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-0 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | We have that:
:$\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$
Consider $\ds 2 \sum_{i \mathop = 1}^n i$.
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{i \mathop = 1}^n i
| r = 2 \paren {1 + 2 + \dotsb + \paren {n - 1} + n}
| c =
}}
{{eqn | r = \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + \paren {n + \pare... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | We have that:
:$\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$
Consider $\ds 2 \sum_{i \mathop = 1}^n i$.
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{i \mathop = 1}^n i
| r = 2 \paren {1 + 2 + \dotsb + \paren {n - 1} + n}
| c =
}}
{{eqn | r = \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + \paren {n + \p... | Closed Form for Triangular Numbers/Direct Proof | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Direct_Proof | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Integer Addition is Commutative",
"Integer Addition is Associative"
] |
proofwiki-1 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{m - 1} \paren {a + i d}
| r = m \paren {a + \frac {m - 1} 2 d}
| c = Sum of Arithmetic Sequence
}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren {a + i d}
| r = \paren {n + 1} \paren {a + \frac n 2 d}
| c = Let $n = m - 1$
}}
{{eqn | l = \sum_{i \matho... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{m - 1} \paren {a + i d}
| r = m \paren {a + \frac {m - 1} 2 d}
| c = [[Sum of Arithmetic Sequence]]
}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren {a + i d}
| r = \paren {n + 1} \paren {a + \frac n 2 d}
| c = Let $n = m - 1$
}}
{{eqn | l = \sum_{i \m... | Closed Form for Triangular Numbers/Proof by Arithmetic Sequence | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Arithmetic_Sequence | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Sum of Arithmetic Sequence"
] |
proofwiki-2 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Proof by induction:
=== Basis for the Induction ===
When $n = 1$, we have:
:$\ds \sum_{i \mathop = 1}^1 i = 1$
Also:
:$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$
This is our base case.
=== Induction Hypothesis ===
:$\forall k \in \N: k \ge 1: \ds \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$
Thi... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Proof by [[Principle of Mathematical Induction|induction]]:
=== Basis for the Induction ===
When $n = 1$, we have:
:$\ds \sum_{i \mathop = 1}^1 i = 1$
Also:
:$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$
This is our [[Definition:Basis for the Induction|base case]].
=== Induction Hypothesis ===
:$\for... | Closed Form for Triangular Numbers/Proof by Induction | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Induction | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Principle of Mathematical Induction",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Summation",
"Closed Form for Triangular Numbers/Proof by Induction",
"Principle of Mathematical Induction"
] |
proofwiki-3 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Triangular numbers are $k$-gonal numbers where $k = 3$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
{{begin-eqn}}
{{eqn | l = T_n
| r = \frac n 2 \paren {\paren {3 - 2} n - 3 + 4}
| c = Closed Form for Polygonal Numbers
}}
{{eqn... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | [[Definition:Triangular Number|Triangular numbers]] are [[Definition:Polygonal Number|$k$-gonal numbers]] where $k = 3$.
From [[Closed Form for Polygonal Numbers]] we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
{{begin-eqn}}
{{eqn | l = T_n
| r = \frac n 2 \paren {\paren {... | Closed Form for Triangular Numbers/Proof by Polygonal Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Polygonal_Numbers | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Definition:Triangular Number",
"Definition:Polygonal Number",
"Closed Form for Polygonal Numbers",
"Closed Form for Polygonal Numbers"
] |
proofwiki-4 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | We have that:
:$\ds \map T n = 1 + 2 + \dotsb + n = \sum_{i \mathop = 1}^n i$
Thus:
{{begin-eqn}}
{{eqn | l = \map T n
| r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + 2 + 1
| c =
}}
{{eqn | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + \paren {n - \paren {n - 2} } + \paren {n - \paren {n - 1} }
... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | We have that:
:$\ds \map T n = 1 + 2 + \dotsb + n = \sum_{i \mathop = 1}^n i$
Thus:
{{begin-eqn}}
{{eqn | l = \map T n
| r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + 2 + 1
| c =
}}
{{eqn | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + \paren {n - \paren {n - 2} } + \paren {n - \paren {n - 1} }
... | Closed Form for Triangular Numbers/Proof by Recursion | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Recursion | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [] |
proofwiki-5 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Let:
{{begin-eqn}}
{{eqn | l = S
| o = :=
| r = \sum_{i \mathop = 1}^n i
}}
{{eqn | l = u_r
| o = :=
| r = r
}}
{{eqn | l = v_r
| o = :=
| r = r \paren {r + 1}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = v_r - v_{r - 1}
| r = r \paren {r + 1} - \paren {r - 1} r
| c =
... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Let:
{{begin-eqn}}
{{eqn | l = S
| o = :=
| r = \sum_{i \mathop = 1}^n i
}}
{{eqn | l = u_r
| o = :=
| r = r
}}
{{eqn | l = v_r
| o = :=
| r = r \paren {r + 1}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = v_r - v_{r - 1}
| r = r \paren {r + 1} - \paren {r - 1} r
| c... | Closed Form for Triangular Numbers/Proof by Telescoping Series | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Telescoping_Series | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [] |
proofwiki-6 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Observe that:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2}
| r = -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2}
}}
{{eqn | r = -\paren {1 - \paren {n + 1}^2}
| c = Telescoping Series
}}
{{eqn | r = \paren {n + 1}^2 - 1
}}
{{end-eqn}}
Moreover, we have:
:$\paren ... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Observe that:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2}
| r = -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2}
}}
{{eqn | r = -\paren {1 - \paren {n + 1}^2}
| c = [[Telescoping Series/Example 1|Telescoping Series]]
}}
{{eqn | r = \paren {n + 1}^2 - 1
}}
{{end-e... | Closed Form for Triangular Numbers/Proof by Telescoping Sum | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_by_Telescoping_Sum | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Telescoping Series/Example 1"
] |
proofwiki-7 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numb... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | From [[Faulhaber's Formula]]:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the [[Defini... | Closed Form for Triangular Numbers/Proof using Bernoulli Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_using_Bernoulli_Numbers | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Faulhaber's Formula",
"Definition:Bernoulli Numbers",
"Number to Power of Zero Falling is One"
] |
proofwiki-8 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | From Binomial Coefficient with One:
:$\forall k \in \Z, k > 0: \dbinom k 1 = k$
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n k
| r = \sum_{k \mathop = 1}^n \binom k 1
| c = Binomial Coefficient with One
}}
{{eqn | r = \binom {n + 1} 2
| c = Sum of k Choose m up to n
}}
{{eqn | r = \frac {\pa... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | From [[Binomial Coefficient with One]]:
:$\forall k \in \Z, k > 0: \dbinom k 1 = k$
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n k
| r = \sum_{k \mathop = 1}^n \binom k 1
| c = [[Binomial Coefficient with One]]
}}
{{eqn | r = \binom {n + 1} 2
| c = [[Sum of k Choose m up to n]]
}}
{{eqn | ... | Closed Form for Triangular Numbers/Proof using Binomial Coefficients | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_using_Binomial_Coefficients | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Binomial Coefficient with One",
"Binomial Coefficient with One",
"Sum of Binomial Coefficients over Upper Index"
] |
proofwiki-9 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Let $\N_n^* = \set {1, 2, 3, \cdots, n}$ be the initial segment of natural numbers.
Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$
Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$
Let $\phi: A \to B$ be the mapping:
:$\map \phi {x, y} = \tuple {y, x}$
By definition of dual ordering, $\phi$ is a bijec... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | Let $\N_n^* = \set {1, 2, 3, \cdots, n}$ be the [[Definition:Initial Segment of One-Based Natural Numbers|initial segment of natural numbers]].
Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$
Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$
Let $\phi: A \to B$ be the [[Definition:Mapping|mapping]]... | Closed Form for Triangular Numbers/Proof using Cardinality of Set | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_using_Cardinality_of_Set | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Definition:Initial Segment of Natural Numbers/One-Based",
"Definition:Mapping",
"Definition:Dual Ordering",
"Definition:Bijection",
"Inclusion-Exclusion Principle",
"Definition:Count",
"Definition:Finite Set",
"Definition:Count",
"Definition:Finite Set",
"Inclusion-Exclusion Principle",
"Defini... |
proofwiki-10 | Closed Form for Triangular Numbers | The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = n^2
| c = Odd Number Theorem
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
| r = n^2 + n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j}... | The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
| r = n^2
| c = [[Odd Number Theorem]]
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
| r = n^2 + n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 1}^n \paren {... | Closed Form for Triangular Numbers/Proof using Odd Number Theorem | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers | https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers/Proof_using_Odd_Number_Theorem | [
"Triangular Numbers",
"Sums of Sequences",
"Closed Forms",
"Closed Form for Triangular Numbers"
] | [
"Definition:Closed Form Expression",
"Definition:Triangular Number"
] | [
"Odd Number Theorem"
] |
proofwiki-11 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From Binomial Coefficient with One:
:$\dbinom n 1 = n$
From Binomial Coefficient with Two:
:$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$
Thus:
{{begin-eqn}}
{{eqn | l = 2 \binom n 2 + \binom n 1
| r = 2 \dfrac {n \paren {n - 1} } 2 + n
| c =
}}
{{eqn | r = n \paren {n - 1} + n
| c =
}}
{{eqn | r = n... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From [[Binomial Coefficient with One]]:
:$\dbinom n 1 = n$
From [[Binomial Coefficient with Two]]:
:$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$
Thus:
{{begin-eqn}}
{{eqn | l = 2 \binom n 2 + \binom n 1
| r = 2 \dfrac {n \paren {n - 1} } 2 + n
| c =
}}
{{eqn | r = n \paren {n - 1} + n
| c =
}}
{... | Sum of Sequence of Squares/Proof by Binomial Coefficients | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Binomial_Coefficients | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Binomial Coefficient with One",
"Binomial Coefficient with Two",
"Sum of Binomial Coefficients over Upper Index"
] |
proofwiki-12 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
When $n = 0$, we see from the definition of vacuous sum that:
:$0 = \ds \sum_{i \mathop = 1}^0 i^2 = \frac {0 \paren 1 \paren 1} 6 = 0$
and so $\map P 0$ holds.
===... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
When $n = 0$, we see from the definition of [[Definition:Vacuous Summation|vacuous sum]] that:... | Sum of Sequence of Squares/Proof by Induction | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Induction | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Summation/Vacuous Summation",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Sum of Sequence of Squares/Proof by Induction",
"Principle of Mathematical Induction"
] |
proofwiki-13 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n 3 i \paren {i + 1}
| r = n \paren {n + 1} \paren {n + 2}
| c = Sum from $1$ to $n$ of $r \paren {r + 1}$
}}
{{eqn | ll= \leadsto
| l = \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i
| r = n \paren {n + 1} \paren {n + 2}
}}
{{eqn | ll= \... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n 3 i \paren {i + 1}
| r = n \paren {n + 1} \paren {n + 2}
| c = [[Sum from 1 to n of r(r+1)|Sum from $1$ to $n$ of $r \paren {r + 1}$]]
}}
{{eqn | ll= \leadsto
| l = \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i
| r = n \paren {n + 1} \... | Sum of Sequence of Squares/Proof by Products of Consecutive Integers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Products_of_Consecutive_Integers | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Sum from 1 to n of r(r+1)",
"Closed Form for Triangular Numbers"
] |
proofwiki-14 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}
| r = \sum_{i \mathop = 1}^n \paren {i^3 + 3 i^2 + 3 i + 1 - i^3}
| c = Binomial Theorem
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i^2 + 3 i + 1}
| c =
}}
{{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}
| r = \sum_{i \mathop = 1}^n \paren {i^3 + 3 i^2 + 3 i + 1 - i^3}
| c = [[Binomial Theorem]]
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i^2 + 3 i + 1}
| c =
}}
{{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \ma... | Sum of Sequence of Squares/Proof by Sum of Differences of Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Sum_of_Differences_of_Cubes | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Binomial Theorem",
"Summation is Linear",
"Closed Form for Triangular Numbers",
"Telescoping Series/Example 2",
"Binomial Theorem"
] |
proofwiki-15 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | :File:Sum of Sequences of Squares.jpg
We can observe from the above diagram that:
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}$
Therefore we have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
| r = \sum_{i \mathop = 1}^n \paren {\sum_{j \ma... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | :[[File:Sum of Sequences of Squares.jpg]]
We can observe from the above diagram that:
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}$
Therefore we have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
| r = \sum_{i \mathop = 1}^n \paren {\sum_... | Sum of Sequence of Squares/Proof by Summation of Summations | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Summation_of_Summations | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"File:Sum of Sequences of Squares.jpg",
"Closed Form for Triangular Numbers"
] |
proofwiki-16 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From Sum of Consecutive Triangular Numbers is Square:
:$(1): \quad n^2 = T_n + T_{n - 1}$
where $T_n$ is the $n$th triangular number.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
| r = 1 + \paren {T_1 + T_2} + \paren {T_2 + T_3} + \paren {T_3 + T_4} + \cdots + \paren {T_{n - 1} + T_n}
| c = fr... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From [[Sum of Consecutive Triangular Numbers is Square]]:
:$(1): \quad n^2 = T_n + T_{n - 1}$
where $T_n$ is the $n$th [[Definition:Triangular Number|triangular number]].
Then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
| r = 1 + \paren {T_1 + T_2} + \paren {T_2 + T_3} + \paren {T_3 + T_4} + \cdots +... | Sum of Sequence of Squares/Proof by Summation of Triangular Numbers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_by_Summation_of_Triangular_Numbers | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Sum of Consecutive Triangular Numbers is Square",
"Definition:Triangular Number",
"Closed Form for Triangular Numbers",
"Sum of Sequence of Triangular Numbers",
"Definition:Common Denominator"
] |
proofwiki-17 | Sum of Sequence of Squares | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numb... | :$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ | From [[Faulhaber's Formula]]:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the [[Defini... | Sum of Sequence of Squares/Proof using Bernoulli Numbers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares | https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares/Proof_using_Bernoulli_Numbers | [
"Sum of Sequence of Squares",
"Square Numbers",
"Sums of Sequences"
] | [] | [
"Faulhaber's Formula",
"Definition:Bernoulli Numbers"
] |
proofwiki-18 | Union is Associative | Set union is associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$ | {{begin-eqn}}
{{eqn | o =
| r = x \in A \cup \paren {B \cup C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = x \in A \lor \paren {x \in B \lor x \in C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in A \lor x \in B} \lor x \in C
| c = Ru... | [[Definition:Set Union|Set union]] is [[Definition:Associative Operation|associative]]:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$ | {{begin-eqn}}
{{eqn | o =
| r = x \in A \cup \paren {B \cup C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = x \in A \lor \paren {x \in B \lor x \in C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in A \lor x \in B} \lor x \in C
| c = [[... | Union is Associative | https://proofwiki.org/wiki/Union_is_Associative | https://proofwiki.org/wiki/Union_is_Associative | [
"Union is Associative",
"Set Union",
"Associative Laws of Set Theory",
"Examples of Associative Operations",
"Direct Proofs"
] | [
"Definition:Set Union",
"Definition:Associative Operation"
] | [
"Rule of Association/Disjunction"
] |
proofwiki-19 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | We start with the algebraic definitions for sine and cosine:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | We start with the algebraic definitions for [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]]:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x... | Pythagoras's Theorem/Algebraic Proof | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Algebraic_Proof | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"Definition:Sine",
"Definition:Cosine",
"Sum of Squares of Sine and Cosine",
"Equivalence of Definitions of Sine and Cosine",
"Definition:Sine",
"Definition:Cosine",
"File:SineCosine.png",
"Sum of Squares of Sine and Cosine"
] |
proofwiki-20 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :400px
Let $ABC$ be a right triangle whose angle $BAC$ is a right angle.
Construct squares $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$.
Construct $AL$ parallel to $BD$ (or $CE$).
Since $\angle BAC$ and $\angle BAG$ are both right angles, from Two Angles making Two Right Angles make Straight Line it follows that $... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Euclid-I-47.png|400px]]
Let $ABC$ be a [[Definition:Right Triangle|right triangle]] whose angle $BAC$ is a [[Definition:Right Angle|right angle]].
[[Construction of Square on Given Straight Line|Construct squares]] $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$.
[[Construction of Parallel Line|Construct $... | Pythagoras's Theorem/Classic Proof | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Classic_Proof | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Euclid-I-47.png",
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Right Angle",
"Construction of Square on Given Straight Line",
"Construction of Parallel Line",
"Definition:Right Angle",
"Two Angles making Two Right Angles make Straight Line",
"Definition:Line/Straight Line",
"Two ... |
proofwiki-21 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | Consider the triangle shown below.
:350px
We can take $4$ copies of this triangle and form them into a square using isometries, specifically rotations and translations.
This new figure is shown below.
:500px
This figure is clearly a square, since all the angles are right angles, and the lines connecting the corners are... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | Consider the [[Definition:Triangle (Geometry)|triangle]] shown below.
:[[File:Pythagoras1-1.png|350px]]
We can take $4$ copies of this [[Definition:Triangle (Geometry)|triangle]] and form them into a [[Definition:Square (Geometry)|square]] using isometries, specifically rotations and translations.
This new figure is... | Pythagoras's Theorem/Proof 1 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_1 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"Definition:Triangle (Geometry)",
"File:Pythagoras1-1.png",
"Definition:Triangle (Geometry)",
"Definition:Quadrilateral/Square",
"File:Pythagoras1-2.png",
"Definition:Quadrilateral/Square",
"Definition:Angle",
"Definition:Right Angle",
"Definition:Area",
"Definition:Area",
"Definition:Quadrilate... |
proofwiki-22 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :400px
We have:
:$\dfrac b c = \dfrac d b$
and:
:$\dfrac a c = \dfrac e a$
using the fact that all the triangles involved are similar.
That is:
:$b^2 = c d$
:$a^2 = c e$
Adding, we now get:
:$a^2 + b^2 = c d + c e = c \paren {d + e} = c^2$
{{qed}} | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras2.png|400px]]
We have:
:$\dfrac b c = \dfrac d b$
and:
:$\dfrac a c = \dfrac e a$
using the fact that all the [[Definition:Triangle (Geometry)|triangles]] involved are [[Definition:Similar Triangles|similar]].
That is:
:$b^2 = c d$
:$a^2 = c e$
Adding, we now get:
:$a^2 + b^2 = c d + c e = c \paren... | Pythagoras's Theorem/Proof 2 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_2 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras2.png",
"Definition:Triangle (Geometry)",
"Definition:Similar Triangles"
] |
proofwiki-23 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :300px
The area of the big square is $c^2$.
It is also equal to $4 \dfrac {a b} 2 + \paren {a - b}^2$.
So:
{{begin-eqn}}
{{eqn | l = c^2
| r = 4 \frac {a b} 2 + \paren {a - b}^2
| c =
}}
{{eqn | r = 2 a b + a^2 - 2 a b + b^2
| c =
}}
{{eqn | r = a^2 + b^2
| c =
}}
{{end-eqn}}
{{qed}} | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras3.png|300px]]
The area of the big square is $c^2$.
It is also equal to $4 \dfrac {a b} 2 + \paren {a - b}^2$.
So:
{{begin-eqn}}
{{eqn | l = c^2
| r = 4 \frac {a b} 2 + \paren {a - b}^2
| c =
}}
{{eqn | r = 2 a b + a^2 - 2 a b + b^2
| c =
}}
{{eqn | r = a^2 + b^2
| c =
}}
... | Pythagoras's Theorem/Proof 3 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_3 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras3.png"
] |
proofwiki-24 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :800px
Dissect the square on the left (which has area $c^2$) as shown.
Rearrange the pieces to make the two squares on the right, with areas $a^2$ and $b^2$.
{{qed}} | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras4.png|800px]]
Dissect the [[Definition:Square (Geometry)|square]] on the left (which has [[Definition:Area|area]] $c^2$) as shown.
Rearrange the pieces to make the two [[Definition:Square (Geometry)|squares]] on the right, with areas $a^2$ and $b^2$.
{{qed}} | Pythagoras's Theorem/Proof 4 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_4 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras4.png",
"Definition:Quadrilateral/Square",
"Definition:Area",
"Definition:Quadrilateral/Square"
] |
proofwiki-25 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :700px
The two squares both have the same area, that is, $\paren {a + b}^2$.
The one on the left has four triangles of area $\dfrac {a b} 2$ and a square of area $c^2$.
The one on the right has four triangles of area $\dfrac {a b} 2$ and two squares: one of area $a^2$ and one of area $b^2$.
Take away the triangles from... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras5.png|700px]]
The two squares both have the same area, that is, $\paren {a + b}^2$.
The one on the left has four triangles of area $\dfrac {a b} 2$ and a square of area $c^2$.
The one on the right has four triangles of area $\dfrac {a b} 2$ and two squares: one of area $a^2$ and one of area $b^2$.
... | Pythagoras's Theorem/Proof 5 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_5 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras5.png"
] |
proofwiki-26 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :500px
We have that $CH = BS = AB = AJ$.
Hence the result follows directly from Pythagoras's Theorem for Parallelograms.
{{qed}} | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras6.png|500px]]
We have that $CH = BS = AB = AJ$.
Hence the result follows directly from [[Pythagoras's Theorem for Parallelograms]].
{{qed}} | Pythagoras's Theorem/Proof 6 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_6 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras6.png",
"Pythagoras's Theorem for Parallelograms"
] |
proofwiki-27 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | :500px
Let $\triangle ABC$ be a right triangle and $h_c$ the altitude from $c$.
We have:
:$\angle CAB \cong \angle DCB$
:$\angle ABC \cong \angle ACD$
Then we have:
:$\triangle ADC \sim \triangle ACB \sim \triangle CDB$
Use the fact that if $\triangle XYZ \sim \triangle X'Y'Z'$ then by Ratio of Areas of Similar Triangl... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | :[[File:Pythagoras7.png|500px]]
Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] and $h_c$ the [[Definition:Altitude of Triangle|altitude]] from $c$.
We have:
:$\angle CAB \cong \angle DCB$
:$\angle ABC \cong \angle ACD$
Then we have:
:$\triangle ADC \sim \triangle ACB \sim \triangle CDB$
Us... | Pythagoras's Theorem/Proof 7 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_7 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras7.png",
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Altitude of Triangle",
"Ratio of Areas of Similar Triangles"
] |
proofwiki-28 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | Let $\Box ABCD$ be an arbitrary rectangle with opposing sides $AB = CD$ and $AD = BC$.
300px
Let $O$ be the point where the diameters of $\Box ABCD$ meet.
By Diagonals of Rectangle are Equal:
:$AC = BD$
By Diameters of Parallelogram Bisect each other:
:$OA = OB = OC = OD$
Let a circle be constructed on center $O$ with ... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | Let $\Box ABCD$ be an arbitrary [[Definition:Rectangle|rectangle]] with [[Definition:Opposite Sides|opposing sides]] $AB = CD$ and $AD = BC$.
[[File:Rect in Circle.png|300px]]
Let $O$ be the point where the [[Definition:Diameter of Circle|diameters]] of $\Box ABCD$ meet.
By [[Diagonals of Rectangle are Equal]]:
:$AC... | Pythagoras's Theorem/Proof 8 | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Proof_8 | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"Definition:Quadrilateral/Rectangle",
"Definition:Polygon/Opposite",
"File:Rect in Circle.png",
"Definition:Circle/Diameter",
"Diagonals of Rectangle are Equal",
"Diameters of Parallelogram Bisect each other",
"Definition:Circle",
"Definition:Circle/Center",
"Definition:Circle/Radius",
"Definition... |
proofwiki-29 | Pythagoras's Theorem | Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$ | thumb
From Perpendicular in Right-Angled Triangle makes two Similar Triangles, we have that $\triangle c'c_{upper}b$ is similar to $\triangle c_{lower}c'a$ is similar to $\triangle abc$
{{improve|probably better to refer to triangles by their vertices than their sides, makes it easier for several reasons}}
Looking at t... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]].
Then:
:$a^2 + b^2 = c^2$ | [[File:Pythagoras short algebraic proof.png|thumb]]
From [[Perpendicular in Right-Angled Triangle makes two Similar Triangles]], we have that $\triangle c'c_{upper}b$ is [[Definition:Similar Triangles|similar]] to $\triangle c_{lower}c'a$ is [[Definition:Similar Triangles|similar]] to $\triangle abc$
{{improve|probab... | Pythagoras's Theorem/Short Algebraic Proof | https://proofwiki.org/wiki/Pythagoras's_Theorem | https://proofwiki.org/wiki/Pythagoras's_Theorem/Short_Algebraic_Proof | [
"Pythagoras's Theorem",
"Right Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse"
] | [
"File:Pythagoras short algebraic proof.png",
"Perpendicular in Right-Angled Triangle makes two Similar Triangles",
"Definition:Similar Triangles",
"Definition:Similar Triangles",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse",
"Definition:Altitude of Triangle",
"Definition:Similar Triangles",
... |
proofwiki-30 | Euclid's Theorem | For any finite set of prime numbers, there exists a prime number not in that set.
{{:Euclid:Proposition/IX/20}} | Let $\mathbb P$ be a finite set of prime numbers.
Consider the number:
:$\ds n_p = \paren {\prod_{p \mathop \in \mathbb P} p} + 1$
Take any $p_j \in \mathbb P$.
We have that:
:$\ds p_j \divides \prod_{p \mathop \in \mathbb P} p$
Hence:
:$\ds \exists q \in \Z: \prod_{p \mathop \in \mathbb P} p = q p_j$
So:
{{begin-eqn}}... | For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]].
{{:Euclid:Proposition/IX/20}} | Let $\mathbb P$ be a [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]].
Consider the number:
:$\ds n_p = \paren {\prod_{p \mathop \in \mathbb P} p} + 1$
Take any $p_j \in \mathbb P$.
We have that:
:$\ds p_j \divides \prod_{p \mathop \in \mathbb P} p$
Hence:
:$\ds \exists q \in \Z: \... | Euclid's Theorem | https://proofwiki.org/wiki/Euclid's_Theorem | https://proofwiki.org/wiki/Euclid's_Theorem | [
"Prime Numbers",
"Euclid's Theorem"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Division Theorem",
"Definition:Prime Number",
"Definition:Composite Number",
"Positive Integer Greater than 1 has Prime Divisor",
"Definition:Prime Number"
] |
proofwiki-31 | Euclid's Theorem | For any finite set of prime numbers, there exists a prime number not in that set.
{{:Euclid:Proposition/IX/20}} | Assume that there are only finitely many prime numbers, and that there is a grand total of $n$ primes.
Then it is possible to define the set of all primes:
: $\mathbb P = \set {p_1, p_2, \ldots, p_n}$
From Euclid's Theorem, however, we can always create a prime which is not in $\mathbb P$.
So we can never create a fini... | For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]].
{{:Euclid:Proposition/IX/20}} | Assume that there are only [[Definition:Finite Set|finitely]] many [[Definition:Prime Number|prime numbers]], and that there is a grand total of $n$ primes.
Then it is possible to define the set of all primes:
: $\mathbb P = \set {p_1, p_2, \ldots, p_n}$
From [[Euclid's Theorem]], however, we can always create a prim... | Euclid's Theorem/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Euclid's_Theorem | https://proofwiki.org/wiki/Euclid's_Theorem/Corollary_1/Proof_1 | [
"Prime Numbers",
"Euclid's Theorem"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Euclid's Theorem",
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Factor",
"Definition:Infinite Set",
"Definition:Prime Number"
] |
proofwiki-32 | Euclid's Theorem | For any finite set of prime numbers, there exists a prime number not in that set.
{{:Euclid:Proposition/IX/20}} | Assume that there are only finitely many prime numbers.
Let $p$ be the largest of these.
Then from Existence of Prime between Prime and Factorial there exists a prime number $q$ such that:
:$p < q \le p! + 1$
So there cannot be such a $p$.
{{qed}} | For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]].
{{:Euclid:Proposition/IX/20}} | Assume that there are only [[Definition:Finite Set|finitely]] many [[Definition:Prime Number|prime numbers]].
Let $p$ be the largest of these.
Then from [[Existence of Prime between Prime and Factorial]] there exists a [[Definition:Prime Number|prime number]] $q$ such that:
:$p < q \le p! + 1$
So there cannot be suc... | Euclid's Theorem/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Euclid's_Theorem | https://proofwiki.org/wiki/Euclid's_Theorem/Corollary_1/Proof_2 | [
"Prime Numbers",
"Euclid's Theorem"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Existence of Prime between Prime and Factorial",
"Definition:Prime Number"
] |
proofwiki-33 | Euclid's Theorem | For any finite set of prime numbers, there exists a prime number not in that set.
{{:Euclid:Proposition/IX/20}} | Let $\mathbb P$ be the set of all prime numbers.
{{AimForCont}} there exists a largest prime number $p_m$.
Then:
:$\mathbb P \subseteq \closedint 1 {p_m} = \set {1, 2, \ldots, p_m}$
and so $\mathbb P$ is a finite set.
By Euclid's Theorem, there exists a prime number $q$ such that $q \notin \mathbb P$.
But that means $q... | For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]].
{{:Euclid:Proposition/IX/20}} | Let $\mathbb P$ be the set of all [[Definition:Prime Number|prime numbers]].
{{AimForCont}} there exists a largest [[Definition:Prime Number|prime number]] $p_m$.
Then:
:$\mathbb P \subseteq \closedint 1 {p_m} = \set {1, 2, \ldots, p_m}$
and so $\mathbb P$ is a [[Definition:Finite Set|finite set]].
By [[Euclid's Th... | Euclid's Theorem/Corollary 2/Proof 1 | https://proofwiki.org/wiki/Euclid's_Theorem | https://proofwiki.org/wiki/Euclid's_Theorem/Corollary_2/Proof_1 | [
"Prime Numbers",
"Euclid's Theorem"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set"
] | [
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Finite Set",
"Euclid's Theorem",
"Definition:Prime Number",
"Definition:Prime Number",
"Proof by Contradiction"
] |
proofwiki-34 | Euclid's Theorem | For any finite set of prime numbers, there exists a prime number not in that set.
{{:Euclid:Proposition/IX/20}} | {{AimForCont}} there exists a largest prime number $p$.
Let $b = p! + 1$.
Let $q$ be a prime number that divides $b$.
Since $p$ is the largest prime number, $q \le p$.
However, no positive integer $d \le p$ is a divisor of $b$.
Hence $q \not \le p$.
Hence the result, by Proof by Contradiction.
{{qed}} | For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]].
{{:Euclid:Proposition/IX/20}} | {{AimForCont}} there exists a largest [[Definition:Prime Number|prime number]] $p$.
Let $b = p! + 1$.
Let $q$ be a [[Definition:Prime Number|prime number]] that [[Definition:Divisor of Integer|divides]] $b$.
Since $p$ is the largest [[Definition:Prime Number|prime number]], $q \le p$.
However, no [[Definition:Posit... | Euclid's Theorem/Corollary 2/Proof 2 | https://proofwiki.org/wiki/Euclid's_Theorem | https://proofwiki.org/wiki/Euclid's_Theorem/Corollary_2/Proof_2 | [
"Prime Numbers",
"Euclid's Theorem"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set"
] | [
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Proof by Contradiction"
] |
proofwiki-35 | Square Root of Prime is Irrational | The square root of a prime number is irrational. | Let $p$ be prime.
{{AimForCont}} that $\sqrt p$ is rational.
Then there exist natural numbers $m$ and $n$ such that:
{{begin-eqn}}
{{eqn | l = \sqrt p
| r = \frac m n
| c =
}}
{{eqn | ll= \leadsto
| l = p
| r = \frac {m^2} {n^2}
| c =
}}
{{eqn | ll= \leadsto
| l = n^2 p
| r =... | The [[Definition:Square Root|square root]] of a [[Definition:Prime Number|prime number]] is [[Definition:Irrational Number|irrational]]. | Let $p$ be [[Definition:Prime Number|prime]].
{{AimForCont}} that $\sqrt p$ is [[Definition:Rational Number|rational]].
Then there exist [[Definition:Natural Numbers|natural numbers]] $m$ and $n$ such that:
{{begin-eqn}}
{{eqn | l = \sqrt p
| r = \frac m n
| c =
}}
{{eqn | ll= \leadsto
| l = p
... | Square Root of Prime is Irrational/Proof 1 | https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational | https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational/Proof_1 | [
"Square Root of Prime is Irrational",
"Prime Numbers",
"Irrational Numbers",
"Square Roots",
"Irrationality Proofs"
] | [
"Definition:Square Root",
"Definition:Prime Number",
"Definition:Irrational Number"
] | [
"Definition:Prime Number",
"Definition:Rational Number",
"Definition:Natural Numbers",
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Even Integer",
"Definition:Square Number",
"Definition:Prime Decomposition",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Proof ... |
proofwiki-36 | Square Root of Prime is Irrational | The square root of a prime number is irrational. | Let $p \in \Z$ be a prime number.
Consider the polynomial:
:$\map P x = x^2 - p$
over the ring of polynomials $\Q \sqbrk X$ over the rational numbers.
From Difference of Two Squares:
:$x^2 - p = \paren {x + \sqrt p} \paren {x - \sqrt p}$
Because $p$ is prime, $\sqrt p$ is not an integer.
From Polynomial which is Irredu... | The [[Definition:Square Root|square root]] of a [[Definition:Prime Number|prime number]] is [[Definition:Irrational Number|irrational]]. | Let $p \in \Z$ be a [[Definition:Prime Number|prime number]].
Consider the [[Definition:Polynomial over Ring in One Variable|polynomial]]:
:$\map P x = x^2 - p$
over the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] $\Q \sqbrk X$ over the [[Definition:Rational Number|rational numbers]].
From ... | Square Root of Prime is Irrational/Proof 2 | https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational | https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational/Proof_2 | [
"Square Root of Prime is Irrational",
"Prime Numbers",
"Irrational Numbers",
"Square Roots",
"Irrationality Proofs"
] | [
"Definition:Square Root",
"Definition:Prime Number",
"Definition:Irrational Number"
] | [
"Definition:Prime Number",
"Definition:Polynomial over Ring/One Variable",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Rational Number",
"Difference of Two Squares",
"Definition:Prime Number",
"Definition:Integer",
"Polynomial which is Irreducible over Integers is Irreducible over Ra... |
proofwiki-37 | Derivative of Exponential Function | Let $\exp$ be the exponential function.
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\exp x}
| r = \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h
| c = {{Defof|Derivative}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h
| c = Exponential of Sum
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\ex... | Let $\exp$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\exp x}
| r = \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h
| c = {{Defof|Derivative}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h
| c = [[Exponential of Sum]]
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac ... | Derivative of Exponential Function/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Exponential_Function | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_1 | [
"Derivative of Exponential Function",
"Derivatives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Exponential of Sum",
"Combination Theorem for Limits of Functions/Real/Multiple Rule",
"Derivative of Exponential at Zero"
] |
proofwiki-38 | Derivative of Exponential Function | Let $\exp$ be the exponential function.
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | We use the fact that the exponential function is the inverse of the natural logarithm function:
:$y = e^x \iff x = \ln y$
{{begin-eqn}}
{{eqn | l = \dfrac {\d x} {\d y}
| r = \dfrac 1 y
| c = Derivative of Natural Logarithm Function
}}
{{eqn | ll= \leadsto
| l = \dfrac {\d y} {\d x}
| r = \df... | Let $\exp$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | We use the fact that the [[Definition:Real Exponential Function|exponential function]] is the [[Definition:Inverse Mapping|inverse]] of the [[Definition:Natural Logarithm|natural logarithm function]]:
:$y = e^x \iff x = \ln y$
{{begin-eqn}}
{{eqn | l = \dfrac {\d x} {\d y}
| r = \dfrac 1 y
| c = [[Deri... | Derivative of Exponential Function/Proof 2 | https://proofwiki.org/wiki/Derivative_of_Exponential_Function | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_2 | [
"Derivative of Exponential Function",
"Derivatives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Definition:Exponential Function/Real",
"Definition:Inverse Mapping",
"Definition:Natural Logarithm",
"Derivative of Natural Logarithm Function",
"Derivative of Inverse Function"
] |
proofwiki-39 | Derivative of Exponential Function | Let $\exp$ be the exponential function.
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\ln e^x}
| r = \map {\frac \d {\d x} } x
| c = Exponential of Natural Logarithm
}}
{{eqn | ll= \leadsto
| l = \frac 1 {e^x} \map {\frac \d {\d x} } {e^x}
| r = 1
| c = Chain Rule for Derivatives, Derivative of Natural Logarithm Function, D... | Let $\exp$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\ln e^x}
| r = \map {\frac \d {\d x} } x
| c = [[Exponential of Natural Logarithm]]
}}
{{eqn | ll= \leadsto
| l = \frac 1 {e^x} \map {\frac \d {\d x} } {e^x}
| r = 1
| c = [[Chain Rule for Derivatives]], [[Derivative of Natural Logarithm F... | Derivative of Exponential Function/Proof 3 | https://proofwiki.org/wiki/Derivative_of_Exponential_Function | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_3 | [
"Derivative of Exponential Function",
"Derivatives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Exponential of Natural Logarithm",
"Derivative of Composite Function",
"Derivative of Natural Logarithm Function",
"Derivative of Identity Function"
] |
proofwiki-40 | Derivative of Exponential Function | Let $\exp$ be the exponential function.
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | This proof assumes the power series definition of $\exp$.
That is, let:
:$\ds \exp x = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$
From Series of Power over Factorial Converges, the interval of convergence of $\exp$ is the entirety of $\R$.
So we may apply Differentiation of Real Power Series to $\exp$ for all $x \in... | Let $\exp$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | This proof assumes the [[Definition:Exponential Function/Real/Power Series Expansion|power series definition of $\exp$]].
That is, let:
:$\ds \exp x = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$
From [[Series of Power over Factorial Converges]], the [[Definition:Interval of Convergence|interval of convergence]] of... | Derivative of Exponential Function/Proof 4 | https://proofwiki.org/wiki/Derivative_of_Exponential_Function | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_4 | [
"Derivative of Exponential Function",
"Derivatives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Definition:Exponential Function/Real/Power Series Expansion",
"Series of Power over Factorial Converges",
"Definition:Interval of Convergence",
"Differentiation of Real Power Series",
"Differentiation of Real Power Series"
] |
proofwiki-41 | Derivative of Exponential Function | Let $\exp$ be the exponential function.
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | This proof assumes the limit definition of $\exp$.
So let:
:$\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$
Let $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
Let:
:$N = \ceiling {\max \set {\size {x_0 - 1}, \size {x_0 + 1} } }$
where $\ceiling {\, \cdot \,}$ denotes the ce... | Let $\exp$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ | This proof assumes the [[Definition:Exponential Function/Real/Limit of Sequence|limit definition of $\exp$]].
So let:
:$\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$
Let $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
Let:
:$N = \ceiling {\max \set {\size {x_0 - 1}, \si... | Derivative of Exponential Function/Proof 5 | https://proofwiki.org/wiki/Derivative_of_Exponential_Function | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_5 | [
"Derivative of Exponential Function",
"Derivatives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Definition:Exponential Function/Real/Limit of Sequence",
"Definition:Ceiling Function",
"Closed Real Interval is Compact Space/Metric Space",
"Definition:Compact Space/Real Analysis",
"Derivative of Composite Function",
"Derivative of Exponential Function/Proof 5/Lemma",
"Derivative of Exponential Func... |
proofwiki-42 | Derivative of Sine Function | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | From the definition of the sine function, we have:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
From Radius of Convergence of Power Series over Factorial, this series converges for all $x$.
From Power Series is Differentiable on Interval of Convergence:
{{begin-eqn}... | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | From the definition of the [[Definition:Sine|sine function]], we have:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
From [[Radius of Convergence of Power Series over Factorial]], this series converges for all $x$.
From [[Power Series is Differentiable on Interval... | Derivative of Sine Function/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Sine_Function | https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_1 | [
"Derivatives of Trigonometric Functions",
"Sine Function",
"Derivative of Sine Function"
] | [] | [
"Definition:Sine",
"Radius of Convergence of Power Series over Factorial",
"Power Series is Differentiable on Interval of Convergence",
"Definition:Cosine"
] |
proofwiki-43 | Derivative of Sine Function | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sin x}
| r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h
| c = {{Defof|Derivative of Real Function at Point}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \cos h + \sin h \cos x - \sin x} h
| c = Sine of Sum
}}
{{eqn | r = \lim... | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sin x}
| r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h
| c = {{Defof|Derivative of Real Function at Point}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \cos h + \sin h \cos x - \sin x} h
| c = [[Sine of Sum]]
}}
{{eqn | r = ... | Derivative of Sine Function/Proof 2 | https://proofwiki.org/wiki/Derivative_of_Sine_Function | https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_2 | [
"Derivatives of Trigonometric Functions",
"Sine Function",
"Derivative of Sine Function"
] | [] | [
"Sine of Sum",
"Combination Theorem for Limits of Functions/Real/Sum Rule",
"Limit of Sinc Function at Zero",
"Limit of (Cosine (X) - 1) over X at Zero"
] |
proofwiki-44 | Derivative of Sine Function | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \dfrac \d {\d x} \sin x
| r = \dfrac \d {\d x} \map \cos {\frac \pi 2 - x}
| c = Cosine of Complement equals Sine
}}
{{eqn | r = \map \sin {\frac \pi 2 - x}
| c = Derivative of Cosine Function and Chain Rule for Derivatives
}}
{{eqn | r = \cos x
| c = Sine of Complemen... | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \dfrac \d {\d x} \sin x
| r = \dfrac \d {\d x} \map \cos {\frac \pi 2 - x}
| c = [[Cosine of Complement equals Sine]]
}}
{{eqn | r = \map \sin {\frac \pi 2 - x}
| c = [[Derivative of Cosine Function]] and [[Chain Rule for Derivatives]]
}}
{{eqn | r = \cos x
| c = [[Sin... | Derivative of Sine Function/Proof 3 | https://proofwiki.org/wiki/Derivative_of_Sine_Function | https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_3 | [
"Derivatives of Trigonometric Functions",
"Sine Function",
"Derivative of Sine Function"
] | [] | [
"Cosine of Complement equals Sine",
"Derivative of Cosine Function",
"Derivative of Composite Function",
"Sine of Complement equals Cosine"
] |
proofwiki-45 | Derivative of Sine Function | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sin x}
| r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h
| c = {{Defof|Derivative of Real Function at Point}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - ... | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sin x}
| r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h
| c = {{Defof|Derivative of Real Function at Point}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - ... | Derivative of Sine Function/Proof 4 | https://proofwiki.org/wiki/Derivative_of_Sine_Function | https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_4 | [
"Derivatives of Trigonometric Functions",
"Sine Function",
"Derivative of Sine Function"
] | [] | [
"Werner Formulas/Cosine by Sine",
"Combination Theorem for Limits of Functions/Real/Multiple Rule",
"Combination Theorem for Limits of Functions/Real/Product Rule",
"Cosine Function is Continuous",
"Limit of Sinc Function at Zero"
] |
proofwiki-46 | Derivative of Sine Function | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map \arcsin x
| r = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }
| c = Arcsine as Integral
}}
{{eqn | ll= \leadsto
| l = \dfrac {\map \d {\map \arcsin y} } {\d y}
| r = \dfrac {\map \d {\ds \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}
}}
{{eqn | r = \dfrac 1 {\sqrt... | :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ | {{begin-eqn}}
{{eqn | l = \map \arcsin x
| r = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }
| c = [[Arcsine as Integral]]
}}
{{eqn | ll= \leadsto
| l = \dfrac {\map \d {\map \arcsin y} } {\d y}
| r = \dfrac {\map \d {\ds \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}
}}
{{eqn | r = \dfrac 1 {\... | Derivative of Sine Function/Proof 5 | https://proofwiki.org/wiki/Derivative_of_Sine_Function | https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_5 | [
"Derivatives of Trigonometric Functions",
"Sine Function",
"Derivative of Sine Function"
] | [] | [
"Arcsine as Integral",
"Derivative of Arcsine Function",
"Definition:Inverse Sine/Real/Arcsine",
"Definition:Bijection",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Inverse of Mapping",
"Definition:Mapping",
"Inverse of Inverse of Bijection",
"Definition:Inverse Mapping",
"Definition:Sin... |
proofwiki-47 | 0.999...=1 | :$0.999 \ldots = 1$ | By Sum of Infinite Geometric Sequence:
:$0.999 \ldots = \dfrac a {1 - r}$
where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.
Since our ratio is less than $1$, then we know that $\ds \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to:
:$\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 - \fra... | :$0.999 \ldots = 1$ | By [[Sum of Infinite Geometric Sequence]]:
:$0.999 \ldots = \dfrac a {1 - r}$
where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.
Since our ratio is less than $1$, then we know that $\ds \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to:
:$\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 ... | 0.999...=1/Proof 1 | https://proofwiki.org/wiki/0.999...=1 | https://proofwiki.org/wiki/0.999...=1/Proof_1 | [
"Number Theory",
"Direct Proofs",
"0.999...=1"
] | [] | [
"Sum of Infinite Geometric Sequence"
] |
proofwiki-48 | 0.999...=1 | :$0.999 \ldots = 1$ | {{begin-eqn}}
{{eqn | l = 0.333 \ldots
| r = 1 / 3
}}
{{eqn | ll= \leadsto
| l = 3 \paren {0.333 \ldots}
| r = 3 \paren {1 / 3}
}}
{{eqn | ll= \leadsto
| l = 0.999 \ldots
| r = 3 / 3
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}} | :$0.999 \ldots = 1$ | {{begin-eqn}}
{{eqn | l = 0.333 \ldots
| r = 1 / 3
}}
{{eqn | ll= \leadsto
| l = 3 \paren {0.333 \ldots}
| r = 3 \paren {1 / 3}
}}
{{eqn | ll= \leadsto
| l = 0.999 \ldots
| r = 3 / 3
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}} | 0.999...=1/Proof 2 | https://proofwiki.org/wiki/0.999...=1 | https://proofwiki.org/wiki/0.999...=1/Proof_2 | [
"Number Theory",
"Direct Proofs",
"0.999...=1"
] | [] | [] |
proofwiki-49 | 0.999...=1 | :$0.999 \ldots = 1$ | Let $c = 0.999 \ldots$
Then:
{{begin-eqn}}
{{eqn | l = c
| r = 0.999 \ldots
}}
{{eqn | ll= \leadsto
| l = 10 c
| r = \paren {9.999 \ldots}
| c = multiplying $c$ by $10$
}}
{{eqn | ll= \leadsto
| l = 10 c - c
| r = \paren {9.999 \ldots} - \paren {0.999 \ldots}
| c = subtractin... | :$0.999 \ldots = 1$ | Let $c = 0.999 \ldots$
Then:
{{begin-eqn}}
{{eqn | l = c
| r = 0.999 \ldots
}}
{{eqn | ll= \leadsto
| l = 10 c
| r = \paren {9.999 \ldots}
| c = multiplying $c$ by $10$
}}
{{eqn | ll= \leadsto
| l = 10 c - c
| r = \paren {9.999 \ldots} - \paren {0.999 \ldots}
| c = subtracti... | 0.999...=1/Proof 3 | https://proofwiki.org/wiki/0.999...=1 | https://proofwiki.org/wiki/0.999...=1/Proof_3 | [
"Number Theory",
"Direct Proofs",
"0.999...=1"
] | [] | [] |
proofwiki-50 | 0.999...=1 | :$0.999 \ldots = 1$ | We begin with the knowledge that:
{{begin-eqn}}
{{eqn | l = \frac 9 9
| r = \frac 1 1 = 1
}}
{{end-eqn}}
Now we divide $9$ by $9$ using the standard process of long division, only instead of stating that $90$ divided by $9$ is $10$, we say that it is "$9$ remainder $9$," yielding the following result:
<pre>
... | :$0.999 \ldots = 1$ | We begin with the knowledge that:
{{begin-eqn}}
{{eqn | l = \frac 9 9
| r = \frac 1 1 = 1
}}
{{end-eqn}}
Now we divide $9$ by $9$ using the standard process of long division, only instead of stating that $90$ divided by $9$ is $10$, we say that it is "$9$ remainder $9$," yielding the following result:
<pre>
... | 0.999...=1/Proof 4 | https://proofwiki.org/wiki/0.999...=1 | https://proofwiki.org/wiki/0.999...=1/Proof_4 | [
"Number Theory",
"Direct Proofs",
"0.999...=1"
] | [] | [] |
proofwiki-51 | 0.999...=1 | :$0.999 \ldots = 1$ | {{begin-eqn}}
{{eqn | n = 1
| l = 0 . \underset n {\underbrace {999 \cdots 9} }
| r = 1 - 0.1^n
| c =
}}
{{eqn | l = 0.999 \cdots
| r = \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}
| c = {{Defof|Real Numbers}}
}}
{{eqn | r = \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \,... | :$0.999 \ldots = 1$ | {{begin-eqn}}
{{eqn | n = 1
| l = 0 . \underset n {\underbrace {999 \cdots 9} }
| r = 1 - 0.1^n
| c =
}}
{{eqn | l = 0.999 \cdots
| r = \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}
| c = {{Defof|Real Numbers}}
}}
{{eqn | r = \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \,... | 0.999...=1/Proof 5 | https://proofwiki.org/wiki/0.999...=1 | https://proofwiki.org/wiki/0.999...=1/Proof_5 | [
"Number Theory",
"Direct Proofs",
"0.999...=1"
] | [] | [
"Sequence of Powers of Number less than One"
] |
proofwiki-52 | Schur-Zassenhaus Theorem | Let $G$ be a finite group and $N$ be a normal subgroup in $G$.
Let $N$ be a Hall subgroup of $G$.
Then there exists $H$, a complement of $N$, such that $G$ is the semidirect product of $N$ and $H$. | The proof proceeds by induction.
By definition, $N$ is a Hall subgroup {{iff}} the index and order of $N$ in $G$ are relatively prime numbers.
Let $G$ be a group whose identity is $e$.
We induct on $\order G$, where $\order G$ is the order of $G$.
We may assume that $N \ne \set e$.
Let $p$ be a prime number dividing $\... | Let $G$ be a [[Definition:Finite Group|finite group]] and $N$ be a [[Definition:Normal Subgroup|normal subgroup]] in $G$.
Let $N$ be a [[Definition:Hall Subgroup|Hall subgroup]] of $G$.
Then there exists $H$, a [[Definition:Complement of Subgroup|complement]] of $N$, such that $G$ is the [[Definition:Inner Semidirec... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
By definition, $N$ is a [[Definition:Hall Subgroup|Hall subgroup]] {{iff}} the [[Definition:Index of Subgroup|index]] and [[Definition:Order of Structure|order]] of $N$ in $G$ are [[Definition:Coprime Integers|relatively prime numbers]].
Let $G... | Schur-Zassenhaus Theorem | https://proofwiki.org/wiki/Schur-Zassenhaus_Theorem | https://proofwiki.org/wiki/Schur-Zassenhaus_Theorem | [
"Group Theory",
"Hall Subgroups"
] | [
"Definition:Finite Group",
"Definition:Normal Subgroup",
"Definition:Hall Subgroup",
"Definition:Complement of Subgroup",
"Definition:Semidirect Product/Inner"
] | [
"Principle of Mathematical Induction",
"Definition:Hall Subgroup",
"Definition:Index of Subgroup",
"Definition:Order of Structure",
"Definition:Coprime/Integers",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Prime Numbe... |
proofwiki-53 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^3
| r = \frac {n^2 \paren {n + 1}^2} 4
| c = Sum of Sequence of Cubes
}}
{{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2
| c =
}}
{{eqn | r = {T_n}^2
| c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}} | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^3
| r = \frac {n^2 \paren {n + 1}^2} 4
| c = [[Sum of Sequence of Cubes]]
}}
{{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2
| c =
}}
{{eqn | r = {T_n}^2
| c = [[Closed Form for Triangular Numbers]]
}}
{{end-eqn}}
{{qed}} | Square of Triangular Number equals Sum of Sequence of Cubes/Proof 1 | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Square_of_Triangular_Number_equals_Sum_of_Sequence_of_Cubes/Proof_1 | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Sum of Sequence of Cubes",
"Closed Form for Triangular Numbers"
] |
proofwiki-54 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^1 i^3
| r = 1^3
| c =
}}
{{eqn | r = 1
| c =
}}
{{eqn | r =... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^1 i^3
... | Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2 | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Square_of_Triangular_Number_equals_Sum_of_Sequence_of_Cubes/Proof_2 | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Closed Form for Triangular Numbers",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2",
"Cube Number as Difference... |
proofwiki-55 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | {{begin-eqn}}
{{eqn | l = \paren {r - 1} r \paren {r + 1}
| r = r \paren {r^2 - 1}
| c = Difference of Two Squares
}}
{{eqn | r = r^3 - r
| c =
}}
{{eqn | ll= \leadsto
| l = r^3
| r = \paren {r - 1} r \paren {r + 1} + r
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{r \mathop = 1}... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | {{begin-eqn}}
{{eqn | l = \paren {r - 1} r \paren {r + 1}
| r = r \paren {r^2 - 1}
| c = [[Difference of Two Squares]]
}}
{{eqn | r = r^3 - r
| c =
}}
{{eqn | ll= \leadsto
| l = r^3
| r = \paren {r - 1} r \paren {r + 1} + r
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{r \mathop ... | Sum of Sequence of Cubes/Proof 7 | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_7 | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Difference of Two Squares",
"Sum from 1 to n of r(r+1)(r+2)",
"Closed Form for Triangular Numbers"
] |
proofwiki-56 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | First, from Closed Form for Triangular Numbers:
:$\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
So:
:$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$
Next we use induction on $n$ to show that:
:$\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
The proof proceeds... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | First, from [[Closed Form for Triangular Numbers]]:
:$\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
So:
:$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$
Next we use [[Principle of Mathematical Induction|induction]] on $n$ to show that:
:$\ds \sum_{i \mathop = 1}^n i^3 = \dfr... | Sum of Sequence of Cubes/Proof by Induction | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_by_Induction | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Closed Form for Triangular Numbers",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Sum of Sequence of Cubes/Proof by Induction",
"Principl... |
proofwiki-57 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | By Nicomachus's Theorem, we have:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
Also by Nicomachus's Theorem, we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.
So if we add them all up together, we get:
{{begin-e... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | By [[Nicomachus's Theorem]], we have:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
Also by [[Nicomachus's Theorem]], we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.
So if we add them all up together, we g... | Sum of Sequence of Cubes/Proof by Nicomachus | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_by_Nicomachus | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Nicomachus's Theorem",
"Nicomachus's Theorem",
"Odd Number Theorem"
] |
proofwiki-58 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | From Closed Form for Triangular Numbers:
:$(1): \quad \ds \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$
From Sum of Sequence of Squares:
:$(2): \quad \ds \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$
Let $\ds \map S n = \sum_{i \mathop = 1}^n i^3$.
Then:
{{be... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | From [[Closed Form for Triangular Numbers]]:
:$(1): \quad \ds \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$
From [[Sum of Sequence of Squares]]:
:$(2): \quad \ds \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$
Let $\ds \map S n = \sum_{i \mathop = 1}^n i^3$.... | Sum of Sequence of Cubes/Proof by Recursion | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_by_Recursion | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Closed Form for Triangular Numbers",
"Sum of Sequence of Squares"
] |
proofwiki-59 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numb... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | From [[Faulhaber's Formula]]:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the [[Defini... | Sum of Sequence of Cubes/Proof using Bernoulli Numbers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_using_Bernoulli_Numbers | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Faulhaber's Formula",
"Definition:Bernoulli Numbers"
] |
proofwiki-60 | Sum of Sequence of Cubes | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | :$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \paren {1 + 2 + 3 + \cdots + N}^2$
{{begin-eqn}}
{{eqn | l = \paren {1 + 2 + 3 + \cdots + N}^2
| r = 1 \times \paren {1 + 2 + 3 + \cdots + N}
| c =
}}
{{eqn | o = +
| r = 2 \times \paren {1 + 2 + 3 + \cdots + N}
| c =
}}
{{eqn | o = +
| r = \cd... | :$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ | :$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \paren {1 + 2 + 3 + \cdots + N}^2$
{{begin-eqn}}
{{eqn | l = \paren {1 + 2 + 3 + \cdots + N}^2
| r = 1 \times \paren {1 + 2 + 3 + \cdots + N}
| c =
}}
{{eqn | o = +
| r = 2 \times \paren {1 + 2 + 3 + \cdots + N}
| c =
}}
{{eqn | o = +
| r = \c... | Sum of Sequence of Cubes/Proof using Multiplication Table | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes | https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes/Proof_using_Multiplication_Table | [
"Sum of Sequence of Cubes",
"Sums of Sequences",
"Sums of Cubes",
"Cube Numbers"
] | [] | [
"Definition:Term of Expression",
"Definition:Matrix/Square Matrix",
"Closed Form for Triangular Numbers",
"Definition:Addition/Sum",
"Definition:Term of Expression",
"Definition:Matrix/Row",
"Definition:Term of Expression",
"Definition:Matrix/Column",
"1+2+...+n+(n-1)+...+1 = n^2"
] |
proofwiki-61 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | {{begin-eqn}}
{{eqn | l = \sin c \sin a \cos B
| r = \cos b - \cos c \cos a
| c = Spherical Law of Cosines
}}
{{eqn | r = \cos b - \cos c \paren {\cos b \cos c + \sin b \sin c \cos A}
| c = Spherical Law of Cosines
}}
{{eqn | r = \cos b \paren {1 - \cos^2 c} - \sin b \sin c \cos c \cos A
| c = r... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | {{begin-eqn}}
{{eqn | l = \sin c \sin a \cos B
| r = \cos b - \cos c \cos a
| c = [[Spherical Law of Cosines]]
}}
{{eqn | r = \cos b - \cos c \paren {\cos b \cos c + \sin b \sin c \cos A}
| c = [[Spherical Law of Cosines]]
}}
{{eqn | r = \cos b \paren {1 - \cos^2 c} - \sin b \sin c \cos c \cos A
... | Analogue Formula for Spherical Law of Cosines/Proof 1 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Analogue_Formula_for_Spherical_Law_of_Cosines/Proof_1 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"Spherical Law of Cosines",
"Spherical Law of Cosines",
"Sum of Squares of Sine and Cosine",
"Spherical Law of Cosines",
"Spherical Law of Cosines",
"Sum of Squares of Sine and Cosine"
] |
proofwiki-62 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :500px
Suppose $c$ is less than $\dfrac \pi 2$.
Let $BA$ be produced to $D$ so that $BD = \dfrac \pi 2$.
Then:
:$AD = \dfrac \pi 2 - c$
and:
:$\angle CAD = pi - A$
Let $C$ and $D$ be joined by an arc of a great circle, denoted $x$.
From the triangle $\sphericalangle DAC$, using the Spherical Law of Cosines:
{{begin-eqn... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :[[File:Spherical-Cosine-Formula-Analog.png|500px]]
Suppose $c$ is less than $\dfrac \pi 2$.
Let $BA$ be [[Definition:Production|produced]] to $D$ so that $BD = \dfrac \pi 2$.
Then:
:$AD = \dfrac \pi 2 - c$
and:
:$\angle CAD = pi - A$
Let $C$ and $D$ be joined by an [[Definition:Arc of Circle|arc]] of a [[Definitio... | Analogue Formula for Spherical Law of Cosines/Proof 2 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Analogue_Formula_for_Spherical_Law_of_Cosines/Proof_2 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"File:Spherical-Cosine-Formula-Analog.png",
"Definition:Production",
"Definition:Circle/Arc",
"Definition:Great Circle",
"Definition:Spherical Triangle",
"Spherical Law of Cosines",
"Definition:Spherical Triangle",
"Spherical Law of Cosines",
"Definition:Point"
] |
proofwiki-63 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :500px
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $O$ be joined to each of $A$, $B$ and $C... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :[[File:Spherical-Cosine-Formula-2.png|500px]]
Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$.
By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$,... | Analogue Formula for Spherical Law of Cosines/Proof 3 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Analogue_Formula_for_Spherical_Law_of_Cosines/Proof_3 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"File:Spherical-Cosine-Formula-2.png",
"Definition:Polygon/Vertex",
"Definition:Spherical Triangle",
"Definition:Sphere/Geometry",
"Definition:Spherical Triangle",
"Definition:Circle/Arc",
"Definition:Great Circle",
"Definition:Great Circle",
"Definition:Circle/Center",
"Definition:Great Circle",
... |
proofwiki-64 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | Let $\triangle ABC$ be embedded in a Cartesian coordinate system by identifying:
:$C := \tuple {0, 0}$
:$B := \tuple {a, 0}$
:400px
Thus by definition of sine and cosine:
:$A = \tuple {b \cos C, b \sin C}$
By the Distance Formula:
:$c = \sqrt {\paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2}$
Hence:
{{begin-eqn}}
{{e... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | Let $\triangle ABC$ be embedded in a [[Definition:Cartesian Coordinate System|Cartesian coordinate system]] by identifying:
:$C := \tuple {0, 0}$
:$B := \tuple {a, 0}$
:[[File:CosineRuleCartesian.png|400px]]
Thus by definition of [[Definition:Sine of Angle|sine]] and [[Definition:Cosine of Angle|cosine]]:
:$A = \tu... | Law of Cosines/Proof 1 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Law_of_Cosines/Proof_1 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"Definition:Cartesian Coordinate System",
"File:CosineRuleCartesian.png",
"Definition:Sine/Definition from Triangle",
"Definition:Cosine/Definition from Triangle",
"Distance Formula",
"Distance Formula",
"Square of Difference",
"Real Multiplication Distributes over Addition",
"Sum of Squares of Sine... |
proofwiki-65 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | Let $\triangle ABC$ be a triangle.
=== Case 1: $AC$ greater than $AB$ ===
Using $AC$ as the radius, we construct a circle whose center is $A$.
Now we extend:
:$CB$ to $D$
:$AB$ to $F$
:$BA$ to $G$
:$CA$ to $E$.
$D$ is joined with $E$, thus:
:300px
Using the Intersecting Chords Theorem we have:
:$GB \cdot BF = CB \cdot ... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
=== Case 1: $AC$ greater than $AB$ ===
Using $AC$ as the radius, we construct a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $A$.
Now we extend:
:$CB$ to $D$
:$AB$ to $F$
:$BA$ to $G$
:$CA$ to $E$.
$D$ is joined ... | Law of Cosines/Proof 2 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Law_of_Cosines/Proof_2 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"Definition:Triangle (Geometry)",
"Definition:Circle",
"Definition:Circle/Center",
"File:CosineRule.png",
"Intersecting Chords Theorem",
"Thales' Theorem",
"Definition:Right Angle",
"Definition:Cosine/Definition from Triangle",
"File:CosineRule2.png",
"Definition:Intersection (Geometry)",
"Defin... |
proofwiki-66 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | === Lemma: Right Triangle ===
{{:Law of Cosines/Right Triangle}}{{qed|lemma}}
=== Acute Triangle ===
{{:Law of Cosines/Proof 3/Acute Triangle}}{{qed|lemma}}
=== Obtuse Triangle ===
{{:Law of Cosines/Proof 3/Obtuse Triangle}}{{qed}} | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | === [[Law of Cosines/Right Triangle|Lemma: Right Triangle]] ===
{{:Law of Cosines/Right Triangle}}{{qed|lemma}}
=== [[Law of Cosines/Proof 3/Acute Triangle|Acute Triangle]] ===
{{:Law of Cosines/Proof 3/Acute Triangle}}{{qed|lemma}}
=== [[Law of Cosines/Proof 3/Obtuse Triangle|Obtuse Triangle]] ===
{{:Law of Cosine... | Law of Cosines/Proof 3 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Law_of_Cosines/Proof_3 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"Law of Cosines/Right Triangle",
"Law of Cosines/Proof 3/Acute Triangle",
"Law of Cosines/Proof 3/Obtuse Triangle"
] |
proofwiki-67 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :500px
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $AD$ be the tangent to the great circle ... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :[[File:Spherical-Cosine-Formula.png|500px]]
Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$.
By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$, $... | Spherical Law of Cosines/Proof 1 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Spherical_Law_of_Cosines/Proof_1 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"File:Spherical-Cosine-Formula.png",
"Definition:Polygon/Vertex",
"Definition:Spherical Triangle",
"Definition:Sphere/Geometry",
"Definition:Spherical Triangle",
"Definition:Circle/Arc",
"Definition:Great Circle",
"Definition:Great Circle",
"Definition:Circle/Center",
"Definition:Great Circle",
... |
proofwiki-68 | Law of Cosines | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :500px
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $O$ be joined to each of $A$, $B$ and $C... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | :[[File:Spherical-Cosine-Formula-2.png|500px]]
Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$.
By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$,... | Spherical Law of Cosines/Proof 2 | https://proofwiki.org/wiki/Law_of_Cosines | https://proofwiki.org/wiki/Spherical_Law_of_Cosines/Proof_2 | [
"Law of Cosines",
"Triangles",
"Cosine Function",
"Named Theorems"
] | [
"Definition:Triangle (Geometry)"
] | [
"File:Spherical-Cosine-Formula-2.png",
"Definition:Polygon/Vertex",
"Definition:Spherical Triangle",
"Definition:Sphere/Geometry",
"Definition:Spherical Triangle",
"Definition:Circle/Arc",
"Definition:Great Circle",
"Definition:Great Circle",
"Definition:Circle/Center",
"Definition:Great Circle",
... |
proofwiki-69 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | As Complex Sine Function is Absolutely Convergent and Complex Cosine Function is Absolutely Convergent, we have:
{{begin-eqn}}
{{eqn | l = \cos z + i \sin z
| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n + 1} } {\paren... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | As [[Complex Sine Function is Absolutely Convergent]] and [[Complex Cosine Function is Absolutely Convergent]], we have:
{{begin-eqn}}
{{eqn | l = \cos z + i \sin z
| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n + 1} ... | Euler's Formula/Proof | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Proof | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Sine Function is Absolutely Convergent/Complex Case",
"Cosine Function is Absolutely Convergent/Complex Case",
"Sum of Absolutely Convergent Series"
] |
proofwiki-70 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | Consider the differential equation:
:$D_z \map f z = i \cdot \map f z$
=== Step 1 ===
We will prove that $z = \cos \theta + i \sin \theta$ is a solution.
{{begin-eqn}}
{{eqn | l = z
| r = \cos \theta + i \sin \theta
| c =
}}
{{eqn | l = \frac {\d z} {\d \theta}
| r = -\sin \theta + i \cos \theta
... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | Consider the [[Definition:Differential Equation|differential equation]]:
:$D_z \map f z = i \cdot \map f z$
=== Step 1 ===
We will prove that $z = \cos \theta + i \sin \theta$ is a [[Definition:Solution to Differential Equation|solution]].
{{begin-eqn}}
{{eqn | l = z
| r = \cos \theta + i \sin \theta
|... | Euler's Formula/Real Domain/Proof 1 | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Real_Domain/Proof_1 | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Definition:Differential Equation",
"Definition:Differential Equation/Solution",
"Derivative of Sine Function",
"Derivative of Cosine Function",
"Linear Combination of Derivatives",
"Definition:Differential Equation/Solution",
"Derivative of Exponential Function",
"Derivative of Composite Function",
... |
proofwiki-71 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | This:
:$e^{i \theta} = \cos \theta + i \sin \theta$
is logically equivalent to this:
:$\dfrac {\cos \theta + i \sin \theta} {e^{i \theta} } = 1$
for every $\theta$.
Note that the left expression is nowhere undefined.
Taking the derivative of this:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d \theta} e^{-i \theta} \paren {\c... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | This:
:$e^{i \theta} = \cos \theta + i \sin \theta$
is [[Definition:Logical Equivalence|logically equivalent]] to this:
:$\dfrac {\cos \theta + i \sin \theta} {e^{i \theta} } = 1$
for every $\theta$.
Note that the left expression is nowhere undefined.
Taking the [[Definition:Derivative of Complex Function|deriva... | Euler's Formula/Real Domain/Proof 2 | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Real_Domain/Proof_2 | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Definition:Logical Equivalence",
"Definition:Derivative/Complex Function",
"Product Rule for Derivatives",
"Derivative of Exponential Function",
"Definition:Constant Mapping"
] |
proofwiki-72 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | It follows from Argument of Product equals Sum of Arguments that the $\map \arg z$ function for all $z \in \C$ satisfies the relationship:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2}$
which means that $\map \arg z$ is a kind of logarithm, in the sense that it satisfies the fundamental property of logarith... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | It follows from [[Argument of Product equals Sum of Arguments]] that the [[Definition:Argument of Complex Number|$\map \arg z$ function]] for all $z \in \C$ satisfies the relationship:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2}$
which means that $\map \arg z$ is a kind of [[Definition:General Logarithm... | Euler's Formula/Real Domain/Proof 3 | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Real_Domain/Proof_3 | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Argument of Product equals Sum of Arguments",
"Definition:Argument of Complex Number",
"Definition:General Logarithm",
"Definition:Real Function",
"Definition:Complex Modulus",
"Definition:Complex Function",
"Definition:Complex Number",
"Derivative of Composite Function",
"Definition:Inverse Mappin... |
proofwiki-73 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | Note that the following proof, as written, only holds for real $\theta$.
Define:
:$\map x \theta = e^{i \theta}$
:$\map y \theta = \cos \theta + i \sin \theta$
Consider first $\theta \ge 0$.
Taking Laplace transforms:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {\map x \theta} } s
| r = \map {\laptrans {e^{i \theta... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | Note that the following proof, as written, only holds for [[Definition:Real Number|real]] $\theta$.
Define:
:$\map x \theta = e^{i \theta}$
:$\map y \theta = \cos \theta + i \sin \theta$
Consider first $\theta \ge 0$.
Taking [[Definition:Laplace Transform|Laplace transforms]]:
{{begin-eqn}}
{{eqn | l = \map {\lap... | Euler's Formula/Real Domain/Proof 4 | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Real_Domain/Proof_4 | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Definition:Real Number",
"Definition:Laplace Transform",
"Laplace Transform of Exponential",
"Linear Combination of Laplace Transforms",
"Laplace Transform of Cosine",
"Laplace Transform of Sine",
"Definition:Laplace Transform",
"Injectivity of Laplace Transform"
] |
proofwiki-74 | Euler's Formula | Let $z \in \C$ be a complex number.
Then:
:$e^{i z} = \cos z + i \sin z$ | As Sine Function is Absolutely Convergent and Cosine Function is Absolutely Convergent, we have:
{{begin-eqn}}
{{eqn | l = \cos \theta + i \sin \theta
| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\p... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$e^{i z} = \cos z + i \sin z$ | As [[Sine Function is Absolutely Convergent]] and [[Cosine Function is Absolutely Convergent]], we have:
{{begin-eqn}}
{{eqn | l = \cos \theta + i \sin \theta
| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n +... | Euler's Formula/Real Domain/Proof 5 | https://proofwiki.org/wiki/Euler's_Formula | https://proofwiki.org/wiki/Euler's_Formula/Real_Domain/Proof_5 | [
"Euler's Formula",
"Euler's Identities",
"Exponential Function",
"Sine Function",
"Cosine Function"
] | [
"Definition:Complex Number"
] | [
"Sine Function is Absolutely Convergent",
"Cosine Function is Absolutely Convergent",
"Sum of Absolutely Convergent Series"
] |
proofwiki-75 | Lagrange's Theorem (Group Theory) | Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Then:
:$\order H$ divides $\order G$
where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.
In fact:
:$\index G H = \dfrac {\order G} {\order H}$
where $\index G H$ is the index of $H$ in $G$.
When $G$ is an infinite group, we can still interp... | Let $G$ be finite.
Consider the mapping $\phi: G \to G / H^l$, defined as:
:$\phi: G \to G / H^l: \map \phi x = x H$
where $G / H^l$ is the left coset space of $G$ modulo $H$.
For every $x H \in G / H^l$, there exists a corresponding $x \in G$, so $\phi$ is a surjection.
From Cardinality of Codomain of Surjection it fo... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\order H$ [[Definition:Divisor of Integer|divides]] $\order G$
where $\order G$ and $\order H$ are the [[Definition:Order of Structure|order]] of $G$ and $H$ respectively.
In fact:
:$\index G H = \d... | Let $G$ be [[Definition:Finite Group|finite]].
Consider the [[Definition:Mapping|mapping]] $\phi: G \to G / H^l$, defined as:
:$\phi: G \to G / H^l: \map \phi x = x H$
where $G / H^l$ is the [[Definition:Left Coset Space|left coset space of $G$ modulo $H$]].
For every $x H \in G / H^l$, there exists a corresponding $... | Lagrange's Theorem (Group Theory)/Proof 1 | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory) | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory)/Proof_1 | [
"Order of Groups",
"Index of Subgroups",
"Subgroups",
"Lagrange's Theorem (Group Theory)"
] | [
"Definition:Finite Group",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Index of Subgroup",
"Definition:Infinite Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Infinite Group",
"Definition:Infinite Gr... | [
"Definition:Finite Group",
"Definition:Mapping",
"Definition:Coset Space/Left Coset Space",
"Definition:Surjection",
"Cardinality of Codomain of Surjection",
"Definition:Finite Set",
"Cosets are Equivalent",
"Definition:Element",
"Left Coset Space forms Partition",
"Number of Elements in Partition... |
proofwiki-76 | Lagrange's Theorem (Group Theory) | Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Then:
:$\order H$ divides $\order G$
where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.
In fact:
:$\index G H = \dfrac {\order G} {\order H}$
where $\index G H$ is the index of $H$ in $G$.
When $G$ is an infinite group, we can still interp... | Let $G$ be a group.
Let $H$ be a subgroup of $G$.
From Cosets are Equivalent, a left coset $y H$ has the same number of elements as $H$, namely $\order H$.
Since left cosets are identical or disjoint, each element of $G$ belongs to exactly one left coset.
From the definition of index of subgroup, there are $\index G H$... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\order H$ [[Definition:Divisor of Integer|divides]] $\order G$
where $\order G$ and $\order H$ are the [[Definition:Order of Structure|order]] of $G$ and $H$ respectively.
In fact:
:$\index G H = \d... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
From [[Cosets are Equivalent]], a [[Definition:Left Coset|left coset]] $y H$ has the same number of [[Definition:Element|elements]] as $H$, namely $\order H$.
Since [[Definition:Left Coset|left cosets]] are [[Congruence Cl... | Lagrange's Theorem (Group Theory)/Proof 2 | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory) | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory)/Proof_2 | [
"Order of Groups",
"Index of Subgroups",
"Subgroups",
"Lagrange's Theorem (Group Theory)"
] | [
"Definition:Finite Group",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Index of Subgroup",
"Definition:Infinite Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Infinite Group",
"Definition:Infinite Gr... | [
"Definition:Group",
"Definition:Subgroup",
"Cosets are Equivalent",
"Definition:Coset/Left Coset",
"Definition:Element",
"Definition:Coset/Left Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Element",
"Definition:Coset/Left Coset",
"Definition:Index of Subgroup",
"Definition:Co... |
proofwiki-77 | Lagrange's Theorem (Group Theory) | Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Then:
:$\order H$ divides $\order G$
where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.
In fact:
:$\index G H = \dfrac {\order G} {\order H}$
where $\index G H$ is the index of $H$ in $G$.
When $G$ is an infinite group, we can still interp... | Follows directly from the Orbit-Stabilizer Theorem applied to Group Action on Coset Space.
{{Qed}} | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\order H$ [[Definition:Divisor of Integer|divides]] $\order G$
where $\order G$ and $\order H$ are the [[Definition:Order of Structure|order]] of $G$ and $H$ respectively.
In fact:
:$\index G H = \d... | Follows directly from the [[Orbit-Stabilizer Theorem]] applied to [[Group Action on Coset Space]].
{{Qed}} | Lagrange's Theorem (Group Theory)/Proof 3 | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory) | https://proofwiki.org/wiki/Lagrange's_Theorem_(Group_Theory)/Proof_3 | [
"Order of Groups",
"Index of Subgroups",
"Subgroups",
"Lagrange's Theorem (Group Theory)"
] | [
"Definition:Finite Group",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Index of Subgroup",
"Definition:Infinite Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Infinite Group",
"Definition:Infinite Gr... | [
"Orbit-Stabilizer Theorem",
"Action of Group on Coset Space is Group Action"
] |
proofwiki-78 | Carathéodory's Theorem (Analysis) | Let $I \subseteq \R$.
Let $c \in I$ be an interior point of $I$.
{{Disambiguate|Definition:Interior Point}}
{{explain|In this case, there appears not to be a definition for "interior point" which appropriately captures the gist of this. It is clearly a point inside a real interval, but the concept has not yet been defi... | === Necessary Condition ===
Suppose $f$ is differentiable at $c$.
Then by definition $\map {f'} c$ exists.
So we can define $\varphi$ by:
:$\map \varphi x = \begin{cases}
\dfrac {\map f x - \map f c} {x - c} & : x \ne c, x \in I \\
\map {f'} c & : x = c
\end{cases}$
Condition $(2)$, that $\varphi$ is continuous at $c$,... | Let $I \subseteq \R$.
Let $c \in I$ be an [[Definition:Interior Point|interior point]] of $I$.
{{Disambiguate|Definition:Interior Point}}
{{explain|In this case, there appears not to be a definition for "interior point" which appropriately captures the gist of this. It is clearly a point inside a real interval, but ... | === Necessary Condition ===
Suppose $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $c$.
Then by definition $\map {f'} c$ exists.
So we can define $\varphi$ by:
:$\map \varphi x = \begin{cases}
\dfrac {\map f x - \map f c} {x - c} & : x \ne c, x \in I \\
\map {f'} c & : x = c
\end{case... | Carathéodory's Theorem (Analysis) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Analysis) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Analysis) | [
"Direct Proofs",
"Real Analysis",
"Differential Calculus"
] | [
"Definition:Interior Point",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Real Function",
"Definition:Continuous Real Function/Point"
] | [
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Continuous Real Function/Point",
"Definition:Continuous Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point"
] |
proofwiki-79 | One-Step Subgroup Test | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subset of $G$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ {{iff}}:
:$(1): \quad H \ne \O$, that is, $H$ is non-empty
:$(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$. | === Sufficient Condition ===
Let $H$ be a subset of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is non-empty is one of the conditions.
It is also noted that the group operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$.
So it remains to show tha... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subset|subset]] of $G$.
Then $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ {{iff}}:
:$(1): \quad H \ne \O$, that is, $H$ is [[Definition:Non-Empty Set|non-empty]]
:$(2): \quad \forall a, b \in H:... | === Sufficient Condition ===
Let $H$ be a [[Definition:Subset|subset]] of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is [[Definition:Non-Empty Set|non-empty]] is one of the conditions.
It is also noted that the [[Definition:Group Operation|group operation]] of $\struct {H, \circ}$ is t... | One-Step Subgroup Test | https://proofwiki.org/wiki/One-Step_Subgroup_Test | https://proofwiki.org/wiki/One-Step_Subgroup_Test | [
"Named Theorems",
"Subgroups"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Non-Empty Set"
] | [
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Group Product/Group Law",
"Definition:Group",
"Axiom:Group Axioms",
"Definition:Non-Empty Set",
"Axiom:Group Axioms",
"Definition:Group",
"Definition:Group",
"Definition:Group"
] |
proofwiki-80 | Two-Step Subgroup Test | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subset of $G$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ {{iff}}:
:$(1): \quad H \ne \O$, that is, $H$ is non-empty
:$(2): \quad a, b \in H \implies a \circ b \in H$
:$(3): \quad a \in H \implies a^{-1} \in H$.
That is, $\struct {H, \circ}$ is a su... | === Necessary Condition ===
Let $H$ be a subset of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is nonempty is one of the conditions.
It is also noted that the group operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$.
So it remains to show that ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subset|subset]] of $G$.
Then $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ {{iff}}:
:$(1): \quad H \ne \O$, that is, $H$ is [[Definition:Non-Empty Set|non-empty]]
:$(2): \quad a, b \in H \implies... | === Necessary Condition ===
Let $H$ be a [[Definition:Subset|subset]] of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is [[Definition:Non-Empty Set|nonempty]] is one of the conditions.
It is also noted that the [[Definition:Group Operation|group operation]] of $\struct {H, \circ}$ is the... | Two-Step Subgroup Test | https://proofwiki.org/wiki/Two-Step_Subgroup_Test | https://proofwiki.org/wiki/Two-Step_Subgroup_Test | [
"Named Theorems",
"Subgroups"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Non-Empty Set",
"Definition:Subgroup",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Group Product/Group Law",
"Definition:Closed under Inversion"... | [
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Group Product/Group Law",
"Definition:Group",
"Axiom:Group Axioms",
"Definition:Non-Empty Set",
"Axiom:Group Axioms",
"Definition:Group",
"Definition:Group",
"Definition:Group",
"Definition:Group"
] |
proofwiki-81 | Fundamental Theorem of Arithmetic | For every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes, uniquely up to the order in which they appear. | In Integer is Expressible as Product of Primes it is proved that every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes.
In Prime Decomposition of Integer is Unique, it is proved that this prime decomposition is unique up to the order of the factors.
{{qed}} | For every [[Definition:Integer|integer]] $n$ such that $n > 1$, $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]], uniquely up to the order in which they appear. | In [[Integer is Expressible as Product of Primes]] it is proved that every [[Definition:Integer|integer]] $n$ such that $n > 1$, $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]].
In [[Prime Decomposition of Integer is Unique]], it is proved tha... | Fundamental Theorem of Arithmetic | https://proofwiki.org/wiki/Fundamental_Theorem_of_Arithmetic | https://proofwiki.org/wiki/Fundamental_Theorem_of_Arithmetic | [
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization",
"Fundamental Theorems"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number"
] | [
"Integer is Expressible as Product of Primes",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Prime Decomposition of Integer is Unique",
"Definition:Prime Decomposition",
"Definition:Unique",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-82 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | As for Euclid's Lemma for Prime Divisors, this can be verified by direct application of general version of Euclid's Lemma for irreducible elements.
{{qed}} | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | As for [[Euclid's Lemma for Prime Divisors]], this can be verified by direct application of [[Euclid's Lemma for Irreducible Elements/General Result|general version of Euclid's Lemma for irreducible elements]].
{{qed}} | Euclid's Lemma for Prime Divisors/General Result/Proof 1 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/General_Result/Proof_1 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Euclid's Lemma for Prime Divisors",
"Euclid's Lemma for Irreducible Elements/General Result"
] |
proofwiki-83 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Proof by induction:
For all $r \in \N_{>0}$, let $\map P r$ be the proposition:
:$\ds p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$
$\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$p ... | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $r \in \N_{>0}$, let $\map P r$ be the [[Definition:Propositional Function|proposition]]:
:$\ds p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$
$\map P 1$ is true, as this just says $p \divides a_1 \impl... | Euclid's Lemma for Prime Divisors/General Result/Proof 2 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/General_Result/Proof_2 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Principle of Mathematical Induction",
"Definition:Propositional Function",
"Euclid's Lemma for Prime Divisors",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Euclid's Lemma for Prime Divisors/General Result/Proof 2",
"Euclid's Lemma for Prime D... |
proofwiki-84 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Let $p \divides n$.
{{AimForCont}}:
:$\forall i \in \set {1, 2, \ldots, r}: p \nmid a_i$
By Prime not Divisor implies Coprime:
:$\forall i \in \set {1, 2, \ldots, r}: p \perp a_i$
By Integer Coprime to all Factors is Coprime to Whole:
:$p \perp n$
By definition of coprime:
:$p \nmid n$
The result follows by Proof by Co... | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Let $p \divides n$.
{{AimForCont}}:
:$\forall i \in \set {1, 2, \ldots, r}: p \nmid a_i$
By [[Prime not Divisor implies Coprime]]:
:$\forall i \in \set {1, 2, \ldots, r}: p \perp a_i$
By [[Integer Coprime to all Factors is Coprime to Whole]]:
:$p \perp n$
By definition of [[Definition:Coprime Integers|coprime]]:
:$... | Euclid's Lemma for Prime Divisors/General Result/Proof 3 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/General_Result/Proof_3 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Prime not Divisor implies Coprime",
"Integer Coprime to all Factors is Coprime to Whole",
"Definition:Coprime/Integers",
"Proof by Contradiction"
] |
proofwiki-85 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | We have that the integers form a Euclidean domain.
Then from Irreducible Elements of Ring of Integers we have that the irreducible elements of $\Z$ are the primes and their negatives.
The result then follows directly from Euclid's Lemma for Irreducible Elements.
{{qed}} | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | We have that the [[Integers are Euclidean Domain|integers form a Euclidean domain]].
Then from [[Irreducible Elements of Ring of Integers]] we have that the [[Definition:Irreducible Element of Ring|irreducible elements]] of $\Z$ are the [[Definition:Prime Number|primes]] and their [[Definition:Negative|negatives]].
... | Euclid's Lemma for Prime Divisors/Proof 1 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/Proof_1 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Integers are Euclidean Domain",
"Irreducible Elements of Ring of Integers",
"Definition:Irreducible Element of Ring",
"Definition:Prime Number",
"Definition:Negative",
"Euclid's Lemma for Irreducible Elements"
] |
proofwiki-86 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Let $p \divides a b$.
Suppose $p \nmid a$.
Then from the definition of prime:
:$p \perp a$
where $\perp$ indicates that $p$ and $a$ are coprime.
Thus from Euclid's Lemma it follows that:
:$p \divides b$
Similarly, if $p \nmid b$ it follows that $p \divides a$.
So:
:$p \divides a b \implies p \divides a$ or $p \divides ... | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Let $p \divides a b$.
Suppose $p \nmid a$.
Then from the definition of [[Definition:Prime Number|prime]]:
:$p \perp a$
where $\perp$ indicates that $p$ and $a$ are [[Definition:Coprime Integers|coprime]].
Thus from [[Euclid's Lemma]] it follows that:
:$p \divides b$
Similarly, if $p \nmid b$ it follows that $p \d... | Euclid's Lemma for Prime Divisors/Proof 2 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/Proof_2 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Prime Number",
"Definition:Coprime/Integers",
"Euclid's Lemma"
] |
proofwiki-87 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Let $p \divides a b$.
Suppose $p \nmid a$.
Then by {{EuclidPropLink|book = VII|prop = 29|title = Prime not Divisor implies Coprime}}:
:$p \perp a$
As $p \divides a b$, it follows by definition of divisor:
:$\exists e \in \Z: e p = a b$
So by {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}... | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Let $p \divides a b$.
Suppose $p \nmid a$.
Then by {{EuclidPropLink|book = VII|prop = 29|title = Prime not Divisor implies Coprime}}:
:$p \perp a$
As $p \divides a b$, it follows by definition of [[Definition:Divisor of Integer|divisor]]:
:$\exists e \in \Z: e p = a b$
So by {{EuclidPropLink|book = VII|prop = 19|t... | Euclid's Lemma for Prime Divisors/Proof 3 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma_for_Prime_Divisors/Proof_3 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-88 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Follows directly from Integers are Euclidean Domain.
{{qed}} | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Follows directly from [[Integers are Euclidean Domain]].
{{qed}} | Euclid's Lemma/Proof 1 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma/Proof_1 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Integers are Euclidean Domain"
] |
proofwiki-89 | Euclid's Lemma | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | Let $a, b, c \in \Z$.
We have that $a \perp b$.
That is:
:$\gcd \set {a, b} = 1$
where $\gcd$ denotes greatest common divisor.
From Bézout's Identity, we may write:
:$a x + b y = 1$
for some $x, y \in \Z$.
Upon multiplication by $c$, we see that:
:$c = c \paren {a x + b y} = c a x + c b y$
Now note that $c a x + c b y$... | Let $a, b, c \in \Z$.
Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]].
Then $a \divides c$. | Let $a, b, c \in \Z$.
We have that $a \perp b$.
That is:
:$\gcd \set {a, b} = 1$
where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
From [[Bézout's Identity]], we may write:
:$a x + b y = 1$
for some $x, y \in \Z$.
Upon multiplication by $c$, we see that:
:$c = c \paren... | Euclid's Lemma/Proof 2 | https://proofwiki.org/wiki/Euclid's_Lemma | https://proofwiki.org/wiki/Euclid's_Lemma/Proof_2 | [
"Euclid's Lemma",
"Divisors",
"Coprime Integers"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Coprime/Integers"
] | [
"Definition:Greatest Common Divisor/Integers",
"Bézout's Identity",
"Definition:Integer Combination",
"Common Divisor Divides Integer Combination"
] |
proofwiki-90 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | There are two cases:
:$(1): \quad$ If $p \divides n$, then $n^p \equiv 0 \equiv n \pmod p$.
:$(2): \quad$ Otherwise, $p \nmid n$.
Then, by Fermat's Little Theorem, $n^{p-1} \equiv 1 \pmod p$.
Multiplying both sides by $n$, then by Congruence of Product we have:
: $n^p \equiv n \pmod p$
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | There are two cases:
:$(1): \quad$ If $p \divides n$, then $n^p \equiv 0 \equiv n \pmod p$.
:$(2): \quad$ Otherwise, $p \nmid n$.
Then, by [[Fermat's Little Theorem]], $n^{p-1} \equiv 1 \pmod p$.
Multiplying both sides by $n$, then by [[Congruence of Product]] we have:
: $n^p \equiv n \pmod p$
{{qed}} | Fermat's Little Theorem/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Corollary_1/Proof_1 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Fermat's Little Theorem",
"Congruence of Product"
] |
proofwiki-91 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Suppose we want to form strings of colored beads, each with exactly $p$ beads in.
Suppose we have an unlimited number of beads to allow unlimited use of each of $n$ colors.
How many different strings can we form?
The answer is $n^p$ as each bead can be chosen in $n$ ways and there are $p$ choices for each string.
Out o... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Suppose we want to form strings of colored beads, each with exactly $p$ beads in.
Suppose we have an unlimited number of beads to allow unlimited use of each of $n$ colors.
How many different strings can we form?
The answer is $n^p$ as each bead can be chosen in $n$ ways and there are $p$ choices for each string.
O... | Fermat's Little Theorem/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Corollary_1/Proof_2 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Division Theorem",
"Definition:Prime Number",
"Definition:Multiplication/Integers"
] |
proofwiki-92 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Consider the integer sequence $n, 2 n, 3 n, \dotsc, \paren {p - 1} n$.
Note that none of these integers is congruent modulo $p$ to any of the others.
If this were the case, we would have $a n \equiv b n \pmod p$ for some $1 \le a < b \le p - 1$.
Then as $\map \gcd {n, p} = 1$, and we can cancel the $n$, we get $a \equi... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Consider the [[Definition:Integer Sequence|integer sequence]] $n, 2 n, 3 n, \dotsc, \paren {p - 1} n$.
Note that none of these [[Definition:Integer|integers]] is [[Definition:Congruence (Number Theory)|congruent modulo $p$]] to any of the others.
If this were the case, we would have $a n \equiv b n \pmod p$ for some ... | Fermat's Little Theorem/Proof 1 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Proof_1 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Integer Sequence",
"Definition:Integer",
"Definition:Congruence (Number Theory)",
"Cancellability of Congruences",
"Euclid's Lemma",
"Definition:Integer",
"Definition:Integer Sequence",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Set",
"Definition:Reduced Residue Syst... |
proofwiki-93 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | By Prime not Divisor implies Coprime:
:$p \nmid n \implies p \perp n$
and Euler's Theorem (Number Theory) can be applied.
Thus:
:$n^{\map \phi p} \equiv 1 \pmod p$
But from Euler Phi Function of Prime Power:
:$\map \phi p = p \paren {1 - \dfrac 1 p} = p - 1$
and the result follows.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | By [[Prime not Divisor implies Coprime]]:
:$p \nmid n \implies p \perp n$
and [[Euler's Theorem (Number Theory)]] can be applied.
Thus:
:$n^{\map \phi p} \equiv 1 \pmod p$
But from [[Euler Phi Function of Prime Power]]:
:$\map \phi p = p \paren {1 - \dfrac 1 p} = p - 1$
and the result follows.
{{qed}} | Fermat's Little Theorem/Proof 2 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Proof_2 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Prime not Divisor implies Coprime",
"Euler's Theorem (Number Theory)",
"Euler Phi Function of Prime Power"
] |
proofwiki-94 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$.
From {{Corollary|Reduced Residue System under Multiplication forms Abelian Group}}, $\struct {\Z'_p, \times}$ forms a group of order $p - 1$ under modulo multiplication.
By Element to Power of Group Order is Identity, we have:... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Let $\struct {\Z'_p, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $p$]].
From {{Corollary|Reduced Residue System under Multiplication forms Abelian Group}}, $\struct {\Z'_p, \times}$ forms a [[Definition:Group|group]] of [[Definition:Order of... | Fermat's Little Theorem/Proof 3 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Proof_3 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Multiplicative Group of Reduced Residues",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Modulo Multiplication",
"Element to Power of Group Order is Identity"
] |
proofwiki-95 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Proof by induction over $n$.
Induction base:
:$1^p \equiv 1 \pmod p$
Induction step:
Assume $n^p \equiv n \pmod p$
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^p
| r = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} \cdot 1^k
| c = Binomial Theorem
}}
{{eqn | q = \forall k: 0 < k < p
| l = {p \choose k}
... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | [[Second Principle of Mathematical Induction|Proof by induction]] over $n$.
Induction base:
:$1^p \equiv 1 \pmod p$
Induction step:
Assume $n^p \equiv n \pmod p$
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^p
| r = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} \cdot 1^k
| c = [[Binomial Theorem]]
}}
{{eqn ... | Fermat's Little Theorem/Proof 4 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Proof_4 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Second Principle of Mathematical Induction",
"Binomial Theorem",
"Binomial Coefficient of Prime"
] |
proofwiki-96 | Fermat's Little Theorem | Let $p$ be a prime number.
Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Let $S$ be a set of n elements, and consider $p$-tuples $\left( a_1, a_2, \dots, a_p \right)$ and consider the Group Action of these p-tuples by $\Z/p\Z$ via cyclic shifts.
For example, if:
:$S = \{ b, r, g \}$
and:
:$p=5$
then $rgbgg$ is equivalent to:
:$grgbg, ggrgb, bggrg, gbggr$.
We use Burnside's Lemma, counting t... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
Then:
:$n^{p - 1} \equiv 1 \pmod p$ | Let $S$ be a set of n elements, and consider $p$-tuples $\left( a_1, a_2, \dots, a_p \right)$ and consider the [[Definition:Group Action|Group Action]] of these p-tuples by $\Z/p\Z$ via cyclic shifts.
For example, if:
:$S = \{ b, r, g \}$
and:
:$p=5$
then $rgbgg$ is equivalent to:
:$grgbg, ggrgb, bggrg, gbggr$.
We us... | Fermat's Little Theorem/Proof 5 | https://proofwiki.org/wiki/Fermat's_Little_Theorem | https://proofwiki.org/wiki/Fermat's_Little_Theorem/Proof_5 | [
"Fermat's Little Theorem",
"Number Theory"
] | [
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Group Action",
"Burnside's Lemma",
"Burnside's Lemma"
] |
proofwiki-97 | De Morgan's Laws (Set Theory) | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let the cardinality $\card I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcap_{i \mathop = 1}^n ... | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let the [[Definition:Cardinality|cardinality]] $\card I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory) | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection/Proof | [
"Set Difference",
"Set Union",
"Set Intersection",
"Relative Complement",
"Set Complement",
"De Morgan's Laws"
] | [] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"In... |
proofwiki-98 | De Morgan's Laws (Set Theory) | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let the cardinality $\size I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcup_{i \mathop = 1}^n ... | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let the [[Definition:Cardinality|cardinality]] $\size I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory) | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union/Proof | [
"Set Difference",
"Set Union",
"Set Intersection",
"Relative Complement",
"Set Complement",
"De Morgan's Laws"
] | [] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Union",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Union is ... |
proofwiki-99 | De Morgan's Laws (Set Theory) | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let $x \in S$ througout.
{{begin-eqn}}
{{eqn | o =
| r = x \in \relcomp S {T_1 \cup T_2}
}}
{{eqn | o = \leadsto
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | o = \leadsto
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eqn... | {{:De Morgan's Laws (Set Theory)/Set Difference}} | Let $x \in S$ througout.
{{begin-eqn}}
{{eqn | o =
| r = x \in \relcomp S {T_1 \cup T_2}
}}
{{eqn | o = \leadsto
| r = x \notin \paren {T_1 \cup T_2}
| c = {{Defof|Relative Complement}}
}}
{{eqn | o = \leadsto
| r = \neg \paren {x \in T_1 \lor x \in T_2}
| c = {{Defof|Set Union}}
}}
{{eq... | De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union/Proof 2 | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory) | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Relative_Complement/Complement_of_Union/Proof_2 | [
"Set Difference",
"Set Union",
"Set Intersection",
"Relative Complement",
"Set Complement",
"De Morgan's Laws"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations",
"De Morgan's Laws (Logic)/Conjunction of Negations",
"Definition:Set Equality/Definition 1"
] |
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