avewright/proofwiki-math huggingface dataset:
Theorem: Right Module induces Left Module over same Ring iff Actions are Commutative

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda $
Then $\struct {G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$ {{iff}}:
:$\forall \lambda, \mu \in R: \forall x \in G: x \circ \paren{ \lambda \times_R \mu} = x \circ \paren {\mu \times_R \lambda}$

Proof:
=== Necessary Condition ===
Let $\struct {G, +_G, \circ'}$ be a left module over $\struct {R, +_R, \times_R}$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ \paren {\lambda \times_R \mu}
      | r = \paren {\lambda \times_R \mu} \circ' x
      | c = Definition of $\circ'$
}}
{{eqn | r = \lambda \circ' \paren {\mu \circ' x}
      | c = {{Left-module-axiom|3}}
}}
{{eqn | r = \paren {x \circ \mu} \circ \lambda
      | c = Definition of $\circ'$
}}
{{eqn | r = x \circ \paren {\mu \times_R \lambda}
      | c = {{Right-module-axiom|3}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Fermat Problem

Let $\triangle ABC$ be a triangle
Let the vertices of $\triangle ABC$ all have angles less than $120 \degrees$.
Let $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ be equilateral triangles constructed on the sides of $ABC$.
Let $AE$, $BF$ and $CG$ be constructed.
Let $P$ be the point at which $AE$, $BF$ and $CG$ meet.
:500px
Then $P$ is the Fermat-Torricelli point of $\triangle ABC$.
If one of vertices of $\triangle ABC$ be of $120 \degrees$ or more, then that vertex is itself the Fermat-Torricelli point of $\triangle ABC$.

Proof:
The sum of the distances will be a minimum when the lines $PA$, $PB$ and $PC$ all meet at an angle of $120 \degrees$.
This is a consequence of the '''Fermat problem''' being a special case of the Steiner tree problem.
Consider the circles which circumscribe the $3$ equilateral triangles $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$.
Consider quadrilaterals formed by $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ along with another point on each of those circumscribing circles.
Because these are cyclic quadrilaterals, the angle formed with these new points is $120 \degrees$.
It follows that $\Box APBG$, $\Box BPCE$ and $\Box APCF$ are those cyclic quadrilaterals.
Hence $\angle APC = \angle APB = \angle BPC = 120 \degrees$ and the result follows.
{{qed}}
Theorem: Left Module over Commutative Ring induces Right Module

Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct {G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$.

Proof:
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring.
From Left Module over Ring Induces Right Module over Opposite Ring, $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$.
{{qed}}
Theorem: Right Module over Commutative Ring induces Left Module

Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct{G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ’ x = x \circ \lambda$
Then $\struct{G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$.

Proof:
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring.
From Right Module over Ring Induces Left Module over Opposite Ring, $\struct{G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$.
{{qed}}
Theorem: Right Ideal is Right Module over Ring/Ring is Right Module over Ring

Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times}$ is a right module over $\struct {R, +, \times}$.

Proof:
From Ring is Ideal of Itself, $R$ is a right ideal.
From Right Ideal is Right Module over Ring, $\struct {R, +, \times}$ is a right module over $\struct {R, +, \times}$.
{{qed}}
Theorem: Left Ideal is Left Module over Ring/Ring is Left Module over Ring

Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times}$ is a left module over $\struct {R, +, \times}$.

Proof:
From Ring is Ideal of Itself, $R$ is a left ideal.
From Left Ideal is Left Module over Ring, $\struct {R, +, \times}$ is a left module over $\struct {R, +, \times}$.
{{qed}}
Theorem: Integration by Inversion

:$\ds \int_0^{+\infty} \map f x \rd x = \int_0^{+\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^{+\infty} \map f x \rd x
      | r = \int_{x \mathop \to 0}^{x \mathop \to +\infty} \map f x \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \map \rd {\frac 1 x}
      | c = Integration by Substitution
}}
{{eqn | r = \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \paren {-\dfrac 1 {x^2} } \rd x
      | c = Power Rule for Derivatives
}}
{{eqn | r = -\int_{x \mathop \to +\infty}^{x \mathop \to 0} \dfrac {\map f {\frac 1 x} } {x^2} \rd x
      | c = simplifying
}}
{{eqn | r = \int_{x \mathop \to 0}^{x \mathop \to +\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x
      | c = Reversal of Limits of Definite Integral
}}
{{eqn | r = \int_0^{+\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x
      | c = {{Defof|Improper Integral}}
}}
{{end-eqn}}
{{qed}}
Category:Definite Integrals
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Theorem: Left Module Does Not Necessarily Induce Right Module over Ring

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct{G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct{G, +_G, \circ'}$ is not necessarily a right module over $\struct {R, +_R, \times_R}$

Proof:
Proof by Counterexample:
Let $\struct {S, +, \times}$ be a ring with unity
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
From Ring of Square Matrices over Ring is Ring, $\struct {\map {\MM_S} 2, +, \times}$ is a ring.
Let:
:$G := \set {\begin {bmatrix} x & 0 \\ y & 0 \end{bmatrix} : x, y \in S}$
Theorem: Right Module Does Not Necessarily Induce Left Module over Ring

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct{G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda$
Then $\struct{G, +_G, \circ'}$ is not necessarily a left module over $\struct {R, +_R, \times_R}$

Proof:
Proof by Counterexample:
Let $\struct {S, +, \times}$ be a ring with unity
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
From Ring of Square Matrices over Ring is Ring, $\struct {\map {\MM_S} 2, +, \times}$ is a ring.
Let:
:$G := \set {\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix} : x, y \in S}$
Theorem: Left Module Does Not Necessarily Induce Right Module over Ring/Lemma

:$G$ is a left ideal of $\struct {\map {\MM_S} 2, +, \times}$.

Proof:
From Test for Left Ideal, the following need to be proved:
:$(1): \quad G \ne \O$
:$(2): \quad \forall \mathop {\mathbf X}, \mathop {\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
:$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop {\mathbf R} \in \map {\MM_S} 2: \mathbf R \times \mathbf J \in G$
Theorem: Right Module Does Not Necessarily Induce Left Module over Ring/Lemma

:$G$ is a right ideal of $\struct {\map {\MM_S} 2, +, \times}$.

Proof:
From Test for Right Ideal, the following need to be proved:
:$(1): \quad G \ne \O$
:$(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
:$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$
Theorem: Ideal is Bimodule over Ring

Let $\struct {R, +, \times}$ be a ring.
Let $J \subseteq R$ be an ideal of $R$.
Let $\circ_l : R \times J \to J$ be the restriction of $\times$ to $R \times J$.
Let $\circ_r : J \times R \to J$ be the restriction of $\times$ to $J \times R$.
Then $\struct {J, +, \circ_l, \circ_r}$ is a bimodule over $\struct {R, +, \times}$.

Proof:
By definition of an ideal, $J$ is both a left ideal and a right ideal.
From Left Ideal is Left Module over Ring then $\struct {J, +, \circ_l}$ is a left module.
From Right Ideal is Right Module over Ring then $\struct {J, +, \circ_r}$ is a right module.
Then:
{{begin-eqn}}
{{eqn | q = \forall x, y \in R: \forall j \in J
      | l = \paren {x \circ_l j} \circ_r y
      | r = \paren {x \times j} \times y
      | c = Definition of $\circ_l$ and $\circ_r$
}}
{{eqn | r = x \times \paren {j \times y}
      | c = {{Ring-axiom|M1}}
}}
{{eqn | r = x \circ_l \paren {j \circ_r y}
      | c = Definition of $\circ_l$ and $\circ_r$
}}
{{end-eqn}}
Hence $\struct {J, +, \circ_l, \circ_r}$ is a bimodule over $\struct {R, +, \times}$ by definition.
{{qed}}
Theorem: Ideal is Bimodule over Ring/Ring is Bimodule over Ring

Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times, \times}$ is a bimodule over $\struct {R, +, \times}$.

Proof:
From Ring is Ideal of Itself and Ideal is Bimodule over Ring, $\struct {R, +, \times, \times}$ is a bimodule over $\struct {R, +, \times}$.
{{qed}}
Theorem: Left Module over Commutative Ring induces Bimodule

Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct{G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct{G, +_G, \circ, \circ'}$ is a bimodule over $\struct {R, +_R, \times_R}$.

Proof:
From Left Module over Commutative Ring induces Right Module, $\struct{G, +_G, \circ'}$ is a right module.
Let $\lambda, \mu \in R$ and $x \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \lambda \circ \paren{x \circ' \mu}
      | r = \lambda \circ \paren{\mu \circ x}
      | c = Definition of $\circ’$
}}
{{eqn | r = \paren {\lambda \circ \mu} \circ x
      | c = {{Module-axiom|3}}
}}
{{eqn | r = \paren {\mu \circ \lambda} \circ x
      | c = Ring product $\circ$ is commutative
}}
{{eqn | r = \mu \circ \paren{\lambda \circ x}
      | c = {{Module-axiom|3}}
}}
{{eqn | r = \paren{\lambda \circ x} \circ' \mu
      | c = Definition of $\circ'$
}}
{{end-eqn}}
Hence $\struct {G, +_G, \circ, \circ'}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition.
{{qed}}
Theorem: Right Module over Commutative Ring induces Bimodule

Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct {G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda $
Then $\struct {G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$.

Proof:
From Right Module over Commutative Ring induces Left Module, $\struct {G, +_G, \circ'}$ is a left module.
Let $\lambda, \mu \in R$ and $x \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \lambda \circ' \paren {x \circ \mu}
      | r = \paren {x \circ \mu} \circ \lambda
      | c = Definition of $\circ’$
}}
{{eqn | r = x \circ \paren {\mu \circ \lambda}
      | c = Right module axiom $(\text {RM} 3)$ : Associativity of Scalar Multiplication
}}
{{eqn | r = x \circ \paren {\lambda \circ \mu}
      | c = Ring product $\circ$ is commutative
}}
{{eqn | r = \paren {x \circ \lambda} \circ \mu
      | c = Right module axiom $(\text {RM} 3)$ : Associativity of Scalar Multiplication
}}
{{eqn | r = \paren {\lambda \circ' x} \circ \mu
      | c = Definition of $\circ'$
}}
{{end-eqn}}
Hence $\struct{G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition.
{{qed}}
Theorem: Diagonals of Kite are Perpendicular

Let $ABCD$ be a kite such that $AC$ and $BD$ are its diagonals.
Then $AC$ and $BD$ are perpendicular.

Proof:
:450px
Let $AC$ and $BD$ meet at $E$.
Consider the triangles $\triangle ABD$ and $\triangle CBD$.
We have that:
:$AB = CB$
:$AD = CD$
:$BD$ is common.
Hence by Triangle Side-Side-Side Congruence, $\triangle ABD$ and $\triangle CBD$ are congruent.
Consider the triangles $\triangle ABE$ and $\triangle CBE$.
We have from the congruence of $\triangle ABD$ and $\triangle CBD$ that:
:$\angle ABE = \angle CBE$
:$AB = CB$
and $BE$ is common.
Hence by Triangle Side-Angle-Side Congruence, $\triangle ABE$ and $\triangle CBE$ are congruent.
We have that $AC$ is a straight line.
We have from the congruence of $\triangle ABE$ and $\triangle CBE$ that:
:$\angle BEC = \angle BEA$
From Two Angles on Straight Line make Two Right Angles, $\angle BEC + \angle BEA$ make two right angles.
Thus:
:$2 \angle BEC = 2 \angle BEA = 2$  right angles
and so:
:$\angle BEC = \angle BEA$ are both right angles.
That is, $AC$ and $BD$ are perpendicular.
{{qed}}
Theorem: One Diagonal of Kite Bisects the Other

Let $ABCD$ be a kite such that:
:$AC$ and $BD$ are its diagonals
:$AB = BC$
:$AD = DC$
Then $BD$ is the perpendicular bisector of $AC$.

Proof:
:450px
Let $AC$ and $BD$ meet at $E$.
From Diagonals of Kite are Perpendicular, $AC$ and $BD$ are perpendicular.
That is:
:$\angle AEB = \angle CEB$
both being right angles.
Consider the triangles $\triangle ABE$ and $\triangle CBE$.
We have that:
:$\angle AEB = \angle CEB$ are both right angles.
:$AB = BC$
:$BE$ is common
Hence by Triangle Right-Angle-Hypotenuse-Side Congruence, $\triangle ABE$ and $\triangle CBE$ are congruent.
Thus:
:$AE = CE$
and it has already been established that $BD$ is perpendicular to $AC$.
Hence the result by definition of perpendicular bisector.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Group/Product Rule

:$f * g : \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.

Proof:
By definition, a topological group is a topological semigroup.
Hence $\struct {G, *, \tau_{_G} }$ is a topological semigroup.
From Product Rule for Continuous Mappings to Topological Semigroup:
:$f * g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Group/Multiple Rule

:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.

Proof:
By definition, a topological group is a topological semigroup.
Hence $\struct {G, *, \tau_{_G}}$ is a topological semigroup.
From Multiple Rule for Continuous Mappings to Topological Semigroup:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ are continuous mappings.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Group/Inverse Rule

:$g^{-1}: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is a continuous mapping.

Proof:
By definition of a topological group:
:$\phi: \struct {G, \tau_G} \to \struct {G, \tau_G}$ such that $\forall x \in G: \map \phi x = x^{-1}$ is a continuous mapping
From Composite of Continuous Mappings is Continuous:
:the composition $\phi \circ g: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is continuous.
Now:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map {\paren {g^{-1} } } x
      | r = \paren {\map g x}^{-1}
      | c = Definition of $g^{-1}$
}}
{{eqn | r = \map \phi {\map g x}
      | c = Definition of $\phi$
}}
{{eqn | r = \map {\paren {\phi \circ g} } x
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
From Equality of Mappings:
:$g^{-1} = \phi \circ g$
The result follows.
{{qed}}
Category:Combination Theorem for Continuous Mappings to Topological Group
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Theorem: Combination Theorem for Continuous Mappings/Topological Semigroup/Product Rule

:$f * g: \struct{S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.

Proof:
Let $f \times g: S \to G \times G$ be the mapping defined by:
:$\forall x \in S : \map {\paren {f \times g} } x = \tuple {\map f x, \map g x}$
From Pointwise Operation is Composite of Operation with Mapping to Cartesian Product:
:$f * g = * \circ \paren {f \times g}$
Let $\tau$ be the product topology on $G \times G$.
From {{Corollary|Continuous Mapping to Product Space}}:
:$f \times g : \struct{S, \tau_{_S}} \to \struct {G \times G, \tau}$ is a continuous mapping.
Now $* : \struct {G \times G, \tau} \to \struct {G, \tau_{_G} }$ is continuous by definition of a topological semigroup.
From Composite of Continuous Mappings is Continuous:
:$* \circ \paren {f \times g}$ is continuous.
The result follows.
{{qed}}
Category:Combination Theorem for Continuous Mappings to Topological Semigroup
26nnkn16qd67z35yn9dr1xzs53h67qk
Theorem: Combination Theorem for Continuous Mappings/Topological Semigroup/Multiple Rule

:$\lambda * f: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is a continuous mapping
:$f * \lambda: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is a continuous mapping.

Proof:
Let $c_\lambda : S \to G$ be the constant mapping defined by:
:$\forall x \in S: \map {c_\lambda} x = \lambda$
From Constant Mapping is Continuous, $c_\lambda$ is continuous.
From Product Rule for Continuous Mappings to Topological Semigroup:
:$c_\lambda * f$ and $f * c_\lambda$ are continuous.
Now:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map {\paren {c_\lambda * f} } x
      | r = \map {c_\lambda} x * \map f x
      | c = Definition of $c_\lambda * f$ 
}}
{{eqn | r = \lambda * \map f x
      | c = Definition of $c_\lambda$ 
}}
{{eqn | r = \map {\paren {\lambda * f} } x
      | c = Definition of $\lambda * f$ 
}}
{{end-eqn}}
From Equality of Mappings:
:$c_\lambda * f = \lambda * f$
Similarly:
:$f * c_\lambda = f * \lambda$
The result follows.
{{qed}}
Category:Combination Theorem for Continuous Mappings to Topological Semigroup
qyu1hk4ol80pu0kqqt1sea7vp1o7va0
Theorem: Volume of Solid of Revolution/Parametric Form

Let $x: \R \to \R$ and $y: \R \to \R$ be real functions defined on the interval $\closedint a b$.
Let $y$ be integrable on the (closed) interval $\closedint a b$.
Let $x$ be differentiable on the (open) interval $\openint a b$.
Let the points be defined:
:$A = \tuple {\map x a, \map y a}$
:$B = \tuple {\map x b, \map y b}$
:$C = \tuple {\map x b, 0}$
:$D = \tuple {\map x a, 0}$
Let the figure $ABCD$ be defined as being bounded by the straight lines $y = 0$, $x = a$, $x = b$ and the curve defined by:
:$\set {\tuple {\map x t, \map y t}: a \le t \le b}$
Let the solid of revolution $S$ be generated by rotating $ABCD$ around the $x$-axis (that is, $y = 0$).
Then the volume $V$ of $S$ is given by:
:$\ds V = \pi \int_a^b \paren {\map y t}^2 \map {x'} t \rd t$

Proof:
{{begin-eqn}}
{{eqn|c = Volume of Solid of Revolution
     |l = V
     |r = \pi \int_a^b \paren {\map f x}^2 \rd x
}}
{{eqn|c = Integration by Substitution/Definite Integral
     |r = \pi \int_a^b \paren {\map y t}^2 \map {x'} t \rd t
}}
{{end-eqn}}
{{qed}}
Theorem: Weight of Body at Earth's Surface

Let $B$ be a body of mass $m$ situated at (or near) the surface of Earth.
Then the weight of $B$ is given by:
:$W = m g$
where $g$ is the value of the acceleration due to gravity at the surface of Earth.

Proof:
The weight of $B$ is the magnitude of the force exerted on it by the influence of the gravitational field it is in.
By Newton's Second Law of Motion, that force is given by:
:$\mathbf W = -m g \mathbf k$
where:
:$g$ is the value of the acceleration due to gravity at the surface of Earth
:$\mathbf k$ is a unit vector directed vertically upwards.
Hence the magnitude of $\mathbf W$ is given by:
:$W = \size {-m g \mathbf k} = m g$
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Ring/Sum Rule

:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological ring, $\struct {R, +, \tau_{_R} }$ is a topological group.
From Product Rule for Continuous Mappings to Topological Group:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Ring/Negation Rule

:$-g : \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological ring:
:$\struct {R, +, \tau_{_R} }$ is a topological group.
From Inverse Rule for Continuous Mappings to Topological Group:
:$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Ring/Multiple Rule

:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological ring:
:$\struct {R, *, \tau_{_R} }$ is a topological semigroup.
From Multiple Rule for Continuous Mappings to Topological Semigroup:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Ring/Product Rule

:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological ring:
:$\struct {R, *, \tau_{_R} }$ is a topological semigroup.
From Product Rule for Continuous Mappings to Topological Semigroup:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Ring/Combined Rule

:$\lambda * f + \mu * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping
:$f * \lambda + g * \mu: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.

Proof:
{{ProofWanted}}
Category:Combination Theorem for Continuous Mappings
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Theorem: Combination Theorem for Continuous Mappings/Topological Division Ring/Sum Rule

:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological division ring, $\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Sum Rule for Continuous Mappings into Topological Ring:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Division Ring/Multiple Rule

:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Multiple Rule for Continuous Mappings to Topological Ring:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Topological Division Ring/Product Rule

:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.

Proof:
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Product Rule for Continuous Mappings to Topological Ring:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
{{qed}}