hcju/proofwikiqa huggingface dataset:
Recall that $X$ is uncountable if there is no injection $X \hookrightarrow \N$. === $(1)$ implies $(2)$ === Suppose there exists an injection $f: X \hookrightarrow \N$. Let $Y \subseteq X$ be uncountable. Then $\hat f: Y \to \N$ defined as: :$\hat f = \set {\map f x: x \in Y}$ is an injection $Y \hookrightarrow N$, which is a contradiction. {{qed|lemma}} === $(2)$ implies $(3)$ === Suppose $\sequence {x_n}_{n \mathop \in \N}$ is a sequence of distinct points such that every $x \in X$ equals $x_n$ for some $n$. Since the $x_n$ are distinct, this $n$ is unique. Let $f: X \to \N$ be defined by $\map f x = n$, where $n$ is the unique natural number such that $x_n = x$. Then $f$ is injective because $n = m$ implies that $x_n = x_m$. So there is an injection $X \hookrightarrow \N$, a contradiction. {{qed|lemma}} === $(3)$ implies $(4)$ === Suppose there exists a surjection $f: \N \to X$. Therefore every $x \in X$ is $\map f n$ for some $n \in \N$. Define $g: X \to \N$ by $\map g x = \inf \set {n \in \N : \map f n = x}$. Then if $\map g {x_1} = \map g {x_2} = n$, we have $\map f n = x_1$ and $\map f n = x_2$. So by the definition of a mapping, $x_1 = x_2$. Therefore $g$ is an injection $X \to \N$, a contradiction. {{qed|lemma}} === $(4)$ implies $(5)$ === Since there exists no surjection $\N \to X$, and a bijection $\N \to X$ must be surjective, there is no such mapping. {{qed|lemma}} === $(5)$ implies $(1)$ === Suppose $f: X \hookrightarrow \N$ is an injection. We essentially relabel the image of $f$ to obtain a bijection, and thus a contradiction. Construct a map $g: X \to \N$ as follows. Let $S_1 = \Img f$ and: :$x_1 = \map {f^{-1} } {\inf \set {n \in \N : n \in S_1} }$ and define $\map f x = 1$. Note that $x$ is a singleton because $f$ is injective. Given $S_n \subseteq \Img f$, let: :$x_n = \map {f^{-1} } {\inf \set {n \in \N : n \in S_n} }$ and set $S_{n + 1} = S_n \setminus x_n$. The mapping: :$n \stackrel {f^{-1} } \longrightarrow x_k \stackrel g \longrightarrow k$ defines a bijection from $\Img f$ to $\N$, so $g$ is injective because $f$ is. Furthermore $g$ assigns each $n \in \N$ to some $x \in X$ (because $X$ is infinite) so $g$ is surjective. Thus $g$ is a bijection, a contradiction. {{qed|lemma}} === $(6)$ implies $(1)$ === {{proof wanted|to do}} {{qed}} Category:Uncountable Sets 951p86012533eucwpxfwrtezutjl1c8
If $R$ has a nontrivial decomposition $R = R_1 \times R_2$ then $\tuple {1_R, 0_R}$ is a non-trivial idempotent element of $R$. {{explain|non-trivial decomposition, non-trivial element}} Conversely suppose there is $0_R, 1_R \ne e \in R$ with $e^2 = e$. Let $R_1 = \ideal e$, the ideal generated by $e$, and $R_2 = R / \ideal e$. Since $e \paren {e - 1_R} = 0_R$, it follows by definition that $e$ is a zero divisor. So by Unit Not Zero Divisor it is not a unit. Therefore, $1 \notin R_1$ and $\ideal e \subsetneqq R$. Also $R_1 \cap R_2 = \set {0_R}$ so the product is direct (that is, the Universal Property for Direct Products is satisfied). Finally we define the "gluing homomorphism" $\phi : R \to R_1 \times R_2$ by :$\phi: a \mapsto \tuple {a e, a + \ideal e}$ which is easily shown to be an isomorphism. {{qed}} Category:Commutative Algebra fh03qu2of6klj77b3msijjski733boh
Let $x = 0_D$. Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero. Thus by definition of strict positivity property: :$\neg \map P {0_D \times 0_D}$ {{qed|lemma}} Now suppose $x \ne 0_D$. One of two cases applies: :$\map P x$ :$\neg \map P x$ Let $\map P x$. Then by definition of (strict) positivity: :$\map P {x \times x}$ Now suppose $\neg \map P x$. Then by the trichotomy law of ordered integral domains: :$\map P {-x}$ Then again by definition: :$\map P {\paren {-x} \times \paren {-x} }$. But by Product of Ring Negatives: :$\paren {-x} \times \paren {-x} = x \times x$ So again: :$\map P {x \times x}$ Hence the result. {{qed}}
$\le$ is a total ordering on $\R$. The trichotomy follows directly from Trichotomy Law. {{Qed}}
We have by definition of the unity that: :$\forall a \in D: 1_D \times a = a = a \times 1_D$ This particularly applies to $1_D$ itself: :$1_D = 1_D \times 1_D$ But then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive: :$\map P {1_D \times 1_D} \implies \map P {1_D}$ {{qed}}
From Integers form Integral Domain we have that $\struct {\Z, +, \times}$ forms an integral domain. From Natural Numbers are Non-Negative Integers we have that the set $\N$ can be considered as a subset of the integers. Then we have that $\struct {\N_{> 0}, +, \times}$ is a (commutative) semiring. So it follows by definition of semiring, in particular the fact that $+$ and $\times$ on $\N_{> 0}$ are closed, that: :$\forall a, b \in \N_{> 0}: a + b \in \N_{> 0}$ :$\forall a, b \in \N_{> 0}: a \times b \in \N_{> 0}$ It follows that we can define a property $P$ on $\Z$ such that: :$\forall x \in \Z: \map P x \iff x \in \N_{> 0}$ Checking that $P$ fulfils the conditions for it to be the (strict) positivity property: : $(1): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a + b}$ Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$. : $(2): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a \times b}$ Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$. : $(3): \quad \forall a \in \Z: \map P a \lor \map P {-a} \lor a = 0$ This follows from the definition of the integers as the inverse completion of the natural numbers. If $a \in \Z - \N$ then $\exists b \in \N: a + b = 0$ and so $a = -b$. Hence the conditions are fulfilled and $\struct {\Z, +, \times}$ forms an ordered integral domain. {{qed}}
This follows directly from Rational Numbers form Ordered Field: The set of rational numbers $\Q$ forms an ordered field under addition and multiplication: $\struct {\Q, +, \times, \le}$. An ordered field is also an ordered integral domain. {{Qed}}
This follows directly from Real Numbers form Ordered Field. The set of real numbers $\R$ forms an ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$. An ordered field is also an ordered integral domain. {{Qed}}
We have that the real numbers $\struct {\R, +, \times}$ form an integral domain. It remains to specify a property $P$ on $\R$ such that: :$(1): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a + b}$ :$(2): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a \times b}$ :$(3): \quad \forall a \in \R: \map P a \lor \map P {-a} \lor a = 0$ We have that the integers $\struct {\Q, +, \times}$ form an ordered integral domain, where $\Q$ denotes the set of rational numbers. Let $P'$ be the (strict) positivity property on $\struct {\Q, +, \times}$. Let us define the property $P$ on $\R$ as: :$\forall a \in \R: \map P a \iff a = \eqclass {\sequence {a_n} } {}: \forall n \in \N: \map {P'} {a_n}$ That is, an element $a = \eqclass {\sequence {a_n} } {}$ has $P$ {{iff}} $a_n$ has (strict) positivity property in $\Q$ for all $n \in \N$. Now let $a = \eqclass {\sequence {a_n} } {}$ and $b = \eqclass {\sequence {b_n} } {}$ such that $\map P a$ and $\map P b$. Then by definition of real addition: :$\eqclass {\sequence {a_n} } {} + \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n + b_n} } {}$ and real multiplication: :$\eqclass {\sequence {a_n} } {} \times \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n \times b_n} } {}$ It can be seen from the definition of (strict) positivity $P'$ on $\Q$ that $\map P {a + b}$ and $\map P {a \times b}$. It can be seen that if $\map P a$ then $\neg \map P {-a}$ and vice versa. Also we note that $\neg \map P 0$ and of course $\neg \map P {-0}$. So the property $P$ we defined fulfils the criteria for the (strict) positivity property. Hence the result. {{qed}}
We have that Numbers of Type Integer a plus b root 2 form Subdomain of Reals. We also have that such numbers are real. The result follows from Real Numbers form Ordered Integral Domain. {{explain|So is a subdomain of an ordered integral domain always an ordered integral domain? Does this need to be proved?}} {{qed}}
Trivially true for $n = 1$, and true by definition for $n = 2$. Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$. Then: :$\displaystyle \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$ :$\displaystyle \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$ Take any $\displaystyle \sum_{i \mathop = 1}^{k + 1} s_i = \sum_{i \mathop = 1}^k s_i + s_{k + 1}$. Then: :$\displaystyle \map P {\sum_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$ so: :$\displaystyle \map P {\sum_{i \mathop = 1}^{k + 1} s_i}$ because the (strict) positivity property holds for all $s_i \in S$. Similarly: :$\displaystyle \map P {\prod_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$ so: :$\displaystyle \map P {\prod_{i \mathop = 1}^{k + 1} s_i}$ The result follows by the Principle of Complete Induction. {{qed}}
Let $\struct {D, +, \times}$ be an integral domain whose order is $n$. {{AimForCont}} there exists a (strict) positivity property $P$ on $\struct {D, +, \times}$. From Unity of Ordered Integral Domain is Strictly Positive, we have that $\map P {1_D}$. Let $\struct {D, +}$ be the additive group of $\struct {D, +, \times}$. From Order of Element Divides Order of Finite Group, the order of $1_D$ is also finite. So: :$\exists m \in \N: m \cdot 1_D = 0_D$ where $0_D$ is the zero of $D$. But then from the General Positivity Rule in Ordered Integral Domain that would mean $\map P {0_D}$ which contradicts the trichotomy law of the positivity property. Further, note that from Inverse Element is Power of Order Less 1 we have that $-1_D = \paren {m - 1} \cdot 1_D$. But from the General Positivity Rule in Ordered Integral Domain $\map P {\paren {m - 1} \cdot 1_D}$ and so $\map P {-1_D}$ which is an even more dramatic way of showing that the trichotomy law of the positivity property is contradicted. Hence an integral domain whose order is finite can not be ordered. {{qed}}
First note that from Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain only when $m$ is prime. So for $m$ composite the result holds. If $m$ is prime, and $\struct {\Z_m, +, \times}$ is therefore an integral domain, its order is finite.  The result follows from Finite Integral Domain cannot be Ordered. {{qed}}
By definition of the strict positivity property: {{:Definition:Strict Positivity Property}} Let us define a relation $<$ on $D$ as: :$\forall a, b \in D: a < b \iff \map P {-a + b}$ Setting $a = 0$: :$\forall b \in D: 0 < b \iff \map P b$ demonstrating that (strictly) positive elements of $D$ are those which are greater than zero. From Relation Induced by Strict Positivity Property is Compatible with Addition we have that $<$ is compatible with $+$. From Relation Induced by Strict Positivity Property is Transitive we have that $<$ is transitive. From Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive we have that $<$ is asymmetric and antireflexive. Thus by definition, $<$ is a strict ordering. Let the relation $\le$ be defined as the reflexive closure of $<$. From Reflexive Closure of Strict Ordering is Ordering we have that $\le$ is an ordering on $D$. From Relation Induced by Strict Positivity Property is Trichotomy, and from the Trichotomy Law (Ordering), we have that $\le$ is a total ordering. Hence the result. {{qed}}
Let $a < b$ and $b < c$. Thus: {{begin-eqn}} {{eqn | o =        | r = a < b, b < c       | c =  }} {{eqn | o = \leadsto       | r = \map P {-a + b}, \map P {-b + c}       | c = Definition of $<$ }} {{eqn | o = \leadsto       | r = \map P {\paren {-a + b} + \paren {-b + c} }       | c = {{Defof|Strict Positivity Property}} }} {{eqn | o = \leadsto       | r = \map P {-a + c}       | c = Properties of $+$ in $D$ }} {{eqn | o = \leadsto       | r = a < c       | c = Definition of $<$ }} {{end-eqn}} And so $<$ is seen to be transitive. {{qed}}
Let $a < b$: {{begin-eqn}} {{eqn | o =        | r = a < b       | c =  }} {{eqn | o = \leadsto       | r = \map P {-a + b}       | c = by definition of $<$ }} {{eqn | o = \leadsto       | r = \map P {-a + b + \paren {-c} + c}       | c = for any $c \in D$ }} {{eqn | o = \leadsto       | r = \map P {\paren {-a + \paren {-c} } + \paren {b + c} }       | c = properties of $+$ in $D$ }} {{eqn | o = \leadsto       | r = \map P {-\paren {a + c} + \paren {b + c} }       | c = properties of $+$ in $D$ }} {{eqn | o = \leadsto       | r = a + c < b + c       | c = by definition of $<$ }} {{eqn | o = \leadsto       | r = c + a < c + b       | c = commutativity of $+$ }} {{end-eqn}} And so $<$ is seen to be compatible with $+$. {{qed}}
If $x < y$ then $\map P {-x + y}$. Hence: {{begin-eqn}} {{eqn | o =        | r = \map P {-x + y}, \, \map P z       | c =  }} {{eqn | o = \leadsto       | r = \map P {z \times \paren {-x + y} }       | c = {{Defof|Strict Positivity Property}} }} {{eqn | o = \leadsto       | r = \map P {-z \times x + z \times y}       | c = Distributivity of $\times$ over $+$ }} {{eqn | o = \leadsto       | r = z \times x < z \times y       | c =  }} {{end-eqn}} The other result follows from the fact that $\times$ is commutative in an integral domain. {{qed}}
Take any $a, b \in D$ and consider $-a + b$. From the trichotomy law of ordered integral domains, exactly one of the following applies: :$(1): \quad \map P {-a + b}$ :$(2): \quad \map P {-\paren {-a + b} }$ :$(3): \quad -a + b = 0$ Taking each of these in turn and taking into account that $\struct {D, +}$ is the additive group of $\struct {D, +, \times}$: {{begin-eqn}} {{eqn | n = 1       | o =        | r = \map P {-a + b}       | c =  }} {{eqn | o = \leadsto       | r = a < b       | c = by definition of $<$ }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 2       | o =        | r = \map P {-\paren {-a + b} }       | c =  }} {{eqn | o = \leadsto       | r = \map P {-b + a}       | c =  }} {{eqn | o = \leadsto       | r = b < a       | c = by definition of $<$ }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 3       | o =        | r = -a + b = 0       | c =  }} {{eqn | o = \leadsto       | r = a = b       | c = adding $a$ to both sides }} {{end-eqn}} {{qed|lemma}} The result has been proved. {{qed}}
From the trichotomy law of ordered integral domains, for all $x \in D$ exactly one of the following applies: :$(1): \quad \map P x$ :$(2): \quad \map P {-x}$ :$(3): \quad x = 0$ Let $a, b \in D: a < b$. Then by definition: :$\map P {-a + b}$ Suppose also that $b < a$. Then by definition: :$\map P {-b + a}$ and so: :$\map P {-\paren {-a + b} }$ But then that contradicts the trichotomy law of ordered integral domains: Either: :$(1): \quad \map P {-a + b}$ or: :$(2): \quad \map P {-\paren {-a + b} }$ So $(1)$ and $(2)$ can not both apply, and so if $a < b$ it is not possible that $b < a$. Thus $<$ is asymmetric. Now suppose $\exists a \in D: a < a$. Then that would mean $\map P {-a + a}$, that is, $\map P {0_D}$. But this also contradicts the conditions of the trichotomy law of ordered integral domains. Hence: :$\forall a \in D: a \not < a$ and so $<$ is antireflexive. {{qed}} Category:Ordered Integral Domains 7wj0qad5d2yz9f13et7dv0yx2k16dak
Let $P$ be the (strict) positivity property on $D$. Let $<$ be the (strict) total ordering defined on $D$ as: :$a < b \iff a \le b \land a \ne b$ From the trichotomy law, exactly one of three possibilities holds for any $ a \in D$: $(1): \quad \map P a$: In this case $0 < a$ and so $\size a = a$. So: :$\map P a \implies \map P {\size a}$ $(2): \quad \map P {-a}$: In this case $a < 0$ and so $\size a = -a$. So: :$\map P {-a} \implies \map P {\size a}$ $(3): \quad a = 0$: In this case $\size a = a$. So: :$a = 0 \implies \size a = 0$. Hence the result. {{qed}}
Let $P$ be the (strict) positivity property on $D$. Let $<$ be the (strict) total ordering defined on $D$ as: :$a < b \iff a \le b \land a \ne b$ Let $N$ be the (strict) negativity property on $D$. Let $a \in D$. If $\map P a$ or $a = 0$ then $a \le \size a$. If $\map N a$ then by Properties of Strict Negativity: $(1)$ and definition of absolute value: :$a < 0 < \size a$ and hence by transitivity $<$ we have: :$a < \size a$ By similar reasoning: :$-a < \size a$ Thus for all $a, b \in D$ we have: :$a \le \size a, b \le \size b$ As $<$ is compatible with $+$, we have: :$a + b \le \size a + \size b$ and: :$-\paren {a + b} = \paren {-a} + \paren {-b} \le \size a + \size b$ But either: :$\size {a + b} = a + b$ or: :$\size {a + b} = -\paren {a + b}$ Hence the result: :$\size {a + b} \le \size a + \size b$ {{qed}}
{{AimForCont}} there exists an element $a \in D$ such that $0 < a < 1$. Let us create the set $S$ of all such elements: :$S = \set {x \in D: 0 < x < 1}$ We know $S$ is not empty because it has already been asserted that $a$ is in it. And all the elements in $S$ are strictly positive by definition. Because $D$ is well-ordered it follows that $S$ has a minimal element, which we will call $m$. Thus we have $0 < m < 1$. Then from Square of Element Less than Unity in Ordered Integral Domain we have: :$0 < m \times m < m$ Thus $m \times m \in S$ is an element of $S$ which is smaller than $m$. But $m$ was supposed to be the minimal element of $S$. From this contradiction we deduce that there can be no such element $a$ such that $0 < a < 1$. {{qed}}
We have that $0 < x < 1$. From Relation Induced by Strict Positivity Property is Compatible with Multiplication: :$0 \times x < x \times x < 1 \times x$ Hence the result. {{qed}} Category:Ordered Integral Domains payuibiqj3z0g90g32oa6vktwe0b6lr
Let $\overline S$ be the set of all elements of $D_{> 0_D}$ that are not in $S$: :$\overline S = D_{> 0_D} \setminus S$ {{AimForCont}} $\overline S$ is not empty. Then as $D$ is well-ordered, it follows that $\overline S$ has a minimal element, which we will call $m$. Then $m - 1 \notin \overline S$. But $1 < m$ as $1 \in S$, and from One Succeeds Zero in Well-Ordered Integral Domain, $1$ is the minimal positive element of $D$. So $0 < m - 1$ and so $m - 1$ is strictly positive. Because $m - 1 \notin \overline S$, it follows that $m - 1 \in S$. By construction of $S$ it follows that $m \in S$. So $m \in S$ and $m \in \overline S$, that is: :$m \notin S$ From this contradiction we deduced that there can therefore be no such $m$. Hence: :$D_{\ge 0_D} \setminus S = \O$. From Set Difference with Superset is Empty Set it follows that $D_{\ge 0_D} \subseteq S$. {{qed}}
By definition, if $a \divides b$ then $\exists d \in D: a \times d = b$. Then $\paren {a \times d} \times c = b \times c$, that is: :$\paren {a \times c} \times d = b \times c$ which follows because $\times$ is commutative and associative in an integral domain. Hence the result. {{qed}}
From Integers form Integral Domain, the concept '''divisibility''' is fully applicable to the integers. Therefore this result follows directly from Unity Divides All Elements. {{qed}}
From Integers form Integral Domain, the integers are an integral domain. Hence we can apply Product of Ring Negatives: :$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$ and Product with Ring Negative: :$\forall n \in \Z: \exists -1 \in \Z: -n = \paren {-1} \times \paren n$ {{qed}}
Let $n > 0$. Then $\size n = n$ and Integer Divides Itself applies. Let $n = 0$. Then Integer Divides Itself holds again. Let $n < 0$. Then $\size n = -n$ and Integer Divides its Negative applies. {{qed}} Category:Divisors Category:Absolute Value Function qwokq3o9x7dlgf3p8phr3bj8v8ig1wr
From Integers form Integral Domain, the concept '''divides''' is fully applicable to the integers. Therefore this result follows directly from Element of Integral Domain Divides Zero. {{qed}}
Let $m \mathrel \backslash n$. From Integer Divides its Negative, we have $-m \divides m$. From Divisor Relation on Positive Integers is Partial Ordering it follows that $-m \divides n$. From Integer Divides its Negative, we have $n \divides -n$. From Divisor Relation on Positive Integers is Partial Ordering it follows that $m \divides -n$. From above, we have $-m \divides n$. From Integer Divides its Negative, we have $n \divides -n$. From Divisor Relation on Positive Integers is Partial Ordering it follows that $-m \divides -n$. The reverse implications follow similarly. {{qed}}
=== Necessary Condition === Let $a = \pm 1$. From Integer Divides Itself we have that $1 \divides 1$. From Integer Divides its Negative we have that $-1 \divides 1$. {{qed|lemma}} === Sufficient Condition === Let $\exists a \in \Z: a \divides 1$. Then $\exists c \in \Z: a c = 1$. From Absolute Value Function is Completely Multiplicative we have that: :$\size a \cdot \size c = \size 1$ Neither $a$ nor $c$ can be zero, from Integers form Integral Domain. So $\size a \ge 1$ and $\size c \ge 1$. But if $\size a > 1$ then: :$\size a \cdot \size c > \size c$ and so: :$\size a \cdot \size c > 1$ So: :$\size a = 1$ that is: :$a = 1$ or $a = -1$ {{qed}}
Let $a, b \in \Z_{> 0}$ such that $a \divides b$ and $b \divides a$. Then: {{begin-eqn}} {{eqn | l = a \divides b       | o = \implies       | r = \size a \le \size b       | c = Absolute Value of Integer is not less than Divisors }} {{eqn | l = b \divides a       | o = \implies       | r = \size b \le \size a       | c = Absolute Value of Integer is not less than Divisors }} {{eqn | o = \leadsto       | r = \size a = \size b }} {{end-eqn}} If we restrict ourselves to the domain of positive integers, we can see: :$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ {{qed}}