \paren {p - 1}^2$ modulo $p$
But:
:$r^2 = \paren {-r}^2$
and so these $p - 1$ [[Definition:Integer|integers]] fall int... | Number of Quadratic Residues of Prime | https://proofwiki.org/wiki/Number_of_Quadratic_Residues_of_Prime | https://proofwiki.org/wiki/Number_of_Quadratic_Residues_of_Prime | [
"Prime Numbers",
"Quadratic Residues"
] | [
"Definition:Odd Prime",
"Definition:Quadratic Residue",
"Definition:Quadratic Residue/Non-Residue",
"Definition:Quadratic Residue",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Integer"
] | [
"Definition:Quadratic Residue",
"Definition:Integer",
"Definition:Square Number",
"Definition:Integer",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Odd Integer",
"Definition:Quadratic Residue",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Integer",
"Definition:In... |
proofwiki-2011 | Cancellability of Congruences | Let $a, b, c, n \in \Z$ be integers.
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Let $c a \equiv c b \pmod n$.
Then we have that $c a - c b = k n$ for some $k \in \Z$ by definition of congruence.
Now $d = \gcd \set {c, n}$, so from Integers Divided by GCD are Coprime we have:
:$\exists r, s \in Z: r \perp s: c = d r, n = d s$
So we substitute for $c$ and $n$ in $c a - c b = k n$:
:$d r a - d r b = ... | Let $a, b, c, n \in \Z$ be [[Definition:Integer|integers]].
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Let $c a \equiv c b \pmod n$.
Then we have that $c a - c b = k n$ for some $k \in \Z$ by definition of [[Definition:Congruence (Number Theory)|congruence]].
Now $d = \gcd \set {c, n}$, so from [[Integers Divided by GCD are Coprime]] we have:
:$\exists r, s \in Z: r \perp s: c = d r, n = d s$
So we substitute for $c$... | Cancellability of Congruences | https://proofwiki.org/wiki/Cancellability_of_Congruences | https://proofwiki.org/wiki/Cancellability_of_Congruences | [
"Modulo Multiplication",
"Modulo Arithmetic",
"Cancellability of Congruences"
] | [
"Definition:Integer"
] | [
"Definition:Congruence (Number Theory)",
"Integers Divided by GCD are Coprime",
"Euclid's Lemma",
"Category:Modulo Multiplication",
"Category:Modulo Arithmetic",
"Category:Cancellability of Congruences"
] |
proofwiki-2012 | Cancellability of Congruences | Let $a, b, c, n \in \Z$ be integers.
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Recall that $c \perp n$ means $\gcd \set {c, n} = 1$.
The result follows directly from Cancellability of Congruences.
{{qed}} | Let $a, b, c, n \in \Z$ be [[Definition:Integer|integers]].
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Recall that $c \perp n$ means $\gcd \set {c, n} = 1$.
The result follows directly from [[Cancellability of Congruences]].
{{qed}} | Cancellability of Congruences/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Cancellability_of_Congruences | https://proofwiki.org/wiki/Cancellability_of_Congruences/Corollary_1/Proof_1 | [
"Modulo Multiplication",
"Modulo Arithmetic",
"Cancellability of Congruences"
] | [
"Definition:Integer"
] | [
"Cancellability of Congruences"
] |
proofwiki-2013 | Cancellability of Congruences | Let $a, b, c, n \in \Z$ be integers.
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | We are given that $c$ and $n$ are coprime.
So:
{{begin-eqn}}
{{eqn | q = \exists x, y \in \Z
| l = c x + n y
| r = 1
| c = Integer Combination of Coprime Integers
}}
{{eqn | ll= \leadsto
| l = 1 - cx
| r = y n
}}
{{eqn | n = 1
| l = c x
| o = \equiv
| r = 1
| rr= \p... | Let $a, b, c, n \in \Z$ be [[Definition:Integer|integers]].
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | We are given that $c$ and $n$ are [[Definition:Coprime Integers|coprime]].
So:
{{begin-eqn}}
{{eqn | q = \exists x, y \in \Z
| l = c x + n y
| r = 1
| c = [[Integer Combination of Coprime Integers]]
}}
{{eqn | ll= \leadsto
| l = 1 - cx
| r = y n
}}
{{eqn | n = 1
| l = c x
| o ... | Cancellability of Congruences/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Cancellability_of_Congruences | https://proofwiki.org/wiki/Cancellability_of_Congruences/Corollary_1/Proof_2 | [
"Modulo Multiplication",
"Modulo Arithmetic",
"Cancellability of Congruences"
] | [
"Definition:Integer"
] | [
"Definition:Coprime/Integers",
"Integer Combination of Coprime Integers",
"Equal Numbers are Congruent",
"Modulo Multiplication is Well-Defined",
"Equal Numbers are Congruent",
"Modulo Multiplication is Well-Defined",
"Modulo Multiplication is Well-Defined"
] |
proofwiki-2014 | Cancellability of Congruences | Let $a, b, c, n \in \Z$ be integers.
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Let:
:$c a \equiv c b \pmod n$
Then by definition of congruence:
:$n \divides k \paren {x - y}$
We have that:
:$c \perp n$
Thus by Euclid's Lemma:
:$n \divides \paren {x - y}$
So by definition of congruence:
:$a \equiv b \pmod n$
{{qed}} | Let $a, b, c, n \in \Z$ be [[Definition:Integer|integers]].
Then:
:$c a \equiv c b \pmod n \iff a \equiv b \pmod {n / d}$
where $d = \gcd \set {c, n}$. | Let:
:$c a \equiv c b \pmod n$
Then by definition of [[Definition:Congruence Modulo Integer|congruence]]:
:$n \divides k \paren {x - y}$
We have that:
:$c \perp n$
Thus by [[Euclid's Lemma]]:
:$n \divides \paren {x - y}$
So by definition of [[Definition:Congruence Modulo Integer|congruence]]:
:$a \equiv b \pmod n$
... | Cancellability of Congruences/Corollary 1/Proof 3 | https://proofwiki.org/wiki/Cancellability_of_Congruences | https://proofwiki.org/wiki/Cancellability_of_Congruences/Corollary_1/Proof_3 | [
"Modulo Multiplication",
"Modulo Arithmetic",
"Cancellability of Congruences"
] | [
"Definition:Integer"
] | [
"Definition:Congruence (Number Theory)/Integers",
"Euclid's Lemma",
"Definition:Congruence (Number Theory)/Integers"
] |
proofwiki-2015 | Solution of Linear Diophantine Equation | The linear Diophantine equation:
:$a x + b y = c$
has solutions {{iff}}:
:$\gcd \set {a, b} \divides c$
where $\divides$ denotes divisibility.
If this condition holds with $\gcd \set {a, b} > 1$ then division by $\gcd \set {a, b}$ reduces the equation to:
:$a' x + b' y = c'$
where $\gcd \set {a', b'} = 1$.
If $x_0, y_... | We assume that both $a$ and $b$ are non-zero, otherwise the solution is trivial.
The first part of the problem is a direct restatement of Set of Integer Combinations equals Set of Multiples of GCD:
The set of all integer combinations of $a$ and $b$ is precisely the set of integer multiples of the GCD of $a$ and $b$:
:$... | The [[Definition:Linear Diophantine Equation|linear Diophantine equation]]:
:$a x + b y = c$
has solutions {{iff}}:
:$\gcd \set {a, b} \divides c$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
If this condition holds with $\gcd \set {a, b} > 1$ then division by $\gcd \set {a, b}$ reduces t... | We assume that both $a$ and $b$ are non-[[Definition:Zero (Number)|zero]], otherwise the solution is trivial.
The first part of the problem is a direct restatement of [[Set of Integer Combinations equals Set of Multiples of GCD]]:
The [[Definition:Set|set]] of all [[Definition:Integer Combination|integer combination... | Solution of Linear Diophantine Equation | https://proofwiki.org/wiki/Solution_of_Linear_Diophantine_Equation | https://proofwiki.org/wiki/Solution_of_Linear_Diophantine_Equation | [
"Linear Diophantine Equations",
"Greatest Common Divisor"
] | [
"Definition:Diophantine Equation/Linear Diophantine Equation",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Zero (Number)",
"Set of Integer Combinations equals Set of Multiples of GCD",
"Definition:Set",
"Definition:Integer Combination",
"Definition:Set of Integer Multiples",
"Definition:Greatest Common Divisor/Integers",
"Euclid's Lemma"
] |
proofwiki-2016 | GCD with Prime | Let $p$ be a prime number.
Then:
$\quad \forall n \in \Z: \gcd \set {n, p} = \begin {cases}
p & : p \divides n \\
1 & : p \nmid n
\end {cases}$ | The only divisors of $p$ are $1$ and $p$ itself by definition.
$\gcd \set {n, p} = p$ {{iff}} $p$ divides $n$.
Hence the result.
{{qed}}
Category:Greatest Common Divisor
Category:Prime Numbers
cp93gzbif27auasul35s4r3x96ixnrr | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
$\quad \forall n \in \Z: \gcd \set {n, p} = \begin {cases}
p & : p \divides n \\
1 & : p \nmid n
\end {cases}$ | The only [[Definition:Divisor of Integer|divisors]] of $p$ are $1$ and $p$ itself by definition.
$\gcd \set {n, p} = p$ {{iff}} $p$ [[Definition:Divisor of Integer|divides]] $n$.
Hence the result.
{{qed}}
[[Category:Greatest Common Divisor]]
[[Category:Prime Numbers]]
cp93gzbif27auasul35s4r3x96ixnrr | GCD with Prime | https://proofwiki.org/wiki/GCD_with_Prime | https://proofwiki.org/wiki/GCD_with_Prime | [
"Greatest Common Divisor",
"Prime Numbers"
] | [
"Definition:Prime Number"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Category:Greatest Common Divisor",
"Category:Prime Numbers"
] |
proofwiki-2017 | Prime Divides Power | Let $p$ be a prime number.
Let $a, n \in \Z$ be integers.
Then $p$ divides $a^n$ {{iff}} $p^n$ divides $a^n$. | === Sufficient Condition ===
Let $p^n \divides a^n$.
We have $p \divides p^n$ as $p^n = p \paren {p^{n - 1} }$.
From the fact that Divisor Relation is Transitive, we have that $p \divides a^n$.
{{qed|lemma}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $a, n \in \Z$ be [[Definition:Integer|integers]].
Then $p$ [[Definition:Divisor of Integer|divides]] $a^n$ {{iff}} $p^n$ [[Definition:Divisor of Integer|divides]] $a^n$. | === Sufficient Condition ===
Let $p^n \divides a^n$.
We have $p \divides p^n$ as $p^n = p \paren {p^{n - 1} }$.
From the fact that [[Divisor Relation is Transitive]], we have that $p \divides a^n$.
{{qed|lemma}} | Prime Divides Power | https://proofwiki.org/wiki/Prime_Divides_Power | https://proofwiki.org/wiki/Prime_Divides_Power | [
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Divisor Relation is Transitive"
] |
proofwiki-2018 | Existence of Laurent Series | Let $z_0 \in \C$ be a complex number.
Let $R \in \R_{>0}$ be a real number.
Let $\map {B'} {z_0, R}$ be the open punctured disk at $z_0$ of radius $R$.
Let $f: \map {B'} {z_0, R} \to \C$ be holomorphic.
Then there exists a sequence $\sequence {a_n}_{n \mathop \in \Z}$ such that:
:$\ds \map f z = \sum_{n \mathop = -\inf... | {{Proofread}}
{{MissingLinks}}
Choose circles $C_1$ and $C_3$ centered at $z_0$
Connect $C_2$ and $C_3$ by path $C_2$ such that $z$ is inside $C_1 + C_2 - C_3 - C_2$ as shown below:
:390px
This curve and its interior are contained in $B'$, so by Cauchy's Integral Formula:
{{begin-eqn}}
{{eqn | l = \map f z
| r = ... | Let $z_0 \in \C$ be a [[Definition:Complex Number|complex number]].
Let $R \in \R_{>0}$ be a [[Definition:Real Number|real number]].
Let $\map {B'} {z_0, R}$ be the [[Definition:Open Punctured Complex Disk|open punctured disk]] at $z_0$ of radius $R$.
Let $f: \map {B'} {z_0, R} \to \C$ be [[Definition:Holomorphic Fu... | {{Proofread}}
{{MissingLinks}}
Choose [[Definition:Circle|circles]] $C_1$ and $C_3$ [[Definition:Center of Circle|centered]] at $z_0$
Connect $C_2$ and $C_3$ by [[Definition:Path (Complex Analysis)|path]] $C_2$ such that $z$ is inside $C_1 + C_2 - C_3 - C_2$ as shown below:
:[[File:Path figure.png|390px]]
This curv... | Existence of Laurent Series | https://proofwiki.org/wiki/Existence_of_Laurent_Series | https://proofwiki.org/wiki/Existence_of_Laurent_Series | [
"Laurent Series",
"Complex Analysis"
] | [
"Definition:Complex Number",
"Definition:Real Number",
"Definition:Open Punctured Complex Disk",
"Definition:Holomorphic Function",
"Definition:Sequence"
] | [
"Definition:Circle",
"Definition:Circle/Center",
"Definition:Path (Complex Analysis)",
"File:Path figure.png",
"Cauchy's Integral Formula",
"Sum of Infinite Geometric Sequence",
"Sum of Infinite Geometric Sequence",
"Definition:Parenthesis",
"Category:Laurent Series",
"Category:Complex Analysis"
] |
proofwiki-2019 | Congruence by Divisor of Modulus | Let $z \in \R$ be a real number.
Let $a, b \in \R$ such that $a$ is congruent modulo $z$ to $b$, that is:
:$a \equiv b \pmod z$
Let $m \in \R$ such that $z$ is an integer multiple of $m$:
:$\exists k \in \Z: z = k m$
Then:
: $a \equiv b \pmod m$ | We are given that $\exists k \in \Z: z = k m$.
Thus:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists k' \in \Z
| l = a
| r = b + k' z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | ll= \leadsto
... | Let $z \in \R$ be a [[Definition:Real Number|real number]].
Let $a, b \in \R$ such that $a$ is [[Definition:Congruence (Number Theory)|congruent modulo $z$]] to $b$, that is:
:$a \equiv b \pmod z$
Let $m \in \R$ such that $z$ is an [[Definition:Integer Multiple|integer multiple]] of $m$:
:$\exists k \in \Z: z = k m$
... | We are given that $\exists k \in \Z: z = k m$.
Thus:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists k' \in \Z
| l = a
| r = b + k' z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | ll= \leadsto... | Congruence by Divisor of Modulus | https://proofwiki.org/wiki/Congruence_by_Divisor_of_Modulus | https://proofwiki.org/wiki/Congruence_by_Divisor_of_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Congruence (Number Theory)",
"Definition:Integral Multiple/Real Numbers",
"Definition:Integer"
] | [
"Definition:Integer"
] |
proofwiki-2020 | Congruence of Sum with Constant | Let $a, b, z \in \R$.
Let $a$ be congruent to $b$ modulo $z$:
: $a \equiv b \pmod z$
Then:
:$\forall c \in \R: a + c \equiv b + c \pmod z$ | Follows directly from the definition of Modulo Addition:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c = given
}}
{{eqn | l = c
| o = \equiv
| r = c
| rr= \pmod z
| c = Congruence Modulo Real Number is Equivalence Relation
}}
{{eqn | ll= \leadsto
... | Let $a, b, z \in \R$.
Let $a$ be [[Definition:Congruence (Number Theory)|congruent to $b$ modulo $z$]]:
: $a \equiv b \pmod z$
Then:
:$\forall c \in \R: a + c \equiv b + c \pmod z$ | Follows directly from the definition of [[Definition:Modulo Addition|Modulo Addition]]:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c = given
}}
{{eqn | l = c
| o = \equiv
| r = c
| rr= \pmod z
| c = [[Congruence Modulo Real Number is Equivalence Rel... | Congruence of Sum with Constant | https://proofwiki.org/wiki/Congruence_of_Sum_with_Constant | https://proofwiki.org/wiki/Congruence_of_Sum_with_Constant | [
"Modulo Addition"
] | [
"Definition:Congruence (Number Theory)"
] | [
"Definition:Modulo Addition",
"Congruence Modulo Real Number is Equivalence Relation",
"Category:Modulo Addition"
] |
proofwiki-2021 | Congruence of Quotient | Let $a, b \in \Z$ and $n \in \N$.
Let $a$ be congruent to $b$ modulo $n$, that is $a \equiv b \pmod n$.
Let $d \in \Z: d > 0$ such that $d$ is a common divisor of $a, b$ and $n$.
Then:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d}$ | By definition of congruence modulo $n$:
:$a = b + k n$
Dividing through by $d$ (which you can do because $d$ divides all three terms), we get:
:$\dfrac a d = \dfrac b d + k \dfrac n d$
from which the result follows directly.
{{qed}} | Let $a, b \in \Z$ and $n \in \N$.
Let $a$ be [[Definition:Congruence (Number Theory)|congruent to $b$ modulo $n$]], that is $a \equiv b \pmod n$.
Let $d \in \Z: d > 0$ such that $d$ is a [[Definition:Common Divisor of Integers|common divisor]] of $a, b$ and $n$.
Then:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d}$ | By definition of [[Definition:Congruence (Number Theory)|congruence modulo $n$]]:
:$a = b + k n$
Dividing through by $d$ (which you can do because $d$ [[Definition:Divisor of Integer|divides]] all three terms), we get:
:$\dfrac a d = \dfrac b d + k \dfrac n d$
from which the result follows directly.
{{qed}} | Congruence of Quotient/Proof 1 | https://proofwiki.org/wiki/Congruence_of_Quotient | https://proofwiki.org/wiki/Congruence_of_Quotient/Proof_1 | [
"Congruence of Quotient",
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Common Divisor/Integers"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-2022 | Congruence of Quotient | Let $a, b \in \Z$ and $n \in \N$.
Let $a$ be congruent to $b$ modulo $n$, that is $a \equiv b \pmod n$.
Let $d \in \Z: d > 0$ such that $d$ is a common divisor of $a, b$ and $n$.
Then:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d}$ | From Congruence by Product of Moduli, we have that:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d} \iff d \dfrac a d \equiv d \dfrac b d \pmod {n / d} \iff a \equiv b \pmod n$
{{qed}} | Let $a, b \in \Z$ and $n \in \N$.
Let $a$ be [[Definition:Congruence (Number Theory)|congruent to $b$ modulo $n$]], that is $a \equiv b \pmod n$.
Let $d \in \Z: d > 0$ such that $d$ is a [[Definition:Common Divisor of Integers|common divisor]] of $a, b$ and $n$.
Then:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d}$ | From [[Congruence by Product of Moduli]], we have that:
:$\dfrac a d \equiv \dfrac b d \pmod {n / d} \iff d \dfrac a d \equiv d \dfrac b d \pmod {n / d} \iff a \equiv b \pmod n$
{{qed}} | Congruence of Quotient/Proof 2 | https://proofwiki.org/wiki/Congruence_of_Quotient | https://proofwiki.org/wiki/Congruence_of_Quotient/Proof_2 | [
"Congruence of Quotient",
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Common Divisor/Integers"
] | [
"Congruence by Product of Moduli"
] |
proofwiki-2023 | Divisibility by 11 | Let $N \in \N$ be expressed as:
:$N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$
Then $N$ is divisible by $11$ {{iff}} $\ds \sum_{r \mathop = 0}^n \paren {-1}^r a_r$ is divisible by $11$.
That is, a divisibility test for $11$ is achieved by alternately adding and subtracting the digits and taking the result modulo $... | As:
:$10 \equiv -1 \pmod {11}$
we have:
:$10^r \equiv \paren {-1}^r \pmod {11}$
from Congruence of Powers.
Thus:
:$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {11}$
from the definition of Modulo Addition.
The result follows.
{{qed}} | Let $N \in \N$ be expressed as:
:$N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$
Then $N$ is [[Definition:Divisor of Integer|divisible]] by $11$ {{iff}} $\ds \sum_{r \mathop = 0}^n \paren {-1}^r a_r$ is [[Definition:Divisor of Integer|divisible]] by $11$.
That is, a divisibility test for $11$ is achieved by alte... | As:
:$10 \equiv -1 \pmod {11}$
we have:
:$10^r \equiv \paren {-1}^r \pmod {11}$
from [[Congruence of Powers]].
Thus:
:$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {11}$
from the definition of [[Definition:Modulo Addition|Modulo Addition]].
The result follows.
{{qed}} | Divisibility by 11 | https://proofwiki.org/wiki/Divisibility_by_11 | https://proofwiki.org/wiki/Divisibility_by_11 | [
"Divisibility Tests",
"11"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Congruence of Powers",
"Definition:Modulo Addition"
] |
proofwiki-2024 | Solutions of Polynomial Congruence | Let $\map P x$ be an integral polynomial.
Let $a \equiv b \pmod n$.
Then:
:$\map P a \equiv \map P b \pmod n$
In particular, $a$ is a solution to the polynomial congruence $\map P x \equiv 0 \pmod n$ {{iff}} $b$ is also. | Let $\map P x = c_m x^m + c_{m - 1} x^{m - 1} + \cdots + c_1 x + c_0$.
Since $a \equiv b \pmod n$, from Congruence of Product and Congruence of Powers, we have:
:$\forall r \in \Z_{>0}: c_r a^r \equiv c_r b^r \pmod n$
From Modulo Addition we then have:
{{begin-eqn}}
{{eqn | l = \map P a
| r = c_m a^m + c_{m - 1} ... | Let $\map P x$ be an [[Definition:Integral Polynomial|integral polynomial]].
Let $a \equiv b \pmod n$.
Then:
:$\map P a \equiv \map P b \pmod n$
In particular, $a$ is a [[Definition:Solution to Polynomial Congruence|solution]] to the [[Definition:Polynomial Congruence|polynomial congruence]] $\map P x \equiv 0 \pm... | Let $\map P x = c_m x^m + c_{m - 1} x^{m - 1} + \cdots + c_1 x + c_0$.
Since $a \equiv b \pmod n$, from [[Congruence of Product]] and [[Congruence of Powers]], we have:
:$\forall r \in \Z_{>0}: c_r a^r \equiv c_r b^r \pmod n$
From [[Definition:Modulo Addition|Modulo Addition]] we then have:
{{begin-eqn}}
{{eqn | l ... | Solutions of Polynomial Congruence | https://proofwiki.org/wiki/Solutions_of_Polynomial_Congruence | https://proofwiki.org/wiki/Solutions_of_Polynomial_Congruence | [
"Polynomial Congruences"
] | [
"Definition:Integral Polynomial",
"Definition:Polynomial Congruence/Solution",
"Definition:Polynomial Congruence"
] | [
"Congruence of Product",
"Congruence of Powers",
"Definition:Modulo Addition",
"Definition:Polynomial Congruence/Solution",
"Definition:Polynomial Congruence",
"Category:Polynomial Congruences"
] |
proofwiki-2025 | Chinese Remainder Theorem | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be pairwise coprime positive integers.
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the system of linear congruences:
{{begin-eqn}}
{{eqn | l = x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
}}
{{eqn | l = x
| o = \equiv
| r = ... | === Existence ===
Let $N_k = \dfrac N {n_k}$ for $k = 1, 2, \ldots, r$.
From Integer Coprime to all Factors is Coprime to Whole:
:$\forall k \in \set {1, 2, \ldots, r}: \gcd \set {N_k, n_k} = \gcd \set {n_1 n_2 \cdots n_{k - 1} n_{k + 1} \cdots n_r, n_k} = 1$
From Solution of Linear Congruence, the linear congruence:
:... | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]] [[Definition:Positive Integer|positive integers]].
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the [[Definition:Simultaneous Linear Congruences|system of linear congruences]]:
{{begin-eq... | === Existence ===
Let $N_k = \dfrac N {n_k}$ for $k = 1, 2, \ldots, r$.
From [[Integer Coprime to all Factors is Coprime to Whole]]:
:$\forall k \in \set {1, 2, \ldots, r}: \gcd \set {N_k, n_k} = \gcd \set {n_1 n_2 \cdots n_{k - 1} n_{k + 1} \cdots n_r, n_k} = 1$
From [[Solution of Linear Congruence]], the [[Defin... | Chinese Remainder Theorem | https://proofwiki.org/wiki/Chinese_Remainder_Theorem | https://proofwiki.org/wiki/Chinese_Remainder_Theorem | [
"Chinese Remainder Theorem",
"Modulo Arithmetic",
"Commutative Algebra",
"Number Theory",
"Named Theorems"
] | [
"Definition:Pairwise Coprime/Integers",
"Definition:Positive/Integer",
"Definition:Simultaneous Congruences/Linear",
"Definition:Simultaneous Congruences/Solution",
"Definition:Unique",
"Definition:Congruence (Number Theory)"
] | [
"Integer Coprime to all Factors is Coprime to Whole",
"Solution of Linear Congruence",
"Definition:Linear Congruence",
"Euclid's Lemma",
"Divisor Divides Multiple",
"Congruent to Zero iff Modulo is Divisor"
] |
proofwiki-2026 | Chinese Remainder Theorem | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be pairwise coprime positive integers.
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the system of linear congruences:
{{begin-eqn}}
{{eqn | l = x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
}}
{{eqn | l = x
| o = \equiv
| r = ... | The mapping $\phi$ is indeed a ring homomorphism, because each canonical projection $\phi_i: A \to A / I_i$ is a ring homomorphism.
The kernel of $\phi$ is given by:
:$\ds \ker \phi = \set {x \in A: \forall i, 1 \le i \le n : x \in I_i} = \bigcap_{1 \mathop \le i \mathop \le n} I_i =: I$
It remains then to be proved th... | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]] [[Definition:Positive Integer|positive integers]].
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the [[Definition:Simultaneous Linear Congruences|system of linear congruences]]:
{{begin-eq... | The mapping $\phi$ is indeed a [[Definition:Ring Homomorphism|ring homomorphism]], because each [[Definition:Quotient Epimorphism|canonical projection]] $\phi_i: A \to A / I_i$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is given by:
:$\ds \k... | Chinese Remainder Theorem (Commutative Algebra)/Proof 1 | https://proofwiki.org/wiki/Chinese_Remainder_Theorem | https://proofwiki.org/wiki/Chinese_Remainder_Theorem_(Commutative_Algebra)/Proof_1 | [
"Chinese Remainder Theorem",
"Modulo Arithmetic",
"Commutative Algebra",
"Number Theory",
"Named Theorems"
] | [
"Definition:Pairwise Coprime/Integers",
"Definition:Positive/Integer",
"Definition:Simultaneous Congruences/Linear",
"Definition:Simultaneous Congruences/Solution",
"Definition:Unique",
"Definition:Congruence (Number Theory)"
] | [
"Definition:Ring Homomorphism",
"Definition:Quotient Epimorphism",
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism",
"Definition:Surjection",
"Definition:Ideal of Ring",
"Definition:Coprime Ideals",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Cartesian Product/Coo... |
proofwiki-2027 | Chinese Remainder Theorem | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be pairwise coprime positive integers.
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the system of linear congruences:
{{begin-eqn}}
{{eqn | l = x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
}}
{{eqn | l = x
| o = \equiv
| r = ... | Consider $\pi$ only as a homomorphism of groups.
Then Chinese Remainder Theorem (Groups) is applicable as Subgroup of Abelian Group is Normal.
It remains to demonstrate that the condition $I_i + I_j = R$ for all $i \ne j$ assumed here is equivalent to:
:$\ds \forall k \le n - 1: I_{k + 1} + \bigcap_{i \mathop = 1}^k I... | Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]] [[Definition:Positive Integer|positive integers]].
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the [[Definition:Simultaneous Linear Congruences|system of linear congruences]]:
{{begin-eq... | Consider $\pi$ only as a [[Definition:Group Homomorphism|homomorphism of groups]].
Then [[Chinese Remainder Theorem (Groups)]] is applicable as [[Subgroup of Abelian Group is Normal]].
It remains to demonstrate that the condition $I_i + I_j = R$ for all $i \ne j$ assumed here is equivalent to:
:$\ds \forall k \le n ... | Chinese Remainder Theorem (Commutative Algebra)/Proof 2 | https://proofwiki.org/wiki/Chinese_Remainder_Theorem | https://proofwiki.org/wiki/Chinese_Remainder_Theorem_(Commutative_Algebra)/Proof_2 | [
"Chinese Remainder Theorem",
"Modulo Arithmetic",
"Commutative Algebra",
"Number Theory",
"Named Theorems"
] | [
"Definition:Pairwise Coprime/Integers",
"Definition:Positive/Integer",
"Definition:Simultaneous Congruences/Linear",
"Definition:Simultaneous Congruences/Solution",
"Definition:Unique",
"Definition:Congruence (Number Theory)"
] | [
"Definition:Group Homomorphism",
"Chinese Remainder Theorem (Groups)",
"Subgroup of Abelian Group is Normal",
"Intersection of Ideals of Ring contains Product"
] |
proofwiki-2028 | Solution to Simultaneous Linear Congruences | Let:
{{begin-eqn}}
{{eqn | l = a_1 x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
| c =
}}
{{eqn | l = a_2 x
| o = \equiv
| r = b_2
| rr= \pmod {n_2}
| c =
}}
{{eqn | o = \vdots
| c =
}}
{{eqn | l = a_r x
| o = \equiv
| r = b_r
| rr= \pmod {n_r}
... | We take the case where $r = 2$.
Suppose $x \in \Z$ satisfies both:
{{begin-eqn}}
{{eqn | l = a_1 x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
| c =
}}
{{eqn | l = a_2 x
| o = \equiv
| r = b_2
| rr= \pmod {n_2}
| c =
}}
{{end-eqn}}
That is, $\exists r, s \in \Z$ such that:... | Let:
{{begin-eqn}}
{{eqn | l = a_1 x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
| c =
}}
{{eqn | l = a_2 x
| o = \equiv
| r = b_2
| rr= \pmod {n_2}
| c =
}}
{{eqn | o = \vdots
| c =
}}
{{eqn | l = a_r x
| o = \equiv
| r = b_r
| rr= \pmod {n_r}
... | We take the case where $r = 2$.
Suppose $x \in \Z$ satisfies both:
{{begin-eqn}}
{{eqn | l = a_1 x
| o = \equiv
| r = b_1
| rr= \pmod {n_1}
| c =
}}
{{eqn | l = a_2 x
| o = \equiv
| r = b_2
| rr= \pmod {n_2}
| c =
}}
{{end-eqn}}
That is, $\exists r, s \in \Z$ such tha... | Solution to Simultaneous Linear Congruences | https://proofwiki.org/wiki/Solution_to_Simultaneous_Linear_Congruences | https://proofwiki.org/wiki/Solution_to_Simultaneous_Linear_Congruences | [
"Modulo Arithmetic"
] | [
"Definition:Simultaneous Congruences",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Integer Combination",
"Definition:Divisor (Algebra)/Integer",
"Definition:Conditional/Necessary Condition",
"Definition:Conditional/Sufficient Condition",
"Bézout's Identity",
"Intersection of Congruence Classes",
"Principle of Mathematical Induction",
"Category:Modulo Arithmetic"
] |
proofwiki-2029 | Second Supplement to Law of Quadratic Reciprocity | $\quad \paren {\dfrac 2 p} = \paren {-1}^{\paren {p^2 - 1} / 8} = \begin {cases}
+1 & : p \equiv \pm 1 \pmod 8 \\
-1 & : p \equiv \pm 3 \pmod 8
\end {cases}$ | Consider the numbers in the set $S = \set {2 \times 1, 2 \times 2, 2 \times 3, \dots, 2 \times \dfrac {p - 1} 2} = \set {2, 4, 6, \dots, p - 1}$.
From Gauss's Lemma:
:$\paren {\dfrac 2 p} = \paren {-1}^n$
where $n$ is the number of elements in $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$.
As... | $\quad \paren {\dfrac 2 p} = \paren {-1}^{\paren {p^2 - 1} / 8} = \begin {cases}
+1 & : p \equiv \pm 1 \pmod 8 \\
-1 & : p \equiv \pm 3 \pmod 8
\end {cases}$ | Consider the numbers in the set $S = \set {2 \times 1, 2 \times 2, 2 \times 3, \dots, 2 \times \dfrac {p - 1} 2} = \set {2, 4, 6, \dots, p - 1}$.
From [[Gauss's Lemma (Number Theory)|Gauss's Lemma]]:
:$\paren {\dfrac 2 p} = \paren {-1}^n$
where $n$ is the number of elements in $S$ whose [[Definition:Least Positive Res... | Second Supplement to Law of Quadratic Reciprocity | https://proofwiki.org/wiki/Second_Supplement_to_Law_of_Quadratic_Reciprocity | https://proofwiki.org/wiki/Second_Supplement_to_Law_of_Quadratic_Reciprocity | [
"Legendre Symbol",
"Named Theorems",
"Law of Quadratic Reciprocity"
] | [] | [
"Gauss's Lemma (Number Theory)",
"Definition:Set of Residue Classes/Least Positive",
"Definition:Element",
"Definition:Set of Residue Classes/Least Positive",
"Definition:Element",
"Definition:Floor Function",
"Definition:Element",
"Definition:Residue Class",
"Definition:Odd Prime",
"Gauss's Lemma... |
proofwiki-2030 | First Supplement to Law of Quadratic Reciprocity | $\quad \paren {\dfrac {-1} p} = \paren {-1}^{\paren {p - 1} / 2} = \begin {cases}
+1 & : p \equiv 1 \pmod 4 \\
-1 & : p \equiv 3 \pmod 4
\end {cases}$ | From Euler's Criterion for Quadratic Residue, and the definition of the Legendre symbol, we have that:
:$\paren {\dfrac a p} \equiv a^{\paren {p - 1} / 2} \pmod p$
The result follows by putting $a = -1$.
{{qed}} | $\quad \paren {\dfrac {-1} p} = \paren {-1}^{\paren {p - 1} / 2} = \begin {cases}
+1 & : p \equiv 1 \pmod 4 \\
-1 & : p \equiv 3 \pmod 4
\end {cases}$ | From [[Euler's Criterion for Quadratic Residue]], and the definition of the [[Definition:Legendre Symbol|Legendre symbol]], we have that:
:$\paren {\dfrac a p} \equiv a^{\paren {p - 1} / 2} \pmod p$
The result follows by putting $a = -1$.
{{qed}} | First Supplement to Law of Quadratic Reciprocity | https://proofwiki.org/wiki/First_Supplement_to_Law_of_Quadratic_Reciprocity | https://proofwiki.org/wiki/First_Supplement_to_Law_of_Quadratic_Reciprocity | [
"Legendre Symbol",
"Named Theorems",
"Law of Quadratic Reciprocity"
] | [] | [
"Euler's Criterion/Quadratic Residue",
"Definition:Legendre Symbol"
] |
proofwiki-2031 | Product of Sums | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | We have that both series are absolutely convergent.
Thus by Manipulation of Absolutely Convergent Series, it is permitted to expand the product as:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{i \mathop \in A} \paren {a_i \sum_{j \mathop \in B} b_j}$
But since $a_i$ is a constan... | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be [[Definition:Absolutely Convergent Series|absolutely convergent]] [[Definition:Sequence|sequences]].
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | We have that both series are [[Definition:Absolutely Convergent Series|absolutely convergent]].
Thus by [[Manipulation of Absolutely Convergent Series]], it is permitted to expand the product as:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{i \mathop \in A} \paren {a_i \sum_{j... | Product of Sums | https://proofwiki.org/wiki/Product_of_Sums | https://proofwiki.org/wiki/Product_of_Sums | [
"Number Theory",
"Analysis"
] | [
"Definition:Absolutely Convergent Series",
"Definition:Sequence"
] | [
"Definition:Absolutely Convergent Series",
"Manipulation of Absolutely Convergent Series",
"Definition:Constant",
"Definition:Summation",
"Category:Number Theory",
"Category:Analysis"
] |
proofwiki-2032 | Product of Sums | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
| r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
| c =
}}
{{eqn | o =
| ro= +
| r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
| c =... | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be [[Definition:Absolutely Convergent Series|absolutely convergent]] [[Definition:Sequence|sequences]].
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
| r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
| c =
}}
{{eqn | o =
| ro= +
| r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
| c ... | Product of Sums of Four Squares/Proof 1 | https://proofwiki.org/wiki/Product_of_Sums | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Proof_1 | [
"Number Theory",
"Analysis"
] | [
"Definition:Absolutely Convergent Series",
"Definition:Sequence"
] | [] |
proofwiki-2033 | Product of Sums | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two quaternions.
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
| r = \size {\mathbf m \mathbf n}
| c = Quaternion Modulus of Product of Qua... | Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be [[Definition:Absolutely Convergent Series|absolutely convergent]] [[Definition:Sequence|sequences]].
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$ | Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two [[Definition:Quaternion|quaternions]].
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
| r = \size {\mathbf m \mathbf n}
| c = [[Quatern... | Product of Sums of Four Squares/Proof 2 | https://proofwiki.org/wiki/Product_of_Sums | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Proof_2 | [
"Number Theory",
"Analysis"
] | [
"Definition:Absolutely Convergent Series",
"Definition:Sequence"
] | [
"Definition:Quaternion",
"Quaternion Modulus of Product of Quaternions"
] |
proofwiki-2034 | Green's Theorem | Let $\Gamma$ be a positively oriented piecewise smooth Jordan curve in $\R^2$.
Let $U = \Int \Gamma$, that is, the interior of $\Gamma$.
Let $A$ and $B$ be functions of $\tuple {x, y}$ defined on an open region containing $U$ and have continuous partial derivatives in such a set.
Then:
:$\ds \oint_\Gamma \paren {A \rd ... | It suffices to demonstrate the theorem for rectangular regions in the $x y$-plane.
The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles.
As the proof is for a recta... | Let $\Gamma$ be a positively oriented piecewise smooth [[Definition:Jordan Curve|Jordan curve]] in $\R^2$.
Let $U = \Int \Gamma$, that is, the [[Definition:Interior of Jordan Curve|interior]] of $\Gamma$.
Let $A$ and $B$ be functions of $\tuple {x, y}$ defined on an [[Definition:Open Region in Plane|open region]] con... | It suffices to demonstrate the theorem for rectangular regions in the [[Definition:X-Y Plane|$x y$-plane]].
The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles.
... | Green's Theorem | https://proofwiki.org/wiki/Green's_Theorem | https://proofwiki.org/wiki/Green's_Theorem | [
"Green's Theorem",
"Integral Calculus"
] | [
"Definition:Jordan Curve",
"Definition:Jordan Curve/Interior",
"Definition:Open Region/Plane",
"Definition:Continuous Real Function",
"Definition:Partial Derivative",
"Definition:Contour Integral"
] | [
"Definition:X-Y Plane",
"Monotone Convergence Theorem (Measure Theory)"
] |
proofwiki-2035 | Gauss-Ostrogradsky Theorem | Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.
Then:
:$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$
wher... | It suffices to prove the theorem for rectangular prisms.
The Riemann-sum nature of the triple integral then guarantees the theorem for arbitrary regions.
{{explain|Link to a result demonstrating the above, or explain it better here}}
Let:
:$R = \set {\tuple {x, y, z}: a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le... | Let $U$ be a [[Definition:Subset|subset]] of $\R^3$ which is [[Definition:Compact Subset of Real Euclidean Space|compact]] and has a [[Definition:Piecewise Smooth|piecewise smooth]] [[Definition:Boundary (Geometry)|boundary]] $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a [[Definition:Smooth Mapping|smooth]] [[Defi... | It suffices to prove the theorem for [[Definition:Rectangular Prism|rectangular prisms]].
The Riemann-sum nature of the triple integral then guarantees the theorem for arbitrary regions.
{{explain|Link to a result demonstrating the above, or explain it better here}}
Let:
:$R = \set {\tuple {x, y, z}: a_1 \le x \le a... | Gauss-Ostrogradsky Theorem/Formal Proof | https://proofwiki.org/wiki/Gauss-Ostrogradsky_Theorem | https://proofwiki.org/wiki/Gauss-Ostrogradsky_Theorem/Formal_Proof | [
"Gauss-Ostrogradsky Theorem",
"Vector Calculus"
] | [
"Definition:Subset",
"Definition:Compact Space/Euclidean Space",
"Definition:Piecewise Smooth",
"Definition:Boundary (Geometry)",
"Definition:Smooth Mapping",
"Definition:Vector Field",
"Definition:Neighborhood (Metric Space)",
"Definition:Unit Normal",
"Definition:Summation",
"Definition:Expansio... | [
"Definition:Cuboid",
"Definition:Area Element"
] |
proofwiki-2036 | Gauss-Ostrogradsky Theorem | Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.
Then:
:$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$
wher... | Let $S$ be the surface of $U$.
By definition, the surface integral of $\mathbf V$ over $S$ is therefore defined as:
:$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$
where $\rd \mathbf S$ is the differential of vector area.
:700px
Let $\d v$ be an element of volume within $U$.
In th... | Let $U$ be a [[Definition:Subset|subset]] of $\R^3$ which is [[Definition:Compact Subset of Real Euclidean Space|compact]] and has a [[Definition:Piecewise Smooth|piecewise smooth]] [[Definition:Boundary (Geometry)|boundary]] $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a [[Definition:Smooth Mapping|smooth]] [[Defi... | Let $S$ be the [[Definition:Surface|surface]] of $U$.
By definition, the [[Definition:Surface Integral|surface integral]] of $\mathbf V$ over $S$ is therefore defined as:
:$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$
where $\rd \mathbf S$ is the [[Definition:Differential of M... | Gauss-Ostrogradsky Theorem/Informal Proof | https://proofwiki.org/wiki/Gauss-Ostrogradsky_Theorem | https://proofwiki.org/wiki/Gauss-Ostrogradsky_Theorem/Informal_Proof | [
"Gauss-Ostrogradsky Theorem",
"Vector Calculus"
] | [
"Definition:Subset",
"Definition:Compact Space/Euclidean Space",
"Definition:Piecewise Smooth",
"Definition:Boundary (Geometry)",
"Definition:Smooth Mapping",
"Definition:Vector Field",
"Definition:Neighborhood (Metric Space)",
"Definition:Unit Normal",
"Definition:Summation",
"Definition:Expansio... | [
"Definition:Surface",
"Definition:Surface Integral",
"Definition:Differential of Mapping",
"Definition:Vector Area",
"File:Gauss-Ostrogradsky-theorem-proof-1.png",
"Definition:Volume",
"Definition:Cuboid",
"Definition:Flux/Total",
"Definition:Vector Quantity",
"Definition:Polyhedron/Face",
"Defi... |
proofwiki-2037 | Implications of Stokes' Theorem | Stokes' Theorem implies all of the following results:
* Classical Stokes' Theorem
* Green's Theorem
* Gauss's Theorem
* Fundamental Theorem of Calculus
* Cauchy's Residue Theorem | === Classical Stokes' Theorem ===
We note that given $f_1, f_2, f_3: \R^3 \to \R$, $f_1 \rd x + f_2 \rd y + f_3 \rd z$ is a $1$-form defined on a surface.
Since we are integrating over the boundary of a surface, which is a $1$-manifold, we may use the general Stokes' theorem:
{{begin-eqn}}
{{eqn | l = \oint_{\partial S... | [[General Stokes' Theorem|Stokes' Theorem]] implies all of the following results:
* [[Classical Stokes' Theorem]]
* [[Green's Theorem]]
* [[Gauss's Theorem]]
* [[Fundamental Theorem of Calculus]]
* [[Cauchy's Residue Theorem]] | === Classical Stokes' Theorem ===
We note that given $f_1, f_2, f_3: \R^3 \to \R$, $f_1 \rd x + f_2 \rd y + f_3 \rd z$ is a [[Definition:Differential Form|$1$-form]] defined on a surface.
Since we are integrating over the boundary of a surface, which is a $1$-manifold, we may use the general Stokes' theorem:
{{begin... | Implications of Stokes' Theorem | https://proofwiki.org/wiki/Implications_of_Stokes'_Theorem | https://proofwiki.org/wiki/Implications_of_Stokes'_Theorem | [
"Calculus"
] | [
"General Stokes' Theorem",
"Kelvin-Stokes Theorem",
"Green's Theorem",
"Gauss-Ostrogradsky Theorem",
"Fundamental Theorem of Calculus",
"Cauchy's Residue Theorem"
] | [
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form",
"Definition:Differential Form"
] |
proofwiki-2038 | Divergence Test | Let $\sequence {a_n}$ be a sequence in $\R$ which does not converge to $0$.
Then $\ds \sum_{i \mathop = 1}^\infty a_n$ diverges. | We know that Terms in Convergent Series Converge to Zero.
This is the contrapositive statement of this theorem.
Thus, the theorem holds by Rule of Transposition.
{{qed}} | Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence in $\R$]] which does not [[Definition:Convergent Series|converge]] to $0$.
Then $\ds \sum_{i \mathop = 1}^\infty a_n$ [[Definition:Divergent Series|diverges]]. | We know that [[Terms in Convergent Series Converge to Zero]].
This is the [[Definition:Contrapositive Statement|contrapositive statement]] of this theorem.
Thus, the theorem holds by [[Rule of Transposition]].
{{qed}} | Divergence Test | https://proofwiki.org/wiki/Divergence_Test | https://proofwiki.org/wiki/Divergence_Test | [
"Divergence Test",
"Divergent Series",
"Convergence Tests",
"Named Theorems",
"Proofs by Contraposition"
] | [
"Definition:Sequence",
"Definition:Convergent Series",
"Definition:Divergent Series"
] | [
"Terms in Convergent Series Converge to Zero",
"Definition:Contrapositive Statement",
"Rule of Transposition"
] |
proofwiki-2039 | Eisenstein's Lemma | Let $p$ be an odd prime.
Let $a \in \Z$ be an odd integer such that $p \nmid a$.
Let $\paren {\dfrac a p}$ be defined as the Legendre symbol.
Then:
:$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$
where:
:$\ds \map \alpha {a, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\frac {k a} p}$
:$\floor {\dfrac... | We will borrow some ideas and techniques from the proof of Gauss's Lemma.
Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.
Let us create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$.
Arranging $S'$ into increasing order, we get:
:$S' = \set {b_1, b_2, \ldots, b_m, c_1... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Let $a \in \Z$ be an [[Definition:Odd Integer|odd integer]] such that $p \nmid a$.
Let $\paren {\dfrac a p}$ be defined as the [[Definition:Legendre Symbol|Legendre symbol]].
Then:
:$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$
where:
:$\ds \map \alpha {... | We will borrow some ideas and techniques from the proof of [[Gauss's Lemma (Number Theory)|Gauss's Lemma]].
Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.
Let us create $S'$ from $S$ by replacing each element of $S$ by its [[Definition:Least Positive Residue|least positive residue]] modulo $p$.
Arranging... | Eisenstein's Lemma | https://proofwiki.org/wiki/Eisenstein's_Lemma | https://proofwiki.org/wiki/Eisenstein's_Lemma | [
"Number Theory"
] | [
"Definition:Odd Prime",
"Definition:Odd Integer",
"Definition:Legendre Symbol",
"Definition:Floor Function"
] | [
"Gauss's Lemma (Number Theory)",
"Definition:Set of Residue Classes/Least Positive",
"Division Theorem",
"Gauss's Lemma (Number Theory)",
"Gauss's Lemma (Number Theory)",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Gauss's Lemma (Numbe... |
proofwiki-2040 | Law of Quadratic Reciprocity | Let $p$ and $q$ be distinct odd primes.
Then:
:$\paren {\dfrac p q} \paren {\dfrac q p} = \paren {-1}^{\dfrac {\paren {p - 1} \paren {q - 1} } 4}$
where $\paren {\dfrac p q}$ and $\paren {\dfrac q p}$ are defined as the Legendre symbol.
An alternative formulation is:
$\paren {\dfrac p q} = \begin{cases}
\quad \paren {\... | Let $p$ and $q$ be distinct odd primes.
Consider the rectangle in the Cartesian $x y$ plane with vertices at $\tuple {0, 0}, \tuple {\dfrac p 2, 0}, \tuple {\dfrac p 2, \dfrac q 2}, \tuple {0, \dfrac q 2}$.
The number of lattice points inside this rectangle is $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2$.
:700px
Consider... | Let $p$ and $q$ be distinct [[Definition:Odd Prime|odd primes]].
Then:
:$\paren {\dfrac p q} \paren {\dfrac q p} = \paren {-1}^{\dfrac {\paren {p - 1} \paren {q - 1} } 4}$
where $\paren {\dfrac p q}$ and $\paren {\dfrac q p}$ are defined as the [[Definition:Legendre Symbol|Legendre symbol]].
An alternative formulatio... | Let $p$ and $q$ be distinct [[Definition:Odd Prime|odd primes]].
Consider the [[Definition:Rectangle|rectangle]] in the [[Definition:Cartesian Plane|Cartesian $x y$ plane]] with [[Definition:Vertex of Polygon|vertices]] at $\tuple {0, 0}, \tuple {\dfrac p 2, 0}, \tuple {\dfrac p 2, \dfrac q 2}, \tuple {0, \dfrac q 2}$... | Law of Quadratic Reciprocity | https://proofwiki.org/wiki/Law_of_Quadratic_Reciprocity | https://proofwiki.org/wiki/Law_of_Quadratic_Reciprocity | [
"Law of Quadratic Reciprocity",
"Legendre Symbol",
"Named Theorems",
"Quadratic Residues"
] | [
"Definition:Odd Prime",
"Definition:Legendre Symbol",
"Law of Quadratic Reciprocity",
"Symbols:Abbreviations/L/LQR"
] | [
"Definition:Odd Prime",
"Definition:Quadrilateral/Rectangle",
"Definition:Cartesian Plane",
"Definition:Polygon/Vertex",
"Definition:Lattice Point/Cartesian Coordinate System",
"Definition:Quadrilateral/Rectangle",
"File:LQR.png",
"Definition:Diameter of Quadrilateral",
"Definition:Equation",
"Def... |
proofwiki-2041 | Manipulation of Exterior Derivative | For the exterior derivative, the following statements are true:
:$\map \d {a b} = a \rd b + \paren {\d a} b$
:$\map \d {a \wedge b} = \d a \wedge b - a \wedge \d b$
where $a \wedge b$ is the wedge product. | {{ProofWanted}}
Category:Topology
55sgp4p9ccfvrwj3rtqotazriywkwoc | For the [[Definition:Exterior Derivative|exterior derivative]], the following statements are true:
:$\map \d {a b} = a \rd b + \paren {\d a} b$
:$\map \d {a \wedge b} = \d a \wedge b - a \wedge \d b$
where $a \wedge b$ is the [[Definition:Wedge Product|wedge product]]. | {{ProofWanted}}
[[Category:Topology]]
55sgp4p9ccfvrwj3rtqotazriywkwoc | Manipulation of Exterior Derivative | https://proofwiki.org/wiki/Manipulation_of_Exterior_Derivative | https://proofwiki.org/wiki/Manipulation_of_Exterior_Derivative | [
"Topology"
] | [
"Definition:Exterior Derivative",
"Definition:Wedge Product"
] | [
"Category:Topology"
] |
proofwiki-2042 | General Stokes' Theorem | Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.
Let the boundary of $X$ be $\partial X$.
Then:
:$\ds \int_{\partial X} \omega = \int_X \rd \omega$
where $\d \omega$ is the exterior derivative of $\omega$. | === Special Case ===
Let there exist a chart:
:$x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$
such that:
:$\map \supp \omega \subseteq V$
where:
:$\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$
{{explain|Yes very good but what *is* it?}}
We may suppose that $V$ is relatively compact.
Th... | Let $\omega$ be a [[Definition:Smooth (Differential Form)|smooth]] $\paren {n - 1}$-[[Definition:Differential Form|form]] with [[Definition:Compact Topological Space|compact]] [[Definition:Support of Real-Valued Function|support]] on a [[Definition:Smooth Manifold|smooth]] $n$-[[Definition:Dimension (Topology)|dimensio... | === Special Case ===
Let there exist a [[Definition:Chart|chart]]:
:$x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$
such that:
:$\map \supp \omega \subseteq V$
where:
:$\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$
{{explain|Yes very good but what *is* it?}}
We may suppose that $V$ ... | General Stokes' Theorem | https://proofwiki.org/wiki/General_Stokes'_Theorem | https://proofwiki.org/wiki/General_Stokes'_Theorem | [
"Stokes' Theorem",
"Integral Calculus",
"Manifolds"
] | [
"Definition:Smooth (Differential Form)",
"Definition:Differential Form",
"Definition:Compact Topological Space",
"Definition:Support of Mapping to Algebraic Structure/Real-Valued Function",
"Definition:Topological Manifold/Smooth Manifold",
"Definition:Dimension (Topology)",
"Definition:Oriented Manifol... | [
"Definition:Chart",
"Definition:Relatively Compact Subspace",
"Definition:Translation Mapping",
"Definition:Tangent Space",
"Fubini's Theorem",
"Fubini's Theorem",
"Definition:Relatively Compact Subspace",
"Definition:Chart"
] |
proofwiki-2043 | Kelvin-Stokes Theorem | Let $S$ be some orientable smooth surface with boundary in $\R^3$.
Let $\mathbf F:\R^3 \to \R^3$ be a vector-valued function with Euclidean coordinate expression:
:$\mathbf F = f_1 \mathbf i + f_2 \mathbf j + f_3 \mathbf k$
where $f_i: \R^3 \to \R$.
Then:
:$\ds \oint_{\partial S} f_1 \rd x + f_2 \rd y + f_3 \rd z = \ii... | Let $\mathbf r:\R^2 \to \R^3, \map {\mathbf r} {s, t}$ be a smooth parametrization of $S$ from some region $R$ in the $st$-plane, so that:
:$\map {\mathbf r} R = S$
and:
:$\map {\mathbf r} {\partial R} = \partial S$
First, we convert the {{LHS}} into a line integral:
{{begin-eqn}}
{{eqn | l = \oint_{\partial S} f_1 \rd... | Let $S$ be some orientable smooth surface with boundary in $\R^3$.
Let $\mathbf F:\R^3 \to \R^3$ be a vector-valued function with Euclidean coordinate expression:
:$\mathbf F = f_1 \mathbf i + f_2 \mathbf j + f_3 \mathbf k$
where $f_i: \R^3 \to \R$.
Then:
:$\ds \oint_{\partial S} f_1 \rd x + f_2 \rd y + f_3 \rd z = \... | Let $\mathbf r:\R^2 \to \R^3, \map {\mathbf r} {s, t}$ be a smooth parametrization of $S$ from some region $R$ in the $st$-plane, so that:
:$\map {\mathbf r} R = S$
and:
:$\map {\mathbf r} {\partial R} = \partial S$
First, we convert the {{LHS}} into a line integral:
{{begin-eqn}}
{{eqn | l = \oint_{\partial S} f_1 \... | Kelvin-Stokes Theorem | https://proofwiki.org/wiki/Kelvin-Stokes_Theorem | https://proofwiki.org/wiki/Kelvin-Stokes_Theorem | [
"Kelvin-Stokes Theorem",
"Stokes' Theorem",
"Integral Calculus"
] | [
"Definition:Area Element"
] | [
"Derivative of Dot Product of Vector-Valued Functions",
"Derivative of Dot Product of Vector-Valued Functions",
"Symmetry of Second Derivatives",
"Definition:Expression",
"Definition:Integration/Integrand",
"Chain Rule for Real-Valued Functions",
"Chain Rule for Real-Valued Functions",
"Green's Theore... |
proofwiki-2044 | Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent | Every even convergent is strictly smaller than every odd convergent. | Let $k \ge 1$.
From Denominators of Simple Continued Fraction are Strictly Positive, $q_k q_{k - 1}>0$.
From Difference between Adjacent Convergents of Simple Continued Fraction:
:$C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
Let $k$ be even.
Then:
:$\exists s \in \N: k = 2 s$
and:
:$C_{2 s} < C_{2... | Every [[Definition:Even Convergent|even convergent]] is [[Definition:Strictly Smaller Real Number|strictly smaller]] than every [[Definition:Odd Convergent|odd convergent]]. | Let $k \ge 1$.
From [[Denominators of Simple Continued Fraction are Strictly Positive]], $q_k q_{k - 1}>0$.
From [[Difference between Adjacent Convergents of Simple Continued Fraction]]:
:$C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
Let $k$ be [[Definition:Even Integer|even]].
Then:
:$\exists... | Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent | https://proofwiki.org/wiki/Even_Convergent_of_Simple_Continued_Fraction_is_Strictly_Smaller_than_Odd_Convergent | https://proofwiki.org/wiki/Even_Convergent_of_Simple_Continued_Fraction_is_Strictly_Smaller_than_Odd_Convergent | [
"Simple Continued Fractions"
] | [
"Definition:Convergent of Continued Fraction/Even",
"Definition:Strictly Smaller Real Number",
"Definition:Convergent of Continued Fraction/Odd"
] | [
"Denominators of Simple Continued Fraction are Strictly Positive",
"Difference between Adjacent Convergents of Simple Continued Fraction",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Convergent of Continued Fraction/Even",
"Definition:Convergent of Continued Fraction/Odd",
"Even Con... |
proofwiki-2045 | Denominators of Simple Continued Fraction are Strictly Increasing | Let $N \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple continued fraction in $\R$ of length $N$.
Let $q_0, q_1, q_2, \ldots$ be its denominators.
Then with the possible exception of $q_0 = q_1$, the sequence $\sequence {q_n}$ is strictly increasing. | By definition of simple continued fraction, all partial denominators of $\sqbrk {a_0, a_1, a_2, \ldots}$ are strictly positive integers, with the possible exception of $a_0$.
So:
:$q_1 = a_1 \ge 1 = q_0$
:$q_2 = a_2 a_1 + 1 \ge a_1 + 1 > a_1 = q_1$
hence:
:$1 = q_0 \le q_1 < q_2$
Suppose $q_k > q_{k - 1} \ge 1$ for som... | Let $N \in \N \cup \set \infty$ be an [[Definition:Extended Natural Number|extended natural number]].
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a [[Definition:Simple Continued Fraction|simple continued fraction]] in $\R$ of [[Definition:Length of Continued Fraction|length]] $N$.
Let $q_0, q_1, q_2, \ldots$ be its [[Def... | By definition of [[Definition:Simple Continued Fraction|simple continued fraction]], all [[Definition:Partial Denominator|partial denominators]] of $\sqbrk {a_0, a_1, a_2, \ldots}$ are [[Definition:Strictly Positive Integer|strictly positive integers]], with the possible exception of $a_0$.
So:
:$q_1 = a_1 \ge 1 = q_0... | Denominators of Simple Continued Fraction are Strictly Increasing | https://proofwiki.org/wiki/Denominators_of_Simple_Continued_Fraction_are_Strictly_Increasing | https://proofwiki.org/wiki/Denominators_of_Simple_Continued_Fraction_are_Strictly_Increasing | [
"Simple Continued Fractions"
] | [
"Definition:Extended Natural Numbers",
"Definition:Simple Continued Fraction",
"Definition:Length of Continued Fraction",
"Definition:Numerators and Denominators of Continued Fraction",
"Definition:Sequence",
"Definition:Strictly Increasing/Sequence"
] | [
"Definition:Simple Continued Fraction",
"Definition:Partial Denominator",
"Definition:Strictly Positive/Integer",
"Principle of Mathematical Induction",
"Definition:Strictly Increasing/Sequence"
] |
proofwiki-2046 | Simple Infinite Continued Fraction Converges to Irrational Number | The value of any simple infinite continued fraction in $\R$ is irrational. | Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple infinite continued fraction.
Note that by Simple Infinite Continued Fraction Converges, a simple infinite continued fraction is indeed convergent, say to $x \in \R$.
Let $p_0, p_1, \ldots$ and $q_0, q_1, \ldots$ be its numerators and denominators.
Let $C_0, C_1, \ldots$ ... | The [[Definition:Value of Infinite Continued Fraction|value]] of any [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]] in $\R$ is [[Definition:Irrational Number|irrational]]. | Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]].
Note that by [[Simple Infinite Continued Fraction Converges]], a [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]] is indeed [[Definition:Convergent Continu... | Simple Infinite Continued Fraction Converges to Irrational Number | https://proofwiki.org/wiki/Simple_Infinite_Continued_Fraction_Converges_to_Irrational_Number | https://proofwiki.org/wiki/Simple_Infinite_Continued_Fraction_Converges_to_Irrational_Number | [
"Simple Continued Fractions",
"Irrational Numbers"
] | [
"Definition:Value of Continued Fraction/Infinite",
"Definition:Simple Continued Fraction/Infinite",
"Definition:Irrational Number"
] | [
"Definition:Simple Continued Fraction/Infinite",
"Simple Infinite Continued Fraction Converges",
"Definition:Simple Continued Fraction/Infinite",
"Definition:Convergent Continued Fraction",
"Definition:Numerators and Denominators of Continued Fraction",
"Definition:Convergent of Continued Fraction",
"Ac... |
proofwiki-2047 | Uniqueness of Analytic Continuation | Let $U \subset V \subset \C$ be open subsets of the complex plane.
Let $V$ be connected.
Suppose:
:$(1): \quad F_1, F_2$ are functions defined on $V$
:$(2): \quad f$ is a function defined on $U$.
Let $F_1$ and $F_2$ be analytic continuations of $f$ to $V$.
Then $F_1 = F_2$. | We have {{hypothesis}} that $F_1$ and $F_2$ are analytic continuations of $f$.
Therefore they are {{afortiori}} analytic functions.
Let $\map g z = \map {F_1} z - \map {F_2} z$.
By various combination laws, $g$ is also an analytic function.
By definition of agreement:
:$\forall z \in U: \map g z = 0$
That is, every ele... | Let $U \subset V \subset \C$ be [[Definition:Open Set (Complex Analysis)|open subsets]] of the [[Definition:Complex Plane|complex plane]].
Let $V$ be [[Definition:Connected Set (Complex Analysis)|connected]].
Suppose:
:$(1): \quad F_1, F_2$ are [[Definition:Complex Function|functions]] defined on $V$
:$(2): \quad f$... | We have {{hypothesis}} that $F_1$ and $F_2$ are [[Definition:Analytic Continuation|analytic continuations]] of $f$.
Therefore they are {{afortiori}} [[Definition:Analytic Complex Function|analytic functions]].
Let $\map g z = \map {F_1} z - \map {F_2} z$.
By various combination laws, $g$ is also an [[Definition:Anal... | Uniqueness of Analytic Continuation | https://proofwiki.org/wiki/Uniqueness_of_Analytic_Continuation | https://proofwiki.org/wiki/Uniqueness_of_Analytic_Continuation | [
"Analytic Continuations"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Complex Number/Complex Plane",
"Definition:Connected Set (Complex Analysis)",
"Definition:Complex Function",
"Definition:Complex Function",
"Definition:Analytic Continuation"
] | [
"Definition:Analytic Continuation",
"Definition:Analytic Function/Complex Plane",
"Definition:Analytic Function/Complex Plane",
"Definition:Agreement/Mappings",
"Definition:Element",
"Definition:Root of Mapping",
"Definition:Open Set/Complex Analysis",
"Definition:Root of Mapping",
"Definition:Isola... |
proofwiki-2048 | Zeroes of Non-Constant Analytic Function are Isolated | Let $U \subset \C$ be some open set.
Let $f$ be an analytic function defined on $U$.
Then either $f$ is a constant function, or the set $\set {z \in U: \map f z = 0}$ is totally disconnected. | Suppose $f$ has no zeroes in $U$.
Then the set described in the theorem is the empty set, and we're done.
So we suppose $\exists z_0 \in U$ such that $\map f {z_0} = 0$.
Since $f$ is analytic, there is a Taylor series for $f$ at $z_0$ which converges for $\cmod {z - z_0} < R$.
Now, since $\map f {z_0} = 0,$ we know $a_... | Let $U \subset \C$ be some [[Definition:Open Set (Complex Analysis)|open set]].
Let $f$ be an [[Definition:Analytic Function|analytic function]] defined on $U$.
Then either $f$ is a [[Definition:Constant Mapping|constant function]], or the set $\set {z \in U: \map f z = 0}$ is [[Definition:Totally Disconnected|total... | Suppose $f$ has no zeroes in $U$.
Then the set described in the theorem is the [[Definition:Empty Set|empty set]], and we're done.
So we suppose $\exists z_0 \in U$ such that $\map f {z_0} = 0$.
Since $f$ is analytic, there is a [[Taylor's Theorem|Taylor series]] for $f$ at $z_0$ which converges for $\cmod {z - z_0... | Zeroes of Non-Constant Analytic Function are Isolated | https://proofwiki.org/wiki/Zeroes_of_Non-Constant_Analytic_Function_are_Isolated | https://proofwiki.org/wiki/Zeroes_of_Non-Constant_Analytic_Function_are_Isolated | [
"Complex Analysis"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Analytic Function",
"Definition:Constant Mapping",
"Definition:Totally Disconnected Space"
] | [
"Definition:Empty Set",
"Taylor's Theorem",
"Definition:Continuous Complex Function",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Totally Disconnected Space",
"Category:Complex Analysis"
] |
proofwiki-2049 | Irrational Number is Limit of Unique Simple Infinite Continued Fraction | Let $x$ be an irrational number.
Then the continued fraction expansion of $x$ is the unique simple infinite continued fraction that converges to $x$. | Follows from:
* Continued Fraction Expansion of Irrational Number Converges to Number Itself
* Simple Infinite Continued Fraction is Uniquely Determined by Limit.
{{Qed}} | Let $x$ be an [[Definition:Irrational Number|irrational number]].
Then the [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]] of $x$ is the [[Definition:Unique|unique]] [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]] that [[Definition:Con... | Follows from:
* [[Continued Fraction Expansion of Irrational Number Converges to Number Itself]]
* [[Simple Infinite Continued Fraction is Uniquely Determined by Limit]].
{{Qed}} | Irrational Number is Limit of Unique Simple Infinite Continued Fraction | https://proofwiki.org/wiki/Irrational_Number_is_Limit_of_Unique_Simple_Infinite_Continued_Fraction | https://proofwiki.org/wiki/Irrational_Number_is_Limit_of_Unique_Simple_Infinite_Continued_Fraction | [
"Simple Continued Fractions",
"Irrational Numbers"
] | [
"Definition:Irrational Number",
"Definition:Continued Fraction Expansion/Real Number",
"Definition:Unique",
"Definition:Simple Continued Fraction/Infinite",
"Definition:Convergent Continued Fraction"
] | [
"Continued Fraction Expansion of Irrational Number Converges to Number Itself",
"Simple Infinite Continued Fraction is Uniquely Determined by Limit"
] |
proofwiki-2050 | Analytic Continuation of Riemann Zeta Function | The Riemann zeta function is meromorphic on $\C$. | The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\set {s: \map \Re s > 0}$ is given by Integral Representation of Riemann Zeta Function in terms of Fractional Part:
:$(1): \quad \ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where $\... | The [[Definition:Riemann Zeta Function|Riemann zeta function]] is [[Definition:Meromorphic Function|meromorphic]] on $\C$. | The (yet to be confirmed) [[Definition:Meromorphic Function|meromorphic]] [[Definition:Analytic Continuation|continuation]] of the [[Definition:Riemann Zeta Function|Riemann zeta function]] to the [[Definition:Half-Plane|half-plane]] $\set {s: \map \Re s > 0}$ is given by [[Integral Representation of Riemann Zeta Funct... | Analytic Continuation of Riemann Zeta Function | https://proofwiki.org/wiki/Analytic_Continuation_of_Riemann_Zeta_Function | https://proofwiki.org/wiki/Analytic_Continuation_of_Riemann_Zeta_Function | [
"Riemann Zeta Function"
] | [
"Definition:Riemann Zeta Function",
"Definition:Meromorphic Function"
] | [
"Definition:Meromorphic Function",
"Definition:Analytic Continuation",
"Definition:Riemann Zeta Function",
"Definition:Half-Plane",
"Integral Representation of Riemann Zeta Function in terms of Fractional Part",
"Definition:Fractional Part",
"Definition:Completed Riemann Zeta Function",
"Definition:Ho... |
proofwiki-2051 | Accuracy of Convergents of Continued Fraction Expansion of Irrational Number | Let $x$ be an irrational number.
Let $(a_0, a_1, \ldots)$ be its continued fraction expansion.
Let $\left \langle {C_n}\right \rangle_{n \geq 0}$ be its sequence of convergents.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Then:
:$\forall k \ge 1: \left\vert{x - \dfrac {p_... | The Continued Fraction Algorithm gives the following system of equations:
{{begin-eqn}}
{{eqn | l = x
| r = \left[{a_0, x_2}\right]
| c =
}}
{{eqn | r = \left[{a_0, a_1, x_3}\right]
| c =
}}
{{eqn | r = \ldots
| c =
}}
{{eqn | r = \left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right]
| c =
... | Let $x$ be an [[Definition:Irrational Number|irrational number]].
Let $(a_0, a_1, \ldots)$ be its [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]].
Let $\left \langle {C_n}\right \rangle_{n \geq 0}$ be its [[Definition:Sequence|sequence]] of [[Definition:Convergent of Cont... | The [[Continued Fraction Algorithm]] gives the following system of equations:
{{begin-eqn}}
{{eqn | l = x
| r = \left[{a_0, x_2}\right]
| c =
}}
{{eqn | r = \left[{a_0, a_1, x_3}\right]
| c =
}}
{{eqn | r = \ldots
| c =
}}
{{eqn | r = \left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right]
| ... | Accuracy of Convergents of Continued Fraction Expansion of Irrational Number | https://proofwiki.org/wiki/Accuracy_of_Convergents_of_Continued_Fraction_Expansion_of_Irrational_Number | https://proofwiki.org/wiki/Accuracy_of_Convergents_of_Continued_Fraction_Expansion_of_Irrational_Number | [
"Continued Fractions"
] | [
"Definition:Irrational Number",
"Definition:Continued Fraction Expansion/Real Number",
"Definition:Sequence",
"Definition:Convergent of Continued Fraction",
"Definition:Numerators and Denominators of Continued Fraction",
"Definition:Convergent of Continued Fraction"
] | [
"Continued Fraction Algorithm",
"Value of Finite Continued Fraction equals Numerator Divided by Denominator",
"Difference between Adjacent Convergents of Simple Continued Fraction",
"Continued Fraction Algorithm"
] |
proofwiki-2052 | Trivial Zeroes of Riemann Zeta Function are Even Negative Integers | Let $s = \sigma + i t$ be a zero of the Riemann zeta function not contained in the critical strip:
:$0 \le \map \Re s \le 1$
Then:
:$s \in \set {-2, -4, -6, \ldots}$
These are called the '''trivial zeros''' of $\zeta$. | {{finish|First it needs to be established that those points are in fact zeroes. This follows directly and trivially from Riemann Zeta Function at Non-Positive Integers, but needs to be made explicit here.}}
First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$.
Therefore, the completed Riemann ... | Let $s = \sigma + i t$ be a [[Definition:Zero of Function|zero]] of the [[Definition:Riemann Zeta Function|Riemann zeta function]] not contained in the [[Definition:Critical Strip|critical strip]]:
:$0 \le \map \Re s \le 1$
Then:
:$s \in \set {-2, -4, -6, \ldots}$
These are called the '''[[Definition:Trivial Zero of ... | {{finish|First it needs to be established that those points are in fact zeroes. This follows directly and trivially from [[Riemann Zeta Function at Non-Positive Integers]], but needs to be made explicit here.}}
First we note that by [[Zeroes of Gamma Function]], $\Gamma$ has no [[Definition:Zero of Function|zeroes]] o... | Trivial Zeroes of Riemann Zeta Function are Even Negative Integers | https://proofwiki.org/wiki/Trivial_Zeroes_of_Riemann_Zeta_Function_are_Even_Negative_Integers | https://proofwiki.org/wiki/Trivial_Zeroes_of_Riemann_Zeta_Function_are_Even_Negative_Integers | [
"Riemann Zeta Function"
] | [
"Definition:Root of Mapping",
"Definition:Riemann Zeta Function",
"Definition:Riemann Zeta Function/Critical Strip",
"Definition:Riemann Zeta Function/Zero/Trivial"
] | [
"Riemann Zeta Function at Non-Positive Integers",
"Zeroes of Gamma Function",
"Definition:Root of Mapping",
"Definition:Completed Riemann Zeta Function",
"Definition:Root of Mapping",
"Functional Equation for Riemann Zeta Function",
"Definition:Prime Number",
"Definition:Absolute Convergence of Produc... |
proofwiki-2053 | Poles of Gamma Function | The gamma function $\Gamma: \C \to \C$ is analytic throughout the complex plane except at $\set {0, -1, -2, -3, \dotsc}$ where it has simple poles. | First we examine the location of the poles.
We examine the Weierstrass form of the Gamma function:
:$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \ds \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$
The factors of the continued product clearly do not tend to zero.
So the continued product is zero ... | The [[Definition:Gamma Function|gamma function]] $\Gamma: \C \to \C$ is [[Definition:Analytic Function|analytic]] throughout the complex plane except at $\set {0, -1, -2, -3, \dotsc}$ where it has [[Definition:Simple Pole|simple poles]]. | First we examine the location of the [[Definition:Simple Pole|poles]].
We examine the [[Definition:Weierstrass Form of Gamma Function|Weierstrass form of the Gamma function]]:
:$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \ds \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$
The [[Definition:Fa... | Poles of Gamma Function | https://proofwiki.org/wiki/Poles_of_Gamma_Function | https://proofwiki.org/wiki/Poles_of_Gamma_Function | [
"Gamma Function"
] | [
"Definition:Gamma Function",
"Definition:Analytic Function",
"Definition:Order of Pole/Simple Pole"
] | [
"Definition:Order of Pole/Simple Pole",
"Definition:Gamma Function/Weierstrass Form",
"Definition:Divisor (Algebra)/Integer",
"Definition:Continued Product",
"Definition:Zero (Number)",
"Definition:Continued Product",
"Definition:Zero (Number)",
"Definition:Divisor (Algebra)/Integer",
"Definition:Ze... |
proofwiki-2054 | Zeroes of Gamma Function | The Gamma function is never equal to $0$. | Suppose $\exists z$ such that $\map \Gamma z = 0$.
We examine the Euler form of the gamma function, which is defined for $\C \setminus \set {0, -1, -2, \ldots}$.
The Euler form, equated with zero, yields
:$\ds 0 = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1} }$
It ... | The [[Definition:Gamma Function|Gamma function]] is never equal to $0$. | Suppose $\exists z$ such that $\map \Gamma z = 0$.
We examine the [[Definition:Gamma Function#Euler Form|Euler form]] of the gamma function, which is defined for $\C \setminus \set {0, -1, -2, \ldots}$.
The Euler form, equated with zero, yields
:$\ds 0 = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \f... | Zeroes of Gamma Function | https://proofwiki.org/wiki/Zeroes_of_Gamma_Function | https://proofwiki.org/wiki/Zeroes_of_Gamma_Function | [
"Gamma Function"
] | [
"Definition:Gamma Function"
] | [
"Definition:Gamma Function",
"Poles of Gamma Function",
"Category:Gamma Function"
] |
proofwiki-2055 | Gamma Difference Equation | :$\map \Gamma {z + 1} = z \, \map \Gamma z$ | Let $z \in \C$, with $\map \Re z > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + 1}
| r = \int_0^\infty t^z e^{-t} \rd t
| c =
}}
{{eqn | r = \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t
| c = Integration by Parts
}}
{{eqn | r = z \int_0^\infty t^{z - 1} e^{-t}... | :$\map \Gamma {z + 1} = z \, \map \Gamma z$ | Let $z \in \C$, with $\map \Re z > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + 1}
| r = \int_0^\infty t^z e^{-t} \rd t
| c =
}}
{{eqn | r = \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t
| c = [[Integration by Parts]]
}}
{{eqn | r = z \int_0^\infty t^{z - 1} e... | Gamma Difference Equation/Proof 1 | https://proofwiki.org/wiki/Gamma_Difference_Equation | https://proofwiki.org/wiki/Gamma_Difference_Equation/Proof_1 | [
"Gamma Function",
"Gamma Difference Equation"
] | [] | [
"Integration by Parts"
] |
proofwiki-2056 | Gamma Difference Equation | :$\map \Gamma {z + 1} = z \, \map \Gamma z$ | {{begin-eqn}}
{{eqn | l = \frac {\map \Gamma {z + 1} } {\map \Gamma z}
| r = \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \fr... | :$\map \Gamma {z + 1} = z \, \map \Gamma z$ | {{begin-eqn}}
{{eqn | l = \frac {\map \Gamma {z + 1} } {\map \Gamma z}
| r = \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \fr... | Gamma Difference Equation/Proof 2 | https://proofwiki.org/wiki/Gamma_Difference_Equation | https://proofwiki.org/wiki/Gamma_Difference_Equation/Proof_2 | [
"Gamma Function",
"Gamma Difference Equation"
] | [] | [
"Combination Theorem for Sequences/Complex/Product Rule",
"Telescoping Product",
"Limit to Infinity of Complex Rational Function"
] |
proofwiki-2057 | Convergents are Best Approximations | Let $x$ be an irrational number.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of its continued fraction expansion.
Let $\dfrac {p_n} {q_n}$ be the $n$th convergent.
Let $\dfrac a b$ be any rational number such that $0 < b < q_{n + 1}$.
Then:
:$\foral... | Let $\dfrac a b$ be a rational number in canonical form such that $b < q_{n + 1}$.
Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds.
Consider the system of equations:
{{begin-eqn}}
{{eqn | l = a
| r = r p_n + s p_{n + 1}
| c =
}}
{{eqn | l = b
| r = r q_... | Let $x$ be an [[Definition:irrational Number|irrational number]].
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the [[Definition:Numerators and Denominators of Continued Fraction|numerators and denominators]] of its [[Definition:Continued Fraction Expansion of Irrational Number|con... | Let $\dfrac a b$ be a [[Definition:Rational Number|rational number]] in [[Definition:Canonical Form of Rational Number|canonical form]] such that $b < q_{n + 1}$.
Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds.
Consider the system of equations:
{{begin-eqn}}
{{eqn | l... | Convergents are Best Approximations | https://proofwiki.org/wiki/Convergents_are_Best_Approximations | https://proofwiki.org/wiki/Convergents_are_Best_Approximations | [
"Continued Fractions"
] | [
"Definition:irrational Number",
"Definition:Numerators and Denominators of Continued Fraction",
"Definition:Continued Fraction Expansion/Real Number",
"Definition:Convergent of Continued Fraction",
"Definition:Rational Number"
] | [
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form",
"Difference between Adjacent Convergents of Simple Continued Fraction",
"Definition:Integer",
"Euclid's Lemma",
"Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent",
"Definition:Convergent of Con... |
proofwiki-2058 | Condition for Rational to be Convergent | Let $x$ be an irrational number.
Let the rational number $\dfrac a b$ satisfy the inequality:
:$\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$
Then $\dfrac a b$ is a convergent of $x$. | {{AimForCont}} $\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, but that $\dfrac a b$ is ''not'' one of the convergents $\dfrac {p_n} {q_n}$ of $x$.
Let $r$ be the unique integer for which $q_r \le b \le q_{r + 1}$.
{{explain|Why does such a unique $r$ exist? Maybe related to Denominators of Simple Continued Fraction are S... | Let $x$ be an [[Definition:Irrational Number|irrational number]].
Let the [[Definition:Rational Number|rational number]] $\dfrac a b$ satisfy the inequality:
:$\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$
Then $\dfrac a b$ is a [[Definition:Convergent of Continued Fraction|convergent]] of $x$. | {{AimForCont}} $\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, but that $\dfrac a b$ is ''not'' one of the [[Definition:Convergent of Continued Fraction|convergents]] $\dfrac {p_n} {q_n}$ of $x$.
Let $r$ be the unique [[Definition:Integer|integer]] for which $q_r \le b \le q_{r + 1}$.
{{explain|Why does such a unique $r$... | Condition for Rational to be Convergent | https://proofwiki.org/wiki/Condition_for_Rational_to_be_Convergent | https://proofwiki.org/wiki/Condition_for_Rational_to_be_Convergent | [
"Continued Fractions"
] | [
"Definition:Irrational Number",
"Definition:Rational Number",
"Definition:Convergent of Continued Fraction"
] | [
"Definition:Convergent of Continued Fraction",
"Definition:Integer",
"Denominators of Simple Continued Fraction are Strictly Increasing",
"Convergents are Best Approximations",
"Triangle Inequality",
"Definition:Integer",
"Definition:Convergent of Continued Fraction",
"Definition:Integer",
"Definiti... |
proofwiki-2059 | Quadratic Irrational is Root of Quadratic Equation | Let $x$ be a quadratic irrational.
Then $x$ is a solution to a quadratic equation with rational coefficients. | Let $x = r + s \sqrt n$.
From Solution to Quadratic Equation, the solutions of $a x^2 + b x + c$ are:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}$
given the appropriate condition on the discriminant.
So if $x = r + s \sqrt n$ is a solution, then so is $x = r - s \sqrt n$.
Hence we have:
{{begin-eqn}}
{{eqn | l = \l... | Let $x$ be a [[Definition:Quadratic Irrational|quadratic irrational]].
Then $x$ is a [[Definition:Root of Polynomial|solution]] to a [[Definition:Quadratic Equation|quadratic equation]] with [[Definition:Rational Number|rational]] [[Definition:Coefficient|coefficients]]. | Let $x = r + s \sqrt n$.
From [[Solution to Quadratic Equation]], the solutions of $a x^2 + b x + c$ are:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}$
given the appropriate condition on the [[Definition:Discriminant of Quadratic Equation|discriminant]].
So if $x = r + s \sqrt n$ is a solution, then so is $x = r... | Quadratic Irrational is Root of Quadratic Equation | https://proofwiki.org/wiki/Quadratic_Irrational_is_Root_of_Quadratic_Equation | https://proofwiki.org/wiki/Quadratic_Irrational_is_Root_of_Quadratic_Equation | [
"Quadratic Equations",
"Quadratic Irrationals"
] | [
"Definition:Quadratic Irrational",
"Definition:Root of Polynomial",
"Definition:Quadratic Equation",
"Definition:Rational Number",
"Definition:Coefficient"
] | [
"Solution to Quadratic Equation",
"Definition:Discriminant of Polynomial/Quadratic Equation",
"Difference of Two Squares",
"Definition:Rational Number",
"Definition:Integer",
"Definition:Rational Number",
"Rational Numbers form Field",
"Category:Quadratic Equations",
"Category:Quadratic Irrationals"... |
proofwiki-2060 | Irrational Number has Periodic Continued Fraction iff Quadratic | Let $x$ be an irrational number.
Then $x$ is a quadratic irrational {{iff}} its continued fraction expansion is periodic. | Follows from:
* Limit of Simple Infinite Periodic Continued Fraction is Quadratic Irrational
* Quadratic Irrational has Periodic Continued Fraction Expansion
{{qed}}
Category:Simple Continued Fractions
Category:Quadratic Irrationals
n1vz0c7bwfzo6t12nrjthsngc91p4en | Let $x$ be an [[Definition:Irrational Number|irrational number]].
Then $x$ is a [[Definition:Quadratic Irrational|quadratic irrational]] {{iff}} its [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]] is [[Definition:Periodic Continued Fraction|periodic]]. | Follows from:
* [[Limit of Simple Infinite Periodic Continued Fraction is Quadratic Irrational]]
* [[Quadratic Irrational has Periodic Continued Fraction Expansion]]
{{qed}}
[[Category:Simple Continued Fractions]]
[[Category:Quadratic Irrationals]]
n1vz0c7bwfzo6t12nrjthsngc91p4en | Irrational Number has Periodic Continued Fraction iff Quadratic | https://proofwiki.org/wiki/Irrational_Number_has_Periodic_Continued_Fraction_iff_Quadratic | https://proofwiki.org/wiki/Irrational_Number_has_Periodic_Continued_Fraction_iff_Quadratic | [
"Simple Continued Fractions",
"Quadratic Irrationals"
] | [
"Definition:Irrational Number",
"Definition:Quadratic Irrational",
"Definition:Continued Fraction Expansion/Real Number",
"Definition:Periodic Continued Fraction"
] | [
"Limit of Simple Infinite Periodic Continued Fraction is Quadratic Irrational",
"Quadratic Irrational has Periodic Continued Fraction Expansion",
"Category:Simple Continued Fractions",
"Category:Quadratic Irrationals"
] |
proofwiki-2061 | Continued Fraction Expansion of Irrational Square Root | Let $n \in \Z$ such that $n$ is not a square.
Then the continued fraction expansion of $\sqrt n$ is of the form:
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
or
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
whe... | We use:
:Expansion of Associated Reduced Quadratic Irrational to establish a series of reduced quadratic irrationals associated to $n$
and then use
:Finitely Many Reduced Associated Quadratic Irrationals to assert this series must repeat (hence be periodic) due to the finite number of such irrationals.
Let:
:$a_0 = \f... | Let $n \in \Z$ such that $n$ is not a [[Definition:Square Number|square]].
Then the [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]] of $\sqrt n$ is of the form:
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
or
:$\sqbrk {a... | We use:
:[[Expansion of Associated Reduced Quadratic Irrational]] to establish a series of [[Definition:Reduced Quadratic Irrational|reduced quadratic irrationals]] [[Definition:Associated Quadratic Irrational|associated]] to $n$
and then use
:[[Finitely Many Reduced Associated Quadratic Irrationals]] to assert this s... | Continued Fraction Expansion of Irrational Square Root | https://proofwiki.org/wiki/Continued_Fraction_Expansion_of_Irrational_Square_Root | https://proofwiki.org/wiki/Continued_Fraction_Expansion_of_Irrational_Square_Root | [
"Continued Fraction Expansion of Irrational Square Root",
"Square Roots",
"Examples of Continued Fractions"
] | [
"Definition:Square Number",
"Definition:Continued Fraction Expansion/Real Number",
"Definition:Periodic Continued Fraction",
"Definition:Integer",
"Definition:Periodic Continued Fraction/Cycle",
"Definition:Palindrome",
"Definition:Periodic Continued Fraction/Cycle/Length",
"Definition:Periodic Contin... | [
"Expansion of Associated Reduced Quadratic Irrational",
"Definition:Quadratic Irrational/Reduced",
"Definition:Quadratic Irrational/Reduced/Associated",
"Finitely Many Reduced Associated Quadratic Irrationals",
"Definition:Quadratic Irrational/Reduced",
"Definition:Quadratic Irrational/Reduced",
"Defini... |
proofwiki-2062 | Solution of Pell's Equation is a Convergent | Let $x = a, y = b$ be a positive solution to Pell's equation $x^2 - n y^2 = 1$.
Then $\dfrac a b$ is a convergent of $\sqrt n$. | Let $a^2 - n b^2 = 1$.
Then we have:
:$\paren {a - b \sqrt n} \paren {a + b \sqrt n} = 1$.
So:
:$a - b \sqrt n = \dfrac 1 {a + b \sqrt n} > 0$
and so $a > b \sqrt n$.
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\sqrt n - \frac a b}
| r = \frac {a - b \sqrt n} b
| c =
}}
{{eqn | r = \frac 1 {b \paren {a + b... | Let $x = a, y = b$ be a [[Definition:Positive Integer|positive]] solution to [[Definition:Pell's Equation|Pell's equation]] $x^2 - n y^2 = 1$.
Then $\dfrac a b$ is a [[Definition:Convergent of Continued Fraction|convergent]] of $\sqrt n$. | Let $a^2 - n b^2 = 1$.
Then we have:
:$\paren {a - b \sqrt n} \paren {a + b \sqrt n} = 1$.
So:
:$a - b \sqrt n = \dfrac 1 {a + b \sqrt n} > 0$
and so $a > b \sqrt n$.
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\sqrt n - \frac a b}
| r = \frac {a - b \sqrt n} b
| c =
}}
{{eqn | r = \frac 1 {b \paren {... | Solution of Pell's Equation is a Convergent | https://proofwiki.org/wiki/Solution_of_Pell's_Equation_is_a_Convergent | https://proofwiki.org/wiki/Solution_of_Pell's_Equation_is_a_Convergent | [
"Pell's Equation",
"Continued Fractions"
] | [
"Definition:Positive/Integer",
"Definition:Pell's Equation",
"Definition:Convergent of Continued Fraction"
] | [
"Condition for Rational to be Convergent",
"Category:Pell's Equation",
"Category:Continued Fractions"
] |
proofwiki-2063 | Convergence of P-Series | Let $p \in \C$ be a complex number.
=== Absolute Convergence if $\map \Re p > 1$ ===
{{:Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1}}
=== Divergence if $0 <\map \Re p \le 1$ ===
{{:Convergence of P-Series/Divergence if p between 0 and 1}}
=== Real Case ===
{{:Convergence of P-Series/Re... | By the Cauchy Integral Test:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges {{iff}} the improper integral $\ds \int_1^\infty \frac {\d t} {t^x}$ exists.
The result follows from Integral to Infinity of Reciprocal of Power of x.
{{qed}} | Let $p \in \C$ be a [[Definition:Complex Number|complex number]].
=== [[Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1|Absolute Convergence if $\map \Re p > 1$]] ===
{{:Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1}}
=== [[Convergence of P-Series/Divergen... | By the [[Cauchy Integral Test]]:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ [[Definition:Convergent Real Sequence|converges]] {{iff}} the [[Definition:Improper Integral|improper integral]] $\ds \int_1^\infty \frac {\d t} {t^x}$ exists.
The result follows from [[Integral to Infinity of Reciprocal of Power of x]]... | Convergence of P-Series/Real/Proof 1 | https://proofwiki.org/wiki/Convergence_of_P-Series | https://proofwiki.org/wiki/Convergence_of_P-Series/Real/Proof_1 | [
"Convergence of P-Series",
"Convergence Tests",
"P-Series"
] | [
"Definition:Complex Number",
"Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1",
"Convergence of P-Series/Divergence if p between 0 and 1",
"Convergence of P-Series/Real"
] | [
"Cauchy Integral Test",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Improper Integral",
"Integral to Infinity of Reciprocal of Power of x"
] |
proofwiki-2064 | Convergence of P-Series | Let $p \in \C$ be a complex number.
=== Absolute Convergence if $\map \Re p > 1$ ===
{{:Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1}}
=== Divergence if $0 <\map \Re p \le 1$ ===
{{:Convergence of P-Series/Divergence if p between 0 and 1}}
=== Real Case ===
{{:Convergence of P-Series/Re... | Let $p = 1$.
Then from Harmonic Series is Divergent the $p$-series diverges.
So let $p > 1$.
We note that the sequence of partial sums is increasing.
Hence it is sufficient to show that they are bounded above.
Let:
:$s_{2^N} := 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dotsb + \dfrac 1 {N^p}$
Then:
{{begin-eqn}}
{{eqn | l... | Let $p \in \C$ be a [[Definition:Complex Number|complex number]].
=== [[Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1|Absolute Convergence if $\map \Re p > 1$]] ===
{{:Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1}}
=== [[Convergence of P-Series/Divergen... | Let $p = 1$.
Then from [[Harmonic Series is Divergent]] the [[Definition:P-Series|$p$-series]] [[Definition:Divergent Series|diverges]].
So let $p > 1$.
We note that the [[Definition:Sequence of Partial Sums|sequence of partial sums]] is [[Definition:Increasing Real Sequence|increasing]].
Hence it is [[Definition:... | Convergence of P-Series/Real/Proof 2 | https://proofwiki.org/wiki/Convergence_of_P-Series | https://proofwiki.org/wiki/Convergence_of_P-Series/Real/Proof_2 | [
"Convergence of P-Series",
"Convergence Tests",
"P-Series"
] | [
"Definition:Complex Number",
"Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1",
"Convergence of P-Series/Divergence if p between 0 and 1",
"Convergence of P-Series/Real"
] | [
"Harmonic Series is Divergent",
"Definition:P-Series",
"Definition:Divergent Series",
"Definition:Series/Sequence of Partial Sums",
"Definition:Increasing/Sequence/Real Sequence",
"Definition:Conditional/Sufficient Condition",
"Definition:Bounded Above Sequence/Real"
] |
proofwiki-2065 | Square Modulo 4 | Let $x \in \Z$ be an integer.
Then $x$ is:
:even {{iff}} $x^2 \equiv 0 \pmod 4$
:odd {{iff}} $x^2 \equiv 1 \pmod 4$ | {{begin-eqn}}
{{eqn | l = x
| r = 2 k
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^2
| r = \paren {2 k}^2
| c =
}}
{{eqn | r = 4 k^2
| c =
}}
{{eqn | o = \equiv
| r = 0
| rr= \pmod 4
| c =
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = x
| r = 2 k ... | Let $x \in \Z$ be an [[Definition:Integer|integer]].
Then $x$ is:
:[[Definition:Even Integer|even]] {{iff}} $x^2 \equiv 0 \pmod 4$
:[[Definition:Odd Integer|odd]] {{iff}} $x^2 \equiv 1 \pmod 4$ | {{begin-eqn}}
{{eqn | l = x
| r = 2 k
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^2
| r = \paren {2 k}^2
| c =
}}
{{eqn | r = 4 k^2
| c =
}}
{{eqn | o = \equiv
| r = 0
| rr= \pmod 4
| c =
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = x
| r = 2 ... | Square Modulo 4 | https://proofwiki.org/wiki/Square_Modulo_4 | https://proofwiki.org/wiki/Square_Modulo_4 | [
"Modulo Arithmetic",
"Square Numbers"
] | [
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] | [] |
proofwiki-2066 | Parity of Smaller Elements of Primitive Pythagorean Triple | Let $\left({x, y, z}\right)$ be a Pythagorean triple, that is, integers such that $x^2 + y^2 = z^2$.
Then $x$ and $y$ are of opposite parity. | From Smaller Elements of Pythagorean Triple not both Odd, $x$ and $y$ are not both odd.
{{AimForCont}} $x$ and $y$ are both even.
Then by definition they have $2$ as a common divisor.
But from Elements of Primitive Pythagorean Triple are Pairwise Coprime this cannot be the case.
So by Proof by Contradiction, $x$ and $y... | Let $\left({x, y, z}\right)$ be a [[Definition:Pythagorean Triple|Pythagorean triple]], that is, [[Definition:Integer|integers]] such that $x^2 + y^2 = z^2$.
Then $x$ and $y$ are of opposite [[Definition:Parity of Integer|parity]]. | From [[Smaller Elements of Pythagorean Triple not both Odd]], $x$ and $y$ are not both [[Definition:Odd Integer|odd]].
{{AimForCont}} $x$ and $y$ are both [[Definition:Even Integer|even]].
Then by definition they have $2$ as a [[Definition:Common Divisor of Integers|common divisor]].
But from [[Elements of Primitiv... | Parity of Smaller Elements of Primitive Pythagorean Triple | https://proofwiki.org/wiki/Parity_of_Smaller_Elements_of_Primitive_Pythagorean_Triple | https://proofwiki.org/wiki/Parity_of_Smaller_Elements_of_Primitive_Pythagorean_Triple | [
"Pythagorean Triples"
] | [
"Definition:Pythagorean Triple",
"Definition:Integer",
"Definition:Parity of Integer"
] | [
"Smaller Elements of Pythagorean Triple not both Odd",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Common Divisor/Integers",
"Elements of Primitive Pythagorean Triple are Pairwise Coprime",
"Proof by Contradiction",
"Definition:Even Integer",
"Definition:Parity of Integer",
"Cat... |
proofwiki-2067 | Dirichlet Series is Analytic | Let $(a_n)$ be sequence of complex numbers.
Let:
:$\ds \map f z = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^z}$
be the associated Dirichlet Series, which is defined at the points where the series converges.
Then $f$ is analytic in every open set such that the sum converges in the set. | {{ProofWanted}}
Category:Dirichlet Series
0jy9evo5zxvsiiapob40ddmut40aecw | Let $(a_n)$ be [[Definition:Complex Sequence|sequence of complex numbers]].
Let:
:$\ds \map f z = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^z}$
be the associated [[Definition:Dirichlet Series|Dirichlet Series]], which is defined at the points where the [[Definition:Series|series]] [[Definition:Convergent Series|con... | {{ProofWanted}}
[[Category:Dirichlet Series]]
0jy9evo5zxvsiiapob40ddmut40aecw | Dirichlet Series is Analytic | https://proofwiki.org/wiki/Dirichlet_Series_is_Analytic | https://proofwiki.org/wiki/Dirichlet_Series_is_Analytic | [
"Dirichlet Series"
] | [
"Definition:Complex Sequence",
"Definition:Dirichlet Series",
"Definition:Series",
"Definition:Convergent Series",
"Definition:Analytic Function",
"Definition:Convergent Series"
] | [
"Category:Dirichlet Series"
] |
proofwiki-2068 | Consecutive Integers are Coprime | $\forall h \in \Z$, $h$ and $h + 1$ have only two common factors: $1$ and $-1$.
That is, consecutive integers are always coprime. | $\gcd \set {h + 1, h} = \gcd \set {h, 1} = \gcd \set {1, 0} = 1$ from the Euclidean Algorithm.
{{qed}} | $\forall h \in \Z$, $h$ and $h + 1$ have only two common [[Definition:Factor (Algebra)|factors]]: $1$ and $-1$.
That is, consecutive [[Definition:Integer|integers]] are always [[Definition:Coprime Integers|coprime]]. | $\gcd \set {h + 1, h} = \gcd \set {h, 1} = \gcd \set {1, 0} = 1$ from the [[Euclidean Algorithm]].
{{qed}} | Consecutive Integers are Coprime/Proof 1 | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime/Proof_1 | [
"Consecutive Integers are Coprime",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Euclidean Algorithm"
] |
proofwiki-2069 | Consecutive Integers are Coprime | $\forall h \in \Z$, $h$ and $h + 1$ have only two common factors: $1$ and $-1$.
That is, consecutive integers are always coprime. | Let $k \in \Z: k \divides h$.
Also assume $k \divides \paren {h + 1}$.
Thus:
:$\exists a, b \in \N: a k = h, b k = \paren {h + 1}$
Then:
:$\paren {h + 1} - h = b k - a k$
and so:
:$1 = \paren {b - a} k$
Since the integers form an integral domain, $\paren {b - a} \in \Z$.
Thus either $k = 1$ and $b - a = 1$, or $k = -1$... | $\forall h \in \Z$, $h$ and $h + 1$ have only two common [[Definition:Factor (Algebra)|factors]]: $1$ and $-1$.
That is, consecutive [[Definition:Integer|integers]] are always [[Definition:Coprime Integers|coprime]]. | Let $k \in \Z: k \divides h$.
Also assume $k \divides \paren {h + 1}$.
Thus:
:$\exists a, b \in \N: a k = h, b k = \paren {h + 1}$
Then:
:$\paren {h + 1} - h = b k - a k$
and so:
:$1 = \paren {b - a} k$
Since the [[Integers form Integral Domain|integers form an integral domain]], $\paren {b - a} \in \Z$.
Thus eith... | Consecutive Integers are Coprime/Proof 2 | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime/Proof_2 | [
"Consecutive Integers are Coprime",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"Integers form Integral Domain"
] |
proofwiki-2070 | Consecutive Integers are Coprime | $\forall h \in \Z$, $h$ and $h + 1$ have only two common factors: $1$ and $-1$.
That is, consecutive integers are always coprime. | A direct application of GCD of Integer with Integer + $n$:
:$\gcd \set {a, a + n} \divides n$
{{qed}} | $\forall h \in \Z$, $h$ and $h + 1$ have only two common [[Definition:Factor (Algebra)|factors]]: $1$ and $-1$.
That is, consecutive [[Definition:Integer|integers]] are always [[Definition:Coprime Integers|coprime]]. | A direct application of [[GCD of Integer with Integer + n|GCD of Integer with Integer + $n$]]:
:$\gcd \set {a, a + n} \divides n$
{{qed}} | Consecutive Integers are Coprime/Proof 3 | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime | https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime/Proof_3 | [
"Consecutive Integers are Coprime",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Coprime/Integers"
] | [
"GCD of Integer with Integer + n"
] |
proofwiki-2071 | Elements of Primitive Pythagorean Triple are Pairwise Coprime | Let $\tuple {x, y, z}$ be a primitive Pythagorean triple.
Then:
:$x \perp y$
:$y \perp z$
:$x \perp z$
That is, all elements of $\tuple {x, y, z}$ are pairwise coprime. | We have that $x \perp y$ by definition.
Suppose there is a prime divisor $p$ of both $x$ and $z$.
That is:
:$\exists p \in \mathbb P: p \divides x, p \divides z$
Then from Prime Divides Power:
:$p \divides x^2, p \divides z^2$
Then from Common Divisor Divides Integer Combination:
:$p \divides \paren {z^2 - x^2} = y^2$
... | Let $\tuple {x, y, z}$ be a [[Definition:Primitive Pythagorean Triple|primitive Pythagorean triple]].
Then:
:$x \perp y$
:$y \perp z$
:$x \perp z$
That is, all [[Definition:Element|elements]] of $\tuple {x, y, z}$ are [[Definition:Pairwise Coprime Integers|pairwise coprime]]. | We have that $x \perp y$ by [[Definition:Primitive Pythagorean Triple|definition]].
Suppose there is a [[Definition:Prime Number|prime]] [[Definition:Divisor of Integer|divisor]] $p$ of both $x$ and $z$.
That is:
:$\exists p \in \mathbb P: p \divides x, p \divides z$
Then from [[Prime Divides Power]]:
:$p \divides x... | Elements of Primitive Pythagorean Triple are Pairwise Coprime | https://proofwiki.org/wiki/Elements_of_Primitive_Pythagorean_Triple_are_Pairwise_Coprime | https://proofwiki.org/wiki/Elements_of_Primitive_Pythagorean_Triple_are_Pairwise_Coprime | [
"Pythagorean Triples",
"Coprime Integers"
] | [
"Definition:Pythagorean Triple/Primitive",
"Definition:Element",
"Definition:Pairwise Coprime/Integers"
] | [
"Definition:Pythagorean Triple/Primitive",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Prime Divides Power",
"Common Divisor Divides Integer Combination",
"Prime Divides Power",
"Definition:Pythagorean Triple/Primitive",
"Category:Pythagorean Triples",
"Category:Coprime Integ... |
proofwiki-2072 | Elements of Primitive Pythagorean Triples Modulo 4 | Let $x \in \Z: x > 2$.
Then $x$ is an element of some primitive Pythagorean triple {{iff}} $x \not \equiv 2 \pmod 4$. | Let $m = k + 1, n = k$ where $k \in \Z: k \ge 1$.
From Consecutive Integers are Coprime, $m \perp n$.
Then we have:
:$m, n \in \Z$ are positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
From Solutions of Pythagorean Equation, these conditions are necessary and... | Let $x \in \Z: x > 2$.
Then $x$ is an element of some [[Definition:Primitive Pythagorean Triple|primitive Pythagorean triple]] {{iff}} $x \not \equiv 2 \pmod 4$. | Let $m = k + 1, n = k$ where $k \in \Z: k \ge 1$.
From [[Consecutive Integers are Coprime]], $m \perp n$.
Then we have:
:$m, n \in \Z$ are [[Definition:Positive Integer|positive integers]]
:$m \perp n$, that is, $m$ and $n$ are [[Definition:Coprime Integers|coprime]]
:$m$ and $n$ are of [[Definition:Parity of Integer... | Elements of Primitive Pythagorean Triples Modulo 4 | https://proofwiki.org/wiki/Elements_of_Primitive_Pythagorean_Triples_Modulo_4 | https://proofwiki.org/wiki/Elements_of_Primitive_Pythagorean_Triples_Modulo_4 | [
"Pythagorean Triples"
] | [
"Definition:Pythagorean Triple/Primitive"
] | [
"Consecutive Integers are Coprime",
"Definition:Positive/Integer",
"Definition:Coprime/Integers",
"Definition:Parity of Integer",
"Solutions of Pythagorean Equation",
"Definition:Biconditional/Semantics of Biconditional/Necessary and Sufficient",
"Definition:Pythagorean Triple/Primitive",
"Definition:... |
proofwiki-2073 | Odd Square Modulo 8 | Let $x \in \Z$ be an odd square.
Then $x \equiv 1 \pmod 8$. | Let $x \in \Z$ be an odd square.
Then $x = n^2$ where $n$ is also odd.
Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$.
Hence:
{{begin-eqn}}
{{eqn | l = x
| r = n^2
| c =
}}
{{eqn | r = \paren {2 k + 1}^2
| c =
}}
{{eqn | r = 4 k^2 + 4 k + 1
| c =
}}
{{eqn | r = 4 k \paren {k + 1} ... | Let $x \in \Z$ be an [[Definition:Odd Integer|odd]] [[Definition:Square Number|square]].
Then $x \equiv 1 \pmod 8$. | Let $x \in \Z$ be an [[Definition:Odd Integer|odd]] [[Definition:Square Number|square]].
Then $x = n^2$ where $n$ is also [[Definition:Odd Integer|odd]].
Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$.
Hence:
{{begin-eqn}}
{{eqn | l = x
| r = n^2
| c =
}}
{{eqn | r = \paren {2 k + 1}^2
... | Odd Square Modulo 8 | https://proofwiki.org/wiki/Odd_Square_Modulo_8 | https://proofwiki.org/wiki/Odd_Square_Modulo_8 | [
"Modulo Arithmetic",
"Odd Squares"
] | [
"Definition:Odd Integer",
"Definition:Square Number"
] | [
"Definition:Odd Integer",
"Definition:Square Number",
"Definition:Odd Integer",
"Definition:Parity of Integer"
] |
proofwiki-2074 | Sum of Consecutive Triangular Numbers is Square | The sum of two consecutive triangular numbers is a square number. | Let $T_{n - 1}$ and $T_n$ be two consecutive triangular numbers.
From Closed Form for Triangular Numbers, we have:
:$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$
:$T_n = \dfrac {n \paren {n + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = T_{n - 1} + T_n
| r = \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2
... | The sum of two consecutive [[Definition:Triangular Number|triangular numbers]] is a [[Definition:Square Number|square number]]. | Let $T_{n - 1}$ and $T_n$ be two consecutive [[Definition:Triangular Number|triangular numbers]].
From [[Closed Form for Triangular Numbers]], we have:
:$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$
:$T_n = \dfrac {n \paren {n + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = T_{n - 1} + T_n
| r = \frac {\paren {n - 1... | Sum of Consecutive Triangular Numbers is Square | https://proofwiki.org/wiki/Sum_of_Consecutive_Triangular_Numbers_is_Square | https://proofwiki.org/wiki/Sum_of_Consecutive_Triangular_Numbers_is_Square | [
"Triangular Numbers",
"Square Numbers",
"Sum of Consecutive Triangular Numbers is Square"
] | [
"Definition:Triangular Number",
"Definition:Square Number"
] | [
"Definition:Triangular Number",
"Closed Form for Triangular Numbers"
] |
proofwiki-2075 | If n is Triangular then so is 9n + 1 | Let $n$ be a triangular number.
Then $9 n + 1$ is also triangular. | Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = 9 n + 1
| r = 9 \frac {k \paren {k + 1} } 2 + 1
| c =
}}
{{eqn | r = \frac {9 k^2 + 9 k + 2} 2
| c =
}}
{{eqn | r = \frac {\paren {3 k + 1} \paren {3 k + 2} } 2
| c =
}}
{{end-eqn}... | Let $n$ be a [[Definition:Triangular Number|triangular number]].
Then $9 n + 1$ is also [[Definition:Triangular Number|triangular]]. | Let $n$ be [[Definition:Triangular Number|triangular]].
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = 9 n + 1
| r = 9 \frac {k \paren {k + 1} } 2 + 1
| c =
}}
{{eqn | r = \frac {9 k^2 + 9 k + 2} 2
| c =
}}
{{eqn | r = \frac {\paren {3 k + 1} \paren {3 k... | If n is Triangular then so is 9n + 1 | https://proofwiki.org/wiki/If_n_is_Triangular_then_so_is_9n_+_1 | https://proofwiki.org/wiki/If_n_is_Triangular_then_so_is_9n_+_1 | [
"Triangular Numbers"
] | [
"Definition:Triangular Number",
"Definition:Triangular Number"
] | [
"Definition:Triangular Number",
"Definition:Triangular Number"
] |
proofwiki-2076 | Sum of Arithmetic Sequence | Let $\sequence {a_k}$ be an arithmetic sequence defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \frac {n - 1} 2 d}
| c =
}}
{{eqn | r = \frac {n \paren {a + l} } 2
... | We have that:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = 2 \paren {a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n... | Let $\sequence {a_k}$ be an [[Definition:Arithmetic Sequence|arithmetic sequence]] defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its [[Definition:Closed-Form Expression|closed-form expression]] is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \... | We have that:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = 2 \paren {a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren... | Sum of Arithmetic Sequence | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence | [
"Sums of Sequences",
"Arithmetic Sequences",
"Sum of Arithmetic Sequence"
] | [
"Definition:Arithmetic Sequence",
"Definition:Closed Form Expression",
"Definition:Arithmetic Sequence/Last Term"
] | [] |
proofwiki-2077 | Sum of Arithmetic Sequence | Let $\sequence {a_k}$ be an arithmetic sequence defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \frac {n - 1} 2 d}
| c =
}}
{{eqn | r = \frac {n \paren {a + l} } 2
... | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = m}^n j
| r = \sum_{j \mathop = 0}^{n - m} \paren {m + j}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = \paren {n - m + 1} \paren {m + \frac {n - m} 2}
| c = Sum of Arithmetic Sequence
}}
{{eqn | r = m \paren {n - m + 1} + \frac 1 2 \paren {... | Let $\sequence {a_k}$ be an [[Definition:Arithmetic Sequence|arithmetic sequence]] defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its [[Definition:Closed-Form Expression|closed-form expression]] is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \... | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = m}^n j
| r = \sum_{j \mathop = 0}^{n - m} \paren {m + j}
| c = [[Translation of Index Variable of Summation]]
}}
{{eqn | r = \paren {n - m + 1} \paren {m + \frac {n - m} 2}
| c = [[Sum of Arithmetic Sequence]]
}}
{{eqn | r = m \paren {n - m + 1} + \frac 1 2 ... | Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 1 | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence/Examples/Sum_of_j_from_m_to_n/Proof_1 | [
"Sums of Sequences",
"Arithmetic Sequences",
"Sum of Arithmetic Sequence"
] | [
"Definition:Arithmetic Sequence",
"Definition:Closed Form Expression",
"Definition:Arithmetic Sequence/Last Term"
] | [
"Translation of Index Variable of Summation",
"Sum of Arithmetic Sequence"
] |
proofwiki-2078 | Sum of Arithmetic Sequence | Let $\sequence {a_k}$ be an arithmetic sequence defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \frac {n - 1} 2 d}
| c =
}}
{{eqn | r = \frac {n \paren {a + l} } 2
... | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = m}^n j
| r = \sum_{j \mathop = 0}^n j - \sum_{j \mathop = 0}^{m - 1} j
| c =
}}
{{eqn | r = \frac {n \paren {n + 1} } 2 - \frac {\paren {m - 1} m} 2
| c = Closed Form for Triangular Numbers
}}
{{end-eqn}} | Let $\sequence {a_k}$ be an [[Definition:Arithmetic Sequence|arithmetic sequence]] defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its [[Definition:Closed-Form Expression|closed-form expression]] is:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}
| r = n \paren {a + \... | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = m}^n j
| r = \sum_{j \mathop = 0}^n j - \sum_{j \mathop = 0}^{m - 1} j
| c =
}}
{{eqn | r = \frac {n \paren {n + 1} } 2 - \frac {\paren {m - 1} m} 2
| c = [[Closed Form for Triangular Numbers]]
}}
{{end-eqn}} | Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 2 | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence | https://proofwiki.org/wiki/Sum_of_Arithmetic_Sequence/Examples/Sum_of_j_from_m_to_n/Proof_2 | [
"Sums of Sequences",
"Arithmetic Sequences",
"Sum of Arithmetic Sequence"
] | [
"Definition:Arithmetic Sequence",
"Definition:Closed Form Expression",
"Definition:Arithmetic Sequence/Last Term"
] | [
"Closed Form for Triangular Numbers"
] |
proofwiki-2079 | Odd Square is Eight Triangles Plus One | Let $n \in \Z$ be an odd integer.
Then $n$ is square {{iff}} $n = 8 m + 1$ where $m$ is triangular. | Follows directly from the identity:
{{begin-eqn}}
{{eqn | l = 8 \frac {k \paren {k + 1} } 2 + 1
| r = 4 k^2 + 4 k + 1
| c =
}}
{{eqn | r = \paren {2 k + 1}^2
| c =
}}
{{end-eqn}}
as follows:
Let $m$ be triangular.
Then from Closed Form for Triangular Numbers:
:$\exists k \in \Z: m = \dfrac {k \paren... | Let $n \in \Z$ be an [[Definition:Odd Integer|odd integer]].
Then $n$ is [[Definition:Square Number|square]] {{iff}} $n = 8 m + 1$ where $m$ is [[Definition:Triangular Number|triangular]]. | Follows directly from the identity:
{{begin-eqn}}
{{eqn | l = 8 \frac {k \paren {k + 1} } 2 + 1
| r = 4 k^2 + 4 k + 1
| c =
}}
{{eqn | r = \paren {2 k + 1}^2
| c =
}}
{{end-eqn}}
as follows:
Let $m$ be [[Definition:Triangular Number|triangular]].
Then from [[Closed Form for Triangular Numbers]]... | Odd Square is Eight Triangles Plus One | https://proofwiki.org/wiki/Odd_Square_is_Eight_Triangles_Plus_One | https://proofwiki.org/wiki/Odd_Square_is_Eight_Triangles_Plus_One | [
"8",
"Triangular Numbers",
"Odd Squares"
] | [
"Definition:Odd Integer",
"Definition:Square Number",
"Definition:Triangular Number"
] | [
"Definition:Triangular Number",
"Closed Form for Triangular Numbers",
"Definition:Odd Integer",
"Definition:Square Number",
"Definition:Odd Integer",
"Definition:Square Number",
"Definition:Odd Integer",
"Definition:Triangular Number"
] |
proofwiki-2080 | Triangular Number Modulo 3 and 9 | Let $n$ be a triangular number.
Then one of the following two conditions applies:
:$n \equiv 0 \pmod 3$
:$n \equiv 1 \pmod 9$ | Let $n = T_r$.
Then from Closed Form for Triangular Numbers:
:$n = \dfrac {r \paren {r + 1} } 2$
It suffices from Euclid's Lemma to investigate the nature of $r \paren {r + 1}$ modulo $3$.
There are three cases to consider:
:$r \equiv 0 \pmod 3$
:$r \equiv 1 \pmod 3$
:$r \equiv 2 \pmod 3$
Let $r \equiv 0 \pmod 3$.
Then... | Let $n$ be a [[Definition:Triangular Number|triangular number]].
Then one of the following two conditions applies:
:$n \equiv 0 \pmod 3$
:$n \equiv 1 \pmod 9$ | Let $n = T_r$.
Then from [[Closed Form for Triangular Numbers]]:
:$n = \dfrac {r \paren {r + 1} } 2$
It suffices from [[Euclid's Lemma]] to investigate the nature of $r \paren {r + 1}$ modulo $3$.
There are three cases to consider:
:$r \equiv 0 \pmod 3$
:$r \equiv 1 \pmod 3$
:$r \equiv 2 \pmod 3$
Let $r \equiv 0 \... | Triangular Number Modulo 3 and 9 | https://proofwiki.org/wiki/Triangular_Number_Modulo_3_and_9 | https://proofwiki.org/wiki/Triangular_Number_Modulo_3_and_9 | [
"Triangular Numbers"
] | [
"Definition:Triangular Number"
] | [
"Closed Form for Triangular Numbers",
"Euclid's Lemma",
"Category:Triangular Numbers"
] |
proofwiki-2081 | Closed Form for Polygonal Numbers | Let $\map P {k, n}$ be the $n$th $k$-gonal number.
The closed-form expression for $\map P {k, n}$ is given by:
{{begin-eqn}}
{{eqn | l = \map P {k, n}
| r = \frac n 2 \paren {\paren {k - 2} n - k + 4}
}}
{{eqn | r = \frac {k - 2} 2 \paren {n^2 - n} + n
}}
{{eqn | r = 2 n + \dfrac 1 2 n k \paren {n - 1} - n^2
}}
{... | By definition of the $n$th $k$-gonal number:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Then:
:$\paren {\paren {k - 2} \paren {j - 1} + 1}$
is an arithmetic sequence.
Its initial term $a$ is $1$, and its common difference $d$ is $k - 2$.
He... | Let $\map P {k, n}$ be the $n$th [[Definition:Polygonal Number|$k$-gonal number]].
The [[Definition:Closed-Form Expression|closed-form expression]] for $\map P {k, n}$ is given by:
{{begin-eqn}}
{{eqn | l = \map P {k, n}
| r = \frac n 2 \paren {\paren {k - 2} n - k + 4}
}}
{{eqn | r = \frac {k - 2} 2 \paren {n^... | By definition of the $n$th [[Definition:Polygonal Number|$k$-gonal number]]:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Then:
:$\paren {\paren {k - 2} \paren {j - 1} + 1}$
is an [[Definition:Arithmetic Sequence|arithmetic sequence]].
... | Closed Form for Polygonal Numbers | https://proofwiki.org/wiki/Closed_Form_for_Polygonal_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Polygonal_Numbers | [
"Polygonal Numbers",
"Closed Forms"
] | [
"Definition:Polygonal Number",
"Definition:Closed Form Expression"
] | [
"Definition:Polygonal Number",
"Definition:Arithmetic Sequence",
"Sum of Arithmetic Sequence"
] |
proofwiki-2082 | Difference Between Adjacent Polygonal Numbers is Triangular Number | Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
:$\map P {k + 1, n} - \map P {k, n} = T_{n - 1}$
where $T_n$ is the $n$th triangular number. | From Closed Form for Polygonal Numbers:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Thus:
{{begin-eqn}}
{{eqn | l = \map P {k + 1, n} - \map P {k, n}
| r = \dfrac n 2 \paren {\paren {\paren {k + 1} - 2} n - \paren {k + 1} + 4} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}
| c =
}}
{{eq... | Let $\map P {k, n}$ be the $n$th [[Definition:Polygonal Number|$k$-gonal number]].
Then:
:$\map P {k + 1, n} - \map P {k, n} = T_{n - 1}$
where $T_n$ is the $n$th [[Definition:Triangular Number|triangular number]]. | From [[Closed Form for Polygonal Numbers]]:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Thus:
{{begin-eqn}}
{{eqn | l = \map P {k + 1, n} - \map P {k, n}
| r = \dfrac n 2 \paren {\paren {\paren {k + 1} - 2} n - \paren {k + 1} + 4} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}
| c =
}... | Difference Between Adjacent Polygonal Numbers is Triangular Number | https://proofwiki.org/wiki/Difference_Between_Adjacent_Polygonal_Numbers_is_Triangular_Number | https://proofwiki.org/wiki/Difference_Between_Adjacent_Polygonal_Numbers_is_Triangular_Number | [
"Polygonal Numbers",
"Triangular Numbers"
] | [
"Definition:Polygonal Number",
"Definition:Triangular Number"
] | [
"Closed Form for Polygonal Numbers"
] |
proofwiki-2083 | Method of Infinite Descent | {{mistake|This technique only works if you can prove that there exists some $n_\gamma$ such that $0 \le n_\gamma < n_\alpha$ for which $\map P {n_\gamma}$ does not work. Otherwise, technically speaking, the proof by Fermat in the historical note is not actually an example of this proof technique. I challenge the studen... | Suppose that $\map P {n_\alpha}$ holds.
Then from the descent step, $\exists n_\beta \in \N_{n_\alpha}: \map P {n_\beta}$.
The descent step then tell us we can deduce a smaller positive solution, $n_\gamma$, such that $\map P {n_\gamma}$ is true and $n_\gamma \in \N_{n_\beta}$.
And again, the descent step tells us we c... | {{mistake|This technique only works if you can prove that there exists some $n_\gamma$ such that $0 \le n_\gamma < n_\alpha$ for which $\map P {n_\gamma}$ does not work. Otherwise, technically speaking, the proof by Fermat in the historical note is not actually an example of this proof technique. I challenge the studen... | Suppose that $\map P {n_\alpha}$ holds.
Then from the descent step, $\exists n_\beta \in \N_{n_\alpha}: \map P {n_\beta}$.
The descent step then tell us we can deduce a smaller positive solution, $n_\gamma$, such that $\map P {n_\gamma}$ is true and $n_\gamma \in \N_{n_\beta}$.
And again, the descent step tells us w... | Method of Infinite Descent/Proof 2 | https://proofwiki.org/wiki/Method_of_Infinite_Descent | https://proofwiki.org/wiki/Method_of_Infinite_Descent/Proof_2 | [
"Method of Infinite Descent",
"Proof Techniques"
] | [
"Definition:Propositional Function",
"Definition:False",
"Definition:True",
"Definition:Natural Numbers",
"Definition:True",
"Definition:True",
"Definition:True"
] | [
"Definition:Bounded Below Set",
"Well-Ordering Principle",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Bounded Below Set",
"Definition:Smallest Element",
"Definition:Empty Set"
] |
proofwiki-2084 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | From the extension to the general Brahmagupta-Fibonacci Identity:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$, for all $m \in \Z$.
The result follows by setting $m = 1$.
{{qed}} | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | From the [[Brahmagupta-Fibonacci Identity/Extension/General|extension to the general Brahmagupta-Fibonacci Identity]]:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$, for all $m \in \Z$.
The result follows by setting $m = 1$.
{{qed}} | Brahmagupta-Fibonacci Identity/Extension/Proof 1 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Extension/Proof_1 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Brahmagupta-Fibonacci Identity/Extension/General"
] |
proofwiki-2085 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
$\map P 1$ is the trivial case:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 1}^1 \paren { {a_j}^2 + {b_j}^2}
| r = {a_1}^2 + ... | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
$\map P 1$ is the trivial case:
{{begin-eqn}}
{{eqn | l = \pro... | Brahmagupta-Fibonacci Identity/Extension/Proof 2 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Extension/Proof_2 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Brahmagupta-Fibonacci Identity",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Brahmagupta-Fibonacci Identity/Extension/Proof 2",
"Brahmagupta-Fibonacci Identity/Extension/Proof 2... |
proofwiki-2086 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | Let $z_j = a_j + i b_j$ for each $j = 1, 2, \ldots, n$.
Let $\ds c + i d = \prod_{j \mathop = 1}^n z_j$.
Then:
{{begin-eqn}}
{{eqn | l = c + i d
| r = \prod_{j \mathop = 1}^n z_j
}}
{{eqn | r = \prod_{j \mathop = 1}^n \paren {a_j + i b_j}
}}
{{end-eqn}}
As $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are i... | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | Let $z_j = a_j + i b_j$ for each $j = 1, 2, \ldots, n$.
Let $\ds c + i d = \prod_{j \mathop = 1}^n z_j$.
Then:
{{begin-eqn}}
{{eqn | l = c + i d
| r = \prod_{j \mathop = 1}^n z_j
}}
{{eqn | r = \prod_{j \mathop = 1}^n \paren {a_j + i b_j}
}}
{{end-eqn}}
As $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ ... | Brahmagupta-Fibonacci Identity/Extension/Proof 3 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Extension/Proof_3 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Definition:Integer",
"Complex Modulus of Product of Complex Numbers/General Result"
] |
proofwiki-2087 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {a c + b d}^2 + \paren {a d - b c}^2
| c =
}}
{{eqn | r = \paren {\paren {a c}^2 + 2 \paren {a c} \paren {b d} + \paren {b d}^2} + \paren {\paren {a d}^2 - 2 \paren {a d} \paren {b c} + \paren {b c}^2}
| c = Square of Sum, Square of Difference
}}
{{eqn | r = a^... | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {a c + b d}^2 + \paren {a d - b c}^2
| c =
}}
{{eqn | r = \paren {\paren {a c}^2 + 2 \paren {a c} \paren {b d} + \paren {b d}^2} + \paren {\paren {a d}^2 - 2 \paren {a d} \paren {b c} + \paren {b c}^2}
| c = [[Square of Sum]], [[Square of Difference]]
}}
{{eqn ... | Brahmagupta-Fibonacci Identity/Proof 1 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Proof_1 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Square of Sum",
"Square of Difference"
] |
proofwiki-2088 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | From the more general version of Brahmagupta-Fibonacci Identity:
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c + n b d}^2 + n \paren {a d - b c}^2$
The result follows by setting $n = 1$.
{{qed}} | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | From the [[Brahmagupta-Fibonacci Identity/General|more general version of Brahmagupta-Fibonacci Identity]]:
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c + n b d}^2 + n \paren {a d - b c}^2$
The result follows by setting $n = 1$.
{{qed}} | Brahmagupta-Fibonacci Identity/Proof 2 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Proof_2 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Brahmagupta-Fibonacci Identity/General"
] |
proofwiki-2089 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | Lagrange's Identity gives:
:$\ds \paren {\sum_{k \mathop = 1}^n {a_k}^2} \paren {\sum_{k \mathop = 1}^n {b_k}^2} - \paren {\sum_{k \mathop = 1}^n a_k b_k}^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \paren {a_i b_j - a_j b_i}^2$
Setting $n = 2$:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{k \mathop = 1}^2 {... | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | [[Lagrange's Identity]] gives:
:$\ds \paren {\sum_{k \mathop = 1}^n {a_k}^2} \paren {\sum_{k \mathop = 1}^n {b_k}^2} - \paren {\sum_{k \mathop = 1}^n a_k b_k}^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \paren {a_i b_j - a_j b_i}^2$
Setting $n = 2$:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{k \mathop = ... | Brahmagupta-Fibonacci Identity/Proof 3 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Proof_3 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Lagrange's Identity"
] |
proofwiki-2090 | Brahmagupta-Fibonacci Identity | Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | Let $z_1 = a - b i, z_2 = c + d i$ be complex numbers.
Let $\cmod z$ denote the complex modulus of a given complex number $z \in \C$.
By definition of complex multiplication:
:$(1): \quad z_1 z_2 = \paren {a c + b d} + \paren {a d - b c} i$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_1 z_2}
| r = \cmod {z_1} \cmod{... | Let $a, b, c, d$ be [[Definition:Number|numbers]].
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$ | Let $z_1 = a - b i, z_2 = c + d i$ be [[Definition:Complex Number|complex numbers]].
Let $\cmod z$ denote the [[Definition:Complex Modulus|complex modulus]] of a given [[Definition:Complex Number|complex number]] $z \in \C$.
By definition of [[Definition:Complex Multiplication|complex multiplication]]:
:$(1): \quad... | Brahmagupta-Fibonacci Identity/Proof 4 | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity | https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity/Proof_4 | [
"Brahmagupta-Fibonacci Identity",
"Sums of Squares",
"Algebra"
] | [
"Definition:Number"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus",
"Definition:Complex Number",
"Definition:Multiplication/Complex Numbers",
"Complex Modulus of Product of Complex Numbers"
] |
proofwiki-2091 | Fermat's Two Squares Theorem | Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands. | === Uniqueness Lemma ===
{{:Fermat's Two Squares Theorem/Uniqueness Lemma}}{{qed|lemma}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Then $p$ can be expressed as the sum of two [[Definition:Square Number|squares]] {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a [[Definition:Prime Number|prime]] of the form $4 k + 1$ as the sum of two [[Definition:Square Number|squares]... | === [[Fermat's Two Squares Theorem/Uniqueness Lemma|Uniqueness Lemma]] ===
{{:Fermat's Two Squares Theorem/Uniqueness Lemma}}{{qed|lemma}} | Fermat's Two Squares Theorem | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem | [
"Fermat's Two Squares Theorem",
"Sums of Squares",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Square Number",
"Definition:Prime Number",
"Definition:Square Number"
] | [
"Fermat's Two Squares Theorem/Uniqueness Lemma"
] |
proofwiki-2092 | Fermat's Two Squares Theorem | Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands. | Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a, b, c, d \in \Z_{>0}$.
We have that:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
| r = \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b
| c = multiplying out and gathering terms
}}
{{eqn | r = p \paren {a b + c d}
| c = as $p = a^2 + b^... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then $p$ can be expressed as the sum of two [[Definition:Square Number|squares]] {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a [[Definition:Prime Number|prime]] of the form $4 k + 1$ as the sum of two [[Definition:Square Number|squares]... | Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a, b, c, d \in \Z_{>0}$.
We have that:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
| r = \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b
| c = multiplying out and gathering terms
}}
{{eqn | r = p \paren {a b + c d}
| c = as $p = a^2 + ... | Fermat's Two Squares Theorem/Uniqueness Lemma/Proof 1 | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem/Uniqueness_Lemma/Proof_1 | [
"Fermat's Two Squares Theorem",
"Sums of Squares",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Square Number",
"Definition:Prime Number",
"Definition:Square Number"
] | [
"Definition:Prime Number",
"Euclid's Lemma for Prime Divisors",
"Brahmagupta-Fibonacci Identity",
"Square of Real Number is Non-Negative",
"Definition:Square/Function"
] |
proofwiki-2093 | Fermat's Two Squares Theorem | Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands. | Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a > b > 0$ and $c > d > 0$.
We are going to show that $a = c$ and $b = d$.
From the two expressions for $p$, we have:
{{begin-eqn}}
{{eqn | l = \paren {a d - b c} \paren {a d + b c}
| r = a^2 d^2 - b^2 c^2
| c = Difference of Two Squares
}}
{{eqn | r = \paren {p ... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then $p$ can be expressed as the sum of two [[Definition:Square Number|squares]] {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a [[Definition:Prime Number|prime]] of the form $4 k + 1$ as the sum of two [[Definition:Square Number|squares]... | Suppose:
:$p = a^2 + b^2 = c^2 + d^2$
where $a > b > 0$ and $c > d > 0$.
We are going to show that $a = c$ and $b = d$.
From the two expressions for $p$, we have:
{{begin-eqn}}
{{eqn | l = \paren {a d - b c} \paren {a d + b c}
| r = a^2 d^2 - b^2 c^2
| c = [[Difference of Two Squares]]
}}
{{eqn | r = \p... | Fermat's Two Squares Theorem/Uniqueness Lemma/Proof 2 | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem/Uniqueness_Lemma/Proof_2 | [
"Fermat's Two Squares Theorem",
"Sums of Squares",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Square Number",
"Definition:Prime Number",
"Definition:Square Number"
] | [
"Difference of Two Squares",
"Euclid's Lemma",
"Brahmagupta-Fibonacci Identity",
"Definition:Contradiction",
"Definition:Divisor (Algebra)/Integer",
"Definition:Common Divisor/Integers",
"Definition:Prime Number",
"Euclid's Lemma"
] |
proofwiki-2094 | Fermat's Two Squares Theorem | Let $p$ be a prime number.
Then $p$ can be expressed as the sum of two squares {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a prime of the form $4 k + 1$ as the sum of two squares is unique except for the order of the two summands. | For $p = 2$ the only possibility is:
:$p = 1^2 + 1^2$
Assume $p \ge 3$.
Let $a \ge b$ and $c \ge d$ satisfy:
:$p = a^2 + b^2 = c^2 + d^2$
As $2 \nmid p$, we have:
:$(1): \quad a > b$ and $c > d$
Observe:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
| r = \paren {a^2 + b^2} c d + \paren {c^2 + d... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then $p$ can be expressed as the sum of two [[Definition:Square Number|squares]] {{iff}} either:
:$p = 2$
or:
:$p \equiv 1 \pmod 4$
The expression of a [[Definition:Prime Number|prime]] of the form $4 k + 1$ as the sum of two [[Definition:Square Number|squares]... | For $p = 2$ the only possibility is:
:$p = 1^2 + 1^2$
Assume $p \ge 3$.
Let $a \ge b$ and $c \ge d$ satisfy:
:$p = a^2 + b^2 = c^2 + d^2$
As $2 \nmid p$, we have:
:$(1): \quad a > b$ and $c > d$
Observe:
{{begin-eqn}}
{{eqn | l = \paren {a c + b d} \paren {a d + b c}
| r = \paren {a^2 + b^2} c d + \paren {c... | Fermat's Two Squares Theorem/Uniqueness Lemma/Proof 3 | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem | https://proofwiki.org/wiki/Fermat's_Two_Squares_Theorem/Uniqueness_Lemma/Proof_3 | [
"Fermat's Two Squares Theorem",
"Sums of Squares",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Square Number",
"Definition:Prime Number",
"Definition:Square Number"
] | [
"Real Number Ordering is Compatible with Multiplication",
"Brahmagupta-Fibonacci Identity"
] |
proofwiki-2095 | Integer as Sum of Two Squares | Let $n$ be a positive integer.
Then:
: $n$ can be expressed as the sum of two squares
{{iff}}:
:each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power. | Let us extract the largest square divisor of $n$, and write:
:$n = m^2 r$
where $r$ is square-free. | Let $n$ be a [[Definition:Positive Integer|positive integer]].
Then:
: $n$ can be expressed as the sum of two [[Definition:Square Number|squares]]
{{iff}}:
:each of its [[Definition:Prime Factor|prime divisors]] of the form $4 k + 3$ (if any) occur to an [[Definition:Even Power|even power]]. | Let us extract the largest [[Definition:Square Number|square]] [[Definition:Divisor of Integer|divisor]] of $n$, and write:
:$n = m^2 r$
where $r$ is [[Definition:Square-Free Integer|square-free]]. | Integer as Sum of Two Squares | https://proofwiki.org/wiki/Integer_as_Sum_of_Two_Squares | https://proofwiki.org/wiki/Integer_as_Sum_of_Two_Squares | [
"Sums of Squares"
] | [
"Definition:Positive/Integer",
"Definition:Square Number",
"Definition:Prime Factor",
"Definition:Power (Algebra)/Even Power"
] | [
"Definition:Square Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Square-Free Integer",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square-Free Integer"
] |
proofwiki-2096 | Square Modulo 8 | Let $x \in \Z$ be an integer.
: If $x$ is even then:
::: $x^2 \equiv 0 \pmod 8$ or $x^2 \equiv 4 \pmod 8$
: If $x$ is odd then:
::: $x^2 \equiv 1 \pmod 8$ | === Proof for Even Integer ===
Let $x \in \Z$ be even.
Then from Square Modulo 4:
: $x^2 \equiv 0 \pmod 4$
Hence there are two possibilities for $x^2$:
: $x^2 \equiv 0 \pmod 8$
: $x^2 \equiv 4 \pmod 8$
The fact that there do exist such squares can be demonstrated by example:
: $2^2 = 4 \equiv 4 \pmod 8$
: $4^2 = 16 \eq... | Let $x \in \Z$ be an [[Definition:Integer|integer]].
: If $x$ is [[Definition:Even Integer|even]] then:
::: $x^2 \equiv 0 \pmod 8$ or $x^2 \equiv 4 \pmod 8$
: If $x$ is [[Definition:Odd Integer|odd]] then:
::: $x^2 \equiv 1 \pmod 8$ | === Proof for Even Integer ===
Let $x \in \Z$ be [[Definition:Even Integer|even]].
Then from [[Square Modulo 4]]:
: $x^2 \equiv 0 \pmod 4$
Hence there are two possibilities for $x^2$:
: $x^2 \equiv 0 \pmod 8$
: $x^2 \equiv 4 \pmod 8$
The fact that there do exist such [[Definition:Square Number|squares]] can be demo... | Square Modulo 8 | https://proofwiki.org/wiki/Square_Modulo_8 | https://proofwiki.org/wiki/Square_Modulo_8 | [
"Modulo Arithmetic",
"Square Numbers"
] | [
"Definition:Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] | [
"Definition:Even Integer",
"Square Modulo 4",
"Definition:Square Number"
] |
proofwiki-2097 | Integer as Sum of Three Squares | Let $r$ be a positive integer.
Then $r$ can be expressed as the sum of three squares {{iff}} it is not of the form:
:$4^n \paren {8 m + 7}$
for some $m, n \in \Z_{\ge 0}$. | === Sufficient Condition ===
Suppose $r$ is not of the form $4^n \paren {8 m + 7}$.
Then we need to show that it can always be expressed as the sum of three squares.
{{ProofWanted}} | Let $r$ be a [[Definition:Positive Integer|positive integer]].
Then $r$ can be expressed as the sum of three [[Definition:Square Number|squares]] {{iff}} it is not of the form:
:$4^n \paren {8 m + 7}$
for some $m, n \in \Z_{\ge 0}$. | === Sufficient Condition ===
Suppose $r$ is not of the form $4^n \paren {8 m + 7}$.
Then we need to show that it can always be expressed as the sum of three [[Definition:Square Number|squares]].
{{ProofWanted}} | Integer as Sum of Three Squares | https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares | https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares | [
"Sums of Squares"
] | [
"Definition:Positive/Integer",
"Definition:Square Number"
] | [
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number",
"Definition:Square Number"
] |
proofwiki-2098 | Product of Sums of Four Squares | Let $a, b, c, d, w, x, y, z$ be numbers.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
| c =
}}
{{eqn | l = =
| o =
| r = \left({a w + b x + c y + d z}\right)^2
| c =
}}
{{eqn | o = +
| r = \left({a x - b w + c z -... | Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
| r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
| c =
}}
{{eqn | o =
| ro= +
| r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
| c =... | Let $a, b, c, d, w, x, y, z$ be [[Definition:Number|numbers]].
Then:
{{begin-eqn}}
{{eqn | o =
| r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
| c =
}}
{{eqn | l = =
| o =
| r = \left({a w + b x + c y + d z}\right)^2
| c =
}}
{{eqn | o = +
| r = ... | Taking each of the squares on the {{RHS}} and multiplying them out in turn:
{{begin-eqn}}
{{eqn | l = \paren {a w + b x + c y + d z}^2
| r = a^2 w^2 + b^2 x^2 + c^2 y^2 + d^2 z^2
| c =
}}
{{eqn | o =
| ro= +
| r = 2 \paren {a b w x + a c w y + a d w z + b c x y + b d x z + c d y z}
| c ... | Product of Sums of Four Squares/Proof 1 | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Proof_1 | [
"Algebra",
"Sums of Squares",
"Product of Sums of Four Squares"
] | [
"Definition:Number"
] | [] |
proofwiki-2099 | Product of Sums of Four Squares | Let $a, b, c, d, w, x, y, z$ be numbers.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
| c =
}}
{{eqn | l = =
| o =
| r = \left({a w + b x + c y + d z}\right)^2
| c =
}}
{{eqn | o = +
| r = \left({a x - b w + c z -... | Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two quaternions.
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
| r = \size {\mathbf m \mathbf n}
| c = Quaternion Modulus of Product of Qua... | Let $a, b, c, d, w, x, y, z$ be [[Definition:Number|numbers]].
Then:
{{begin-eqn}}
{{eqn | o =
| r = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)
| c =
}}
{{eqn | l = =
| o =
| r = \left({a w + b x + c y + d z}\right)^2
| c =
}}
{{eqn | o = +
| r = ... | Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two [[Definition:Quaternion|quaternions]].
Then:
{{begin-eqn}}
{{eqn | l = \size {\mathbf m} \size {\mathbf n}
| r = \size {\mathbf m \mathbf n}
| c = [[Quatern... | Product of Sums of Four Squares/Proof 2 | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Proof_2 | [
"Algebra",
"Sums of Squares",
"Product of Sums of Four Squares"
] | [
"Definition:Number"
] | [
"Definition:Quaternion",
"Quaternion Modulus of Product of Quaternions"
] |
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