nyuuzyou/wikis huggingface dataset:
Limit to Infinity of Summation of Euler Phi Function over Square

== Theorem ==
:$\ds \lim_{n \mathop \to \infty} \dfrac {\map \Phi n} {n^2} = \dfrac 3 {\pi^2}$
where:
:$\map \Phi n = \ds \sum_{k \mathop = 1}^n \map \phi k$
:$\map \phi k$ is the Euler $\phi$ function of $k$.
Numerically, this evaluates to:
:$\dfrac 3 {\pi^2} \approx 0 \cdotp 30396 35509 \ldots$
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* 
* : Chapter $\text V$. Euler's $\phi$-Function, Generalizations, Farey Series
* : $0,30396 35509 \ldots$
Category:Euler Phi Function
Limit to Infinity of Summation of Euler Phi Function over Square/Proof 1

== Theorem ==
== Proof ==
== Sources ==
* : $\text {XVI}$. THE ARITHMETICAL FUNCTIONS $\map \phi n$, $\map d n$, $\map \sigma n$, $\map r n$: $18.5$ The average order of $\map \phi n$
Category:Limit to Infinity of Summation of Euler Phi Function over Square
Limit to Infinity of Summation of Euler Phi Function over Square/Proof 2

== Theorem ==
== Proof ==
We find the probability that the two random numbers not less than $n$ is coprime.
We pick randomly $x, y \in \set {1, \dots, n}$.
There are $n^2$ ways to do this.
There are three cases:
If $x > y$, for each $x$, there are $\map \phi x$ values of $y$ so that $x \perp y$.
Since $1 \le x \le n$, there are a total of $\map \phi 1 + \dots + \map \phi n = \map \Phi n$ valid values of $y$.
If $x < y$, the result is similar: there are $\map \Phi n$ valid values of $x$.
If $x = y$, $x \perp y$ only if $x = y = 1$.
Therefore the probability that the two random numbers not less than $n$ is coprime is:
:$\dfrac {\map \Phi n + \map \Phi n + 1} {n^2}$
Taking limit to infinity and using Probability of Two Random Integers having no Common Divisor:
Category:Limit to Infinity of Summation of Euler Phi Function over Square
Limit to Infinity of x minus Gamma of Reciprocal of x

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \paren {x - \map \Gamma {\dfrac 1 x} } = \gamma$
where:
:$\Gamma$ denotes the $\Gamma$ (Gamma) function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
== Historical Note ==
== Sources ==
* : $0,57721 56649 \ldots$
Category:Gamma Function
Category:Euler-Mascheroni Constant
Limit to Infinity of x minus Gamma of Reciprocal of x/Historical Note

== Historical Note on Limit to Infinity of x minus Gamma of Reciprocal of x ==
 and , in their of $1983$, attribute this to a $1976$ result of , but this has not been corroborated.
== Sources ==
* : $0,57721 56649 \ldots$
Category:Historical Notes
Limit to Negative Infinity of Distribution Function of Finite Signed Borel Measure

== Theorem ==
Let $\mu$ be a finite signed Borel measure on $\R$.
Let $F_\mu$ be the distribution function of $\mu$. 
Then: 
:$\ds \lim_{x \mathop \to -\infty} \map {F_\mu} x = 0$
where $\ds \lim_{x \mathop \to -\infty}$ denotes the limit at $-\infty$.
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
From Decomposition of Distribution Function of Finite Signed Borel Measure, we have: 
:$F_\mu = F_{\mu^+} - F_{\mu^-}$
where $F_{\mu^+}$ and $F_{\mu^-}$ are the distribution functions of $\mu^+$ and $\mu^-$ respectively.
From Limit of Distribution Function of Finite Borel Measure at Negative Infinity, we have: 
:$\ds \lim_{x \mathop \to -\infty} \map {F_{\mu^+} } x = 0$
and:
:$\ds \lim_{x \mathop \to -\infty} \map {F_{\mu^-} } x = 0$
From Properties of Limit at Minus Infinity of Real Function: Difference Rule, we have that: 
:$\ds \lim_{x \mathop \to -\infty} \paren {\map {F_{\mu^+} } x - \map {F_{\mu^-} } x} = 0 - 0 = 0$
So:
:$\ds \lim_{x \mathop \to -\infty} \map {F_\mu} x = 0$
== Sources ==
* : $4.4$: Functions of Finite Variation
Category:Distribution Function of Finite Signed Borel Measure
Limit with Epsilon Powers of 2

== Theorem ==
Let $\openint a b$ be an open real interval.
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a real function.
Let $L \in \R$.
Suppose that:
:$\forall n > 0 \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $
Then the limit of $f$ exists as $x$ tends to $c$, and is equal to $L$.
== Proof ==
Denote by $\map P {2^{-n},\delta}$ the proposition considered in the theorem exposition:
:$\forall \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $
Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in the definition of limit:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$.
Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.
From the Axiom of Archimedes:
:$\forall \epsilon > 0: \exists n \in \N: n > \dfrac 1 \epsilon$.
As $2^n > n$ for all $n$, we have that $2^n > \dfrac 1 \epsilon$.
By Ordering of Reciprocals, $2^{-n} < \epsilon$.
Therefore $\size {\map f x - L} < \epsilon$ is implied by $\size {\map f x - L} < 2^{-n}$.
Thus $\map P {\epsilon,\delta}$ is a weaker statement $\map P {2^{-n},\delta}$ 
== Also see ==
* Limit with Rational Epsilon and Delta
Category:Limits of Real Functions
Limit with Rational Epsilon and Delta

== Theorem ==
Let $\openint a b$ be an open real interval.
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a real function.
Let $L \in \R$.
Suppose that:
:$\forall \epsilon > 0 \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Then the limit of $f$ exists as $x$ tends to $c$, and is equal to $L$.
== Proof ==
Recall the definition of a limit of a real function:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.
Suppose that $\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ in the interval of interest:
:$\forall \epsilon \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Let $\epsilon_0 > 0$ be a real number.
From Between two Real Numbers exists Rational Number, there is some rational $\epsilon_0'$ such that $0 < \epsilon_0' < \epsilon_0$.
Then $\size {\map f x - L} < \epsilon_0'$ implies $\size {\map f x - L} < \epsilon_0$.
Thus $\map P {\epsilon,\delta}$ holding for all rational $\epsilon$ implies $\map P {\epsilon,\delta}$ for all real $\epsilon$.
Now suppose $\map P {\epsilon,\delta}$ has been established for some rational $\epsilon$ and some real $\delta_0$.
From Between two Real Numbers exists Rational Number, there is some rational $\delta_0'$ such that $\size {x-c} < \delta_0' < \delta_0$.
Then $\size {x-c} < \delta_0$ is implied by $\size {x-c} < \delta_0'$
This means $\map P {\epsilon,\delta_0'}$ implies $\map P {\epsilon,\delta_0}$ holds.
We conclude:
:$\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ and $\delta$ in the interval of interest
is a stronger statement than:
:$\map P {\epsilon,\delta}$ holds for all real $\epsilon$ and $\delta$ in the interval of interest.
== Also see ==
* Limit with Epsilon Powers of 2
Category:Limits of Real Functions
Limiting Area of Polygon with given Perimeter

== Theorem ==
Let $\PP$ be the set of plane geometric figures with perimeter $L$.
The element of $P$ with the largest area is the circle of radius $\dfrac L {2 \pi}$ which has area $\dfrac {L^2} {4 \pi}$.
== Proof ==
== Historical Note ==
== Sources ==
* : $6$
* : $6$
Category:Plane Geometry
Limits of Real and Imaginary Parts

== Theorem ==
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 $\cmod {z - z_0} < \delta$:
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:
The proof for equation $(2)$ with imaginary parts follows when $\Re$ is replaced by $\Im$ in the equations above.
Category:Limits of Complex Functions
Limsup Squeeze Theorem

== Theorem ==
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let:
:$(1): \quad \forall n \ge n_0: \size {x_n} \le y_n$
:$(2): \quad \ds \limsup_{n \mathop \to \infty} \paren {y_n} = 0$, where $\limsup$ denotes the limit superior.
Then:
:$\ds \lim_{n \mathop \to \infty} x_n = 0$
== Direct Proof ==
Since $\size {x_n} \ge 0$, we have that $y_n \ge 0$.
Therefore, we know:
:$\ds 0 \le \liminf_{n \mathop \to \infty} \paren {y_n} \le \limsup_{n \mathop \to \infty} \paren {y_n}$
where $\liminf$ denotes the limit inferior.
So:
:$\ds \liminf_{n \mathop \to \infty} \paren {y_n} = \limsup_{n \mathop \to \infty} \paren {y_n} = 0$
by the Squeeze Theorem.
Thus:
:$\ds \lim_{n \mathop \to \infty} \paren {y_n} = 0$
but:
:$\ds 0 \le \size {x_n} \le y_n \implies \lim_{n \mathop \to \infty} \size {x_n} = 0$
Therefore:
:$\ds \lim_{n \mathop \to \infty} \paren {-\size {x_n} } = 0$
Then since $-\size {x_n} \le x_n \le \size {x_n}$, it follows by the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} x_n = 0$
Category:Analysis
Category:Direct Proofs
Category:Named Theorems
Limsup and Liminf are Limits of Bounds

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let $\ds \overline l = \limsup_{n \mathop \to \infty} x_n$ be the limit superior and $\ds \liminf_{n \mathop \to \infty} x_n$ the limit inferior of $\sequence {x_n}$.
Then:
:$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
:$\ds \underline l = \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\inf_{k \mathop \ge n} x_k}$
== Proof ==
First we show that:
:$\ds \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
Let $M_n = \ds \sup_{k \mathop \ge n} x_k$.
By Supremum of Subset, the sequence $\sequence {M_n}$ decreases, for:
:$m \ge n \implies \set {k \in \N: k \ge m} \subseteq \set {k \in \N: k \ge n}$
A lower bound for $\sequence {x_n}$ is also a lower bound for $\sequence {M_n}$.
So from the Monotone Convergence Theorem (Real Analysis) it follows that $\sequence {M_n}$ converges.
Suppose $M_n \to M$ as $n \to \infty$.
From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $\sequence {x_n}$.
Let $L$ be the set of all numbers which are the limit of some subsequence of $\sequence {x_n}$.
Let $l \in L$ be arbitrary.
Let $\sequence {x_{n_r} }$ be a convergent subsequence of $\sequence {x_n}$ such that $x_{n_r} \to l$ as $r \to \infty$.
Then $\forall n_r \ge n: x_{n_r} \le M_n$.
Hence $l \le M_n$ by Lower and Upper Bounds for Sequences and hence (from the same theorem) $l \le M$.
This is true for all $l \in L$, so $\overline l = \limsup_{n \mathop \to \infty} x_n \le M$.
Now, from Terms of Bounded Sequence Within Bounds, we have that:
:$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$
Thus $\overline l + \epsilon$ is an upper bound for $\set {x_k: k \ge n}$.
So $M \le M_n \le \overline l + \epsilon$.
Thus from Real Plus Epsilon, $M \le \overline l$.
Thus we conclude that $M = \overline l$ and hence:
:$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$
:$\ds \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$
can be proved using a similar argument.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.15 \ (6)$
Category:Limits of Sequences
Lindelöf's Lemma

== Theorem ==
Let $C$ be a set of open real sets.
Let $S \subseteq \R$ be a subset of the real numbers that is covered by $C$.
Then there exists a countable subset of $C$ that covers $S$.
== Proof ==
=== Lemma ===
We have that $S$ is covered by $C$.
This means that $S$ is a subset of $\ds \bigcup_{O \mathop \in C} O$.
From the lemma:
:$\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$
where $D$ is a countable subset of $C$.
Hence $S$ is also a subset of $\ds \bigcup_{O \mathop \in D} O$.
In other words, $S$ is covered by $D$.
That is, $S$ is covered by a countable subset of $C$.
Hence the result.
Category:Real Analysis
Category:Lindelöf's Lemma
Lindelöf's Lemma/Lemma 1

== Theorem ==
Let $C$ be a set of open real sets.
Then there is a countable subset $D$ of $C$ such that:
:$\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$
== Proof ==
=== Lemma ===
Let $\ds U = \bigcup_{O \mathop \in C} O$.
Let $x$ be an arbitrary point in $U$.
Since $U$ is the union of the sets in $C$, the point $x$ belongs to a set in $C$.
Name such a set $O_x$.
Since $O_x$ is open, $O_x$ contains an open interval $I_x$ that contains $x$.
By Between two Real Numbers exists Rational Number, a rational number exists between the left hand endpoint of $I_x$ and $x$.
Also, a rational number exists between $x$ and the right hand endpoint of $I_x$.
Form an open interval $R_x$ that has two such rational numbers as endpoints.
All in all:
By the Lemma, $\set {R_x: x \in U}$ is countable as $\set {R_x: x \in U}$ is a set of open intervals with rational numbers as endpoints.
By Countable Set equals Range of Sequence, the countability of $\set {R_x: x \in U}$ means that there exists a sequence $\sequence {R^i}_{i \mathop \in N}$ where:
:$N$ is a subset of $\N$
:$\set {R_x: x \in U}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.
From this follows:
:$\set {R_x: x \in U} = \set {R^i: i \in N}$ as $\set {R^i: i \in N}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.
Two sequences that differ only by one of them having duplicates, have the same range.
Therefore, it is possible to require that $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.
Now, let $i$ be an arbitrary natural number in $N$.
Let $R^i$ be an element in $\set {R^i: i \in N}$.
There is an $x$ in $U$ such that:
:$R_x = R^i$ as $\set {R_x: x \in U} = \set {R^i: i \in N}$
Also, we know that a set $O_x$ in $C$ exists such that:
:$R_x \subset O_x$
The uniqueness of the elements of N makes it possible to define a mapping $\chi$ that sends $i$ to $x$.
This allows us to define, for every $i$ in $N$:
:$O^i = O_x$ where $x = \map \chi i$
We find, for every $i$ in $N$ and $x = \map \chi i$:
:$O^i \in C$ as $O_x \in C$
:$R^i \subset O^i$ as $R_x \subset O_x$ and $R_x = R^i$ and $O_x = O^i$
Every $R^i$ where $i \in N$ is uniquely determined by $i$ as $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.
Therefore, it is possible to define a mapping from $\set {R^i: i \in N}$ to $\set {O^i: i \in N}$ that sends $R^i$ to $O^i$ for every $i$ in $N$.
Image of countable set under mapping is countable.
Therefore, $\set {O^i: i \in N}$ is countable as $\set {R^i: i \in N}$ is countable.
$\set {O^i: i \in N}$ is a subset of $C$ as $O^i \in C$ for every $i \in N$.
Therefore, $\set {O^i: i \in N}$ is a countable subset of $C$.
We find by focusing on $R_x$ for an $x$ in $U$:
So, $\set {R_x: x \in U}$ covers $U$.
We find by focusing on $O^i$ for an $i$ in $N$:
So, $D = \set {O^i: i \in N}$ satisfies the proposition of the theorem.
== Sources ==
* : Ch. $2$: Section $5$, proposition $9$.
Category:Lindelöf's Lemma
Lindelöf's Lemma/Lemma 2

== Lemma ==
Let $R$ be a set of real intervals with rational numbers as endpoints.
Let every interval in $R$ be of the same type of which there are four: $\openint \ldots \ldots$, $\closedint \ldots \ldots$, $\hointr \ldots \ldots$, and $\hointl \ldots \ldots$.
Then $R$ is countable.
== Proof ==
By Rational Numbers are Countably Infinite, the rationals are countable.
By Subset of Countably Infinite Set is Countable, a subset of the rationals is countable.
The endpoint of an interval in $R$ is characterized by a rational number as every interval in $R$ is of the same type.
Therefore, the set consisting of the left hand endpoints of every interval in $R$ is countable.
Also, the set consisting of the right hand endpoints of every interval in $R$ is countable.
The cartesian product of countable sets is countable.
Therefore, the cartesian product of the sets consisting of the respectively left hand and right hand endpoints of every interval in $R$ is countable.
A subset of this cartesian product is in one-to-one correspondence with $R$.
This subset is countable by Subset of Countably Infinite Set is Countable.
By Set which is Equivalent to Countable Set is Countable:
:$R$ is countable.
Category:Lindelöf's Lemma
Lindelöf Hypothesis

== Hypothesis ==
The Lindelöf hypothesis is a conjecture about the rate of growth of the Riemann zeta function on the critical line that is implied by the Riemann Hypothesis.
It states that:
:$\forall \epsilon \in \R_{> 0}: \map \zeta {\dfrac 1 2 + i t} \text{ is } \map \OO {t^\epsilon}$
as $t \to \infty$ (see big-O notation).
Since $\epsilon$ can always be replaced by a smaller value, we can also write the conjecture as:
:$\forall \epsilon \in \R_{> 0}: \map \zeta {\dfrac 1 2 + i t} \text{ is } \map o {t^\epsilon}$
as $t \to \infty$ (see little-o notation).
Category:Unproven Hypotheses
Category:Riemann Zeta Function
Lindelöf Property is Preserved under Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is a Lindelöf space, then $T_B$ is also a Lindelöf space.
== Proof ==
Let $T_A$ be a Lindelöf space.
Take an open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\set {\phi^{-1} \sqbrk U: U \in \UU}$ 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:
:$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} }, \ldots} = \set {U_1, \ldots, U_n, \ldots} \subseteq \UU$
is a countable subcover of $\UU$ on $T_B$.
== Also see ==
* Compactness Properties Preserved under Continuous Surjection
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Lindelöf Spaces
Category:Continuous Mappings
Category:Surjections
Lindelöf T3 Space is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ space which is also Lindelöf.
Then $T$ is paracompact.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Lindelöf Spaces
Category:T3 Spaces
Category:Paracompact Spaces
Line Joining Centers of Two Circles Touching Externally

== Theorem ==
== Proof ==
Let the circles $ABC$ and $ADE$ touch externally at $A$.
Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$.
We are to show that the straight line joining $F$ to $G$ passes through $A$.
:
Suppose, as in the diagram above, that it does not, and that it were possible for it to pass through $C$ and $D$, as $FCDG$.
(It is clear that the diagram does not have $F$ and $G$ as the actual centers of these circles - it is the point of this proof to demonstrate that this would not be possible.)
Join $AF$ and $AG$.
We have that:
* Since $F$ is the center of $ABC$, then $FA$ and $FC$ are both radii of $ABC$, and so $FA = FC$.
* Since $G$ is the center of $ADE$, then $GA$ and $GD$ are both radii of $ADE$, and so $GA = GD$.
So $FA + AG = FC + GD$.
So all of $FCDG$ is greater than $FA + AG$.
But from Sum of Two Sides of Triangle Greater than Third Side $FA + AG$ is greater than $FCDG$.
Hence we have a contradiction, and so $FG$ has to go through point $A$, the point of contact of the two circles.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Line Joining Centers of Two Circles Touching Internally

== Theorem ==
Let two circles touch internally.
Then the straight line joining their centers passes through the point where they touch.
== Proof ==
Let the circles $ABC$ and $ADE$ touch internally at $A$.
Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$.
We are to show that the straight line joining $F$ to $G$ passes through $A$.
:
Suppose, as in the diagram above, that it does not.
Let $FG$ fall on $H$ instead.
It will also pass through $D$ on its way, which lies on circle $ADE$.
Join $AF$ and $AG$.
From Sum of Two Sides of Triangle Greater than Third Side $AG + GF$ is greater than $AF$.
Therefore $AG + GF$ is greater than $FH$ as $F$ is the center and both $AF$ and $FH$ are radii.
So, subtract $FG$ from both $AG + GF$ and $FH$.
It follows that $AG$ is greater than $GH$, and hence also greater than $GD$.
But as $G$ is the center of $ADE$, it follows that $AG$ is supposed to be equal to $GD$.
So the $AG + GF$ can not be greater than $AF$ and so must be coincident with it.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be two parallel straight lines.
Let $AB$ be at right angles to the plane of reference.
Then $CD$ needs to be demonstrated to be also at right angles to the plane of reference.
Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$.
Let $BD$ be joined.
Then by :
:$AB$, $CD$ and $BD$ are all in the same plane.
Let $DE$ be drawn in the plane of reference at right angles to $BD$ and equal to $AB$.
Let $BE$, $AE$ and $AD$ be joined.
We have that $AB$ is at right angles to the plane of reference.
Therefore from , $AB$ is at right angles to all the straight lines which meet it and are in the plane of reference.
Therefore each of $\angle ABD$ and $\angle ABE$ is a right angle.
We have that the straight line $BD$ falls on the parallels $AB$ and $CD$.
Therefore from Parallelism implies Supplementary Interior Angles $\angle ABD$ and $\angle CDB$ are equal to two right angles.
But $\angle ABD$ is a right angle.
Therefore $\angle CDB$ is also a right angle.
Therefore $CD$ is at right angles to $BD$.
We have that $AB = DE$, and that $BD$ is common.
The two sides $AB$ and $BD$ of $\triangle ABD$ are equal to the two sides $ED$ and $DB$ of $\triangle EBD$.
Also $\angle ABD = \angle EDB$, as they are both right angles.
Therefore $AD = BE$.
We have that:
:$AB = DE$
and:
:$BE = AD$
Thus the two sides $AB$ and $BE$ of $\triangle ABE$ are equal to the two sides $ED$ and $DA$ of $\triangle EAD$.
We have that $AE$ is common.
So:
: $\angle ABE = \angle EDA$
But $\angle ABE$ is a right angle.
Therefore $\angle EDA$ is a right angle.
Therefore $ED$ is at right angles to $AD$.
But $ED$ is also at right angles to $DB$.
Therefore by :
:$ED$ is at right angles to the plane through $DA$ and $BD$.
But from :
:$DC$ is in the plane through $DA$ and $BD$ inasmuch as $AB$ and $BD$ are in the same plane which $AB$ and $BD$ are.
Also $DC$ is in the same plane which $AB$ and $BD$ are.
Therefore $ED$ is at right angles to $DC$.
So $CD$ is also at right angles to $DE$.
But $CD$ is also at right angles to $BD$.
Therefore $CD$ is set up at right angles to the straight lines $DE$ and $DB$ which cut one another, at $D$.
So from :
:$CD$ is also at right angles to the plane through $DE$ and $DB$ are.
But the plane through $DE$ and $DB$ is the plane of reference.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Lines
Category:Perpendiculars
Category:Planes