nyuuzyou/wikis huggingface dataset:
Relative Sizes of Components of Ratios

== Theorem ==
That is, if $a : b = c : d$ then:
:$a > c \implies b > d$
:$a = c \implies b = d$
:$a < c \implies b < d$
== Proof ==
Let a first magnitude $A$ have the same ratio to a second $B$ as a third $C$ has to a fourth $D$.
:
Let $A > C$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.
But $A : B = C : D$.
So from Relative Sizes of Proportional Magnitudes $C : D > C : B$.
But from Relative Sizes of Magnitudes on Unequal Ratios $D > B$.
In a similar way it can be shown that:
: if $A = C$ then $B = D$
: if $A < C$ then $B < D$
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relative Sizes of Definite Integrals

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$, where $a < b$.
If:
:$\forall t \in \closedint a b: \map f t \le \map g t$
then:
:$\ds \int_a^b \map f t \rd t \le \int_a^b \map g t \rd t$
Similarly, if:
:$\forall t \in \closedint a b: \map f t < \map g t$
then:
:$\ds \int_a^b \map f t \rd t < \int_a^b \map g t \rd t$
== Proof ==
Suppose that $\forall t \in \closedint a b: \map f t \le \map g t$.
From the Fundamental Theorem of Calculus, $g - f$ has a primitive on $\closedint a b$.
Let $H$ be such a primitive.
Then:
:$\forall t \in \closedint a b: D_t \map H t = \map g t - \map f t \ge 0$
By Derivative of Monotone Function, it follows that $H$ is increasing on $\closedint a b$.
Thus:
:$\map H b \ge \map H a$
Hence:
:$\ds \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$
The proof for the second case is similar.
== Also see ==
* Sign of Function Matches Sign of Definite Integral
== Sources ==
* : $\S 13.23$
Category:Integral Calculus
Category:Inequalities
Relative Sizes of Elements in Perturbed Proportion

== Theorem ==
That is, let:
:$a : b = e : f$
:$b : c = d : e$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$
== Proof ==
Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio.
Let the proportion of them be perturbed, that is:
:$A : B = E : F$
:$B : C = D : E$
Let $A > C$.
Then we need to show that $D > F$.
:
We have that $A > C$.
So from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.
But $A : B = E : F$, and $C : B = E : D$
So from Relative Sizes of Proportional Magnitudes $E : F > E : D$.
But from Relative Sizes of Magnitudes on Unequal Ratios $F < D$ and so $D > F$.
Similarly we can prove that $A = C \implies D = F$ and $A < C \implies D < F$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relative Sizes of Magnitudes on Unequal Ratios

== Theorem ==
That is:
:$a : c > b : c \implies a > b$
:$c : b > c : a \implies b < a$
== Proof ==
Let $A$ have to $C$ a greater ratio than $B$ has to $C$.
:
Suppose $A = B$.
Then from Ratios of Equal Magnitudes $A : C = B : C$.
But by hypothesis $A : C > B : C$, so $A \ne B$.
Suppose $A < B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $A : C < B : C$.
But by hypothesis $A : C > B : C$.
Therefore it must be that $A > B$.
Let $C$ have to $B$ a greater ratio than $C$ has to $A$.
Suppose $B = A$.
Then from Ratios of Equal Magnitudes $C : B = C : A$.
But by hypothesis $C : B > C : A$, so $B \ne A$.
Suppose $B > A$.
Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $C : B < C : A$.
But by hypothesis $C : B > C : A$.
Therefore it must be that $B < A$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relative Sizes of Platonic Solids in Same Sphere

== Summary ==
== Proof ==
Let:
:$\mathscr D$ be a regular dodecahedron
:$\mathscr I$ be a regular icosahedron
:$\mathscr C$ be a cube
which are inscribed in a given sphere.
Let $\map e {\mathscr C}, \map e {\mathscr D}, \map e {\mathscr I}$ be the edge (that is, the side) of $\mathscr C, \mathscr D, \mathscr I$ respectively.
Let $\map s {\mathscr C}, \map s {\mathscr D}, \map s {\mathscr I}$ be the area of the surfaces of $\mathscr C, \mathscr D, \mathscr I$ respectively.
Let $\map v {\mathscr C}, \map v {\mathscr D}, \map v {\mathscr I}$ be the volumes of $\mathscr C, \mathscr D, \mathscr I$ respectively.
Let $AB$ be cut at $C$ in extreme and mean ratio such that $AC$ is the greater segment.
From :
:$\map e {\mathscr C}^2 : \map e {\mathscr I}^2 = \paren {PQ^2 + PR^2} : \paren {PQ^2 + QR^2}$ 
where:
:$PQ$ is the radius of the circle which circumscribes the faces of both $\mathscr D$ and $\mathscr I$
:$R$ is the point at which $PQ$ has been cut in extreme and mean ratio such that $PR$ is the greater segment.
Thus from :
:$AB : AC = PQ : PR$
Thus:
:$\paren {PQ^2 + PR^2} : \paren {PQ^2 + QR^2} = \paren {AB^2 + AC^2} : \paren {AB^2 + BC^2}$
Hence:
:$\map e {\mathscr C} : \map e {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
and so $(1)$ is shown to be true.
From :
:$\map s {\mathscr D} : \map s {\mathscr I} = \map e {\mathscr C} : \map e {\mathscr D}$
and so $(2)$ is shown to be true.
From :
:$\map e {\mathscr C} : \map e {\mathscr D} = \map v {\mathscr D} : \map v {\mathscr I}$
But from $(2)$ above:
:$\map s {\mathscr D} : \map s {\mathscr I} = \map e {\mathscr C} : \map e {\mathscr D}$
Thus from :
:$\map v {\mathscr D} : \map v {\mathscr I} = \map s {\mathscr D} : \map s {\mathscr I}$
and so $(3)$ is shown to be true.
From $(1)$ above:
:$\map e {\mathscr C} : \map e {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
From :
:$\map e {\mathscr C} : \map e {\mathscr D} = \map v {\mathscr D} : \map v {\mathscr I}$
Thus from :
:$\map v {\mathscr D} : \map v {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
and so $(4)$ is shown to be true.
Result $(3)$ appears to have been included in Comparison of the Dodecahedron with the Icosahedron by . This work is now lost.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
* : Chapter $\text {A}.6$: Apollonius (ca. B.C.)
Category:Platonic Solids
Relative Sizes of Proportional Magnitudes

== Theorem ==
That is:
:$a : b = c : d, c : d > e : f \implies a : b > e : f$
== Proof ==
Let a first magnitude $A$ have to a second $B$ the same ratio as a third $C$ to a fourth $D$.
Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.
:
We have that $C : D > E : F$.
From , there will be some equimultiples of $C, E$ and other arbitrary equimultiples of $D, F$ such that the multiple of $C$ is in excess of the multiple of $D$, while the multiple of $E$ is not in excess of the multiple of $F$.
Let these equimultiples be taken.
Let $G, H$ be equimultiples of $C, E$, and $K, L$ be other arbitrary equimultiples of $D, F$, so that $G > K$ but $H \le L$.
Whatever multiple $G$ is of $C$, let $M$ be also that multiple of $A$.
Also, whatever multiple $K$ is of $D$, let $N$ be also that multiple of $B$.
Now we have that $A : B = C : D$ and of $A, C$ equimultiples $M, G$ have been taken.
We also have that of $B, D$ other arbitrary equimultiples $N, K$ have been taken.
Therefore:
: $M > N \implies G > K$
: $M = N \implies G = K$
: $M < N \implies G < K$
from .
But $G > K$ and so $M > N$.
But $H \le L$, and:
: $M, H$ are equimultiples $A, E$
: $N, L$ are other, arbitrary equimultiples $B, F$.
Therefore from , $A : B > E : F$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relative Sizes of Ratios on Unequal Magnitudes

== Theorem ==
That is:
:$a > b \implies a : c > b : c$
:$a > b \implies c : a < c : b$
== Proof ==
Let $AB, C$ be unequal magnitudes, and let $AB$ be greater.
Let $D$ be another arbitrary magnitude.
We are to show that $AB$ has to $D$ a greater ratio than $C$ has to $D$, and $D$ has to $C$ a greater ratio than it has to $AB$.
We have that $AB > C$, so let $BE = C$.
Then from , the lesser of the magnitudes $AE, EB$, if multiplied, will eventually be greater than $D$.
There are two cases.
;First, let $AE < EB$.
:
Let $AE$ be multiplied by some number, and let $FG$ be a multiple of it which is greater than $D$.
Then whatever multiple $FG$ is of $AE$, let $GH$ be made the same multiple of $EB$ and $K$ of $C$.
Let $L := 2D, M := 3D, N := 4D \ldots$ until one of these multiples is greater than $K$.
Suppose $M \le K$ while $N > K$.
WE have that $FG$ is the same multiple of $AE$ that $GH$ is of $EB$.
So from Multiplication of Numbers is Left Distributive over Addition, $FG$ is the same multiple of $AE$ that $FHK$ is of $AB$.
But $FG$ is the same multiple of $AE$ that $K$ is of $C$.
Therefore $FH, K$ are equimultiples of $AB, C$.
Again, we have that $GH$ is the same multiple of $EB$ that $K$ is of $C$, and $EB = C$.
Therefore $GH = K$.
But $K \ge M$ and so $GH \ge M$.
Also, $FG > D$ so $FH > D + M$.
But $D + M = N$ by the construction of $M$ and $N$, and $FH > M + D$.
So $FH > N$ while $K \le N$.
Also, $FH, K$ are equimultiples of $AB, C$ while $N$ is a multiple of some arbitrary $D$.
So from Ratios of Equal Magnitudes $AB$ has a greater ratio to $D$ than $C$ has to $D$.
Next, note that with the same construction, we can show similarly that $N > K$ while $N \le FH$.
Also we have that $N$ is a multiple of $D$, while $FH, K$ are other equimultiples of $AB, C$.
Therefore $D$ has a greater ratio to $C$ than $D$ has to $AB$.
;Second, let $AE > EB$.
From , $EB$, if multiplied, will eventually be greater than $D$.
:
Let it be so multiplied, and let $GH$ be a multiple of $EB$ and greater than $D$.
Whatever multiple $GH$ is of $EB$, let $FG$ be made the same multiple of $AE$, and $K$ of $C$.
Then we can prove similarly that $FH, K$ are equimultiples of $AB, C$.
Similarly, let $M, N$ be consecutive multiples of $D$ such that $M \le FG$ and $N > FG$.
But $GH > D$, therefore $FH > D + M$, that is, $FH > N$.
Now $K \le N$ inasmuch as $FG$ also, which is greater than $GH$, that is, than $K$, is not in excess of $N$.
Hence the result.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relative Sizes of Sides of Acute Triangle

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be an acute triangle, and so by definition, $\angle ABC$ is acute.
Construct a perpendicular $AD$ from $BC$.
Then the square on $AC$ is less than those on $AB$ and $BC$ by twice the rectangle contained by $BC$ and $BD$.
The proof is as follows.
From Square of Difference, squares on $BC$ and $BD$ equal the square on $DC$ plus twice the rectangle contained by $BC$ and $BD$.
Add the square on $AD$ to each.
So the squares on $BC$, $BD$ and $AD$ together equal the squares on $DC$ and $AD$ plus twice the rectangle contained by $BC$ and $BD$.
But by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $BD$ because $\angle ADB$ is a right angle.
Also by Pythagoras's Theorem, the square on $AC$ equals the squares on $AD$ and $DC$ because $\angle ADC$ is a right angle.
So the squares on $AB$ and $BC$ equals the square on $AC$ plus twice the rectangle contained by $BC$ and $BD$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Triangles
Relative Sizes of Sides of Obtuse Triangle

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be an obtuse triangle where $\angle BAC$ is obtuse.
Produce $CA$ past $A$.
Construct a perpendicular $DB$ from $CA$ produced.
Then the square on $BC$ is greater than those on $BA$ and $AC$ by twice the rectangle contained by $CA$ and $AD$.
The proof is as follows.
From Square of Sum, the square on $DC$ equals the squares on $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$.
Add the square on $DB$ to each.
So the squares on $CD$ and $DB$ together equal the squares on $DB$, $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$.
But by Pythagoras's Theorem, the square on $CB$ equals the squares on $CD$ and $DB$ because $\angle BDC$ is a right angle.
Also by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $DB$ because $\angle BDA$ is a right angle.
So the square on $CB$ equals the squares on $CA$ and $AB$ plus twice the rectangle contained by $CA$ and $AD$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Triangles
Relative Sizes of Successive Ratios

== Theorem ==
That is, let:
:$a : b = d : e$
:$b : c = e : f$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$
== Proof ==
Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two are in the same ratio:
:$A : B = D : E$
:$B : C = E : F$
and let $A > C$ ex aequali.
We need to show that $D > F$.
Similarly, we need to show that $A = C \implies D = F$ and $A < C \implies D < F$.
:
Since $A > C$ we have from Relative Sizes of Ratios on Unequal Magnitudes that $A : B > C : B$.
But $A : B = D : E$ and $C : B = F : E$.
Therefore from Relative Sizes of Proportional Magnitudes $D : E > F : E$.
But from Relative Sizes of Magnitudes on Unequal Ratios it follows that $D > F$.
Similarly we can show that $A = C \implies D = F$ and $A < C \implies D < F$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Relatively Compact Subspace/Examples/Open Unit Interval in Itself

== Example of Relatively Compact Subspace ==
The open unit interval $\openint 0 1$ is not a relatively compact subspace of $\openint 0 1$ itself.
== Proof ==
From Open Real Interval is not Compact, $\openint 0 1$ is not a compact space no matter what topological space it is embedded in.
From Underlying Set of Topological Space is Clopen, $\openint 0 1$ is both closed and open in $\openint 0 1$.
From Set is Closed iff Equals Topological Closure:
:$\openint 0 1 = \cl {\openint 0 1}$
The result follows by definition of relatively compact subspace.
== Sources ==
* : $5$: Compact spaces: $5.4$: Properties of compact spaces
Category:Relatively Compact Subspaces
Relatively Compact Subspace/Examples/Open Unit Interval in Reals

== Example of Relatively Compact Subspace ==
The open unit interval $\openint 0 1$ is a relatively compact subspace of the real number line $\R$.
== Proof ==
From Closure of Open Real Interval is Closed Real Interval, the closure of $\openint 0 1$ is $\closedint 0 1$.
From Closed Bounded Subset of Real Numbers is Compact, $\closedint 0 1$ is compact in $\R$.
The result follows by definition of relatively compact subspace.
== Sources ==
* : $5$: Compact spaces: $5.4$: Properties of compact spaces
Category:Relatively Compact Subspaces
Relativisation is Standard Model

== Theorem ==
Let $P$ be a well-formed formula.
Let $A$ be a finite set such that $x \in A$ $x$ is a free variable in $P$.
Then:
:$A \subseteq B \implies \paren {B \models P \iff P^B}$
== Proof ==
The proof proceeds by Induction on Well-Formed Formulas of $P$.
$P$ must be of the form:
:$x \in y$
:$\paren {Q \land R}$
:$\neg Q$
or:
:$\forall x: Q$
for some propositions $Q$ and $R$.
=== Atoms ===
Let $P$ be of the form $x \in y$.
Then:
:$A = \set {x, y}$
By definition of standard structure:
:$B \models P \iff \paren {x \in y \land x \in B \land y \in B}$
By definition of relativisation:
:$P^B \iff x \in y$
If $A \subseteq B$, then:
:$x \in B \land y \in B$
and the two statements are equivalent.
=== Inductive Step for $\neg$ ===
Let $P$ be of the form $\neg Q$ and that the statement holds for $Q$.
Then the free variables in $Q$ are precisely those in $P$.
=== Inductive Step for $\implies$ ===
Suppose $P$ is of the form $\paren {Q \implies R}$.
Suppose, further, that the statement holds for $Q$ and $R$.
The free variables of $Q$ and $R$ have to all be members of $A$, and thus are members of $B$ since $A \subseteq B$.
=== Inductive Step for $\forall x:$ ===
Let $P$ be of the form:
:$\forall x: Q$
Let the statement hold for $Q$.
Then the free variables of $Q$ are either in $A$ or $x$.
Furthermore, $x \notin A$ because:
:$x \in A \implies x$ is a free variable in $\forall x: Q$
which is a contradiction.
== Sources ==
* : $\S 12.5$
Category:Inner Model Theory
Reliability in terms of Lifetime Distribution Function

== Theorem ==
Let $D$ be a device whose reliability as a function of time is $\map {\mathrm R} t$.
Then:
:$\map {\mathrm R} t = 1 - \map {\mathrm F} t$
where $\map {\mathrm F} t$ is the lifetime distribution function of $D$ $t$.
== Proof ==
== Sources ==
* 
* 
Category:Reliability
Remainder is Arithmetically Definable

== Theorem ==
Let $\operatorname{rem}: \N^2 \to \N$ be defined as:
:$\map \rem {n, m} = \begin{cases}
\text{the remainder when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$
where the $\text{remainder}$ is as defined in the Division Theorem:
:If $n = m q + r$, where $0 \le r < m$, then $r$ is the remainder.
Then there exists a $\Sigma_1$ WFF of $3$ free variables:
:$\map \phi {r, n, m}$
such that:
:$r = \map \rem {n, m} \iff \N \models \map \phi {\sqbrk r, \sqbrk n, \sqbrk m}$
where $\sqbrk a$ denotes the unary representation of $a \in \N$.
== Proof ==
Define:
:$\map \phi {r, n, m} := \paren {m = 0 \land r = 0} \lor \paren {m \ne 0 \land r < m \land \exists q: n = \paren {m \times q} + r}$
Suppose $m = 0$.
Then, only the first case:
:$m = 0 \land r = 0$
can possibly hold.
Therefore, the formula holds $r = 0$, which matches the definition.
Suppose $m \ne 0$.
Then, only the second case:
:$m \ne 0 \land r < m \land \exists q: n = \paren {m \times q} + r$
can possibly hold.
But:
:$r < m \land \exists q: n = \paren {m \times q} + r$
is the definition of $r$.
Therefore, in all cases:
:$r = \map \rem {n, m} \iff \N \models \map \rem {\sqbrk r, \sqbrk n, \sqbrk m}$
Finally, that $\phi$ is $\Sigma_1$ follows from:
:Conjunction of Existential Quantifier
:Existential Quantifier Distributes over Disjunction
Remainder is Primitive Recursive

== Theorem ==
Let $m, n \in \N$ be natural numbers.
Let us define the function $\operatorname{rem}: \N^2 \to \N$:
:$\map \rem {n, m} = \begin{cases}
\text{the remainder when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$
where the $\text{remainder}$ is as defined in the Division Theorem:
:If $n = m q + r$, where $0 \le r < m$, then $r$ is the remainder.
Then $\rem$ is primitive recursive.
== Proof ==
Let $\map \rem {n, m} = r$.
We see that as $n$ increases by $1$, then so does $r$, except when $n = m-1$ in which case increasing $n$ by $1$ makes $r$ go to $0$.
We also take into account the case where $m = 0$:
So we can define $\rem$ by cases:
:$\map \rem {n, m} = \begin{cases}
0 & : \map \rem {n - 1, m} = m - 1 \text { or } m = 0 \text { or } n = 0\\
\map \rem {n - 1, m} + 1 & : \text {otherwise} \\
\end{cases}$
We remind ourselves of the following primitive recursive functions:
* Signum function $\sgn$
* Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$
* Cut-off subtraction $m \, \dot - \, n$.
We have:
* $\map \sgn m = 1 \iff m \ne 0$
* $\map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} } = 1 \iff \map \rem {n, m} \ne m \, \dot - \, 1$.
So:
:$\map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} } = 1 \iff m > 0 \land \map \rem {n, m} \ne m \, \dot - \, 1$.
So we see that:
:$\map \rem {n + 1, m} = \paren {\map \rem {n, m} + 1} \, \map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} }$
So $\rem$ is obtained by primitive recursion (over the first variable, which is allowed by Permutation of Variables of Primitive Recursive Function) from the primitive recursive functions:
* Signum function $\sgn$
* Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$
* Cut-off subtraction $m \, \dot - \, n$
* Addition
* Multiplication.
Thus we can use Definition by Cases is Primitive Recursive and it follows that $\rem$ is primitive recursive.
Category:Primitive Recursive Functions
Remainder of Fibonacci Number Divided by Fibonacci Number is Plus or Minus Fibonacci Number

== Theorem ==
Let $F_n$ and $F_m$ be Fibonacci numbers.
By the Division Theorem, let:
:$F_n = q F_m + r$
where:
:$q \in \Z$
:$r \in \Z: 0 \le r < \size {F_m}$
Then either $r$ or $\size {F_m} - r$, or both, is a Fibonacci number.
== Proof ==
Follows directly from Residue of Fibonacci Number Modulo Fibonacci Number.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $32$
Category:Fibonacci Numbers
Remainder on Division is Least Positive Residue

== Theorem ==
Let $a, b \in \Z$ be integers such that $a \ge 0$ and $b \ne 0$.
Let $r$ be the remainder resulting from the operation of integer division of $a$ by $b$:
$a = q b + r, 0 \le r < \size b$
Then $r$ is equal to the least positive residue of $a \pmod b$.
== Theorem ==
By definition of least positive residue:
:$a = q b + r \iff r \equiv a \pmod b$
for some $q \in \Z$.
By the Division Theorem, there exists a $q$ such that:
:$0 \le r < \size b$
which is precisely the definition of the least positive residue of $a \pmod b$.
== Sources ==
* : $\S 2.5$. Congruence of integers: Example $39$
Category:Modulo Arithmetic
Removable Discontinuity/Examples/Example 1

== Example of Removable Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x^2 - 1} {x - 1}$
Then $f$ has a removable discontinuity at $x = 1$.
In this case the removable discontinuity may be removed by defining $\map f 1$ to equal $2$.
== Sources ==
* 
Category:Examples of Removable Discontinuities
Removable Discontinuity/Examples/Example 2

== Example of Removable Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} x \map \sin {\dfrac 1 x} & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Then $f$ has a removable discontinuity at $x = 0$.
In this case the removable discontinuity may be removed by redefining $\map f 0$ to equal $0$.
== Sources ==
* 
* 
Category:Examples of Removable Discontinuities
Removable Singularity/Examples/Sine of z over z

== Example of Removable Singularity ==
Let $f: \C \setminus \set 0 \to \C$ be the complex function defined as:
:$\map f z = \dfrac {\sin z} z$
Then $f$ has a removable singularity at the point $z = 0$.
== Proof ==
We are given that $f: \C \setminus \set 0 \to \C$ is the complex function:
:$\map f z = \dfrac {\sin z} z$
By Limit of Sine of X over X at Zero:
:$\ds \lim_{z \mathop \to 0} \frac {\sin z} z = 1$
So $f$ is holomorphic
Hence $f$ has a removable singularity at the point $z = 0$.
== Sources ==
* 
* 
Category:Examples of Removable Singularities
Removable Singularity at Infinity implies Constant Function

== Theorem == 
Let $f : \C \to \C$ be an entire function.
Let $f$ have an removable singularity at $\infty$. 
Then $f$ is constant.
 
== Proof ==
We are given that $f$ has a removable singularity at $\infty$.
By Riemann Removable Singularities Theorem, $f$ must be bounded in a neighborhood of $\infty$.
That is, there exists a real number $M > 0$ such that:
:$\forall z \in \set {z : \cmod z > r}: \cmod {\map f z} \le M$
for some real $r \ge 0$.
However, by Continuous Function on Compact Space is Bounded, $f$ is also bounded on $\set {z: \cmod z \le r}$. 
As $\set {z: \cmod z > r} \cup \set {z: \cmod z \le r} = \C$, $f$ is therefore bounded on $\C$.
Therefore by Liouville's Theorem, $f$ is constant.
Category:Removable Singularities
Category:Entire Functions
Category:Constant Mappings
Renaming Mapping/Examples/Projection of Plane onto x-axis

== Example of Renaming Mapping ==
Let $P$ denote the Cartesian plane.
Let $X$ denote the $x$-axis of $P$.
Let $\pi_x: P \to X$ be the perpendicular projection of $P$ onto $X$.
Then:
:the equivalence relation $\RR_\pi$ induced by $\pi_x$ is:
::$p_1 \mathrel {\RR_\pi} p_2 \iff p_1$ and $p_2$ are on the same vertical line
:the quotient set $P / \RR_\pi$ of $P$ determined by $\RR_\pi$ is the set of points of the $x$-axis
:the equivalence class $\eqclass p {\RR_f}$ of $p$ under $\RR_f$ is the $x$-coordinate of $p$.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
Category:Quotient Sets