avewright/proofwiki-math huggingface dataset:
Theorem: Stone Space of Boolean Lattice is Topological Space

Let $\struct {B, \preceq, \wedge, \vee}$ be a Boolean lattice.
Let $\struct {U, \tau}$ be the Stone space of $B$.
Then $\struct {U, \tau}$ is a topological space.

Proof:
The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form
:$\set {x \in U: b \in x}$
for some $b \in B$, where $U$ is the set of all ultrafilters on $B$.
By Union from Synthetic Basis is Topology, it suffices to show that $Q$ is a synthetic basis.
First, we must show that $Q$ is a cover of $U$.
Let $x_0 \in U$ be arbitrary.
By filter axiom $\paren 1$, $x_0 \ne \O$.
Thus, there is some $b_0 \in x_0$.
Then, $x_0 \in \set {x \in U: b_0 \in x} \in Q$.
As $x_0$ was arbitrary, it follows that $Q$ is a cover of $U$.
Hence, synthetic basis axiom $\paren {\text B 1}$ holds for $Q$.
{{qed|lemma}}
Let $P, Q \in Q$ be arbitrary.
Then for some $b, c \in B$:
:$P = \set {x \in U: b \in x}$
:$Q = \set {y \in U: c \in y}$
Let $R = \set {z \in U : b \wedge c \in z}$
We will show that:
:$P \cap Q = R$
Let $z \in P \cap Q$ be arbitrary.
Then by the definitions of $P$ and $Q$:
:$b \in z$
:$c \in z$
By definition of lattice, $\wedge$ denotes the meet operator.
Therefore:
:$b \wedge c \preceq b$
:$b \wedge c \preceq c$
:$\forall d: (d \preceq b) \land (d \preceq c) \implies (d \preceq b \wedge c)$
Therefore, by filter axioms $\paren 2$ and $\paren 3$, $b \wedge c \in z$.
As $z \in P \cap Q$ was arbitrary, it follows that $P \cap Q \subseteq R$.
Now, let $v \in R$ be arbitrary.
Then $b \wedge c \in v$.
As stated above:
:$b \wedge c \preceq b$
:$b \wedge c \preceq c$
Therefore, by filter axiom $\paren 3$:
:$b \in v$
:$c \in v$
Thus $v \in P \cap Q$.
Since this holds for each $v \in R$, $R \subseteq P \cap Q$.
From the above, it follows that $P \cap Q = R$ by the definition of set equality.
Therefore, synthetic basis axiom $\paren {\text B 2}$ holds for $Q$ by taking $\AA = \set {P \cap Q}$.
{{qed}}
Category:Stone Spaces
he3tqc816ateounn766bn0izjkxkkcq
Theorem: Ring Without Unity may have Quotient Ring with Unity

Let $\struct {R, +, \circ}$ be a ring.
Let $I$ be an ideal of $R$.
Let $\struct {R / I, +, \circ}$ be the associated quotient ring.
Then $\struct {R / I, +, \circ}$ may have a unity even if $\struct {R, +, \circ}$ has not.

Proof:
Consider the external direct product of rings $\Z \oplus 2 \Z$.
From Integer Multiples form Commutative Ring, $2 \Z$ does not admit a unity.
By Unity of External Direct Sum of Rings, neither does $\Z \oplus 2 \Z$.
Now consider the ideal $\set 0 \times 2 \Z$ of $\Z \oplus 2 \Z$.
We have for all $a \in \Z$ and $b, c \in 2 \Z$ that:
:$\tuple {0, b} - \tuple {0, c} = \tuple {0, b - c}$
:$\tuple {a, b} \cdot \tuple {0, c} = \tuple {0, b \cdot c}$
so by the Test for Ideal, indeed $\set 0 \times 2 \Z$ is an ideal in $\Z \oplus 2 \Z$.
By Quotient Ring of External Direct Sum of Rings, we have:
:$\paren {\Z \oplus 2 \Z} / \paren {\set 0 \times 2 \Z} \cong \paren {Z / \set 0} \oplus \paren {2 \Z / 2 \Z}$
By Quotient Ring by Null Ideal and Quotient Ring Defined by Ring Itself is Null Ring, this last ring is isomorphic to $\Z \times \set 0 \cong \Z$.
Since $\Z$ has a unity, this construction provides an example of the required kind.
{{qed}}
Theorem: Epsilon-Function Differentiability Condition

Let $f: D \to \R$ be a continuous function, where $D \subseteq \R$ is an open set.
Let $x \in \R$.
Then $f$ is differentiable at $x$ {{iff}} there exist $\alpha \in \R$ and $r \in \R_{>0}$ such that for all $h \in \openint {-r} r \setminus \set 0$:
:$\map f {x + h} = \map f x + h \paren {\alpha + \map \epsilon h}$
where:
:$\openint {-r} r$ denotes an open interval of $0$
:$\epsilon: \openint {-r} r \setminus \set 0 \to \R$ is a real function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
If the conditions are true, then:
:$\alpha = \map {f'} x$
If $f$ is continuously differentiable at $x$, then $\epsilon$ is a continous function.

Proof:
=== Necessary Condition ===
Assume that $f$ is differentiable in $x$.
By definition of open set, there exists $r \in \R_{>0}$ such that $\openint {x - r} {x + r} \subseteq D$.
Define $\epsilon: \openint {-r} r \setminus \set 0 \to \R$ by:
:$\map \epsilon h = \dfrac {\map f {x + h} - \map f x} h - \map {f'} x$
If $h \in \openint {-r} r \setminus \set 0$, then:
:$x + h \in \openint {x - r} {x + r} \setminus \set x \subseteq D$
so $\epsilon$ is well-defined.
As $f$ is differentiable at $x$, it follows that:
:$\ds \lim_{h \mathop \to 0} \map \epsilon h = \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h - \map {f'} x = \map {f'} x - \map {f'} x = 0$
If we put:
:$\alpha = \map {f'} x$
it follows that for any $h \in \openint {-r} r \setminus \set 0$:
:$\map f {x + h} = \map f x + h \paren {\alpha + \map \epsilon h}$
{{qed|lemma}}
If $f$ is continuously differentiable, we can choose $r \in \R_{>0}$ such that $f'$ is continuous in $\openint {-r} r$.
From Continuity of Composite Mapping: Corollary, it follows that $\epsilon$ will be continuous in $\openint {-r} r \setminus \set 0$.
{{qed|lemma}}
Theorem: Von Neumann Hierarchy is Cumulative

Let $x$ and $y$ be ordinals such that $x < y$.
Then:
:$\map V x \subsetneqq \map V y$

Proof:
By Von Neumann Hierarchy Comparison:
:$(1): \quad \map V x \in \map V y$
By $(1)$ and the {{Axiom-link|Foundation}}:
:$\map V x \ne \map V y$
Furthermore, by $(1)$ and  Von Neumann Hierarchy is Supertransitive:
{{finish}}
Category:Von Neumann Hierarchy
5n724nfrxdy07evgcu0uyykv75e34di
Theorem: Limits of Real and Imaginary Parts

Let $f: D \to \C$ be a complex function, where $D \subseteq \C$.
Let $z_0 \in D$ be a complex number.
Suppose $f$ is continuous at $z_0$.
Then:
:$(1): \quad \ds \lim_{z \mathop \to z_0} \map \Re {\map f z} = \map \Re {\lim_{z \mathop \to z_0} \map f z}$
:$(2): \quad \ds \lim_{z \mathop \to z_0} \map \Im {\map f z} = \map \Im {\lim_{z \mathop \to z_0} \map f z}$
where:
:$\map \Re {\map f z}$ denotes the real part of $\map f z$
:$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.

Proof:
By definition of continuity:
:$\forall \epsilon > 0: \exists \delta > 0: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$
Given $\epsilon > 0$, we find $\delta > 0$ so for all $z \in \C$ with $\cmod {z - z_0} < \delta$:
{{begin-eqn}}
{{eqn | l = \epsilon
      | o = >
      | r = \cmod {\map f z - \map f {z_0} }
}}
{{eqn | o = \ge
      | r = \cmod {\map \Re {\map f z - \map f {z_0} } }
      | c = Modulus Larger than Real Part
}}
{{eqn | r = \cmod {\map \Re {\map f z} - \map \Re {\map f {z_0} } }
      | c = Addition of Real and Imaginary Parts
}}
{{end-eqn}}
It follows that:
:$\forall \epsilon > 0: \exists \delta > 0: \cmod {z - z_0} < \delta \implies \cmod {\map \Re {\map f z} - \map \Re {\map f {z_0} } } < \epsilon$
Then equation $(1)$ is proven by:
{{begin-eqn}}
{{eqn | l = \lim_{z \mathop \to z_0} \map \Re {\map f z}
      | r = \map \Re {\map f {z_0} }
      | c = {{Defof|Limit of Complex Function}}
}}
{{eqn | r = \map \Re {\lim_{z \mathop \to z_0} \map f z}
      | c = {{Defof|Continuous Complex Function}}
}}
{{end-eqn}}
The proof for equation $(2)$ with imaginary parts follows when $\Re$ is replaced by $\Im$ in the equations above.
{{qed}}
Category:Limits of Complex Functions
tskdi71dr95z9gsu0ink1c1r7hr1ck2
Theorem: Odd Number Theorem/Corollary

A recurrence relation for the square numbers is:
:$S_n = S_{n - 1} + 2 n - 1$

Proof:
{{begin-eqn}}
{{eqn | l = S_n
      | r = \sum_{j \mathop = 1}^n \paren {2 j - 1}
      | c = Odd Number Theorem
}}
{{eqn | r = \sum_{j \mathop = 1}^{n - 1} \paren {2 j - 1} + \paren {2 n - 1}
      | c = {{Defof|Summation}}
}}
{{eqn | r = S_{n - 1} + \paren {2 n - 1}
      | c = Odd Number Theorem
}}
{{end-eqn}}
{{qed}}
Category:Odd Number Theorem
e5imnpnznujgeroh2tc2wl2ur5udrfg
Theorem: Effect of Sequence of Elementary Row Operations on Determinant

Let $\hat o_1, \ldots, \hat o_m$ be a finite sequence of elementary row operations. 
Here, $\hat o_i$ denotes an elementary row operation on a square matrix of order $n$ over a commutative ring with unity $\struct {R, +, \circ}$.
Here, $i \in \set {1, \ldots, m}$.
Then there exists $c \in R$ such that for all square matrices of order $n$ $\mathbf A$ over $R$:
:$\map \det {\mathbf A} = c \map \det {\mathbf A'}$
where $\mathbf A'$ is the square matrix of order $n$ that results from applying the elementary row operations $\hat o_1, \ldots, \hat o_m$ on $\mathbf A$.

Proof:
Proof by induction on $m$, the number of elementary row operations in the sequence $\hat o_1, \ldots, \hat o_m$.
Theorem: Every Filter has Adherent Point implies Every Ultrafilter Converges

Let $T = \struct {S, \tau}$ be a topological space.
Let every filter on $S$ have a adherent point in $S$.
Then every ultrafilter on $S$ converges to a point in $S$.

Proof:
{{Recall|Ultrafilter on Set|ultrafilter}}
{{:Definition:Ultrafilter on Set/Definition 1}}
Let $T = \struct {S, \tau}$ be such that every filter on $S$ has a adherent point in $S$.
Let $\FF$ be an arbitrary ultrafilter on $S$.
We have {{hypothesis}} that $\FF$ has a adherent point $x \in S$.
By Adherent Point of Filter iff Superfilter Converges, there exists a filter $\FF'$ on $S$ which converges to $x$ satisfying $\FF \subseteq \FF'$.
By definition of ultrafilter:
:$\FF$ is an ultrafilter on $S$ {{iff}} whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$
Therefore:
:$\FF = \FF'$
Thus $\FF$ converges to $x$.
As $\FF$ was arbitrary, it follows that:
:every ultrafilter on $S$ converges to a point in $S$.
{{qed}}
Theorem: Every Ultrafilter Converges implies Every Filter has Adherent Point

Let $T = \struct {S, \tau}$ be a topological space.
Let every ultrafilter on $S$ be convergent.
Then every filter on $S$ has a adherent point.

Proof:
Let $\FF$ be a  filter on $S$.
By the Ultrafilter Lemma, there exists an ultrafilter $\FF'$ such that $\FF \subseteq \FF'$.
We have {{hypothesis}} that $\FF'$ converges to some $x \in S$.
This, by Adherent Point of Filter iff Superfilter Converges, implies that $x$ is a adherent point of $\FF$.
{{qed}}
{{BPI|Ultrafilter Lemma}}
Theorem: Left Distributive and Commutative implies Distributive

Let $\struct {S, \circ, *}$ be an algebraic structure.
Let the operation $\circ$ be left distributive over the operation $*$.
Let $\circ$ be commutative.
Then $\circ$ is distributive over $*$.

Proof:
Let $a, b, c \in S$.
Then
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ c
      | r = c \circ \paren {a * b}
      | c = $\circ$ is commutative
}}
{{eqn | r = \paren {c \circ a} * \paren {c \circ b}
      | c = $\circ$ is left distributive over $*$
}}
{{eqn | r = \paren {a \circ c} * \paren {b \circ c}
      | c = $\circ$ is commutative
}}
{{end-eqn}}
So $\circ$ is right distributive over $*$.
Since $\circ$ is both left distributive and right distributive over $*$, it is distributive over $*$.
{{qed}}
Category:Commutativity
Category:Distributive Operations
c3m6ykmplcactwrchd5w1kw3tb750mq
Theorem: Filter on Product Space Converges iff Projections Converge

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.
Let $\FF \subset \powerset X$ be a filter on $X$.
Then $\FF$ converges {{iff}} for all $i \in I$ the image filter $\map {\pr_i} \FF$ converges.

Proof:
=== Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$
Thus
:$\forall i \in I: \map {\pr_i} \FF$ converges.
Theorem: Tychonoff's Theorem for Hausdorff Spaces

Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact.

Proof:
First assume that $X$ is compact. 
From Projection from Product Topology is Continuous, the projections:
:$\pr_i : X \to X_i$ 
are continuous.
From Continuous Image of Compact Space is Compact, it follows that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
From Topological Space is Compact iff Every Ultrafilter Converges it is enough to show that every ultrafilter on $X$ converges.
Thus let $\FF$ be an ultrafilter on $X$.
From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.
Each $X_i$ is compact by assumption.
So by Topological Space is Compact iff Every Ultrafilter Converges, each $\map {\pr_i} \FF$ converges.
From Filter on Product of Hausdorff Spaces Converges iff Projections Converge, $\FF$ converges.
So, as $\FF$ was arbitrary, $X$ is compact.
{{qed}}
{{BPI|Topological Space is Compact iff Every Ultrafilter Converges}}
However, Topological Space is Compact iff Every Ultrafilter Converges only invokes the Ultrafilter Lemma, so this theorem is safe to use when proving the Boolean Prime Ideal Theorem from the Ultrafilter Lemma.
Theorem: Filter on Product of Hausdorff Spaces Converges iff Projections Converge

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces where $I$ is an arbitrary index set.
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.
Let $\FF \subset \powerset X$ be a filter on $X$.
Then $\FF$ converges {{iff}} for each $i \in I$ the image filter $\map {\pr_i} \FF$ converges.

Proof:
=== Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$.
Thus, for each $i \in I$, $\map {\pr_i} \FF$ converges.
Theorem: Positive Real has Real Square Root

Let $x \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\exists y \in \R: x = y^2$

Proof:
Let $f: \R \to \R$ be defined as:
:$\forall x \in \R: \map f x = x^2$
We have that $f$ is the pointwise product of the identity mapping with itself.
By Product Rule for Continuous Real Functions and Identity Mapping is Continuous, $f$ is continuous.
By Power Function is Unbounded Above:
:$\exists q \in \R: \map f q > x$
Then:
:$0^2 = 0 \le x$
By the Intermediate Value Theorem:
:$\exists y \in \R: 0 < y < q: y^2 = x$
{{qed}}
Category:Real Numbers
s2gk1xsm90w1sdljs6mgljr427djprb
Theorem: Join is Idempotent

Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is idempotent.

Proof:
Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee a
      | r = \sup \set {a, a}
      | c = {{Defof|Join (Order Theory)|Join}}
}}
{{eqn | r = \sup \set a
      | c = {{Defof|Set}}
}}
{{eqn | r = a
      | c = Supremum of Singleton
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Product of Positive Strictly Increasing Mappings is Strictly Increasing

Let $A$ be an ordered set.
Let $B$ be an ordered field.
Let $f, g: A \to B$ be strictly increasing mappings with positive values.
Let $h: A \to B$ be defined by $\map h x = \map f x \map g x$.
Then $h$ is strictly increasing.

Proof:
Let $x, y \in A$ such that $x < y$.
If $\map h x = 0$.
By the definition of strictly increasing:
:$\map f y > \map f x \ge 0$
and:
:$\map g y > \map g x \ge 0$
So:
:$\map h y > 0 = \map h x$
If $\map h x \ne 0$, then $\map h x > 0$
so:
:$\map f x > 0$
and:
:$\map g x > 0$
Also:
:$\map f x < \map f y$
and:
:$\map g x < \map g y$
by the definition of strictly increasing.
Because $\map g x > 0$:
:$\map f x \map g x < \map f y \map g x$
Because $\map f y > 0$:
:$\map f y \map g x < \map f y \map g y$
By transitivity:
:$\map h x = \map f x \map g x < \map f y \map g y = \map h y$
{{qed}}
{{expand|$B$ ordered ring?}}
{{expand|Generalize by induction to finite products}}
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Theorem: Supremum of Singleton

Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\sup \set a = a$
where $\sup$ denotes supremum.

Proof:
Since $a \preceq a$, $a$ is an upper bound of $\set a$.
Let $b$ be another upper bound of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
:$\sup \set a = a$
as desired.
{{qed}}
Theorem: Infimum of Singleton

Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\inf \set a = a$
where $\inf$ denotes infimum.

Proof:
Since $a \preceq a$, $a$ is a lower bound for $\set a$.
Let $b$ be another lower bound for $\set a$.
Then necessarily $b \preceq a$.
It follows that indeed:
:$\inf \set a = a$
as desired.
{{qed}}
Theorem: Meet is Idempotent

Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is idempotent.

Proof:
Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge a
      | r = \inf \set {a, a}
      | c = {{Defof|Meet (Order Theory)}}
}}
{{eqn | r = \inf \set a
      | c = {{Defof|Set}}
}}
{{eqn | r = a
      | c = Infimum of Singleton
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: GCD from Prime Decomposition

Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
      | r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
      | r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
      | l = p_i
      | o = \divides
      | r = a
      | c = 
}}
{{eqn | lo= \lor
      | l = p_i
      | o = \divides
      | r = b
}}
{{end-eqn}}
That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $a$ or $b$.
Then:
:$\gcd \set {a, b} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$
where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.

Proof:
Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.
Let $d \divides a$.
Then:
:$d$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le h_i \le k_i$
:$d \divides a \iff \forall i: 1 \le i \le r, 0 \le h_i \le l_i$
So:
:$d \divides a \land d \divides b \iff \forall i: 1 \le i \le r, 0 \le h_i \le \min \set {k_i, l_i}$
For $d$ to be at its greatest, we want the largest possible exponent for each of these primes.
So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\min \set {k_i, l_i}$.
Hence the result:
:$\gcd \set {a, b} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$
{{Qed}}
Theorem: LCM from Prime Decomposition

Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
      | r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
      | r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
      | l = p_i
      | o = \divides
      | r = a
      | c = 
}}
{{eqn | lo= \lor
      | l = p_i
      | o = \divides
      | r = b
}}
{{end-eqn}}
That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $a$ or $b$.
Then:
:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
where $\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$.

Proof:
{{begin-eqn}}
{{eqn | l = \lcm \set {a, b}
      | r = \frac {a b} {\gcd \set {a, b} }
      | c = Product of GCD and LCM
}}
{{eqn | r = \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }
      | c = GCD from Prime Decomposition
}}
{{eqn | r = \frac {\paren {p_1^{k_1 + l_1} p_2^{k_2 + l_2} \cdots p_r^{k_r + l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }
      | c = 
}}
{{eqn | r = p_1^{k_1 + l_1 - \min \set {k_1, l_1} } p_2^{k_2 + l_2 - \min \set {k_2, l_2} } \cdots p_r^{k_r + l_r - \min \set {k_r, l_r} }
      | c = 
}}
{{eqn | r = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \cdots p_r^{\max \set {k_r, l_r} }
      | c = Sum Less Minimum is Maximum
}}
{{end-eqn}}
{{Qed}}
Theorem: LCM from Prime Decomposition

Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
      | r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
      | r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
      | l = p_i
      | o = \divides
      | r = a
      | c = 
}}
{{eqn | lo= \lor
      | l = p_i
      | o = \divides
      | r = b
}}
{{end-eqn}}
That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $a$ or $b$.
Then:
:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
where $\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$.

Proof:
Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.
Let $a \divides m$.
Then:
:$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$
:$a \divides l \iff \forall i: 1 \le i \le r, 0 \le l_i \le h_i$
So:
:$a \divides m \land b \divides m \iff \forall i: 1 \le i \le r, 0 \le \max \set {k_i, l_i} \le h_i$
For $m$ to be at its smallest, we want the smallest possible exponent for each of these primes.
So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\max \set {k_i, l_i}$.
Hence the result:
:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
{{Qed}}
Theorem: Product of Positive Element and Element Greater than One

Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity $1_R$ and zero $0_R$.
Let $x, y \in R$.
Suppose that $x > 0_R$ and $y > 1_R$.
Then $x \circ y > x$ and $y \circ x > x$.

Proof:
{{begin-eqn}}
{{eqn | l = y
      | o = >
      | r = 1_R
}}
{{eqn | l = x \circ y
      | o = > 
      | r = x \circ 1_R
      | c = $x > 0_R$, Properties of Ordered Ring: $(6)$
}}
{{eqn | l = x \circ y
      | o = >
      | r = x
      | c = {{Defof|Unity of Ring}}
}}
{{end-eqn}}
A similar argument shows that $y \circ x > x$.
{{qed}}
Category:Ordered Rings
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Theorem: Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element

Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
:$\circ^n x \ge x$

Proof:
The proof proceeds by induction:
If $n = 1$, then $\circ^n x = x$.
So:
:$\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
:$\circ^n x > 1$
By Product of Positive Element and Element Greater than One:
:$x \circ \paren {\circ^n x} > x$
Hence:
:$\circ^{n + 1} x \ge x$
{{qed}}
Category:Ordered Rings
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Theorem: Directed Set has Strict Successors iff Unbounded Above

Let $\struct {S, \le}$ be a directed set.
Then every element of $S$ has a strict successor in $S$ {{iff}} $S$ has no upper bound in $S$.

Proof:
=== Necessary Condition ===
Suppose that each element of $S$ has a strict successor in $S$.
If $x$ is any element of $S$, then $x$ has a strict successor.
Therefore, $x$ is not an upper bound of $S$.
{{qed|lemma}}
Theorem: Complex Numbers are Uncountable

The set of complex numbers $\C$ is uncountably infinite.

Proof:
For all $r \in \R$, we have $r = r + 0 i \in C$.
Thus the set of real numbers $\R$ can be considered a subset of $\C$.
As the Real Numbers are Uncountable, it follows from Sufficient Conditions for Uncountability, proposition $(1)$, that $\C$ is uncountably infinite.
{{qed}}
Category:Complex Numbers
Category:Set Theory
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