hcju/proofwikiqa huggingface dataset:
Let $T_A$ be a Lindelöf space. Take an open cover $\mathcal U$ of $T_B$. From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \mathcal U}$ is a cover of $S_A$. But $\phi$ is continuous, and for all $U \in \mathcal U$, $U$ is open in $T_B$. It follows that $\forall U \in \mathcal U: \phi^{-1} \sqbrk U$ is open in $T_A$. So $\set {\phi^{-1} \sqbrk U: U \in \mathcal U}$ is an open cover of $T_A$. $T_A$ is Lindelöf , so we take a countable subcover: :$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n}, \ldots}$ We have that $\phi$ is surjective. So from Surjection iff Right Inverse: :$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$ So: :$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} }, \ldots} = \set {U_1, \ldots, U_n, \ldots} \subseteq \mathcal U$ is a countable subcover of $\mathcal U$ on $T_B$. {{qed}}
Let $\phi$ be surjective, continuous and open. Let $T_A$ be first countable. Let $b \in S_B$. Since $\phi$ is surjective there is a point $a \in S_A$ such that: :$\map \phi a = b$ From the first-countability of $T_A$, there is a local base $\BB$, say, of $a$ which is countable. Let $\BB = \set {V_n: n \in \N}$. We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$. Let $U$ be an open set of $T_B$ that contains $b$. As $b = \map \phi a$ we have that: :$a \in \phi^{-1} \sqbrk U$ From the continuity of $\phi$, we have that $\phi^{-1} \sqbrk U$ is open. As $\BB$ is a local base, there is an open set $V_n \subseteq \phi^{-1} \sqbrk U$ such that $a \in V_n$. $\phi$ is surjective, so from Surjection iff Right Inverse we have that: :$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$ So, applying $\phi$ to $V_n$, from Image of Subset under Relation is Subset of Image: Corollary 2 we obtain: :$\phi \sqbrk {V_n} \subseteq U$ such that $b \in \phi \sqbrk {V_n}$. This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$. Thus, $T_B$ is first countable. {{qed}}
Let $\phi$ be surjective, continuous and open. Let $T_A$ be second-countable. By definition of second-countability $T_A$ has a countable basis, $\BB$, say. Let $\BB = \set {V_n: n \in \N}$. We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$. Let $U$ be an open set of $T_B$. $\phi$ is continuous, so $\phi^{-1} \sqbrk U$ is open in $T_A$. As $\BB$ is a base for $T_A$, there exists an open set $V_n \subseteq \phi^{-1} \sqbrk U$. $\phi$ is surjective, so from Surjection iff Right Inverse we have that: :$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$ So, applying $\phi$ to $V_n$, from Image of Subset under Relation is Subset of Image: Corollary 2 we obtain: :$\phi \sqbrk {V_n} \subseteq U$. This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$. Thus, $T_B$ is second-countable. {{qed}}
From the definition of the Arens-Fort space: $S$ is the the Cartesian product $\Z_{\ge 0} \times \Z_{\ge 0}$ of the set of all positive integers: : $S = \left\{{0, 1, 2, \ldots}\right\} \times \left\{{0, 1, 2, \ldots}\right\}$ We have by definition that $\Z_{\ge 0}$ is countable. From Cartesian Product of Countable Sets is Countable it follows that $S = \Z_{\ge 0} \times \Z_{\ge 0}$ is likewise countable. {{qed}} Category:Arens-Fort Space dw9gl2mqfqzwale79sg8xyunyij39xz
Assume that $T$ is connected. Let $H \subset S$ be a clopen set such that $\O \ne H \ne S$. Then $H$ and $\relcomp S H$ are open sets whose union is $S$. Thus $\set {H \mid \relcomp S H}$ form a partition of $S$. By definition, $T$ is not connected, which is a contradiction. Then such $H$ does not exist. Assume that there exists no proper subset of $S$ which is clopen in $T$. Suppose $T$ is not connected (disconnected). By definition, there is a partition $\set {A \mid B}$ of $S$. Then $\relcomp S A = B$ is open, and $A$ is closed. Thus $A$ is clopen and $\O \ne A \ne S$, which is a contradiction. Finally, $T$ must be connected. {{qed}} Category:Connected Spaces 41enp8q0h5bhoaned763mj5cq7x7gbz
If $p$ is irreducible, it is proven. Suppose $p$ is not irreducible. Then $p = r_1 r_2$ where neither $r_1$ nor $r_2$ are units. If $r_1$ and $r_2$ are irreducible, then the proof is complete. If $r_1$ is not irreducible, then $r_1 = r_{11} r_{12}$ where neither $r_{11}$ nor $r_{12}$ are units. If we continue, we get a proper inclusion of ideals $\ideal r \subset \ideal {r_1} \subset \ideal {r_{11} } \subset \cdots \subset R$. If this process finishes in a finite number of steps, the proof is complete. {{AimForCont}} the process does ''not'' complete in a finite number of steps. Thus the chain $\ideal r \subset \ideal {r_1} \subset \ideal {r_{11} } \subset \cdots \subset R$ is infinite. From Principal Ideal Domain fulfills Ascending Chain Condition, this cannot happen. Thus, the process ends in a finite number of steps. {{qed}} Category:Factorization Category:Ideal Theory Category:Principal Ideal Domains 05ldswkzlatcaniyeaqrq59aonytg7i
Let $I_1\subseteq I_2\subseteq I_3\subseteq \ldots$ be an ascending chain of ideals. Build $\displaystyle I = \bigcup_{i \mathop = 1}^\infty I_i$.  $I$ is an ideal. Since $R$ is a principal ideal domain, $I = \left({a}\right)$ for some $a \in R$. Now, since $a \in I$, there is some $n$ such that $a \in I_n$. Thus $\left({a}\right) \subseteq I_n$. By definition $I_n \subset I = \left({a}\right)$, and so $I_n = I$. Thus: : $\forall m \ge n: I_m = I$ {{MissingLinks|Still some links to be made, I'll get to it later.}} {{qed}} {{Proofread}} Category:Ideal Theory Category:Principal Ideal Domains 74ixkrt5agvvk4h1rgv6paap3v38g51
Let $z \in X$. Let $i \in i$. Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$. Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.  For all $j \in I$ let: ::$Z_j = \begin{cases} X_i & i = j \\ \set{z_j} & j \ne i \end{cases}$ === Lemma 1 === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 1/Lemma 1}}{{qed|lemma}} From Product Space of Subspaces is Subspace of Product Space , $\struct {Y_i, \upsilon_i}$ is a product space. Consider the projection: :$p_i: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$ From Projection from Product of Family is Injection iff Other Factors are Singletons, $p_i$ is injective. From Projection from Product of Family is Surjective, $p_i$ is surjective.   From  Projection from Product Topology is Continuous:General Result, $p_i$ is continuous. From  Projection from Product Topology is Open:General Result, $p_i$ is open. Thus, by definition, we have that $p_i$ is a homeomorphism. Consider the projection: :$\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$ and the restriction: :$\pr_i {\restriction_{Y_i} }: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$ === Lemma 2 === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 1/Lemma 2}}{{qed|lemma}} Thus $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism. {{qed}}
Let $z \in X$. Let $i \in I$.  Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$. Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.  Let $p_i = \pr_i {\restriction_{Y_i}}$. Note that by definitions of a restriction and a projection then: :$\forall y \in Y_i: \map {p_i} y = y_i$  === $p_i$ is an injection === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 2/Injection}}{{qed|lemma}} === $p_i$ is a surjection === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 2/Surjection}}{{qed|lemma}} === $p_i$ is a continuous mapping === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 2/Continuous Mapping}}{{qed|lemma}} === $p_i$ is an open mapping === {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 2/Open Mapping}}{{qed}}
By definition of Fort space, $H$ is finite or contains $p$. Consider the set of sets defined by: :$\DD = \set {S \setminus \set z: z \notin H}$ Because $H$ is finite, $S \setminus H$ is countably infinite. From its method of construction, $\DD$ has the same cardinality as $S \setminus H$ and so is countably infinite. We have that the elements of $\DD$ are all open sets. Thus, by definition, any set of the form $\displaystyle H \subseteq \bigcap_{V \mathop \in \DD} V$ is a $G_\delta$ set. Let $r \in H$. Then: :$\forall z \in S: z \notin H: r \in S \setminus \set z$ Thus by definition of $\DD$: :$\displaystyle H \subseteq \bigcap_{V \mathop \in \DD} V$ Let $r \notin H$. Then: :$\exists z \in S: r \notin S \setminus \set z \in \DD$ So: :$\displaystyle \relcomp S H \subseteq \relcomp S {\bigcap_{V \mathop \in \DD} V}$ and so: :$\displaystyle H = \bigcap_{V \mathop \in \DD} V$ Hence the result. {{qed}} Category:Fort Spaces Category:G-Delta Sets tsr9midx3m0flbt3f7ui7hyy9fwln3y
This is a special case of Integral Closure is Subring. We have an extension of commutative rings with unity, $\Z \subseteq K$, and $\mathbb A$ is the integral closure of $\Z$ in $K$. The theorem says that $\mathbb A$ is a subring of $K$. {{qed}} Category:Algebraic Number Theory pqgxmkfti5damhqprvs3ge08y1xdjwg
Let $AB$ be that aliquot part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted. We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted. :500px Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$. Then from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}}, whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $GF$. But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$, by hypothesis. Therefore, whatever aliquot part $AB$ is of $GF$, the same aliquot part is it of $CD$ also. Therefore $GF = CD$. Let $CF$ be subtracted from each. Therefore $GC = FD$. We have that whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $CG$ Therefore whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $FD$. But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$. Therefore the remainder $EB$ is the same aliquot part of the remainder $FD$ that the whole $AB$ is of the whole $CD$. {{qed}}
A direct application of the Distributive Property: {{begin-eqn}} {{eqn | l = \frac 1 n b - \frac 1 n d       | r = \frac 1 n b + \frac 1 n \paren {-d}       | c =  }} {{eqn | r = \frac 1 n \paren {b + \paren {-d} }       | c =  }} {{eqn | r = \frac 1 n \paren {b - d}       | c =  }} {{end-eqn}} {{qed}}
Let $f$ be a minimal polynomial of $\alpha$ over $K$. By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$.  Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$. By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\deg$ denotes degree. By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \sqbrk x$ such that: :$g = q f + r$ and $\map \deg r < \map \deg f$. Suppose $\map \deg r > 0$. Then evaluating both sides of the equation above at $\alpha$, we obtain $\map r \alpha = 0$. This contradicts the minimality of the degree of $f$. Thus, $r$ is constant and equal to $0$. We have now shown that $f$ divides all polynomials in $K \sqbrk x$ which vanish at $\alpha$. By the monic restriction, it then follows that $f$ is unique. {{qed}} Category:Minimal Polynomials ov202rzdlzmtxpay7d6n1y9di5oau8u
Let $\map f x$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$. From Minimal Polynomial is Unique we have that $\map f x$ is unique. Since $K$ is a field, we may assume that the coefficient of $x^n$ is $1$. {{AimForCont}} that $f$ is not irreducible. Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$. By definition of $f$ as the minimal polynomial in $\alpha$: :$0 = \map f \alpha = \map g \alpha \, \map h \alpha$ Since $L$ is a field, it is an integral domain. Therefore, as $\map g \alpha \, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$. This contradicts the minimality of the degree of $f$. Hence the result, by Proof by Contradiction. {{qed}}
=== Definition $1$ implies Definition $2$ === Let $\alpha \in L$ be an arbitrary element. Let $\sigma: L \mapsto \overline K$ be an arbitrary embedding of $L$ fixing $K$. We wish to show that $\sigma \left({\alpha}\right)\in L$. Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$, which exists because $L / K$ is algebraic. Since $\sigma$ fixes $K$, $\sigma \left({\alpha}\right)$ must also be a root of $m_\alpha$. By our assumption, $\alpha \in L$ implies that all roots of $m_\alpha$ are in $L$ and consequently $\sigma \left({\alpha}\right) \in L$. {{qed|lemma}} === Definition $2$ implies Definition $1$ === Again, let $\alpha \in L$ and let $m_\alpha \in K \left[{x}\right]$ be its minimal polynomial over $K$. We must show that for every root $\beta$ of $m_\alpha$, there exists an embedding $\sigma_\beta$, of $L$ in $\overline K$ such that $\sigma_\beta \left({\alpha}\right) = \beta$.  Consider the intermediate field $K \left[{\alpha}\right]\subset L$. By Abstract Model of Algebraic Extensions, we have an automorphism $\tau_\beta$ for each root $\beta$ of $m_\alpha$ such that $\tau_\beta \left({\alpha}\right) = \beta$ and $\tau_\beta$ fixes $K$ By Extension of Isomorphisms, each $\tau_\beta$ can be extended to an embedding $\sigma_\beta$ of $L$ in $\overline K$ such that: : $\sigma_\beta \restriction_{K \left[{\alpha}\right]} = \tau_\beta$ By our assumption, $\sigma_\beta \left({L}\right) = L$ for each $\beta$. Consequently, every root of $m_\alpha$ is in $L$. {{qed}} Category:Field Extensions 8s4tqfzy86n9r9bm2wj6knflf2k473r
Let $\phi: K[x] \to K [\alpha]$ be the homomorphism which fixes $K$ and maps $x$ to $\alpha$. The kernel of $\phi$ consists of all polynomials which vanish at $\alpha$, which is precisely the ideal $\langle m_\alpha\rangle$. The result follows by the First Fundamental Theorem on Ring Homomorphisms. {{qed}} Category:Field Extensions av97dw9oef35ouryrxbxz5e0nxfkjas
Let $U \in \tau$ be any open set of $T$. Consider any $x \in U$. Then, by definition of local basis, there exists a $B_x \in \mathcal B_x$ such that $B_x \subseteq U$. Also by definition of local basis, $x \in B_x$. So by Set Union Preserves Subsets: : $U = \displaystyle \bigcup_{x \mathop \in U} \left\{{x}\right\} \subseteq \bigcup_{x \mathop \in U} B_x$ Also, since $B_x \subseteq U$ for all $x$, by Union is Smallest Superset: General Result: : $\displaystyle \bigcup_{x \mathop \in U} B_x \subseteq U$ Thus by definition of set equality: : $\displaystyle U = \bigcup_{x \mathop \in U} B_x$ So, by definition, $\displaystyle \mathcal B = \bigcup_{x \mathop \in X} \mathcal B_x$ is a basis for the topology $\tau$. {{qed}}
Let the (natural) number $AB$ be the same aliquant part of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted. We need to show that $EB$ is also the same aliquant part of the remainder $FD$ that the whole $AB$ is of the whole $CD$. :400px Let $GH = AB$. Then whatever aliquant part $GH$ is of $CD$, the same aliquant part also is $AE$ of $CF$. Let $GH$ be divided into the aliquant parts of $CD$, namely $GK + KH$, and $AE$ into the aliquant parts of $CF$, namely $AL + LE$. Thus the multitude of $GK, KH$ will be equal to the multitude of $AL, LE$. We have that whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $AL$ of $CF$. We also have that $CD > CF$. Therefore $GK > AL$. Now let $GM = AL$. Then whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $GM$ of $CF$. Therefore from Subtraction of Divisors obeys Distributive Law the remainder $MK$ is of the same aliquot part of the remainder $FD$ that the whole $GK$ is of the whole $CD$. Again, we have that whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $EL$ of $CF$. We also have that $CD > CF$. Therefore $HK > EL$. Let $KN = EL$. Then whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $KN$ of $CF$. Therefore from Subtraction of Divisors obeys Distributive Law the remainder $NH$ is of the same aliquot part of the remainder $FD$ that the whole $KH$ is of the whole $CD$. But $MK$ was proved to be the same aliquot part of $FD$ that $GK$ is of $CD$. Therefore $MK + NH$ is the same aliquant part of $DF$ that $HG$ is of $CD$. But $MK + NH = EB$ and $HG = BA$. Therefore $EB$ is the same aliquant part of $FD$ that $AB$ is of $CD$. {{Qed}}
A direct application of the Distributive Property: {{begin-eqn}} {{eqn | l = \frac m n b - \frac m n d       | r = \frac m n b + \frac m n \left({-d}\right)       | c =  }} {{eqn | r = \frac m n \left({b + \left({-d}\right)}\right)       | c =  }} {{eqn | r = \frac m n \left({b - d}\right)       | c =  }} {{end-eqn}} {{qed}}
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$, and another $D$ be the same aliquot part of another, $EF$, that $A$ is of $BC$. We need to show that whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$. :200px We have that whatever aliquot part $A$ is of $BC$, the same aliquot part $D$ is of $EF$. Therefore, as many numbers as there are in $BC$ equal to $A$, so many also are there in $EF$ equal to $D$. Let $BC$ be divided into the numbers equal to $A$, namely $BG, GC$. Let $EF$ be divided into the numbers equal to $D$, namely $EH, HF$. Thus the multitude of $BG, GC$ equals the multitude of $EH, HF$. We have that $BG = GC$ and $EH = HF$. Therefore whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part or the same aliquant part is $GC$ of $HF$ also. So, from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}} and {{EuclidPropLink|book = VII|prop = 6|title = Multiples of Divisors obey Distributive Law}}, we have that whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part also, or the same aliquant part, is the sum $BC$ of the sum $EF$. But $BG = A$ and $EH = D$. So whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$ also. {{qed}} {{Euclid Note|9|VII}}
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$, and another $DE$ be the same aliquant part of another, $F$. We need to show that whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part or aliquot part is $C$ of $F$. :250px We have that whatever aliquant part $AB$ is of $C$, the same aliquant part also is $DE$ of $F$. So as many aliquant parts of $C$ as there are in $AB$, the same aliquant parts also is $DE$ of $F$. Let $AB$ be divided into the aliquant parts of $C$, that is, $AG, GB$, and $DE$ into the  aliquant parts of $F$, that is, $DH, HE$. Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$. Now we have that whatever aliquot part $AG$ is of $C$, the same aliquot part also is $DH$ of $F$. From {{EuclidPropLink|book = VII|prop = 9|title = Alternate Ratios of Equal Fractions}}, whatever aliquot part $AG$ is of $DH$, the same aliquot part or the same aliquant part is $C$ of $F$ also. For the same reason, whatever part $GB$ is of $HE$, the same aliquot part or the same aliquant part is $C$ of $F$ also. So from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}} and {{EuclidPropLink|book = VII|prop = 6|title = Multiples of Divisors obey Distributive Law}}, whatever  aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part also, or the same aliquot part, is $C$ of $F$. {{qed}} {{Euclid Note|10|VII}}
As the whole $AB$ is to the whole $CD$, so let the $AE$ subtracted be to $CF$ subtracted. We need to show that $EB : FD = AB : CD$. :200px We have that :$AB : CD = AE : CF$. So from {{EuclidDefLink|VII|20|Proportional}} 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: : {{EuclidPropLink|book = VII|prop = 7|title = Subtraction of Divisors obeys Distributive Law}} and: : {{EuclidPropLink|book = VII|prop = 8|title = Subtraction of Multiples of Divisors obeys Distributive Law}} $EB$ is the same aliquot part or aliquant part of $FD$ that $AB$ is of $CD$. So by {{EuclidDefLink|VII|20|Proportional}} $EB : FD = AB : CD$. {{qed}} {{Euclid Note|11|VII}}
Let $A, B, C, D$ be as many numbers as we please in proportion, so that $A : B = C : D$. We need to show that $A : B = A + C : B + D$. :200px We have that $A : B = C : D$. So from {{EuclidDefLink|VII|20|Proportional}}, whatever aliquot part or aliquant part $A$ is of $B$, the same aliquot part or aliquant part is $C$ of $D$ also. Therefore from: : {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}} and: : {{EuclidPropLink|book = VII|prop = 6|title = Multiples of Divisors obey Distributive Law}} $A + C$ is the same aliquot part or aliquant part of $C + D$ that $A$ is of $B$. So from {{EuclidDefLink|VII|20|Proportional}}, $A : B = A + C : B + D$. {{qed}} {{Euclid Note|12|VII}}
Let the four (natural) numbers $A, B, C, D$ be proportional so that $A : B = C : D$. We need to show that $A : C = B : D$. :250px We have that $A : B = C : D$. So from {{EuclidDefLink|VII|20|Proportional}} 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 {{EuclidPropLink|book = VII|prop = 10|title = Multiples of Alternate Ratios of Equal Fractions}}, whatever aliquot part or aliquant part $A$ is of $C$, the same aliquot part or aliquant part is $B$ of $D$. Therefore from {{EuclidDefLink|VII|20|Proportional}} $A : C = B : D$. {{Qed}} {{Euclid Note|13|VII}}
The Galois group $\Gal {L / K}$ of $L / K$ is defined as: :$\Gal {L / K} = \left\{ {\sigma: L \to L: \sigma}\right.$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\left.\right\}$ We have that $\Gal {L / K}$ is a subset of the automorphism group on $L / K$. We initially note that the Identity Mapping is Automorphism which trivially fixes $K$. Thus $\Gal {L / K}$ is not the empty set. First, to show that $\Gal {L / K}$ is closed under composition of mappings. Let $\sigma, \tau \in \Gal {L / K}$. Let $\rho = \sigma \circ \tau$. If $k \in K$, then: :$\map \rho k = \map \sigma {\map \tau k} = \map \sigma k = k$ and so $\rho$ fixes $K$. By Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\rho$ is an automorphism. Thus, $\Gal {L / K}$ is closed under composition. Next, by Inverse of Algebraic Structure Isomorphism is Isomorphism, $\sigma^{-1}$ is an automorphism, and: :$\forall k \in K: \map \sigma k = k \implies \map {\sigma^{-1} } k = k$ Thus, $\sigma^{-1} \in \Gal {L / K}$. The result follows from the Two-Step Subgroup Test. That is, $\Gal {L / K}$ is a group. {{qed}} Category:Galois Theory tddbfh2xy7mx1e97xp6kkh6oi3pyi8n
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 $A : C = D : F$. :400px We have that $A : B = D : E$. So from {{EuclidPropLink|book=VII|prop=13|title=Proportional Numbers are Proportional Alternately}}, it follows that $A : D = B : E$. Similarly, we have $B : C = E : F$. So again from {{EuclidPropLink|book=VII|prop=13|title=Proportional Numbers are Proportional Alternately}}, it follows that $B : E = C : F$. Putting them together, we get $A : D = C : F$. Finally, again from {{EuclidPropLink|book=VII|prop=13|title=Proportional Numbers are Proportional Alternately}}, it follows that $A : C = D : F$. {{qed}} {{Euclid Note|14|VII}}