hcju/proofwikiqa huggingface dataset:
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 = m^2       | c =  }} {{end-eqn}} Any prime in the prime factorizations of $n^2$ and $m^2$ must occur an even number of times because they are squares. Thus, $p$ must occur in the prime factorization of $n^2 p$ an odd number of times. Therefore, $p$ occurs as a factor of $m^2$ an odd number of times, a contradiction. So $\sqrt p$ must be irrational. {{qed}}
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 Irreducible over Integers is Irreducible over Rationals it follows that $\paren {x + \sqrt p}$ and $\paren {x + \sqrt p}$ do not have rational coefficients. That is: :$\sqrt p$ is not rational. Hence the result. {{qed}}
{{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 {\exp x \paren {\exp h - 1} } h }} {{eqn | r = \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}       | c = Multiple Rule for Limits of Functions, as $\exp x$ is constant }} {{eqn | r = \exp x       | c = Derivative of Exponential at Zero }} {{end-eqn}} {{qed}}
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 = \dfrac 1 {1 / y}       | c = Derivative of Inverse Function }} {{eqn | r = y }} {{eqn | r = e^x }} {{end-eqn}} {{qed}}
{{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, Derivative of Identity Function }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {e^x}       | r = e^x       | c = multiply both sides by $e^x$ }} {{end-eqn}} {{qed}}
This proof assumes the series definition of $\exp$. That is, let: :$\displaystyle \exp x = \sum_{k \mathop = 0}^\infty \frac{x^k}{k!}$ From Series of Power over Factorial Converges, the interval of convergence of $\exp$ is the entirety of $\R$. So we may apply Differentiation of Power Series to $\exp$ for all $x \in \R$. Thus we have: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \exp x       | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} }} {{eqn | r = \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1}       | c = Differentiation of Power Series, with $n = 1$ }} {{eqn | r = \sum_{k \mathop = 1}^\infty \frac {x^{k - 1} } {\paren {k - 1}!} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} }} {{eqn | r = \exp x }} {{end-eqn}} Hence the result. {{qed}}
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 ceiling function. From Closed Real Interval is Compact in Metric Space, $I$ is compact. From Chain Rule for Derivatives: :$\dfrac \d {\d x} \map {f_n} x = \dfrac n {n + x} \map {f_n} x$ === Lemma === {{:Derivative of Exponential Function/Proof 5/Lemma}}{{qed|lemma}} From the lemma: :$\forall x \in I: \sequence {\dfrac \d {\d x} \map {f_{n + N} } x}$ is increasing Hence, from Dini's Theorem, $\sequence {\dfrac \d {\d x} f_{n + N} }$ is uniformly convergent on $I$. Therefore, for $x \in I$: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \exp x       | r = \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_n} x }} {{eqn | r = \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_{n + N} } x       | c = Tail of Convergent Sequence }} {{eqn | r = \lim_{n \mathop \to \infty} \frac \d {\d x} \map {f_{n + N} } x       | c = Derivative of Uniformly Convergent Sequence of Differentiable Functions }} {{eqn | r = \lim_{n \mathop \to \infty} \frac n {n + x} \map {f_n} x       | c = from above }} {{eqn | r = \lim_{n \mathop \to \infty} \map {f_n} x       | c = Combination Theorem for Sequences }} {{eqn | r = \exp x }} {{end-eqn}} In particular: :$\dfrac \d {\d x} \exp x_0 = \exp x_0$ {{qed}}
From the definition of the sine function, we have: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ 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}} {{eqn | l = \map {\frac \d {\d x} } {\sin x}       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}       | c =  }} {{end-eqn}} The result follows from the definition of the cosine function. {{qed}}
{{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_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} + \sin h \cos x} h       | c = collecting terms containing $\map \sin x$ and factoring }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} } h + \lim_{h \mathop \to 0} \frac {\sin h \cos x} h       | c = Sum Rule for Limits of Functions }} {{eqn | r = \map \sin x \times 0 + 1 \times \cos x       | c = Limit of Sine of X over X and Limit of (Cosine (X) - 1) over X }} {{eqn | r = \cos x }} {{end-eqn}} {{qed}}
{{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 Complement equals Cosine }} {{end-eqn}} {{qed}}
{{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} - \frac h 2} } h }} {{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cos {x + \frac h 2} \map \sin {\frac h 2} } h       | c = Simpson's Formula for Cosine by Sine }} {{eqn | r = \lim_{h \mathop \to 0} \map \cos {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}       | c = Multiple Rule for Limits of Functions and Product Rule for Limits of Functions }} {{eqn | r = \cos x \times 1       | c = Cosine Function is Continuous and Limit of Sine of X over X }} {{eqn | r = \cos x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arcsin x       | r = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }       | c = Arcsin as an Integral }} {{eqn | ll= \leadsto       | l = \dfrac {\map \d {\map \arcsin y} } {\d y}       | r = \dfrac {\map \d {\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y} }} {{eqn | r = \dfrac 1 {\sqrt {1 - y^2} }       | c = Corollary to Fundamental Theorem of Calculus: First Part }} {{end-eqn}} Note that we get the same answer as Derivative of Arcsine Function. By definition of real $\arcsin$ function, $\arcsin$ is bijective on its domain $\closedint {-1} 1$. Thus its inverse is itself a mapping. From Inverse of Inverse of Bijection, its inverse is the $\sin$ function. {{begin-eqn}} {{eqn | l = \dfrac {\map \d {\sin \theta} } {\d \theta}       | r = 1 / \dfrac 1 {\sqrt {1 - \sin^2 \theta} }       | c = Derivative of Inverse Function }} {{eqn | r = \pm \sqrt {1 - \sin^2 \theta}       | c = Positive in Quadrant $\text I$ and Quadrant $\text {IV}$, Negative in Quadrant $\text {II}$ and Quadrant $\text {III}$ }} {{eqn | l = \dfrac {\map \d {\sin \theta} } {\d \theta}       | r = \cos \theta }} {{end-eqn}} {{qed}}
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 $\displaystyle \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to: :$\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 - \frac 1 {10} } = \dfrac {\frac 9 {10} } {\frac 9 {10} } = 1$ {{qed}}
{{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, \, 1 - 0.1^3, \, \cdots} } {}       | c = from $(1)$ }} {{eqn | r = \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {}       | c =  }} {{eqn | r = 1 - 0       | c = Sequence of Powers of Number less than One, {{Defof|Real Numbers}} }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
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 $\order N$. Let $\Syl p N$ be the set of Sylow $p$-subgroups of $N$. By the First Sylow Theorem: :$\Syl p N \ne \O$ Let: : $P \in \Syl p N$ : $G_0$ be the normalizer in $G$ of $P$ : $N_0 = N \cap G_0$. By Frattini's Argument: :$G = G_0 N$ By the Second Isomorphism Theorem for Groups and thence Lagrange's Theorem (Group Theory), it follows that: : $N_0$ is a Hall subgroup of $G_0$ : $\index {G_0} {N_0} = \index G H$ Suppose $G_0 < G$. Then by induction applied to $N_0$ in $G_0$, we find that $G_0$ contains a complement $H \in N_0$. We have that: :$\order H = \index {G_0} {N_0}$ and so $H$ is also a complement to $N$ in $G$. So we may assume that $P$ is normal in $G$ (that is: $G_0 < G$). {{explain|What is the significance of $G_0 < G$ here? It has already been stated above. What is its purpose at this point in the argument?}} Let $Z \paren P$ be the center of $P$. By: :Center is Characteristic Subgroup :$P$ is normal in $G$ :Characteristic Subgroup of Normal Subgroup is Normal $Z \paren P$ is also normal in $G$. Let $Z \paren P = N$. Then there exists a long exact sequence of cohomology groups: :$0 \to H^1 \paren {G / N, P^N} \to H^1 \paren {G, P} \to H^1 \paren {N, P} \to H^2 \paren {G / N, P} \to H^2 \paren {G, P}$ which splits as desired. {{explain|The link leads to "exact sequence". An explanation is needed as to what a "long exact sequence" is.}} {{explain|The definition of $P^N$ in this context}} {{explain|The definition of "splits" in this context}} Otherwise: :$Z \paren P \ne N$ In this case $N / Z \paren P$ is a normal (Hall) subgroup of $G / Z \paren P$. By induction: :$N / Z \paren P$ has a complement $H / Z \paren P$ in $E // Z \paren P$. {{explain|The meaning of $E // Z \paren P$ and definition of $E$ in this context.}} {{Explain|Although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}} Let $G_1$ be the preimage of $H // Z \paren P$ in $G$ (under the equivalence relation). {{explain|Under what equivalence relation?}} Then: :$\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N} \times \order {Z \paren P}$ {{explain|The definition of $K$ in $\order {G_1} {{=}} \order {K / Z \paren P} \times \order {Z \paren P} {{=}} \order {G / N}\ \times \order {Z \paren P}$.}} Therefore, $Z \paren P$ is normal Hall subgroup of $G_1$. By induction, $Z \paren P$ has a complement in $G_1$ and is also a complement of $N$ in $G$. {{Explain|Again, although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}} {{qed}}
First, from Closed Form for Triangular Numbers: :$\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ So: :$\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$ Next we use induction on $n$ to show that: :$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ === Basis for the Induction === $\map P 1$ is the case: :$1^3 = \dfrac {1 \paren {1 + 1}^2} 4$ {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^1 i^3       | r = 1^3       | c =  }} {{eqn | r = \dfrac {1^2 \paren {1 + 1}^2} 4       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^3       | r = \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3       | c =  }} {{eqn | r = \frac {k^2 \paren {k + 1}^2} 4 + \paren {k + 1}^3       | c = Induction Hypothesis }} {{eqn | r = \frac {k^4 + 2 k^3 + k^2} 4 + \frac {4 k^3 + 12 k^2 + 12 k + 4} 4       | c =  }} {{eqn | r = \frac {k^4 + 6 k^3 + 13 k^2 + 12 k + 4} 4       | c =  }} {{eqn | r = \frac {\paren {k + 1}^2 \paren {k + 2}^2} 4       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
By Nicomachus's Theorem, we have: :$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$ 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-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^3       | r = \sum_{\stackrel {1 \mathop \le i \mathop \le n^2 \mathop + n \mathop - 1} {i \text { odd} } } i       | c = ... the sum of all the odd numbers up to $n^2 + n - 1$ }} {{eqn | r = \sum_{i \mathop = 1}^{\frac {n^2 \mathop + n} 2} 2 i - 1       | c = ... that is, the first $\dfrac {n^2 + n} 2$ odd numbers }} {{eqn | r = \paren {\frac {n^2 + n} 2}^2       | c = Odd Number Theorem }} {{end-eqn}} Hence the result. {{qed}}
From Closed Form for Triangular Numbers: : $(1): \quad \displaystyle \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$ From Sum of Sequence of Squares: : $(2): \quad \displaystyle \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$ Let $\displaystyle \map S n = \sum_{i \mathop = 1}^n i^3$. Then: {{begin-eqn}} {{eqn | l = \map S n       | r = n^3 + \paren {n - 1}^3 + \paren {n - 2}^3 + \cdots + 2^3 + 1^3       | c = }} {{eqn | r = \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n - k}^3       | c = }} {{eqn | r = \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n^3 - 3 n^2 k + 3 n k^2 - k^3}       | c = }} {{eqn | r = n^4 - 3 n^2 \cdot \map A {n - 1} + 3 n \cdot \map B {n - 1} - \map S {n - 1}       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = \map S n       | r = n^4 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3n \cdot \frac{n \paren {n - 1} \paren {2 n - 1} } 6 - \map S {n - 1}       | c = substituting from $(1)$ and $(2)$ }} {{eqn | n = 4       | l = \map S n       | r = n^3 + \map S {n - 1}       | c = recursive definition }} {{eqn | ll= \leadsto       | l = 2 \map S n       | r = n^4 + n^3 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3 n \cdot \frac {n \paren {n - 1} \paren {2 n - 1} } 6       | c = adding $(3)$ and $(4)$ }} {{eqn | ll= \leadsto       | r = \frac {n^2 \paren {n + 1}^2} 2       | c = simplification }} {{eqn | ll= \leadsto       | l = \map S n       | r = \frac {n^2 \paren {n + 1}^2} 4       | c = }} {{end-eqn}} {{qed}}
{{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}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$ === 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 = \paren {\frac {1 \paren {1 + 1} } 2}^2       | c =  }} {{eqn | r = {T_1}^2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k i^3 = {T_k}^2$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^3       | r = \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3       | c =  }} {{eqn | r = {T_k}^2 + \paren {k + 1}^3       | c = Induction Hypothesis }} {{eqn | r = {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2}       | c = Cube Number as Difference between Squares of Triangular Numbers }} {{eqn | r = {T_{k + 1} }^2       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{> 0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$ {{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 $p = 3$: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^3       | r = \frac {n^{3 + 1} } {3 + 1} + \sum_{k \mathop = 1}^3 \frac {B_k \, 3^{\underline {k - 1} } \, n^{3 - k + 1} } {k!}       | c =  }} {{eqn | r = \frac {n^4} 4 + \frac {B_1 \, 3^{\underline 0} \, n^3} {1!} + \frac {B_2 \, 3^{\underline 1} \, n^2} {2!} + \frac {B_3 \, 3^{\underline 2} \, n^1} {3!}       | c =  }} {{eqn | r = \frac {n^4} 4 + \frac 1 2 \frac {n^3} {1!} + \frac 1 6 \frac {3 n^2} {2!} + 0 \frac {3 \times 2 n} {3!}       | c = {{Defof|Bernoulli Numbers}} and {{Defof|Falling Factorial}} }} {{eqn | r = \frac {n^4} 4 + \frac {n^3} 2 + \frac {n^2} 4       | c = simplifying }} {{eqn | r = \frac {n^2 \left({n + 1}\right)^2} 4       | c = after algebra }} {{end-eqn}}
We aim to demonstrate that the "Sum of Cubes" is the "Square of the Sum" using simple Multiplication Tables. On the right hand side of the equation, the "Square of the Sum" will be represented by a two dimensional array. The sum of all numbers contained inside the two dimensional (n x n) array is the "Square of the Sum". For example: :$ \paren {1 + 2}^2 = 1 + 2 + 2 + 4$ :$\begin{array}{r|cccccccccc} \paren {1 + 2}^2 & 1 & 2 \\ \hline 1  & 1 & 2 \\ 2  & 2 & 4  \\ \end{array}$ :$ \paren {1 + 2 + 3}^2 = 1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 + 9$ :$\begin{array}{r|cccccccccc} \paren {1 + 2 + 3}^2 & 1 & 2 & 3 \\ \hline 1  & 1 & 2 & 3 \\ 2  & 2 & 4 & 6 \\ 3  & 3 & 6 & 9 \\ \end{array}$ :$ \cdots $ :$ \paren {1 + 2 + 3 + \cdots + N}^2 = $ :$\begin{array}{r|cccccccccc} \paren {\sum_{i \mathop = 1}^n i}^2 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ \hline 1  & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ 2  & 2 & 4 & 6 & 8 & 10 & 12 & \cdots & 2N  \\ 3  & 3 & 6 & 9 & 12 & 15 & 18 & \cdots & 3N \\ 4  & 4 & 8 & 12 & 16 & 20 & 24 & \cdots & 4N \\ 5  & 5 & 10 & 15 & 20 & 25 & 30 & \cdots & 5N \\ 6  & 6 & 12 & 18 & 24 & 30 & 36 & \cdots & 6N \\ \cdots  & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ N  & N & 2N & 3N & 4N & 5N & 6N & \cdots & N^2 \\ \end{array}$ From Closed Form for Triangular Numbers: :$\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ So the sum of the numbers in the two dimensional n x n array (i.e. "Square of the Sum") is: :$\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$ So where is the "Sum of the Cubes"? Consider the sum of the set of all numbers in each outermost square edge above of length $1$, $2$, $3$, $\ldots$. We have: {{begin-eqn}} {{eqn | l = 1       | r = \paren 2 \paren 1 \paren 1 - 1^2 = 1^3       | c =  }} {{eqn | l = 2 + 4 + 2       | r = \paren 2 \paren 2 \paren {1 + 2} - 2^2 = 2^3       | c =  }} {{eqn | l = 3 + 6 + 9 + 6 + 3       | r = \paren 2 \paren 3 \paren {1 + 2 + 3} - 3^2 = 3^3       | c =  }} {{eqn | o = \cdots }} {{eqn | l = n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n       | r = \paren 2 \paren n \paren {1 + 2 + \cdots + n} - n^2       | c =  }} {{eqn | r = 2 n \sum_{i \mathop = 1}^n i - n^2       | c =  }} {{eqn | r = 2 n \frac {n \paren {n + 1} } 2 - n^2       | c = Closed Form for Triangular Numbers }} {{eqn | r = n^3 + n^2 - n^2       | c =  }} {{eqn | r = n^3       | c =  }} {{end-eqn}} Therefore,  The "Square of the Sum" (i.e. the sum of all of the numbers in an n x n multiplication table) is identical to  The "Sum of the Cubes" (i.e. the sum of the numbers on each incremental outer square-edge is a cube, sum up the outer edges), or: :$\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ {{qed}}
Let $\triangle ABC$ be embedded in a Cartesian coordinate system by identifying: :$C := \tuple {0, 0}$ :$B := \tuple {a, 0}$ :400px Thus by definition of sine and cosine: :$A = \tuple {b \cos C, b \sin C}$ By the Distance Formula: :$c = \sqrt {\paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2}$ Hence: {{begin-eqn}} {{eqn | l = c^2       | r = \paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2       | c = squaring both sides of Distance Formula }} {{eqn | r = b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C       | c = Square of Difference }} {{eqn | r = a^2 + b^2 \paren {\sin^2 C + \cos^2 C} - 2 a b \cos C       | c = Real Multiplication Distributes over Addition }} {{eqn | r = a^2 + b^2 - 2 a b \cos C       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}}