hcju/proofwikips huggingface dataset:
Between two Rational Numbers exists Irrational Number/Lemma 2
== Lemma for Between two Rational Numbers exists Irrational Number ==  Let $\alpha \in \Q$ and $\beta \in \R \setminus \Q$.  Then:  :$\alpha + \beta \in \R \setminus \Q$  == Proof ==  $\alpha + \beta \in \Q$. By the definition of rational numbers: :$\exists p, q \in \Z: \alpha + \beta = \dfrac p q$ Thus: :$\beta = \dfrac p q - \alpha$ By Rational Subtraction is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$. Therefore: :$\alpha + \beta \in \R \setminus \Q$
Divisibility by 9/Corollary
== Corollary to Divisibility by 9 ==  A number expressed in decimal notation is divisible by $3$  the sum of its digits is divisible by $3$. That is: :$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $3$  : :$a_0 + a_1 + \ldots + a_n$ is divisible by $3$.  == Proof == From Divisibility by 9 we have that: :$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $3^2$  : :$a_0 + a_1 + \ldots + a_n$ is divisible by $3^2$. So: {{begin-eqn}} {{eqn | l = \paren {a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n}       | o = \equiv       | r = \paren {a_0 + a_1 + a_2 + \cdots + a_n}       | rr= \pmod {3^2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n}       | o = \equiv       | r = \paren {a_0 + a_1 + a_2 + \cdots + a_n}       | rr= \pmod 3       | c = Congruence by Divisor of Modulus }} {{end-eqn}}  == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Divisibility by 9/Proof 1|next = Binomial Theorem/Integral Index}}: $\S 2.6$. Algebra of congruences: Example $41$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Integer as Sum of Three Squares|next = Vinogradov's Theorem/Corollary 2}}: $3$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Integer as Sum of Three Squares|next = Vinogradov's Theorem/Corollary 2}}: $3$
Square Root/Examples/2
== Decimal Expansion ==  The decimal expansion of the square root of $2$ starts: :$\sqrt 2 \approx 1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$  {{OEIS|A002193}} == Also known as == :$\sqrt 2$ is sometimes known as '''{{AuthorRef|Pythagoras of Samos|Pythagoras}}'s constant''', for {{AuthorRef|Pythagoras of Samos}}. == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Euler's Number/Limit of Sequence|next = Square Root/Examples/3}}: $\S 1$: Special Constants: $1.3$ * {{BookReference|Theory and Problems of Differential and Integral Calculus|1972|Frank Ayres, Jr.|author2 = J.C. Ault|ed = SI|edpage = SI Edition|prev = Definition:Real Number|next = Definition:Pi/Decimal Expansion}}: Chapter $1$: Variables and Functions: The Set of Real Numbers * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Square Root of 2 is Irrational|next = Pi is Irrational}}: $1$: Complex Numbers: The Real Number System: $4$ * {{BookReference|Les Nombres Remarquables|1983|François Le Lionnais|author2 = Jean Brette|prev = Number of Primes of Form n^2 + 1/Historical Note|next = Length of Diagonal of Unit Square}}: $1,41421 35623 73 \ldots$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Doubling the Cube/Cissoid of Diocles/Historical Note|next = Length of Diagonal of Unit Square}}: $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Doubling the Cube/Cissoid of Diocles/Historical Note|next = Length of Diagonal of Unit Square}}: $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Square Root/Examples/5
== Decimal Expansion ==  The decimal expansion of $\sqrt 5$ starts: :$\sqrt 5 \approx 2 \cdotp 23606 \, 79774 \, 99789 \, 6964 \ldots$  {{OEIS|A002163}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square Root of 3|next = Cube Root of 2}}: $\S 1$: Special Constants: $1.5$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Wallis's Number/Historical Note|next = Natural Logarithm of 10}}: $2 \cdotp 236 \, 067 \ldots$
Axiom:Euclid's Second Postulate/Production
== Definition ==  The process of extending a straight line segment is called '''production'''. Such a straight line segment is said to be '''produced'''. The line segment can be extended in either direction, thus producing an infinite half-line, or in both directions, thus producing an infinite straight line.  {{EuclidPostulateStatement|Second|To produce a finite straight line continuously in a straight line.}} == Sources == * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Proclus Lycaeus|next = Definition:Product|entry = produce}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Proclus Lycaeus|next = Definition:Producer's Risk|entry = produce}}
Sine and Cosine are Periodic on Reals/Corollary
{{mergeto|Cosine of Angle plus Straight Angle]] and [[Sine of Angle plus Straight Angle}} == Corollary to Sine and Cosine are Periodic on Reals ==  :$\map \cos {x + \pi} = -\cos x$ :$\map \sin {x + \pi} = -\sin x$ :$\cos x$ is strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ :$\sin x$ is strictly positive on the interval $\openint 0 \pi$ and strictly negative on the interval $\openint \pi {2 \pi}$  == Proof == From Sine and Cosine are Periodic on Reals: :$\map \sin {x + \dfrac \pi 2} = \cos x$ :$\map \cos {x + \dfrac \pi 2} = -\sin x$ Thus: :$\map \sin {x + \pi} = \map \cos {x + \dfrac \pi 2} = -\sin x$ :$\map \cos {x + \pi} = -\map \sin {x + \dfrac \pi 2} = -\cos x$ It follows directly that: :$\forall x \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}: \cos x \ge 0$ Hence: :$\forall x \in \closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}: \cos x \le 0$ The result for $\sin x$ follows similarly, or we can use: :$\map \sin {x + \dfrac \pi 2} = \cos x$
Sum of Squares of Sine and Cosine/Corollary 2
== Corollary to Sum of Squares of Sine and Cosine == For all $x \in \C$:  :$\csc^2 x - \cot^2 x = 1 \quad \text {(when $\sin x \ne 0$)}$  where $\csc$, $\cot$ and $\sin$ are cosecant, cotangent and sine respectively. == Proof  == When $\sin x \ne 0$: {{begin-eqn}} {{eqn | l = \sin^2 x + \cos^2 x       | r = 1       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = 1 + \frac {\cos^2 x} {\sin^2 x}       | r = \frac 1 {\sin^2 x}       | c = dividing both sides by $\sin^2 x$, as $\sin x \ne 0$ }} {{eqn | ll= \leadsto       | l = 1 + \cot^2 x       | r = \csc^2 x       | c = {{Defof|Cotangent}} and {{Defof|Cosecant}} }} {{eqn | ll= \leadsto       | l = \csc^2 x - \cot^2 x       | r = 1       | c = rearranging }} {{end-eqn}}  == Also defined as == This result can also be reported as: :$\csc^2 x = 1 + \cot^2 x \quad \text{(when $\sin x \ne 0$)}$ or: :$\cot^2 x = \csc^2 x - 1 \quad \text{(when $\sin x \ne 0$)}$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Difference of Squares of Secant and Tangent|next = Shape of Sine Function}}: $\S 5$: Trigonometric Functions: $5.21$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Difference of Squares of Secant and Tangent|next = Sine Function is Odd}}: $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
Kernel of Linear Transformation is Orthocomplement of Range of Adjoint/Corollary
== Corollary to Kernel of Linear Transformation is Orthocomplement of Range of Adjoint == Let $H$ be a Hilbert space.  Let $A \in \map B H$ be a normal operator. Then: :$\ker A = \paren {\Rng A}^\perp$  where: :$A^*$ denotes the adjoint of $A$ :$\ker A$ is the kernel of $A$ :$\Rng A$ is the range of $A$ :$\perp$ signifies orthocomplementation == Proof == {{begin-eqn}} {{eqn | l = \ker A       | r = \paren {\Rng {A^*} }^\perp       | c = Kernel of Linear Transformation is Orthocomplement of Range of Adjoint }} {{eqn | ll= \leadsto       | l = \ker A^*       | r = \paren {\Rng {A^*} }^\perp       | c = Kernel of Normal Operator is Kernel of Adjoint }} {{eqn | ll= \leadsto       | l = \ker A^{**}       | r = \paren {\Rng {A^{**} } }^\perp       | c = substituting $A^*$ for $A$ }} {{eqn | ll= \leadsto       | l = \ker A       | r = \paren {\Rng A}^\perp       | c = Double Adjoint is Itself }} {{end-eqn}}
Bisection of Angle in Cartesian Plane/Corollary
== Corollary to Bisection of Angle in Cartesian Plane == Let $\theta$ be the angular coordinate of a point $P$ in a polar coordinate plane. Let $QOR$ be a straight line that bisects the angle $\theta$.  If $\theta$ is in quadrant I or quadrant II, then the angular coordinates of $Q$ and $R$ are in quadrant I and quadrant III. If $\theta$ is in quadrant III or quadrant IV, then the angular coordinates of $Q$ and $R$ are in quadrant II and quadrant IV.  == Proof == :500px From Bisection of Angle in Cartesian Plane, the angular coordinates of $Q$ and $R$ are $\dfrac \theta 2$ and $\pi + \dfrac \theta 2$. , let $\angle Q = \dfrac \theta 2$ and $\angle R = \pi + \dfrac \theta 2$. Let $\theta$ be in quadrant I or quadrant II. Then $0 < \theta < \pi$. Dividing each term in the inequality by $2$: :$0 < \dfrac \theta 2 < \dfrac \pi 2$ and so $Q$ lies in quadrant I. Adding $\pi$ to each expression in the inequality: :$\pi < \pi + \dfrac \theta 2 < \dfrac {3 \pi} 2$ and so $R$ lies in quadrant I. {{qed|lemma}} Let $\theta$ be in quadrant III or quadrant IV. Then $\pi < \theta < 2 \pi$. Dividing each term in the inequality by $2$: :$\dfrac \pi 2 < \dfrac \theta 2 < \pi$ and so $Q$ lies in quadrant II. Adding $\pi$ to each expression in the inequality: :$\dfrac {3 \pi} 2 < \pi + \dfrac \theta 2 < 2 \pi$ and so $R$ lies in quadrant IV.
Shape of Cosecant Function/Graph
Shape of Cosecant Function/Graph 0 28287 155043 2013-08-05T21:22:52Z Prime.mover 59 Created page with "== Theorem ==  The graph of the cosecant function appears as:   :800px </onlyinclude..." wikitext text/x-wiki == Theorem == The graph of the cosecant function appears as:  :800px  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Graph of Secant Function|next=Sine Function is Odd}}: $\S 5$: Trigonometric Functions: $5.27$
Tangent of Sum/Corollary
== Corollary to Tangent of Sum ==  :$\map \tan {a - b} = \dfrac {\tan a - \tan b} {1 + \tan a \tan b}$  where $\tan$ is tangent. == Proof == {{begin-eqn}} {{eqn | l = \map \tan {a - b}       | r = \frac {\tan a + \map \tan {-b} } {1 - \tan a \, \map \tan {-b} }       | c = Tangent of Sum }} {{eqn | r = \frac {\tan a - \tan b} {1 + \tan a \tan b}       | c = Tangent Function is Odd }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Tangent of Sum|next = Cotangent of Sum}}: $\S 5$: Trigonometric Functions: $5.36$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Tangent of Sum|next = Definition:Hyperbolic Sine}}: $2$: Functions, Limits and Continuity: The Elementary Functions: $4$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Tangent of Sum|next = Prosthaphaeresis Formulas|entry = compound angle formulae|subentry = in trigonometry}}
Cotangent of Sum/Corollary
== Corollary to Cotangent of Sum ==  :$\map \cot {a - b} = \dfrac {\cot a \cot b + 1} {\cot b - \cot a}$  where $\cot $ is cotangent. == Proof == {{begin-eqn}} {{eqn | l = \map \cot {a - b}       | r = \frac {\cot a \, \map \cot {-b} - 1} {\cot a + \map \cot {-b} }       | c = Cotangent of Sum }} {{eqn | r = \frac {-\cot a \cot b - 1} {\cot a - \cot b}       | c = Cotangent Function is Odd }} {{eqn | r = \frac {\cot a \cot b + 1} {\cot b - \cot a}       | c = multiplying numerator and denominator by $-1$ and rearranging }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cotangent of Sum|next = Trigonometric Functions of Negative Angle}}: $\S 5$: Trigonometric Functions: $5.37$
Double Angle Formulas/Cosine/Corollary 1
== Corollary to Double Angle Formula for Cosine  ==  :$\cos 2 \theta = 2 \cos^2 \theta - 1$  where $\cos$ denotes cosine. == Proof  == {{begin-eqn}} {{eqn | l = \cos 2 \theta       | r = \cos^2 \theta - \sin^2 \theta       | c = Double Angle Formula for Cosine }} {{eqn | r = \cos^2 \theta - \paren {1 - \cos^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 2 \cos^2 \theta - 1 }} {{end-eqn}}  == Also see == * Square of Cosine: $\cos^2 \theta = \dfrac {1 + \cos 2 \theta} 2$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Double Angle Formulas/Cosine/Corollary 2|next = Double Angle Formula for Tangent}}: $\S 5$: Trigonometric Functions: $5.39$
Double Angle Formulas/Cosine/Corollary 2
== Corollary to Double Angle Formula for Cosine ==  :$\cos 2 \theta = 1 - 2 \sin^2 \theta$  where $\cos$ and $\sin$ denote cosine and sine respectively. == Proof == {{begin-eqn}} {{eqn | l = \cos 2 \theta       | r = \cos^2 \theta - \sin^2 \theta       | c = Double Angle Formula for Cosine }} {{eqn | r = \paren {1 - \sin^2 \theta} - \sin^2 \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 1 - 2 \sin^2 \theta }} {{end-eqn}}  == Also see == * Square of Sine: $\sin^2 \theta = \dfrac {1 - \cos 2 \theta} 2$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Double Angle Formula for Cosine|next = Double Angle Formulas/Cosine/Corollary 1}}: $\S 5$: Trigonometric Functions: $5.39$
Double Angle Formulas/Sine/Corollary
== Corollary to Double Angle Formula for Sine ==  :$\sin 2 \theta = \dfrac {2 \tan \theta} {1 + \tan^2 \theta}$  where $\sin$ and $\tan$ denote sine and tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \sin 2 \theta       | r = 2 \sin \theta \cos \theta       | c = Double Angle Formula for Sine }} {{eqn | r = 2 \sin \theta \cos \theta \frac {\cos \theta} {\cos \theta}       | c =  }} {{eqn | r = 2 \tan \theta \cos^2 \theta       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {2 \tan \theta} {\sec^2 \theta}       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \frac {2 \tan \theta} {1 + \tan^2 \theta}       | c = Difference of Squares of Secant and Tangent }} {{end-eqn}}  == Sources == * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Half Angle Formula for Sine|next = Tangent Half-Angle Substitution for Cosine|entry = half-angle formula}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Tangent Half-Angle Substitution Formulas|next = Tangent Half-Angle Substitution for Cosine|entry = $t$-formulae}}
Tangent Half-Angle Substitution for Cosine
== Corollary to Double Angle Formula for Cosine ==  :$\cos 2 \theta = \dfrac {1 - \tan^2 \theta} {1 + \tan^2 \theta}$  where $\cos$ and $\tan$ denote cosine and tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \cos 2 \theta       | r = \cos^2 \theta - \sin^2 \theta       | c = Double Angle Formula for Cosine }} {{eqn | r = \paren {\cos^2 \theta - \sin^2 \theta} \frac {\cos^2 \theta}{\cos^2 \theta}       | c =  }} {{eqn | r = \paren {1 - \tan^2 \theta} \cos^2 \theta       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {1 - \tan^2 \theta} {\sec^2 \theta}       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \frac {1 - \tan^2 \theta} {1 + \tan^2 \theta}       | c = Difference of Squares of Secant and Tangent }} {{end-eqn}}  == Sources == * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Tangent Half-Angle Substitution for Sine|next = Double Angle Formula for Tangent|entry = half-angle formula}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Tangent Half-Angle Substitution for Sine|next = Mathematician:Thales of Miletus|entry = $t$-formulae}}
Quintuple Angle Formulas
Quintuple Angle Formulas 0 28391 155470 2013-08-07T19:50:49Z Prime.mover 59 Created page with "== Theorem ==  === Quintuple Angle Formula for Sine === {{:Quintuple Angle Formulas/Sine}}  === Quintuple Angle Formulas/Cosin..." wikitext text/x-wiki == Theorem ==  === Quintuple Angle Formula for Sine === {{:Quintuple Angle Formulas/Sine}} === Quintuple Angle Formula for Cosine === {{:Quintuple Angle Formulas/Cosine}} === Quintuple Angle Formula for Tangent === {{:Quintuple Angle Formulas/Tangent}}  where $\sin, \cos, \tan$ denote sine, cosine and tangent respectively.
Sum of Geometric Sequence/Proof 4/Lemma
== Lemma ==  Let $n \in \N_{>0}$. Then: :$\displaystyle \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$  == Proof == Proof by induction on $n$: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \paren {1 - x} \sum_{i \mathop = 1}^{1 - 1} x^i       | r = 1 \paren {1 - x} }} {{eqn | r = 1 - x }} {{eqn | r = 1 - x^1 }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show 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 our induction hypothesis: :$\displaystyle \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i = 1 - x^k$ Then we need to show: :$\displaystyle \paren {1 - x} \sum_{i \mathop = 0}^k x^i = 1 - x^{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {1 - x} \sum_{i \mathop = 0}^k x^i       | r = \paren {1 - x} \paren {x^k + \sum_{i \mathop = 0}^{k - 1} x^i} }} {{eqn | r = x^k - x^{k + 1} + \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i }} {{eqn | r = x^k - x^{k + 1} + 1 - x^k       | c = from the induction hypothesis }} {{eqn | r = 1 - x^{k + 1}       | c = gathering terms }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
General Distributivity Theorem/Lemma 1
== Lemma ==  Let $\left({R, \circ, *}\right)$ be a ringoid. Then for every sequence $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$: :$\displaystyle \left({\sum_{j \mathop = 1}^n a_j}\right) * b = \sum_{j \mathop = 1}^n \left({a_j * b}\right)$  where: :$\displaystyle \sum_{j \mathop = 1}^n a_j$ is the composite $a_1 \circ a_2 \circ \cdots \circ a_n$ :$n$ is a strictly positive integer: $n \in \Z_{> 0}$. == Proof == The proof proceeds by the Principle of Mathematical Induction. Recall that as $\left({R, \circ, *}\right)$ is a ringoid, $*$ is distributive over $\circ$: :$\forall a, b, c \in S: \left({a \circ b}\right) * c = \left({a * c}\right) \circ \left({b * c}\right)$ For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left({\sum_{j \mathop = 1}^n a_j}\right) * b = \sum_{j \mathop = 1}^n \left({a_j * b}\right)$ === Basis for the Induction === $P(1)$ is true, as this just says: :$a_1 * b = a_1 * b$ $P(2)$ is the case: {{begin-eqn}} {{eqn | l = \left({\sum_{j \mathop = 1}^2 a_j}\right) * b       | r = \left({a_1 \circ a_2}\right) * b       | c = Definition of Composite }} {{eqn | r = \left({a_1 * b}\right) \circ \left({a_2 * b}\right)       | c = $*$ is distributive over $\circ$ as $\left({R, \circ, *}\right)$ is a ringoid }} {{eqn | r = \sum_{j \mathop = 1}^2 \left({a_j * b}\right)       | c = Definition of Composite }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \left({\sum_{j \mathop = 1}^k a_j}\right) * b = \sum_{j \mathop = 1}^k \left({a_j * b}\right)$ Then we need to show: :$\displaystyle \left({\sum_{j \mathop = 1}^{k + 1} a_j}\right) * b = \sum_{j \mathop = 1}^{k + 1} \left({a_j * b}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({\sum_{j \mathop = 1}^{k + 1} a_j}\right) * b       | r = \left({\left({\sum_{j \mathop = 1}^k a_j}\right) \circ a_{k+1} }\right) * b       | c = Definition of Composite }} {{eqn | r = \left({\left({\sum_{j \mathop = 1}^k a_j}\right) * b}\right) \circ \left({a_{k+1} * b}\right)       | c = Basis for the Induction }} {{eqn | r = \left({\sum_{j \mathop = 1}^k \left({a_j * b}\right)}\right) \circ \left({a_{k+1} * b}\right)       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{k + 1} \left({a_j * b}\right)       | c = Associativity of $\circ$ in $\left({R, \circ, *}\right)$ }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{> 0}: \left({\sum_{j \mathop = 1}^k a_j}\right) * b = \sum_{j \mathop = 1}^k \left({a_j * b}\right)$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = General Commutativity Theorem|next = General Distributivity Theorem/Lemma 2}}: $\S 18$: Theorem $18.8$
Integer Multiplication Distributes over Addition/Corollary
== Corollary to Integer Multiplication Distributes over Addition ==  The operation of multiplication on the set of integers $\Z$ is distributive over subtraction: : $\forall x, y, z \in \Z: x \times \left({y - z}\right) = \left({x \times y}\right) - \left({x \times z}\right)$ : $\forall x, y, z \in \Z: \left({y - z}\right) \times x = \left({y \times x}\right) - \left({z \times x}\right)$  == Proof == {{begin-eqn}} {{eqn | l = x \times \left({y - z}\right)       | r = x \times \left({y + \left({- z}\right)}\right)       | c = {{Defof|Integer Subtraction}} }} {{eqn | r = x \times y + x \times \left({- z}\right)       | c =  }} {{eqn | r = x \times y + \left({- \left({x \times z}\right)}\right)       | c = Product with Ring Negative }} {{eqn | r = x \times y - x \times z       | c = {{Defof|Integer Subtraction}} }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \left({y - z}\right) \times x       | r = x \times \left({y - z}\right)       | c = Integer Multiplication is Commutative }} {{eqn | r = x \times y - x \times z       | c = from above }} {{eqn | r = y \times z - z \times x       | c = Integer Multiplication is Commutative }} {{end-eqn}}
Divisors of Product of Coprime Integers/Corollary
== Corollary to Divisors of Product of Coprime Integers ==  Let $p$ be a prime. Let $p \divides b c$, where $b \perp c$. Then $p \divides b$ or $p \divides c$, but not both.  == Proof == From the main result, $p = r s$, where $r \divides b$ and $s \divides c$. But as $p$ is prime, either: :$r = 1$ and $s = p$, or: :$r = p$ and $s = 1$. So $p \divides b$ or $p \divides c$. But $p$ can not divide both, as $b \perp c$.