nyuuzyou/wikis huggingface dataset:
Property of Group Automorphism which Fixes Identity Only/Corollary 1

== Theorem ==
Let $G$ be a finite group whose identity is $e$.
Let $\phi: G \to G$ be a group automorphism.
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only fixed element of $\phi$ is $e$.
Then:
:$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$
== Proof ==
Let $\psi: G \to G$ be the mapping:
:$\forall x \in G: \map \psi x = x^{-1} \, \map \phi x$
From Property of Group Automorphism which Fixes Identity Only:
:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$
That is, $\psi$ is an injection.
From Injection from Finite Set to Itself is Surjection, it follows that $\psi$ is a surjection.
That is:
:$\forall x \in G: \exists g \in G: x = \map \psi g$
which is a restatement of the hypothesis.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $26$
Category:Property of Group Automorphism which Fixes Identity Only
Property of Group Automorphism which Fixes Identity Only/Corollary 2

== Corollary to Property of Group Automorphism which Fixes Identity Only ==
Let $G$ be a finite group whose identity is $e$.
Let $\phi: G \to G$ be a group automorphism.
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only fixed element of $\phi$ is $e$.
Let:
:$\phi^2 = I_G$
where $I_G$ denotes the identity mapping on $G$.
Then:
:$\forall g \in G: \map \phi g = g^{-1}$
== Proof ==
Let $g \in G$.
Then:
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $26$
Category:Property of Group Automorphism which Fixes Identity Only
Property of Group Automorphism which Fixes Identity Only/Corollary 3

== Corollary to Property of Group Automorphism which Fixes Identity Only ==
Let $G$ be a finite group whose identity is $e$.
Let $\phi: G \to G$ be a group automorphism.
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only fixed element of $\phi$ is $e$.
Let:
:$\phi^2 = I_G$
where $I_G$ denotes the identity mapping on $G$.
Then $G$ is an abelian group of odd order.
== Proof ==
Let $s, t \in G$.
Then:
 $G$ is of even order.
Then from Even Order Group has Order 2 Element:
But this contradicts the condition on $\phi$:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
Hence there is no such element of $G$ of order $2$.
Thus $G$ must be of odd order.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $26$
Category:Property of Group Automorphism which Fixes Identity Only
Property of Increasing Mapping on Ordinals

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $K_{II}$ denote the class of all limit ordinals.
Let $F$ be a mapping defined on $\On$ satisfying the following conditions:
Then:
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Exercise $5.4$
Category:Ordinals
Category:Increasing Mappings
Property of being Totally Ordered is of Finite Character

== Theorem ==
Let $P$ be the property of sets defined as:
:$\forall x: \map P x$ denotes that $x$ is totally ordered under a relation $\RR$.
Then $P$ is of finite character.
That is:
:$x$ is totally ordered under $\RR$
:
:every finite subset of $x$ is totally ordered under $\RR$.
== Proof ==
=== Sufficient Condition ===
Let $x$ be totally ordered under $\RR$.
Let $y \subseteq x$.
From Restriction of Total Ordering is Total Ordering it follows that $y$ is also a totally ordered under $\RR$.
This holds in particular if $y$ is a finite set.
Hence every finite subset of $x$ is totally ordered under $\RR$.
=== Necessary Condition ===
Let every finite subset of $x$ be totally ordered under $\RR$.
Let $a, b, c \in x$ be arbitrary.
As $\set a$ is totally ordered under $\RR$:
:$a \mathrel \RR a$
As $a$ is arbitrary, it follows that:
:$\forall a \in x: a \mathrel \RR a$
and so $\RR$ is reflexive on $x$.
As $\set {a, b}$ is totally ordered under $\RR$:
:$a \mathrel \RR b \land b \mathrel \RR a \implies a = b$
As $a$ and $b$ are arbitrary, it follows that:
:$\forall a, b \in x: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$
and so $\RR$ is antisymmetric on $x$.
As $\set {a, b, c}$ is totally ordered under $\RR$:
:$a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$
As $a$, $b$ and $c$ are arbitrary, it follows that:
:$\forall a, b, c \in x: a \mathrel \RR b \land b \mathrel \RR c \implies a = c$
and so $\RR$ is transitive on $x$.
As $\set {a, b}$ is totally ordered under $\RR$:
:$a \mathrel \RR b \lor b \mathrel \RR a$
As $a$ and $b$ are arbitrary, it follows that:
:$\forall a, b \in x: a \mathrel \RR b \lor b \mathrel \RR a$
and so $\RR$ is connected on $x$.
So $\RR$ is reflexive, antisymmetric, transitive and connected on $x$.
That is, $x$ is totally ordered under $\RR$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.7$
Category:Total Orderings
Category:Finite Character
Proportion is Equivalence Relation

== Theorem ==
Proportion is an equivalence relation.
== Proof ==
:Proportion is Reflexive: $\forall x \in \R: x \propto x$
:Proportion is Symmetric: $\forall x, y \in \R: x \propto y \implies y \propto x$
:Proportion is Transitive: $\forall x, y, z \in \R: x \propto y \land y \propto z \implies x \propto z$
The result follows from the definition of an equivalence relation.
Category:Proportion
Category:Examples of Equivalence Relations
Proportion is Reflexive

== Theorem ==
Proportion is a reflexive relation.
That is, every real variable is proportional to itself:
:$\forall x \in \R: x \propto x$
== Proof ==
Let $x$ be arbitrary.
Then $x = 1 \times x$.
The result follows from the definition of reflexivity and proportion.
Category:Proportion
Category:Examples of Reflexive Relations
Proportion is Symmetric

== Theorem ==
Proportion is a symmetric relation.
That is:
:$\forall x, y \in \R: x \propto y \implies y \propto x$
== Proof ==
Let $x, y$ be arbitrary.
Let $x$ be proportional to $y$:
:$x \propto y$
Then by definition:
The result follows from the definition of symmetry and proportion.
Category:Proportion
Category:Examples of Symmetric Relations
Proportion is Transitive

== Theorem ==
Proportion is a transitive relation.
That is:
:$\forall x, y, z \in \R: x \propto y \land y \propto z \implies x \propto z$
== Proof ==
Let $x, y, z$ be arbitrary.
Let $x$ be proportional to $y$ and $y$ to $z$:
:$x \propto y \land y \propto z$
Then by definition:
:$\exists j, k \ne 0: x = j \times y \land y = k \times z$
Substituting $k \times z$ for $y$:
:$x = \paren {j \times k} \times z$
so $j \times k$ is the desired constant of proportion.
The result follows from the definition of transitivity and proportion.
Category:Proportion
Category:Examples of Transitive Relations
Proportion of Numbers is Transitive

== Theorem ==
== Proof ==
Let there be as many (natural) numbers as we please, $A, B, C$, and others equal to them in multitude, $D, E, F$, which taken two and two are in the same ratio, so that:
:$A : B = D : E$
:$B : C = E : F$
We need to show that $A : C = D : F$.
:
We have that $A : B = D : E$.
So from , it follows that $A : D = B : E$.
Similarly, we have $B : C = E : F$.
So again from , it follows that $B : E = C : F$.
Putting them together, we get $A : D = C : F$.
Finally, again from , it follows that $A : C = D : F$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Proportion of Power

== Theorem ==
Let $x$ and $y$ be proportional.
Let $n \in \Z$. 
Then $x^n \propto y^n$.
== Proof ==
Let $x \propto y$.
Then $\exists k \ne 0: x = k \times y$ by the definition of proportion.
Raising both sides of this equation to the $n$th power:
so $k^n$ is the desired constant of proportion.
The result follows from the definition of proportion.
Category:Proportion
Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms

== Theorem ==
=== Lemma ===
== Proof ==
:
Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$.
From :
:let each of $ABCG$ and $DEFH$ be divided into two equal tetrahedra which are similar to the whole and two equal prisms.
It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of all the prisms in $ABCG$ to all the prisms in $DEFH$.
We have that:
:$BO = OC$
and:
:$AL = LC$
Therefore:
:$LO \parallel AB$
and $\triangle ABC$ is similar to $\triangle LOC$.
For the same reason, $\triangle DEF$ is similar to $\triangle RVF$.
We have that:
:$BC = 2 \cdot CO$
and:
:$EF = 2 \cdot FV$
Therefore:
:$BC : CO = EF : FV$
So from :
:$\triangle LOC : \triangle ABC = \triangle DEF : \triangle RFV$
Therefore from :
:$\triangle ABC : \triangle DEF = \triangle LOC : \triangle RFV$
But from :
:the ratio of $\triangle LOC$ to $\triangle RFV$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
Therefore the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
But from :
:the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite equals the ratio of the prism of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite.
Therefore, by :
:the ratio of the two prisms of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite and in which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite are to the two prisms of which the the parallelogram to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite and in which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
Similarly from :
:the two tetrahedra $PMNG$ and $STUH$ can be divided into two equal tetrahedra which are similar and two equal prisms.
As the ratio of the base $PMN$ is to the base $STU$, so will the ratio of the two prisms in the tetrahedron $PMNG$ be to the two prisms in the tetrahedron $STUH$.
But:
:$\triangle PMN : \triangle STU = \triangle LOC : \triangle RFV$
Therefore:
:$\triangle ABC : \triangle DEF$ equals the ratio of the four prisms in the tetrahedron $ABCG$ to the four prisms in the tetrahedron $DEFH$.
Similarly, if the remaining tetrahedra are divided into two tetrahedra and two prisms, then, as $\triangle ABC$ is to $\triangle DEF$, so will all the prisms in $ABCG$ be to all the prisms in $DEFH$.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Category:Prisms
Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms/Lemma

== Lemma to Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms ==
:
== Proof ==
Let perpendiculars be drawn from $G$ and $H$ to the bases $\triangle ABC$ and $\triangle DEF$.
These are equal because, by definition, the heights of $ABCG$ and $DEFH$ are equal.
We have that the straight lines $GC$ and the perpendicular through $G$ are cut by the parallel planes $ABC$ and $PMN$.
So by :
:they will be cut in the same ratios.
We have that $GC$ is bisected by the plane $PMN$ at $N$.
Therefore the perpendicular through $G$ will also be bisected by the plane $PMN$.
For the same reason, the perpendicular through $H$ will be bisected by the plane $STU$.
We have that the perpendicular through $G$ and $H$ are equal.
Thefore the perpendiculars from $\triangle PMN$ and $\triangle STU$ to $\triangle ABC$ and $\triangle DEF$ are also equal.
Therefore the prisms in which the $\triangle LOC$ and $\triangle RVF$ are bases, and $\triangle PMN$ and $\triangle STU$ are opposites are of equal height.
So from :
:the parallelepipeds formed from these prisms are of equal height.
Therefore their halves, that is, the prisms described, are in the same ratio as $\triangle LOC$ is to $\triangle RVF$.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Category:Prisms
Proportional Magnitudes are Proportional Alternately

== Theorem ==
That is:
:$a : b = c : d \implies a : c = b : d$
== Proof ==
Let $A, B, C, D$ be four proportional magnitudes, so that as $A$ is to $B$, then so is $C$ to $D$.
We need to show that as $A$ is to $C$, then $B$ is to $D$.
:
Let equimultiples $E, F$ be taken of $A, B$.
Let other arbitrary equimultiples $G, H$ be taken of $C, D$.
We have that $E$ is the same multiple of $A$ that $F$ is of $B$.
So from Ratio Equals its Multiples we have that $A : B = E : F$
But $A : B = C : D$.
So from Equality of Ratios is Transitive it follows that $C : D = E : F$.
Similarly, we have that $G, H$ are equimultiples of $C, D$.
So from Ratio Equals its Multiples we have that $C : D = G : H$
So from Equality of Ratios is Transitive it follows that $E : F = G : H$.
But from Relative Sizes of Components of Ratios:
:$E > G \implies F > H$
:$E = G \implies F = H$
:$E < G \implies F < H$
Now:
:$E, F$ are equimultiples of $A, B$
:$G, H$ are equimultiples of $C, D$.
Therefore from :
:$A : C = B : D$
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Proportional Magnitudes have Proportional Remainders

== Theorem ==
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$.
=== Porism ===
== Proof ==
As the whole $AB$ is to the whole $CD$, so let the part $AE$ subtracted be to the part $CF$ subtracted.
That is:
:$AB : CD = AE : CF$
We need to show that $EB : FD = AB : CD$.
:
We have that :$AB : CD = AE : CF$.
So from Proportional Magnitudes are Proportional Alternately we have that $BA : AE = DC : CF$.
From Magnitudes Proportional Compounded are Proportional Separated we have that $BA : EA = DF : CF$.
From Proportional Magnitudes are Proportional Alternately: $BE : DF = EA : FC$.
But by hypothesis $AE : CF = AB : CD$.
So by Equality of Ratios is Transitive $EB : FD = AB : CD$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Proportional Magnitudes have Proportional Remainders/Porism

== Porism to Proportional Magnitudes have Proportional Remainders ==
== Proof ==
:
From :
: $EB : FD = AB : CD$
From :
: $AB : BE = CD : FD$
Therefore magnitudes compounded are proportional.
But it was also proved that:
:$BA : AE = DC : CF$
and this is convertendo.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Proportional Numbers are Proportional Alternately

== Theorem ==
== Proof ==
Let the four (natural) numbers $A, B, C, D$ be proportional so that $A : B = C : D$.
We need to show that $A : C = B : D$.
:
We have that $A : B = C : D$.
So from we have that whatever aliquot part or aliquant part $A$ is of $B$, the same aliquot part or aliquant part is $C$ of $D$.
So from , whatever aliquot part or aliquant part $A$ is of $C$, the same aliquot part or aliquant part is $B$ of $D$.
Therefore from $A : C = B : D$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Proportional Numbers have Proportional Differences

== Theorem ==
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$.
== Proof ==
As the whole $AB$ is to the whole $CD$, so let the $AE$ subtracted be to $CF$ subtracted.
We need to show that $EB : FD = AB : CD$.
:
We have that :$AB : CD = AE : CF$.
So from we have that whatever aliquot part or aliquant part $AB$ is of $CD$, the same aliquot part or aliquant part is $AE$ of $CF$.
So from:
: 
and:
: 
$EB$ is the same aliquot part or aliquant part of $FD$ that $AB$ is of $CD$.
So by $EB : FD = AB : CD$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Propositional Tautology is Tautology in Predicate Logic

== Theorem ==
Let $\LL_1$ be the language of predicate logic.
Let $\LL_0$ be the language of propositional logic.
Let the basic WFFs of $\LL_1$ be the vocabulary of $\LL_0$.
Let $\mathbf A$ be a $\mathrm{BI}$-tautology of $\LL_0$, by considering the basic subformulas of $\mathbf A$ as part of the vocabulary of $\LL_0$.
Then $\mathbf A$ is a $\mathrm{PL_A}$-tautology of $\LL_1$.
== Proof ==
We proceed by the Principle of Structural Induction on the bottom-up specification of $\LL_1$, applied to $\mathbf A$.
Define $\map {v_F} {\mathbf A} = F$ for all basic WFFs $\mathbf A$ as a boolean interpretation for $\LL_0$.
Consider the case $\mathbf A$ is formed by either $\paren{ \mathbf W ~ \PP_n }$ or $\paren{ \mathbf W ~ \forall\exists }$.
Then $\mathbf A$ is basic, and therefore:
:$\map {v_F} {\mathbf A} = F$
meaning that $\mathbf A$ is not a $\mathrm{BI}$-tautology.
Consider next the case $\mathbf A$ is formed by either $\paren{ \mathbf W ~ \neg }$ or $\paren{ \mathbf W ~ \lor, \land, \Rightarrow, \Leftrightarrow }$.
That is, $\mathbf A = \neg \mathbf B$ or $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ being one of $\lor, \land, \Rightarrow, \Leftrightarrow$.
Suppose $\mathbf A$ is not a $\mathrm{PL_A}$-tautology.
Then there are a structure $\AA$ and an assignment $\sigma$ such that:
:$\AA, \sigma \not\models \mathbf A$
Then define the boolean interpretation $v_{\AA, \sigma}$ on basic WFFs $\mathbf B$ by:
:$\map {v_{\AA, \sigma} } {\mathbf B} = T$ $\AA, \sigma \models \mathbf B$
which is to say:
:$\map {v_{\AA, \sigma} } {\mathbf B} = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma$
where $\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma$ is the value of $\mathbf A$ in $\AA$ under $\sigma$.
Hence, by definition of boolean interpretation:
And similarly, for $\circ$ being one of $\lor, \land, \Rightarrow, \Leftrightarrow$:
Since by hypothesis $\AA, \sigma \not\models \mathbf A$, it follows that:
:$\map {v_{\AA, \sigma} } {\mathbf A} = F$
and $\mathbf A$ is not a $\mathrm{BI}$-tautology.
The induction case for $\paren{ \mathbf W ~ \neg }$ and $\paren{ \mathbf W ~ \lor, \land, \Rightarrow, \Leftrightarrow }$ follows by transposition.
Hence the result by the Principle of Structural Induction.
== Sources ==
* : $\text{II}.9$ Tautologies: Exercise $\text{II}.9.3$
Category:Boolean Interpretations
Category:Model Theory for Predicate Logic
Propositiones ad Acuendos Juvenes/Problems

== Problems from by ==
=== $1$: De Limace ===
=== $2$: De Viro Ambulante in Via ===
=== $3$: De Duobus Proficistentibus Visis Ciconiis ===
=== $4$: De Homine et Equis in Campo Pascentibus ===
=== $5$: De Emptore in $\text C$ Denariis ===
=== $6$: De Duobis Negotiatoribus $\text C$ Solidos Communis Habentibus ===
=== $7$: De Disco Pesante Libras $\text {XXX}$ ===
=== $8$: De Cupa ===
=== $9$: De Sago ===
=== $10$: De Linteo ===
=== $11$: De Duobus Hominibus Singulas Sorores Accipientibus ===
=== $12$: De Quodam Patrefamilias et Tribus Filiis Eius ===
=== $13$: De Rege et de Eius Exercitu ===
=== $14$: De Bove ===
=== $15$: De Homine ===
=== $16$: De Duobus Hominibus Boves Ducentibus ===
=== $17$: De Tribus Fratribus Singulas Habentibus Sorores ===
=== $18$: De Lupo et Capra et Fasciculo Cauli ===
=== $19$: De Viro et Muliere Ponderantibus Plaustrum ===
=== $20$: De Ericiis ===
=== $21$: De Campo et Ovibus in Eo Locandis ===
=== $22$: De Campo Fastigioso ===
=== $23$: De Campo Quadrangulo ===
=== $24$: De Campo Triangulo ===
=== $25$: De Campo Rotundo ===
=== $26$: De Campo et Cursu Canis ac Fugo Leporis ===
=== $27$: De Civitate Quadrangula ===
=== $28$: De Civitate Triangula ===
=== $29$: De Civitate Rotunda ===
=== $30$: De Basilica ===
=== $31$: De Canava ===
=== $32$: De Quodam Patrefamilias Distribuente Annonam ===
=== $33$: De Alio Patrefamilias Erogante Suae Familiae Annonam ===
=== $34$: Altera de Patrefamilias Partiente Familiae Suae Annonam ===
=== $35$: De Obitu Cuiusdam Patrisfamilias ===
=== $36$: De Salutatione Cuiusdam Senis ad Puerum ===
=== $37$: De Quodam Homine Volenti Aedificare Domum ===
=== $38$: De Quodam Emptore in Animalibus Centum ===
=== $39$: De Quodam Emptore in Oriente ===
=== $40$: De Homine et Ovibus in Monte Pascentibus ===
=== $41$: De Sode et Scrofa ===
=== $42$: De Scala Habente Gradus Centum ===
=== $43$: De Porcis ===
=== $44$: De Salutatione Pueri ad Patrem ===
=== $45$: De Columba ===
=== $46$: De Sacculo ab Homine Invento ===
=== $47$: De Episcopo qui Jussit $\text {XII}$ Panes in Clero Dividi ===
=== $48$: De Homine qui Obviavit Scolaris ===
=== $49$: De Carpentariis ===
=== $50$: De Vino in Vasculum ===
=== $51$: De Vino in Vasculis a Quodam Patre Distributo ===
=== $52$: De Homine Patrefamilias ===
=== $53$: De Homine Patrefamilias Monasterii $\text {XII}$ Monachorum ===
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/10 - De Linteo

== by : Problem $10$ ==
;De Linteo: A Linen Problem
:I have a piece of linen, $60$ cubits long and $40$ cubits wide.
:I want to cut it into portions, each of which will be $6$ cubits long and $4$ wide.
:How many portions can be made of it?
== Solution ==
:$100$.
== Proof ==
We have that:
:$6$ goes into $60$ a total of $10$ times
:$4$ goes into $40$ also a total of $10$ times.
Thus the cloth can be cut into $10 \times 10 = 100$ pieces that are $6$ cubits long and $4$ cubits wide.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/11 - De Duobus Hominibus Singulas Sorores Accipientibus

== by : Problem $11$ ==
;De Duobus Hominibus Singulas Sorores Accipientibus: Two men marrying each other's sister
:If two men take the other's sister in marriage,
:what is the relationship between their sons?
=== Variant 1 ===
=== Variant 2 ===
== Solution ==
Cousins, twice over.
:$(1): \quad$ The sister of the father is the aunt, who is the mother of the cousin
and:
:$(2): \quad$ The brother of the mother is the uncle, who is the father of the cousin.
== Historical Note ==
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $79$
Category:Propositiones ad Acuendos Juvenes
Category:Strange Family Problems
Propositiones ad Acuendos Juvenes/Problems/11 - De Duobus Hominibus Singulas Sorores Accipientibus/Historical Note

== Historical Note on by : Problem $11$: Variant $2$ ==
According to , this problem, along with two related problems (no pun intended) originating from which he also documents, are the earliest of the strange family type.
There are contemporary examples of such:
:At 52, Rolling Stones bassist Bill Wyman married 18-year-old Mandy Smith but divorced after a year.
: Bill's 30-year-old son then, Stephen married Mandy's mother, age 46.
: Therefore, Bill became the father-in-law of his ex mother-in-law
::as well as the stepgrandfather of his former wife.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/11 - De Duobus Hominibus Singulas Sorores Accipientibus/Variant 1

== by : Problem $\text {11 a}$ ==
;De Duobus Hominibus Singulas Matres Accipientibus: Two men marrying each other's mother
:If two men take the other's mother in marriage,
:what is the relationship between their sons?
== Solution ==
Each son is simultaneously the nephew and the uncle of the other.
Let the people involved be:
:man $1$ and mother $1$
:man $2$ and mother $2$
:Let son $1$ be the son of man $1$ and mother $2$.
:Let son $2$ be the son of man $2$ and mother $1$.
Then:
:$(1): \quad$ Son $1$ and man $2$ are both the sons of mother $2$, which makes son $1$ the uncle of son $2$
and at the same time:
:$(2): \quad$ Son $2$ and man $1$ are both the sons of mother $1$, which makes son $1$ the nephew of son $2$
and vice versa.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Strange Family Problems