hcju/proofwikiqa huggingface dataset:
We have that: :$\displaystyle \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$ Consider $\displaystyle 2 \sum_{i \mathop = 1}^n i$. 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 + \paren {n - 1} + \dotsb + 2 + 1}       | c =  }} {{eqn | r = \paren {1 + n} + \paren {2 + \paren {n - 1} } + \dotsb + \paren {\paren {n - 1} + 2} + \paren {n + 1}       | c = Integer Addition is Commutative, Integer Addition is Associative }} {{eqn | r = \paren {n + 1}_1 + \paren {n + 1}_2 + \dotsb + \paren {n + 1}_n       | c =  }} {{eqn | r = n \paren {n + 1}       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = 2 \sum_{i \mathop = 1}^n i       | r = n \paren {n + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop = 1}^n i       | r = \frac {n \paren {n + 1} } 2       | c =  }} {{end-eqn}} {{qed}}
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: :$\paren {i + 1}^2 - i^2 = 2 i + 1$ And also: :$\paren {n + 1}^2 - 1 = n^2 + 2 n$ Combining these results, we obtain: {{begin-eqn}} {{eqn | l = n + 2 \sum_{i \mathop = 1}^n i       | r = n^2 + 2 n }} {{eqn | ll= \leadsto       | l = 2 \sum_{i \mathop = 1}^n i       | r = n \left({n + 1}\right) }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop = 1}^n i       | r = \frac {n \paren {n + 1} } 2 }} {{end-eqn}} This concludes the proof. {{qed}}
Proof by induction: === Basis for the Induction === When $n = 1$, we have: :$\displaystyle \sum_{i \mathop = 1}^1 i = 1$ Also: :$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$ This is our base case. === Induction Hypothesis === :$\forall k \in \N: k \ge 1: \displaystyle \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$ This is our induction hypothesis. It is to be demonstrated that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \frac {\paren {k + 1} \paren {k + 2} } 2$ === Induction Step === This is our induction step: Consider $n = k + 1$. By the properties of summation: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \sum_{i \mathop = 1}^k i + k + 1$ Using the induction hypothesis this can be simplified to: {{begin-eqn}} {{eqn | l = \frac {k \paren {k + 1} } 2 + k + 1       | r = \frac {k \paren {k + 1} + 2 k + 2} 2       | c =  }} {{eqn | r = \frac {k^2 + 3 k + 2} 2       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {\paren {k + 1} + 1} } 2       | c =  }} {{end-eqn}} Hence the result by induction. {{qed}}
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 {\left({n+1}\right) n} 2       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
{{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 \mathop = 0}^n i       | r = \paren {n + 1} \paren {\frac n 2}       | c = Let $a = 0$ and $d = 1$ }} {{eqn | l = 0 + \sum_{i \mathop = 1}^n i       | r = \frac {n \paren {n + 1} } 2       | c =  }} {{eqn | l = \sum_{i \mathop = 1}^n i       | r = \frac {n \paren {n + 1} } 2       | c =  }} {{end-eqn}} {{qed}}
Triangular numbers are $k$-gonal numbers where $k = 3$. From Closed Form for Polygonal Numbers we have that: :$P \left({k, n}\right) = \dfrac n 2 \left({\left({k - 2}\right) n - k + 4}\right)$ Hence: {{begin-eqn}} {{eqn | l = T_n       | r = \frac n 2 \left({\left({3 - 2}\right) n - 3 + 4}\right)       | c = Closed Form for Polygonal Numbers }} {{eqn | r = \frac n 2 \left({n + 1}\right)       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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 bijection: :$(1): \quad \size A = \size B$ We have: {{begin-eqn}} {{eqn | l = A \cup B       | r = \set {\tuple {a, b}: a, b \in \N_n^*} }} {{eqn | l = A \cap B       | r = \set {\tuple {a, b}: a = b: a, b \in \N_n^*} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \size A + \size B       | r = \size {A \cup B} + \size {A \cap B}       | c = Inclusion-Exclusion Principle }} {{eqn | r = n^2 + n       | c = Count of a finite set }} {{end-eqn}} Combined with $\left({1}\right)$ this yields: :$\size A = \dfrac {n^2 + n} 2 = \dfrac {n \paren {n + 1} } 2$ It remains to prove that: :$T_n = \size A$ {{begin-eqn}} {{eqn | l = T_n       | r = \sum_{i \mathop = 1}^n i       | c = Definition of $T_n$ }} {{eqn | r = \sum_{i \mathop \in \N_n^*} i       | c =  }} {{eqn | r = \sum_{i \mathop \in \N_n^*} \size {\set {a: a \in \N_i^*} }       | c = Count of Finite Set,{{Defof|Initial Segment of One-Based Natural Numbers}} }} {{eqn | r = \size {\set {\tuple {a, i} :a\in \N_i^*, i \in \N_n^*} }       | c = Inclusion-Exclusion Principle, sets are mutually exclusive as their second argument in the ordered pair are different }} {{eqn | r = \size {\set {\tuple {a, b}: a \le b, a, b \in \N_n^*} }       | c = Change of Variable, {{Defof|Initial Segment of One-Based Natural Numbers}} }} {{eqn | r = \size A       | c = Definition of $A$ }} {{end-eqn}} {{qed}}
{{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}       | r = n \paren {n + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 1}^n j       | r = \frac {n \paren {n + 1} } 2       | c =  }} {{end-eqn}} {{qed}}
From Sum of Powers of Positive Integers: {{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 numbers. Setting $p = 1$: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^1       | r = 1 + 2 + \cdots + n       | c =  }} {{eqn | r = \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!}       | c =  }} {{eqn | r = \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1       | c = {{Defof|Bernoulli Numbers}}, Number to Power of Zero Falling is One }} {{eqn | r = \frac {n^2 + n} 2       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ When $n = 0$, we see from the definition of vacuous sum that: :$0 = \displaystyle \sum_{i \mathop = 1}^0 i^2 = \frac {0 \paren 1 \paren 1} 6 = 0$ and so $\map P 0$ holds. === Basis for the Induction === When $n = 1$: :$\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$ Now, we have: :$\displaystyle \frac {n \paren {n + 1} \paren {2 n + 1} } 6 = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 1} } 6 = \frac 6 6 = 1$ and $\map P 1$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k i^2 = \frac {k \paren {k + 1} \paren {2 k + 1} } 6$ Then we need to show: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6$ === Induction Step === This is our induction step: Using the properties of summation, we have: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \sum_{i \mathop = 1}^k i^2 + \paren {k + 1}^2$ We can now apply our induction hypothesis, obtaining: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^2       | r = \frac {k \paren {k + 1} \paren {2 k + 1} } 6 + \paren {k + 1}^2       | c =  }} {{eqn | r = \frac {k \paren {k + 1} \paren {2 k + 1} + 6 \paren {k + 1}^2} 6       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k \paren {2 k + 1} + 6 \paren {k + 1} } } 6       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {2 k^2 + 7 k + 6} } 6       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n 3 i \left({i + 1}\right)       | r = n \left({n + 1}\right) \left({n + 2}\right)       | c = Sum of Sequence of Products of Consecutive Integers }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i       | r = n \left({n + 1}\right) \left({n + 2}\right) }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n 3 i^2       | r = n \left({n + 1}\right) \left({n + 2}\right) - 3 \frac {n \left({n + 1}\right) } 2       | c = Closed Form for Triangular Numbers }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n i^2       | r = \frac {n \left({n + 1}\right) \left({n + 2}\right)} 3 - \frac {n \left({n + 1}\right) } 2       | c =  }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n i^2       | r = \frac {2 n \left({n + 1}\right) \left({n + 2}\right) - 3 n \left({n + 1}\right) } 6       | c =  }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n i^2       | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6       | c =  }} {{end-eqn}} {{qed}}
Observe that: : $3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$ That is: : $(1): \quad 6 T_i = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right)$ Then: {{begin-eqn}} {{eqn | l = n^2       | r = \frac {n^2 + n + n^2 - n} 2 }} {{eqn | r = \frac {n \left({n + 1}\right) + n \left({n - 1}\right)} 2 }} {{eqn | r = \frac {n \left({n + 1}\right)} 2 + \frac {n \left({n - 1}\right)} 2 }} {{eqn | r = T_n + T_{n-1} }} {{end-eqn}}   where $T_n$ is the $n$th Triangular number. Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^2       | r = 1 + \left({T_1 + T_2}\right) + \left({T_2 + T_3}\right) + \left({T_3 + T_4}\right) + \cdots + \left({T_{n - 1} + T_n}\right) }} {{eqn | r = 1 + 2 T_2 + 2 T_3 + 2 T_4 + \cdots + 2 T_{n - 1} + T_n }} {{eqn | r = 1 - T_1 - T_n + 2 \left({T_1 + T_2 + T_3 + T_4 + \cdots + T_n}\right) }} {{eqn | r = 2 \left({\sum_{i \mathop = 1}^n T_i} \right) - \frac {n \left({n + 1}\right)} 2 }} {{eqn | r = 2 \left({\frac {n \left({n + 1}\right) \left({n + 2}\right)} 6}\right) - \frac {n \left({n + 1}\right)} 2       | c = Telescoping Series from $(1)$ above }} {{eqn | r = \frac {2 n \left({n^2 + 3 n + 2}\right) - \left({3 n^2 + 3 n}\right)} 6 }} {{eqn | r = \frac {2 n^3 + 3 n^2 + n} 6 }} {{eqn | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 }} {{end-eqn}} {{qed}}
:File:Sum of Sequences of Squares.jpg We can observe from the above diagram that: : $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$ Therefore we have:  {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^2       | r = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \left( {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i - 1} j} \right)       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \left({\frac {n \left({n + 1}\right)} 2 - \frac {i \left({i - 1}\right)} 2}\right)       | c =  }} {{eqn | ll= \implies       | l = 2 \sum_{i \mathop = 1}^n i^2        | r = n^2 \left({n + 1}\right) - \sum_{i \mathop = 1}^n i^2 + \sum_{i \mathop = 1}^n i        | c =  }} {{eqn | ll= \implies       | l = 3 \sum_{i \mathop = 1}^n i^2       | r = n^2 \left({n + 1}\right) + \sum_{i \mathop = 1}^n i        | c =  }} {{eqn | ll= \implies       | l = 3 \sum_{i \mathop = 1}^n i^2        | r = n^2 \left({n + 1}\right) + \frac {n \left({n + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | ll= \implies       | l = 6 \sum_{i \mathop = 1}^n i^2        | r = 2 n^2 \left({n + 1}\right) + n \left({n + 1}\right)       | c =  }} {{eqn | r = n \left({n + 1}\right) \left({2 n + 1}\right)       | c =  }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n i^2        | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right)       | r = \sum_{i \mathop = 1}^n \left({i^3 + 3 i^2 + 3 i + 1 - i^3}\right)       | c = Binomial Theorem }} {{eqn | r = \sum_{i \mathop = 1}^n \left({3 i^2 + 3 i + 1}\right)       | c = }} {{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1       | c = Summation is Linear }} {{eqn | r = 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n       | c = Closed Form for Triangular Numbers }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right)       | r = \left({n + 1}\right)^3 - n^3 + n^3 - \left({n - 1}\right)^3 + \left({n - 1}\right)^3 - \cdots + 2^3 - 1^3       | c = Definition of Summation }} {{eqn | r = \left({n + 1}\right)^3 - 1^3       | c = Telescoping Series: Example 2 }} {{eqn | r = n^3 + 3n^2 + 3n + 1 - 1       | c = Binomial Theorem }} {{eqn | r = n^3 + 3n^2 + 3n       | c =  }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = 3 \sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n       | r = n^3 + 3 n^2 + 3 n }} {{eqn | ll= \implies       | l = 3 \sum_{i \mathop = 1}^n i^2       | r = n^3 + 3 n^2 + 3 n - 3 \frac {n \left({n + 1}\right)} 2 - n }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^2       | r = \frac 1 3 \left({n^3 + 3 n^2 + 3 n - 3 \frac {n \left({n + 1}\right)}2 - n}\right) }} {{eqn | r = \frac 1 3 \left({n^3 + 3 n^2 + 3 n - \frac {3 n^2} 2 - \frac {3 n} 2 - n}\right) }} {{eqn | r = \frac 1 3 \left({n^3 + \frac {3 n^2} 2 + \frac n 2}\right) }} {{eqn | r = \frac 1 6 n \left({2 n^2 + 3 n + 1}\right) }} {{eqn | r = \frac 1 6 n \left({n + 1}\right) \left({2 n + 1}\right) }} {{end-eqn}} {{qed}}
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^2       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^2       | r = \sum_{i \mathop = 1}^n \paren {2 \binom i 2 + \binom i 1}       | c =  }} {{eqn | r = 2 \binom {n + 1} 3 + \binom {n + 1} 2       | c = Sum of Binomial Coefficients over Upper Index }} {{eqn | r = \frac {2 \paren {n + 1} n \paren {n - 1} } {3 \times 2 \times 1} + \frac {\paren {n + 1} n} 2       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6       | c = after algebra }} {{end-eqn}} {{qed}}
From Sum of Powers of Positive Integers: {{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 numbers. Setting $p = 2$: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^2       | r = 1^2 + 2^2 + \cdots + n^2       | c =  }} {{eqn | r = \frac {n^{2 + 1} } {2 + 1} + \frac {B_1 \, p^{\underline 0} \, n^2} {1!} + \frac {B_2 \, p^{\underline 1} \, n^1} {2!}       | c =  }} {{eqn | r = \frac {n^3} 3 + B_1 n^2 + B_2 n       | c = {{Defof|Falling Factorial}} and simplification }} {{eqn | r = \frac {n^3} 3 + \frac {n^2} 2 + \frac n 6       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6       | c = after algebra }} {{end-eqn}} Hence the result. {{qed}}
{{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 = Rule of Association: Disjunction }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {A \cup B} \cup C       | c = {{Defof|Set Union}} }} {{end-eqn}} Therefore: :$x \in A \cup \paren {B \cup C}$ {{iff}} $x \in \paren {A \cup B} \cup C$ Thus it has been shown that: : $A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$ {{qed}}
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 easily seen to be straight. Now to calculate the area of this figure. On the one hand, we can add up the area of the component parts of the square. Specifically, we can add up the $4$ triangles and the inner square. Thus we calculate the area of the large square to be: :$4 \paren {\dfrac 1 2 a b} + c^2 = 2 a b + c^2$ On the other hand, we can similarly calculate the area of the large square to be: :$\paren {a + b}^2 = a^2 + 2 a b + b^2$ Now these two expressions have to be equal, since they both represent the area of the square. Thus: :$a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$ {{qed}}
: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}}
We start with the algebraic definitions for sine and cosine: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ From these, we derive the proof that $\cos^2 x + \sin^2 x = 1$. Then from the Equivalence of Definitions of Sine and Cosine, we can use the geometric interpretation of sine and cosine: :400px :$\sin \theta = \dfrac {\text{Opposite}} {\text{Hypotenuse}}$ :$\cos \theta = \dfrac {\text{Adjacent}} {\text{Hypotenuse}}$ Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page. Thus: {{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = 1       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \paren {\frac a c}^2 + \paren {\frac b c}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = a^2 + b^2       | r = c^2       | c = multiplying both sides by $c^2$ }} {{end-eqn}} {{qed}}
: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 $CA$ is in a straight line with $AG$. For the same reason $BA$ is in a straight line with $AH$. We have that $\angle DBC = \angle FBA$, because both are right angles. We add $\angle ABC$ to each one to make $\angle FBC$ and $\angle DBA$. By common notion 2, $\angle FBC = \angle DBA$. By Triangle Side-Angle-Side Equality, $\triangle ABD = \triangle FBC$. We have that the parallelogram $BDLM$ is on the same base $BD$ and between the same parallels $BD$ and $AL$ as the triangle $\triangle ABD$. So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $BDLM$ is twice the area of $\triangle ABD$. Similarly, we have that the parallelogram $ABFG$ (which happens also to be a square) is on the same base $FB$ and between the same parallels $FB$ and $GC$ as the triangle $\triangle FBC$. So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $ABFG$ is twice the area of $\triangle FBC$. So $BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$. By the same construction, we have that $CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$. But $BDLM + CELM$ is the whole of the square $BDEC$. Therefore the area of the square $BDEC$ is equal to the combined area of the squares $ABFG$ and $ACKH$. {{qed}}
: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}}
:500px We have that $CH = BS = AB = AJ$. Hence the result follows directly from Pythagoras's Theorem for Parallelograms. {{qed}}
: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: :$\dfrac {\paren {XYZ} } {\paren {X' Y' Z'} } = \dfrac {XY^2} {X'Y'^2} = \dfrac {h_z^2} {h_{z'}^2} = \dfrac {t_z^2} {t_{z'}^2} = \ldots$ where $\paren {XYZ}$ denotes the area of $\triangle XYZ$. {{explain|What do $t_z$, $t_{z'}$ mean? A diagram would help this last statement to make sense here.}} This gives us: :$\dfrac {\paren {ADC} } {\paren {ACB} } = \dfrac {AC^2} {AB^2}$ and :$\dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2}$ Taking the sum of these two equalities we obtain: :$\dfrac {\paren {ADC} } {\paren {ACB} } + \dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2} + \dfrac {AC^2} {AB^2}$ Thus: :$\dfrac {\overbrace {\paren {ADC} + \paren {CDB} }^{\paren {ACB} } } {\paren {ACB} } = \dfrac {BC^2 + AC^2} {AB^2}$ Thus: :$AB^2 = BC^2 + AC^2$ as desired. {{qed}}
Let $\mathbb P$ be a finite set of prime numbers. Consider the number: :$\displaystyle n_p = \paren {\prod_{p \mathop \in \mathbb P} p} + 1$ Take any $p_j \in \mathbb P$. We have that: : $\displaystyle p_j \divides \prod_{p \mathop \in \mathbb P} p$ Hence: :$\displaystyle \exists q \in \Z: \prod_{p \mathop \in \mathbb P} p = q p_j$ So: {{begin-eqn}} {{eqn | l = n_p       | r = q p_j + 1       | c = Division Theorem }} {{eqn | ll= \leadsto       | l = n_p       | o = \perp       | r = p_j       | c = {{Defof|Coprime Integers}} }} {{end-eqn}} So $p_j \nmid n_p$. There are two possibilities: :$(1): \quad n_p$ is prime, which is not in $\mathbb P$.  :$(2): \quad n_p$ is composite. But from Positive Integer Greater than 1 has Prime Divisor, it must be divisible by ''some'' prime. That means it is divisible by a prime which is not in $\mathbb P$. So, in either case, there exists at least one prime which is not in the original set $\mathbb P$ we created. {{qed}} {{Euclid Note|20|IX}}