Datasets:
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"
] |
End of preview. Expand in Data Studio
ProofWiki Math Problems and Solutions
This dataset pairs theorem statements (problems) with their formal proofs (solutions) extracted from ProofWiki.
Fields
| Column | Description |
|---|---|
id |
Stable row identifier |
title |
Theorem title |
problem |
Theorem statement with wikilinks resolved |
solution |
Proof text with wikilinks resolved |
problem_wikitext |
Raw MediaWiki wikitext for the theorem |
solution_wikitext |
Raw MediaWiki wikitext for the proof |
proof_title |
Proof page title |
theorem_url |
Source URL for the theorem |
proof_url |
Source URL for the proof |
categories |
ProofWiki categories |
theorem_references |
Linked pages referenced in the theorem |
proof_references |
Linked pages referenced in the proof |
Source and license
- Source dump:
https://proofwiki.org/xmldump/latest.xml.gz - ProofWiki content is licensed under CC BY-SA 3.0
- Current build size: 23,640 theorem-proof pairs
Usage
from datasets import load_from_disk
ds = load_from_disk("data/proofwiki_hf")
print(ds["train"][0]["problem"])
print(ds["train"][0]["solution"])
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