avewright/proofwiki-math huggingface dataset:
Theorem: Equation of Conic Section/Cartesian Form

The general conic section can be expressed in Cartesian coordinates in the form:
:$a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0$
for some $a, b, c, f, g, h \in \R$.

Proof:
By definition, a conic section is the set of points of intersection between a cone and a plane.
Let $P = \tuple {\alpha, \beta, \gamma}$ be the apex of the cone.
Let $Q = \tuple {x, y, z}$ be a point of intersection between the plane and the cone.
From Equation of Right Circular Cone, we have:
:$(1): \quad \paren {x - \alpha}^2 + \paren {y - \beta}^2 = \eta \paren {z - \gamma}^2$
for some (strictly) positive real number $\eta$.
From Equation of Plane, we have:
:$A x + B y + C z + D = 0$
for some real numbers $A, B, C, D$.
Let $C \ne 0$.
Then:
:$z = - D - \dfrac {A x + B y} C$
Hence, plugging in $z$ into $(1)$:
:$\paren {x - \alpha}^2 + \paren {y - \beta}^2 = \eta \paren {\dfrac {A x + B y} C + D + \gamma}^2$
Opening up the brackets and making the corresponding notation of constants, we get the desired result.
{{qed|lemma}}
Let $C = 0$.
Then:
:$A x + B y + D = 0$
In order for the plane to be well-defined, $A$ or $B$ must be non-zero.
{{WLOG}}, since the cone equation $(1)$ is symmetric in $x$ and $y$, we can consider $B \ne 0$ (otherwise interchange coordinates).
Then:
:$y = - D - \dfrac {A x} B$
Hence, plugging  $y$ into $(1)$:
:$\paren {x - \alpha}^2 + \paren {-D - \dfrac {A x} B - \beta}^2 = \eta \paren {z - \gamma}^2$
Opening up the brackets and making the corresponding notation of constants and coordinates ($z \leftrightarrow y$), we get the desired result.
{{qed}}
Theorem: Graph of Quadratic describes Parabola

The locus of the equation defining a quadratic:
:$y = a x^2 + b x + c$
describes a parabola.

Proof:
Consider the focus-directrix property of a parabola $P$.
Let the focus of $P$ be the point $\tuple {0, f}$ on a Cartesian plane.
Let the directrix of $P$ be the straight line $y = -d$.
Let $\tuple {x, y}$ be an arbitrary point on $P$.
Then by the focus-directrix property:
:$y + d = \sqrt {\paren {x - k}^2 + \tuple {y - f}^2}$
where:
:$y + d$ is the distance from $\tuple {x, y}$ to the straight line $y = -d$
:$\sqrt {\paren {x - k}^2 + \paren {y - f}^2}$ is the distance from $\tuple {x, y}$ to the point $\tuple {k, f}$ by the Distance Formula.
Hence:
{{begin-eqn}}
{{eqn | l = \paren {y + d}^2
      | r = \paren {x - k}^2 + \paren {y - f}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2 + 2 y d + d^2
      | r = x^2 - 2 k x + k^2 + y^2 - 2 f y + f^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 y \paren {f + d}
      | r = x^2 - 2 k x + f^2 + k^2 - d^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac 1 {2 \paren {f + d} } x^2 - \frac k {\paren {f + d} } x + \frac {f - d} 2
      | c = 
}}
{{end-eqn}}
This is in the form $y = a x^2 + b^2 + c$.
By setting $k$, $f$ and $d$ appropriately in terms of $a$, $b$ and $c$, the specific focus and directrix can be appropriately positioned.
{{qed}}
Category:Parabolas
Category:Quadratic Equations
Category:Graph of Quadratic describes Parabola
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Theorem: Graph of Quadratic describes Parabola/Corollary 1

The locus of the equation of the square function:
:$y = x^2$
describes a parabola.

Proof:
This is a particular instance of Graph of Quadratic describes Parabola, where:
:$y = a x^2 + b x + c$
is the equation of a parabola.
The result follows by setting $a = 1, b = 0, c = 0$.
{{qed}}
Theorem: Graph of Quadratic describes Parabola/Corollary 2

The locus of the equation of the square root function on the non-negative reals:
:$\forall x \in \R_{\ge 0}: \map f x = \sqrt x$
describes half of a parabola.

Proof:
From {{Corollary|Graph of Quadratic describes Parabola|1}}, where:
:$y = x^2$
is the equation of a parabola.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = x^2$
From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$.
Also we have from Positive Real Number has Two Square Roots:
:$\forall x \in \R: \paren {-x}^2 = x^2$
Thus it is necessary to apply a bijective restriction upon $f$.
Let $g: \R_{\ge 0} \to \R_{\ge 0}$ be the bijective restriction of $f$ to $\R_{\ge 0} \times \R_{\ge 0}$:
:$\forall x \in \R_{\ge 0}: \map g x = x^2$
From Inverse of Bijection is Bijection, $g^{-1}: \R_{\ge 0} \to \R_{\ge 0}$ is also a bijection.
By definition:
:$\forall x \in \R_{\ge 0}: \map {g^{-1} } x = +\sqrt x$
Then from Graph of Inverse Mapping, the graph of $g^{-1}$ is the same as the graph of $g$, reflected in the line $x = y$.
As the graph of $f$ is a parabola, the graph of $g$ is also a parabola, but because of the restriction to $\R_{\ge 0}$, just half of it.
Thus the graph of $g^{-1}$ is also half a parabola.
{{qed}}
Theorem: Natural Numbers under Addition form Commutative Semigroup

The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is a commutative semigroup.

Proof:
By Naturally Ordered Semigroup Exists, there exists a naturally ordered semigroup.
Consider the natural numbers $\N$ defined as the naturally ordered semigroup.
From the definition of the naturally ordered semigroup, $\struct {\N, +}$ is {{apriori}} a semigroup.
By Naturally Ordered Semigroup Axioms imply Commutativity, it follows that $\struct {\N, +}$ is a commutative semigroup.
{{qed}}
Theorem: Gödel's Incompleteness Theorems/First/Corollary

If $T$ is both consistent and complete, it does not contain minimal arithmetic.

Proof:
This is simply the contrapositive of Gödel's First Incompleteness Theorem.
{{qed}}
{{Namedfor|Kurt Friedrich Gödel}}
Theorem: Monomorphism that is Split Epimorphism is Split Monomorphism

Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$.
Let $f: C \to D$ be a morphism in $\mathbf C$ such that $f$ is a monomorphism and a split epimorphism.
Then $f: C \to D$ is a split monomorphism.

Proof:
Let $g: D \to C$ be the right inverse of $f$:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of split epimorphism.
Therefore:
:$f \circ g \circ f = \operatorname{id}_D \circ f = f \circ \operatorname{id}_C$
by the property of the identity morphism.
Since $f$ is left cancellable, by the definition of monomorphism, we have:
:$g \circ f = \operatorname{id}_C$
Hence $f$ is a split monomorphism with left inverse $g$.
{{qed}}
Category:Morphisms
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Theorem: Epimorphism that is Split Monomorphism is Split Epimorphism

Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a epimorphism and a split monomorphism.
Then $f: C \to D$ is a split epimorphism.

Proof:
{{explain|What is a Dual proof}}
Dual proof of Monomorphism that is Split Epimorphism is Split Monomorphism.
{{qed}}
Category:Epimorphisms
Category:Monomorphisms (Category Theory)
k41tqj6wiaco8bl898w5naxuwmhw8kz
Theorem: Westwood's Puzzle

:500px
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.

Proof:
Construct the perpendicular from $E$ to $AC$, and call its foot $G$.
Let $K$ be the intersection of $IE$ and $AC$.
Let $L$ be the intersection of $EH$ and $AC$.
:500px
First we have:
{{begin-eqn}}
{{eqn | n = 1
      | l = \angle CKI
      | r = \angle EKG
      | c = Two Straight Lines make Equal Opposite Angles
}}
{{eqn | l = \angle EGK
      | r = \text {Right Angle}
      | c = Tangent to Circle is Perpendicular to Radius
}}
{{eqn | l = \angle KIC
      | r = \text {Right Angle}
      | c = as $IF \perp CD$
}}
{{eqn | n = 2
      | ll= \therefore
      | l = \angle EGK
      | r = \angle KIC
      | c = {{EuclidPostulateLink|Fourth}}
}}
{{eqn | l = IC
      | r = EJ
      | c = Opposite Sides and Angles of Parallelogram are Equal
}}
{{eqn | l = EJ
      | r = EG
      | c = as both are radii of the same circle
}}
{{eqn | n = 3
      | ll= \therefore
      | l = IC
      | r = EG
      | c = {{EuclidCommonNotionLink|1}}
}}
{{eqn | ll= \therefore
      | l = \Area \triangle IKC
      | r = \Area \triangle GKE
      | c = Triangle Angle-Angle-Side Congruence: $(1)$, $(2)$ and $(3)$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | n = 4
      | l = \angle HLA
      | r = \angle GLE
      | c = Two Straight Lines make Equal Opposite Angles
}}
{{eqn | l = \angle EGL
      | r = \text {Right Angle}
      | c = Tangent to Circle is Perpendicular to Radius
}}
{{eqn | l = \angle AHL
      | r = \text {Right Angle}
      | c = as $HJ \perp AD$
}}
{{eqn | n = 5
      | ll= \therefore
      | l = \angle EGL
      | r = \angle AHL
      | c = {{EuclidPostulateLink|Fourth}}
}}
{{eqn | l = HA
      | r = EF
      | c = Opposite Sides and Angles of Parallelogram are Equal
}}
{{eqn | l = EF
      | r = EG
      | c = as both are radii of the same circle
}}
{{eqn | n = 6
      | ll= \therefore
      | l = HA
      | r = EG
      | c = {{EuclidCommonNotionLink|1}}
}}
{{eqn | ll= \therefore
      | l = \Area \triangle HAL
      | r = \Area \triangle GEL
      | c = Triangle Angle-Angle-Side Congruence: $(4)$, $(5)$ and $(6)$
}}
{{end-eqn}}
Finally:
{{begin-eqn}}
{{eqn | l = \frac {\Area \Box ABCD} 2
      | r = \frac {AD \cdot CD} 2
      | c = Area of Parallelogram
}}
{{eqn | r = \Area \triangle ADC
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \Area \triangle HAL + \Area \triangle IKC + \Area \Box DHLKI
}}
{{eqn | r = \Area \triangle GEL + \Area \triangle GKE + \Area \Box DHLKI
}}
{{eqn | r = \Area \Box DHEI
}}
{{end-eqn}}
{{qed}}
Theorem: Westwood's Puzzle

:500px
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.

Proof:
The crucial geometric truth to note is that:
:$CJ = CG, AG = AF, BF = BJ$
This follows from the fact that:
:$\triangle CEJ \cong \triangle CEG$, $\triangle AEF \cong \triangle AEG$ and $\triangle BEF \cong \triangle BEJ$
This is a direct consequence of the point $E$ being the center of the incircle of $\triangle ABC$.
Then it is just a matter of algebra.
Let $AF = a, FB = b, CJ = c$.
{{begin-eqn}}
{{eqn | l = \paren {a + b}^2 + \paren {b + c}^2
      | r = \paren {a + c}^2
      | c = Pythagoras's Theorem
}}
{{eqn | ll= \leadsto
      | l = a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2
      | r = a^2 + 2 a c + c^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b + b^2 + b c
      | r = a c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b + b^2 + b c + a c
      | r = 2 a c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a + b} \paren {b + c}
      | r = 2 a c
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Vector Cross Product Operator is Bilinear

Let $\mathbf u$, $\mathbf v$ and $\mathbf w$ be vectors in a vector space $\mathbf V$ of $3$ dimensions:
{{begin-eqn}}
{{eqn | l = \mathbf u
      | r = u_i \mathbf i + u_j \mathbf j + u_k \mathbf k
}}
{{eqn | l = \mathbf v
      | r = v_i \mathbf i + v_j \mathbf j + v_k \mathbf k
}}
{{eqn | l = \mathbf w
      | r = w_i \mathbf i + w_j \mathbf j + w_k \mathbf k
}}
{{end-eqn}}
where $\left({\mathbf i, \mathbf j, \mathbf k}\right)$ is the standard ordered basis of $\mathbf V$.
Let $c$ be a real number.
Then:
: $\left({c \mathbf u + \mathbf v}\right) \times \mathbf w = c \left({ \mathbf u \times \mathbf w}\right) + \mathbf v \times \mathbf w$

Proof:
{{begin-eqn}}
{{eqn | l = \left({c \mathbf u + \mathbf v}\right) \times \mathbf w
      | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ c u_i + v_i & c u_j + v_j & c u_k + v_k \\ w_i & w_j & w_k \end{vmatrix}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ c u_i & c u_j & c u_k \\ w_i & w_j & w_k \end{vmatrix} + \begin{vmatrix} \mathbf i& \mathbf j & \mathbf k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{vmatrix}
      | c = Determinant as Sum of Determinants
}}
{{eqn | r = c \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ u_i & u_j & u_k \\ w_i & w_j & w_k \end{vmatrix} + \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{vmatrix}
      | c = Determinant with Row Multiplied by Constant
}}
{{eqn | r = c \left({\mathbf u \times \mathbf w}\right) + \mathbf v \times \mathbf w
      | c = {{Defof|Vector Cross Product}}
}}
{{end-eqn}}
{{qed}}
Category:Vector Cross Product
bhw03ywmrmsthq70e9dmqa9snur0fg6
Theorem: Natural Numbers have No Proper Zero Divisors

Let $\N$ be the natural numbers.
Then for all $m, n \in \N$:
:$m \times n = 0 \iff m = 0 \lor n = 0$
That is, $\N$ has no proper zero divisors.

Proof:
=== Necessary Condition ===
Suppose that $n = 0$ or $m = 0$.
Then from Zero is Zero Element for Natural Number Multiplication:
:$m \times n = 0$
{{qed|lemma}}
Theorem: Diagonals of Rhombus Bisect Angles

Let $OABC$ be a rhombus.
Then:
:$(1): \quad OB$ bisects $\angle AOC$ and $\angle ABC$
:$(2): \quad AC$ bisects $\angle OAB$ and $\angle OCB$
400px

Proof:
{{WLOG}}, we will only prove $OB$ bisects $\angle AOC$.
We have:
{{begin-eqn}}
{{eqn | l = OA
      | r = OC
      | c = {{Defof|Rhombus}}
}}
{{eqn | l = BA
      | r = BC
      | c = {{Defof|Rhombus}}
}}
{{eqn | l = OB
      | r = OB
      | c = Common Side
}}
{{eqn | ll = \leadsto
      | l = \triangle OAB
      | o = \cong
      | r = \triangle OCB
      | c = SSS
}}
{{end-eqn}}
Comparing corresponding angles gives:
:$\angle AOB = \angle COB$
hence $OB$ bisects $\angle AOC$.
{{qed}}
Theorem: Diagonals of Rhombus Bisect Angles

Let $OABC$ be a rhombus.
Then:
:$(1): \quad OB$ bisects $\angle AOC$ and $\angle ABC$
:$(2): \quad AC$ bisects $\angle OAB$ and $\angle OCB$
400px

Proof:
{{WLOG}}, we will only prove $OB$ bisects $\angle AOC$.
Let the position vectors of $A$, $B$ and $C$ with respect to $O$ be $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
By definition of rhombus, we have:
{{begin-eqn}}
{{eqn | n = a
      | l = \mathbf a + \mathbf c
      | r = \mathbf b
      | c = Parallelogram Law
}}
{{eqn | n = b
      | l = \norm {\mathbf a}
      | r = \norm {\mathbf c}
      | c = 
}}
{{end-eqn}}
From the above we have:
{{begin-eqn}}
{{eqn | l = \cos \angle \mathbf a, \mathbf b
      | r = \frac {\mathbf a \cdot \mathbf b} {\norm {\mathbf a} \norm {\mathbf b} }
      | c = {{Defof|Dot Product|subdef = Real Euclidean Space}}
}}
{{eqn | r = \frac {\mathbf a \cdot \paren {\mathbf a + \mathbf c} } {\norm {\mathbf a} \norm {\mathbf b} }
      | c = from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$
}}
{{eqn | r = \frac {\mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf c} {\norm {\mathbf a} \norm {\mathbf b} }
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \frac { {\norm {\mathbf a} }^2 + \mathbf a \cdot \mathbf c} {\norm {\mathbf a} \norm {\mathbf b} }
      | c = Dot Product of Vector with Itself
}}
{{eqn | r = \frac { {\norm {\mathbf c} }^2 + \mathbf a \cdot \mathbf c} {\norm {\mathbf c} \norm {\mathbf b} }
      | c = from $(b)$ above: $\norm {\mathbf a} = \norm {\mathbf c}$
}}
{{eqn | r = \frac {\mathbf c \cdot \mathbf c + \mathbf a \cdot \mathbf c} {\norm {\mathbf c} \norm {\mathbf b} }
      | c = Dot Product of Vector with Itself
}}
{{eqn | r = \frac {\mathbf c \cdot \left({\mathbf a + \mathbf c}\right)} {\norm {\mathbf c} \norm {\mathbf b} }
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \frac {\mathbf c \cdot \mathbf b} {\norm {\mathbf c} \norm {\mathbf b} }
      | c = from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$
}}
{{eqn | r = \cos \angle \mathbf c, \mathbf b
      | c = {{Defof|Dot Product|subdef = Real Euclidean Space}}
}}
{{end-eqn}}
By definition of dot product, the angle between the vectors is between $0$ and $\pi$.
From Shape of Cosine Function, cosine is injective on this interval.
Hence:
:$\angle \mathbf a, \mathbf b = \angle \mathbf c, \mathbf b$
The result follows.
{{qed}}
Theorem: Summation Formula for Polygonal Numbers

Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$

Proof:
We have that:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Proof by induction:
For all $n \in \N_{>0}$, let $\map \Pi n$ be the proposition:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
Theorem: Relative Prime Modulo Tensor is Zero

Let $p \in \Z_{>0}$ and $q \in \Z_{>0}$ be positive coprime integers.
Let $\Z / p \Z$ and $\Z / q \Z$ be $\Z$-modules.
{{explain|It is not a good idea to use the same notation for both a ring and a module. Either $\Z / p \Z$ is a ring or it is a module. Please consider taking the advice in the explain template at the bottom of this page.}}
Then:
:$\Z / p \Z \otimes_\Z \Z / q\Z = 0$
where $\otimes_\Z$ denotes tensor product over integers.

Proof:
By Bézout's Identity there exists $a, b \in \Z$ such that $a p + b q = 1$.
Then for $s \otimes_\Z t \in \Z / p \Z \otimes \Z / q \Z$:
{{begin-eqn}}
{{eqn | l = s \otimes t
      | r = (s \left({a p + b q}\right)) \otimes t
      | c = $s = s \cdot 1$
}}
{{eqn | r = (s a p + s b q) \otimes t
      | c = {{Module-axiom|2}}
}}
{{eqn | r = s b q \otimes t + s a p \otimes t
      | c = Equality in tensor product
}}
{{eqn | r = s b \otimes q t + s a p \otimes t
      | c = Equality in tensor product
}}
{{eqn | r = 0
      | c = by Tensor with Zero Element is Zero in Tensor and the fact that $qt = 0$ in $\Z_q$ and $sap=0$ in $\Z_p$
}}
{{end-eqn}}
{{qed}}
{{explain|Notation needs to be tightened up. When the fundamental stuff like this is being addressed, it is important to explain exactly which operation of which component of the module is being used: the scalar product, or the group operation, or whichever of the ring operations. It is inadequate to use mere concatenation, as it is never completely clear which elements of which structure each one is. As for the rest of this proof, it is still impenetrably vague. The reason for the last line is a complete mystery -- the actual definition of $0$ also needs to be stated.}}
Theorem: Tensor Product is Module

Let $R$ be a ring.
Let $M$ be a $R$-right module.
Let $N$ be a $R$-left module.
Then:
:$\ds T = \bigoplus_{s \mathop \in M \times N} R s$
is a left module.

Proof:
=== Axiom 1 ===
Let $x, y \in T$ with $x = \family {s_i}_{i \mathop \in I}$ and $y = (t_i)_{i \mathop \in I}$.
Let $\lambda\in R$.
Then:
{{begin-eqn}}
{{eqn | l = \lambda \circ \paren {x + y}
      | r = \lambda \circ (\family {s_i}_{i \mathop \in I} + \family {t_i}_{i \mathop \in I})
      | c = Definition of elements in direct sum
}}
{{eqn | r = \lambda \circ \family {s_i + t_i}_{i \mathop \in I}
      | c = Addition in direct sum
}}
{{eqn | r = \family {\lambda \circ s_i + \lambda \circ t_i}_{i \mathop \in I}
      | c = $R$-action in direct sum
}}
{{eqn | r = \family {\lambda \circ s_i}_{i \mathop \in I} + \family {\lambda \circ t_i}_{i \mathop \in I}
      | c = Addition in direct sum
}}
{{eqn | r = \lambda \circ \family {s_i}_{i \mathop \in I} + \lambda \circ \family {t_i}_{i \mathop \in I}
      | c = $R$-action in direct sum
}}
{{eqn | r = \lambda \circ x + \lambda \circ y
      | c = Definition of elements in direct sum
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Supremum of Subset of Real Numbers is Arbitrarily Close

Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be a supremum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A: b − x < \epsilon$

Proof:
Note that $A$ is non-empty as the empty set does not admit a supremum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
:$\forall x \in A: b − x \ge \epsilon$
Then:
:$\forall x \in A: b − \epsilon \ge x$
and so $b − \epsilon$ would be an upper bound of $A$ which is less than $b$.
But since $b$ is a supremum of $A$ there can be no such $b − \epsilon$.
From that contradiction it follows that:
:$\exists x \in A: b − x < \epsilon$
{{qed}}
Theorem: Tensor with Zero Element is Zero in Tensor

Let $R$ be a ring.
Let $M$ be a right $R$-module.
Let $N$ be a left $R$-module.
Let $M \otimes_R N$ denote their tensor product.
Then:
:$0\otimes_R n = m \otimes_R 0 = 0 \otimes_R 0$
is the zero in $M \otimes_R N$.

Proof:
Let $m \in M$ and $n \in N$
Then
{{begin-eqn}}
{{eqn | l = m \otimes_R n
      | r = \paren {m + 0} \otimes_R n
      | c = {{Group-axiom|2}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R n
      | c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R \paren {n + 0}
      | c = {{Group-axiom|2}}
}}
{{eqn | r = m \otimes_R n + m\otimes_R 0
      | c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R \paren {n + 0}
      | c = {{Group-axiom|2}}
}}
{{eqn | r = m \otimes_R n + m \otimes_R 0 + 0 \otimes_R n + 0 \otimes_R 0
      | c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R 0
      | c = {{Defof|Tensor Equality}}
}}
{{end-eqn}}
Hence $0 \otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module.
{{qed}}
Category:Tensor Algebra
Category:Homological Algebra
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Theorem: Primitive of Arcsecant of x over a/Formulation 1

:$\ds \int \arcsec \frac x a \rd x = \begin {cases}
x \arcsec \dfrac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
x \arcsec \dfrac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = x
      | c = Primitive of Constant
}}
{{end-eqn}}
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \arcsec \frac x a \rd x
      | r = x \arcsec \frac x a - \int x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arcsec \frac x a - a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arcsec \frac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
{{begin-eqn}}
{{eqn | l = \int \arcsec \frac x a \rd x
      | r = x \arcsec \frac x a - \int x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arcsec \frac x a + a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arcsec \frac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}