nyuuzyou/wikis huggingface dataset:
Terminal Object is Unique

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $1$ and $1'$ be two terminal objects of $\mathbf C$.
Then there is a unique isomorphism $u: 1 \to 1'$.
Hence, terminal objects are unique up to unique isomorphism.
== Proof ==
Consider the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,0em>*+{1} = "M",
<0em,0em> *+{1'}= "N",
<0em,-4em>*+{1} = "M2",
<4em,-4em>*+{1'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_1},
"N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v},
"N";"N2" **@{-} ?>*@{>} ?*!/_1em/{\operatorname{id}_{1'}},
"M2";"N2"**@{-} ?>*@{>} ?*!/^.6em/{u},
\end{xy}$
It commutes, as each of the morphisms in it points to a terminal object, and hence is unique.
Thus, $v$ is an inverse to $u$, and so $u$ is an isomorphism.
== Also see ==
* Initial Object is Unique
== Sources ==
* : $\S 2.2$: Proposition $2.10$
Category:Category Theory
Terminal Speed of Body under Fall Retarded Proportional to Speed

== Theorem ==
Let $B$ be a body falling in a gravitational field.
Let $B$ be falling through a medium which exerts a resisting force $k \mathbf v$ upon $B$ which is proportional to the speed of $B$ relative to the medium.
Then the terminal speed of $B$ is given by:
:$v = \dfrac {g m} k$
== Proof ==
Let $B$ start from rest.
From Motion of Body Falling through Air, the differential equation governing the motion of $B$ is given by:
:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \dfrac {\d \mathbf s} {\d t}$
Dividing through by $m$ and setting $c = \dfrac k m$ gives:
:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \dfrac {\d \mathbf s} {\d t}$
By definition of velocity:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v$
and so:
When $t = 0$ we have that $\mathbf v = 0$ and so:
:$\mathbf c_2 = \mathbf g$
Hence by taking magnitudes:
:$v = \dfrac g c \paren {1 - e^{-c t} }$
Since $c > 0$ it follows that $v \to \dfrac g c$ as $t \to \infty$.
Thus in the limit:
:$v = \dfrac g c = \dfrac {g m} k$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Category:Terminal Speed
Category:Mechanics
Terminal Speed of Body under Fall Retarded Proportional to Square of Speed

== Theorem ==
Let $B$ be a body falling in a gravitational field.
Let $B$ be falling through a medium which exerts a resisting force of magnitude $k v^2$ upon $B$ which is proportional to the square of the speed of $B$ relative to the medium.
Then the terminal speed of $B$ is given by:
:$v = \sqrt {\dfrac {g m} k}$
== Proof ==
Let $B$ start from rest.
The differential equation governing the motion of $B$ is given by:
:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \paren {\dfrac {\d \mathbf s} {\d t} }^2$
Dividing through by $m$ and setting $c = \dfrac k m$ gives:
:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \paren {\dfrac {\d \mathbf s} {\d t} }^2$
By definition of velocity:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v^2$
for some constant $c$.
and so taking magnitudes of the vector quantities:
We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$.
Hence:
Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$.
Thus in the limit:
:$v = \sqrt {\dfrac g c} = \sqrt {\dfrac {g m} k}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems: Problem $1$
Category:Terminal Speed
Category:Mechanics
Terms in Convergent Series Converge to Zero

== Theorem ==
Let $\sequence {a_n}$ be a sequence in any of the standard number fields $\Q$, $\R$, or $\C$.
Suppose that the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges in any of the standard number fields $\Q$, $\R$, or $\C$.
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = 0$
== Proof ==
Let $\ds s = \sum_{n \mathop = 1}^\infty a_n$.
Then $\ds s_N = \sum_{n \mathop = 1}^N a_n \to s$ as $N \to \infty$.
Also, $s_{N - 1} \to s$ as $N \to \infty$.
Thus:
Hence the result.
== Also see ==
* Terms in Uniformly Convergent Series Converge Uniformly to Zero
== Sources ==
* : $\S 4.3$. Series
* : $\S 6.9$
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.1$
Category:Series
Category:Convergent Series
Terms of Bounded Sequence Within Bounds

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let the limit superior of $\sequence {x_n}$ be $\overline l$.
Let the limit inferior of $\sequence {x_n}$ be $\underline l$.
Then:
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$
== Proof ==
=== Upper Bound ===
First we show that:
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
 this proposition were to be false.
That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \ge \overline l + \epsilon$.
From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \ge \overline l + \epsilon$.
This would contradict the definition of $\overline l$.
=== Lower Bound ===
Next, in the same way, we show that:
: $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$
 this proposition were to be false.
That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \le \underline l + \epsilon$.
From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \le \underline l + \epsilon$.
This would contradict the definition of $\underline l$.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.15 \ (4)$
Category:Limits of Sequences
Test Function/Examples/Exponential of One over x Squared minus One

== Example of Test Function ==
Let $\phi : \R \to \R$ be a real function with support on $x \in \closedint {-1} 1$ such that:
:$\map \phi x = \begin {cases} \map \exp {\dfrac 1 {x^2 - 1} } & : \size x < 1 \\ 0 & : \size x \ge 1 \end {cases}$
Then $\phi$ is a test function.
== Proof ==
Consider a real function $f : \R \to \R$ such that:
:$\map f x = \begin {cases}
\map \exp {-\dfrac 1 x} & : x > 0 \\
0 & : x \le 0 
\end {cases}$
From Nth Derivative of Exponential of Minus One over x:
:$\dfrac {\d^n} {\d x^n} \map \exp {-\dfrac 1 x} = \dfrac {\map {P_n} x} {x^{2 n} } \map \exp {-\dfrac 1 x}$
where $\map {P_n} x$ is a real polynomial of degree $n$.
Then the can be rewritten in terms of at most $n + 1$ terms of the form $\dfrac 1 {x^m} \map \exp {-\dfrac 1 x}$ where $m \in \N$.
Let us take the limit $x \to 0$ from the right:
Therefore:
:$\ds \lim_{x \mathop \to 0^+} \frac {\map {P_n} x} {x^{2 n} } \map \exp {-\frac 1 x} = 0$
By construction:
:$\ds \map \phi x = \map f {1 - x^2} = \begin {cases}
\map \exp {-\dfrac 1 {1 - x^2} } & : 1 - x^2 > 0 \\
0 & : 1 - x^2 \le 0 
\end {cases}$
where:
 
 
Furthermore:
where $y = 1 - x^2$ and $\map {M_3} x$ is a real polynomial of degree $3$.
Similarly, any higher derivative of $\map \phi x$ will have $\map {M_k} x$ instead of $\map {P_n} x$ with $k \ge n$.
Thus:
:$\dfrac {\d^n} {\d x^n} \map \phi x = \dfrac {\map {M_k} x} {y^{2 n} } \map \exp {-\dfrac 1 y}$
Consequently:
Analogously:
:$\ds \lim_{x \mathop \to 1^-} \frac {\map {M_k} x} {y^{2 n} } \map \exp {-\frac 1 y} = 0$
Since outside of the support we have that $\map \phi x = 0$, the limit coming from outside is also $0$.
Therefore, $\map \phi x$ is smooth at the boundaries of its support.
Also:
:$\forall x \in \openint {-1} 1 : \map \phi x \in \map {C^\infty} \R$
By definition, $\phi$ is a test function.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Examples of Test Functions
Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space

== Theorem ==
Let $\map \DD {\R^d}$ be the test function space.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition of test functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of test functions over $\C$.
Then $\struct {\map \DD {\R^d}, +, \, \cdot \,}_\C$ is a vector space.
== Proof ==
Let $f, g, h \in \map \DD {\R^d}$ be test functions with the compact support $K$.
Let $\lambda, \mu \in \C$.
Let $\map 0 x$ be a real-valued function such that:
:$\map 0 x : \R^d \to 0$.
Let us use real number addition and multiplication.
$\forall x \in \R^d$ define pointwise addition as:
:$\map {\paren {f + g}} x := \map f x +_\C \map g x$.
Define pointwise scalar multiplication as:
:$\map {\paren {\lambda \cdot f}} x := \lambda \times_\C \map f x$
Let $\map {\paren {-f} } x := -\map f x$.
=== Closure Axiom ===
By Sum Rule for Continuous Complex Functions, $f + g \in \map \DD {\R^d}$
=== Commutativity Axiom ===
By Pointwise Addition on Complex-Valued Functions is Commutative, $f + g = g + f$
=== Associativity Axiom ===
By Pointwise Addition is Associative, $\paren {f + g} + h = f + \paren {g + h}$.
=== Identity Axiom ===
=== Inverse Axiom ===
=== Distributivity over Scalar Addition ===
=== Distributivity over Vector Addition ===
=== Associativity with Scalar Multiplication ===
=== Identity for Scalar Multiplication ===
== Also see ==
*Complex Vector Space is Vector Space
*Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Examples of Vector Spaces
Category:Functional Analysis
Test Function with Vanishing Partial Derivative

== Theorem ==
Let $\phi \in \map \DD {\R^2}$ be a test function such that:
:$\tuple {x, y} \stackrel \phi {\longrightarrow} \map \phi {x, y}$
Suppose $\phi$ is a solution to the following partial differential equation:
:$\dfrac {\partial \phi} {\partial x} = 0$
Then $\phi$ is identically $0$.
== Proof ==
$\dfrac {\partial \phi} {\partial x} = 0$ implies that:
:$\forall x \in \R : \map \phi {x, y} = \map C y$
where $C : \R \to \C$ is a complex-valued function.
By definition, $\phi$ is a test function.
Hence, $\phi$ must have a compact support $\Omega \subset \R^2$.
Let $\map {B^-_\epsilon} 0 \subset \R^2$ be a closed ball in Euclidean space such that:
:$\Omega \subseteq \map {B^-_\epsilon} 0$
By Closed Ball in Euclidean Space is Compact, $\map {B^-_\epsilon} 0$ is compact.
Then $\map {B^-_\epsilon} 0$ also qualifies as a compact support of $\phi$.
By definition of a test function:
:$\forall y \in \R : \size y > \epsilon \implies \map \phi {x, y} = 0$
But for each $y \in \R$ we have that $\map \phi {x, y} = \map C y$ is a constant.
By smoothness of test functions, this constant has to be the same for all $y \in \R$.
Hence, $\map \phi {x, y} = 0$.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Examples of Test Functions
Category:Partial Differentiation
Test for Ideal

== Theorem ==
Let $J$ be a subset of a ring $\struct {R, +, \circ}$.
Then $J$ is an ideal of $\struct {R, +, \circ}$ these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$
== Proof ==
=== Necessary Condition ===
Let $J$ be an ideal of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being an ideal.
=== Sufficient Condition ===
Suppose conditions $(1)$ to $(3)$ hold.
As $r \in J \implies r \in R$, if $(3)$ holds for $J$, then $J$ is closed under $\circ$ and condition $(3)$ of Subring Test holds.
Thus, $J$ is a subring of $R$.
As $(3)$ defines the condition for $J$, being a subring, to be an ideal, the result holds.
So $J$ is an ideal of $\struct {R, +, \circ}$.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $34$
* : $\S 2.2$: Homomorphisms: Definition $2.5$
* : $\S 58.4$ Ideals
Category:Ideal Theory
Test for Left Ideal

== Theorem ==
Let $J$ be a subset of a ring $\struct {R, +, \circ}$.
Then $J$ is an left ideal of $\struct {R, +, \circ}$ these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall j \in J, r \in R: r \circ j \in J$
== Proof ==
=== Necessary Condition ===
Let $J$ be a left ideal of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being a left ideal.
=== Sufficient Condition ===
Suppose conditions $(1)$ to $(3)$ hold.
Conditions $(1)$ and $(2)$ satisfy the criteria for the One-Step Subgroup Test, thus $J$ is a subgroup of $\struct {R, +}$.
As $(3)$ defines the condition for $J$ to be a left ideal, the result follows.
Category:Ideal Theory
Test for Right Ideal

== Theorem ==
Let $J$ be a subset of a ring $\struct {R, +, \circ}$.
Then $J$ is a right ideal of $\struct {R, +, \circ}$ these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall j \in J, r \in R: j \circ r \in J$
== Proof ==
=== Necessary Condition ===
Let $J$ be a right ideal of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being a right ideal.
=== Sufficient Condition ===
Suppose conditions $(1)$ to $(3)$ hold.
Conditions $(1)$ and $(2)$ satisfy the criteria for the One-Step Subgroup Test, thus $J$ is a subgroup of $\struct {R, +}$.
As $(3)$ defines the condition for $J$ to be a right ideal, the result follows.
Category:Ideal Theory
Test for Submonoid

== Theorem ==
To show that $\struct {T, \circ}$ is a submonoid of a monoid $\struct {S, \circ}$, we need to show that:
:$(1): \quad T \subseteq S$
:$(2): \quad \struct {T, \circ}$ is a magma (that is, that it is closed)
:$(3): \quad \struct {T, \circ}$ has an identity.
== Proof ==
From Subsemigroup Closure Test, $(1)$ and $(2)$ are sufficient to show that $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Demonstrating the presence of an identity is then sufficient to show that it is a monoid.
Category:Monoids
Tests for Finite Set Equality

== Theorem ==
Let $S$ and $T$ be finite sets.
Let $S \subseteq T$.
:$(a): \quad \card S = \card T$
:$(b): \quad \card {T \setminus S} = 0$
:$(c): \quad T \setminus S = \O$
:$(d): \quad T = S$
== Proof ==
We have:
Category:Set Equality
Category:Finite Sets
Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights

== Theorem ==
== Proof ==
:
Let $ABCG$ be a tetrahedron whose base is $\triangle ABC$ and whose apex is $G$.
Let $DEFH$ be a tetrahedron whose base is $\triangle DEF$ and whose apex is $H$.
Let $ABCG$ and $DEFH$ be equal in volume.
It is to be demonstrated that the areas of their bases is reciprocally proportional to their heights.
Let the parallelepipeds $BGML$ and $EHQP$ be completed.
We have that $ABCG$ is equal to $DEFH$.
From:
:
and
:
it follows that:
:$BGML = 6 \cdot ABCG$
and:
:$EHQP = 6 \cdot DEFH$
Therefore:
:$BGML = EHQP$
But from :
:$BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$
where:
:$BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively
:$h \left({EHQP}\right)$ denotes the height of $EHQP$
:$h \left({BGML}\right)$ denotes the height of $BGML$.
But from :
:$BM : EQ = \triangle ABC : \triangle DEF$
Therefore by :
:$\triangle ABC : \triangle DEF = h \left({EHQP}\right) : h \left({BGML}\right)$
But:
:$h \left({EHQP}\right) = h \left({DEFH}\right)$
:$h \left({BGML}\right) = h \left({ABCG}\right)$
where:
:$h \left({ABCG}\right)$ denotes the height of $ABCG$
:$h \left({DEFH}\right)$ denotes the height of $DEFH$.
Therefore:
:$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$
That is the bases of $ABCG$ and $DEFH$ are reciprocally proportional to their heights.
Let $ABCG$ and $DEFH$ be tetrahedra whose bases are reciprocally proportional to their heights:
:$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$
It is to be shown that $ABCG$ is equal to $DEFH$.
Let the parallelepipeds $BGML$ and $EHQP$ be completed.
We have that:
:$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$
while:
:$\triangle ABC : \triangle DEF = BM : EQ$
where $BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively.
Therefore by :
:$BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$
From :
:$BGML = EHQP$
But from:
:
and
:
it follows that:
:$BGML = 6 \cdot ABCG$
and:
:$EHQP = 6 \cdot DEFH$
Therefore:
:$ABCG = DEFH$
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Tetrahedral Number as Sum of Squares

== Theorem ==
:$H_n = \ds \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2$
where $H_n$ denotes the $n$th tetrahedral number.
=== Examples ===
== Proof ==
Let $n$ be even such that $n = 2 m$.
We have:
Let $n$ be odd such that $n = 2 m + 1$.
We have:
Category:Tetrahedral Numbers
Category:Square Numbers
Category:Tetrahedral Number as Sum of Squares
Tetrahedral Numbers which are Sum of 2 Tetrahedral Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$680$
:
:The Dictionary
::$680$
== Mistake ==
:$680$ is the smallest tetrahedral number to be the sum of two tetrahedral numbers: $680 = 120 + 560$.
== Correction ==
In fact it is the $2$nd.
The smallest is $20$:
:$20 = 10 + 10$
However, it would be correct to state this as:
:$680$ is the smallest tetrahedral number to be the sum of two distinct tetrahedral numbers: $680 = 120 + 560$.
== Sources ==
* : $680$
* : $680$
Category:Mistakes/Curious and Interesting Numbers
Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms

== Theorem ==
== Proof ==
:
Let $ABCD$ be a tetrahedron whose base is $ABC$ and whose apex is $D$.
It is to be demonstrated that $ABCD$ can be divided into two equal tetrahedra which are similar to $ABCD$ and two equal prisms which form more than half $ABCD$.
Let $AB, BC, CA, A, DB, DC$ be bisected at the points $E, F, G, H, K, L$.
Let $HE, EG, GH, HK, KL, LH, KF, FG$ be joined.
We have that:
:$AE = EB$
:$AH = DH$
From :
:$EH \parallel DB$
For the same reason:
:$HK \parallel AB$
Therefore $HEBK$ is a parallelogram.
Therefore from :
:$HK = EB$
But $EB = EA$.
Therefore:
:$AE = HK$
But also $AH = HD$.
Thus:
:$EA = KH$
:$AH = HD$
and:
:$\angle EAH = KHD$
Therefore from :
:$\triangle AEH = \triangle HKD$
For the same reason:
:$\triangle AHG = \triangle HLD$
From :
:$\angle EHG = \angle KDL$
We have:
:$EH = KD$
:$HG = DL$
and:
:$\angle EHG = KDL$
Therefore from :
:$\triangle EHG = \triangle KDL$
For the same reason:
:$\triangle AEG = \triangle HKL$
Therefore from :
:the pyramid of which $AEG$ is the base and $H$ the apex is equal and similar to the pyramid of which $HKL$ is the base and $D$ the apex.
We have that $HK$ is parallel to $AB$, which is one of the sides of $\triangle ADB$.
From :
:$\triangle ADB$ is equiangular with $\triangle DHK$
and their sides are proportional.
Therefore by :
:$\triangle ADB$ is similar to $\triangle DHK$.
For the same reason:
:$\triangle DBC$ is similar to $\triangle DKL$
and:
:$\triangle ADC$ is similar to $\triangle DLH$
From :
:$\angle BAC = \angle KHL$
We also have:
:$BA : AC = KH : HL$
and so:
:$\triangle ABC = \triangle HKL$
Therefore the pyramid of which $ABC$ is the base and $D$ the apex is similar to the pyramid of which $HKL$ is the base and $D$ the apex.
But we have that the pyramid of which $AEG$ is the base and $H$ the apex is equal and similar to the pyramid of which $HKL$ is the base and $D$ the apex.
Therefore each of the pyramids $AEGH$ and $HKLD$ is similar to the pyramid $ABCD$.
We have that $BF = FC$.
We also have that the parallelogram $EFGH$ is double the triangle $GFC$.
From :
:$(1): \quad$ the prism contained by $\triangle BKF$, $\triangle EHG$ and the three parallelograms $EBFG, EBKH, HKFG$ is equal to the prism contained by $\triangle GFC$, $\triangle HKL$ and the three parallelograms $KFCL, LCGH, HKFG$.
By joining the straight lines $EF$ and $EK$, it can be seen that the prism in which $EBFG$ is the "base" and the straight line $HK$ is opposite it is greater than pyramid $EBFK$.
But the pyramid $EBFK$ equals pyramid $AEHG$ as they are contained by equal and similar planes.
Thus each of the two prisms in $(1)$ above is greater than each of the pyramids $AEGH$ and $HKLD$.
Thus these two prisms together are greater than the two pyramids $AEGH$ and $HKLD$ together,
But the whole pyramid $ABCD$ consists of these two prisms and these two pyramids.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Category:Prisms
Thabit's Rule

== Theorem ==
Let $n$ be a positive integer such that:
are all prime.
Then:
:$\tuple {2^n a b, 2^n c}$
forms an amicable pair.
== Proof ==
Let $r = 2^n a b, s = 2^n c$.
Let $\map {\sigma_1} k$ denote the divisor sum of an integer $k$.
From Divisor Sum of Power of 2:
:$\map {\sigma_1} {2^n} = 2^{n + 1} - 1$
From Divisor Sum of Prime Number:
From Divisor Sum Function is Multiplicative:
and:
Thus it is seen that:
:$\map {\sigma_1} r = \map {\sigma_1} s$
Now we have:
and so it is seen that:
:$r + s = \map {\sigma_1} r = \map {\sigma_1} s$
Hence the result, by definition of amicable pair.
== Also see ==
* Definition:Thabit Pair
== Historical Note ==
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Thabit's Rule/Historical Note

== Historical Note on Thabit's Rule ==
Thabit's Rule has been discovered and rediscovered several times during the course of history.
The first person on record to discuss it was , in his .
The Thabit pair $\tuple {17 \,296, 18 \, 416}$ was discovered by , who may well have also rediscovered this rule.
In $1636$ it was rediscovered by , who also rediscovered the Thabit pair $\tuple {17 \,296, 18 \, 416}$.
At around the same time, also rediscovered it, and also rediscovered the Thabit pair $\tuple {9 \, 363 \, 584, 9 \, 437 \, 056}$.
These three remain the only known Thabit pairs.
== Sources ==
* : $220$
* : $220$
Category:Historical Notes
Thabit Pair/Examples

== Examples of Thabit Pairs ==
=== $220$ and $284$ ===
=== $17 \, 296$ and $18 \, 416$ ===
=== $9 \, 363 \, 584$ and $9 \, 437 \, 056$ ===
These are the only Thabit pairs known.
Category:Amicable Pairs
Thabit Pair/Examples/17,296-18,416

== Example of Thabit Pair ==
$17 \,296$ and $18 \, 416$ form a Thabit pair.
== Proof ==
Let $n = 4$.
Then we have:
Each of $a, b, c$ are prime.
Thus:
Hence by Thabit's Rule, $17 \, 296$ and $18 \, 416$ form a Thabit pair.
== Historical Note ==
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Thabit Pair/Examples/17,296-18,416/Historical Note

== Historical Note on Thabit pair $17 \,296$ and $18 \,416$ ==
Most of the literature on the subject states that the Thabit pair $\tuple {17 \,296, 18 \, 416}$ was discovered by .
However, as it is generated by Thabit's Rule for the accessibly low number $n = 4$, it is implausible to suppose that it had not in fact been discovered previously by himself.
Hence it would be more accurate to say that rediscovered' it.
In $1636$ it was again rediscovered, this time by , who also rediscovered Thabit's Rule.
== Sources ==
* : $220$
* : $220$
Category:Historical Notes
Thabit Pair/Examples/220-284

== Example of Thabit Pair ==
$220$ and $284$ form a Thabit pair.
== Proof ==
From Amicable Pair: $220$ and $284$:
:$\map {\sigma_1} {220} = \map {\sigma_1} {284} = 504 = 220 + 284$
and so by definition are amicable.
Let $n = 2$.
Then we have:
Each of $a, b, c$ are prime.
Thus:
and so $220$ and $284$ form a Thabit pair.
Category:Amicable Pairs
Thabit Pair/Examples/9,363,584-9,437,056

== Example of Thabit Pair ==
$9 \, 363 \, 584$ and $9 \, 437 \, 056$ form a Thabit pair.
== Proof ==
Let $n = 7$.
Then we have:
Each of $a, b, c$ are prime.
Thus:
Hence by Thabit's Rule, $9 \, 363 \, 584$ and $9 \, 437 \, 056$ form a Thabit pair.
== Historical Note ==
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Thabit Pair/Examples/9,363,584-9,437,056/Historical Note

== Historical Note on Thabit pair $9 \, 363 \, 584$ and $9 \, 437 \, 056$ ==
The Thabit pair $\tuple {9 \, 363 \, 584, 9 \, 437 \, 056}$ was originally discovered by , as a result of applying Thabit's Rule using $n = 7$.
It was rediscovered by , who also rediscovered Thabit's Rule.
== Sources ==
* : $220$
* : $220$
Category:Historical Notes