nyuuzyou/wikis huggingface dataset:
Capelli's Sum

== Theorem ==
:$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$x^{\overline k}$ denotes $x$ to the $k$ rising.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
:$\ds \paren {x + y}^{\overline m} = \sum_k \binom m k x^{\overline k} y^{\overline {m - k} }$
from which it is to be shown that:
:$\ds \paren {x + y}^{\overline {m + 1} } = \sum_k \binom {m + 1} k x^{\overline k} y^{\overline {m + 1 - k} }$
=== Induction Step ===
This is the induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{> 0}: \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $33$
Category:Rising Factorials
Capture-Recapture Sampling/Examples/Arbitrary Example 1

== Example of Capture-Recapture Sampling ==
Let $25$ squirrels be captured, marked and released.
On a later date, let $40$ squirrels be captured and inspected.
Let $5$ of those be ones which were marked on the first capture exercise.
Then a fair estimate of the entire population of squirrels is $200$.
== Proof ==
Let:
:$n_1$ be the number of squirrels captured, marked and released on the first round.
:$n_2$ be the number of squirrels captured on the second round.
:$m$ be the number of marked squirrels in the sample captured on the second round.
Then we have:
== Sources ==
* 
* 
Category:Examples of Capture-Recapture Sampling
Carathéodory's Theorem (Analysis)

== Theorem ==
Let $I \subseteq \R$.
Let $c \in I$ be an interior point of $I$.
Let $f : I \to \R$ be a real function.
Then $f$ is differentiable at $c$ :
:There exists a real function $\varphi : I \to \R$ that is continuous at $c$ and satisfies:
:$(1): \quad \forall x \in I: \map f x - \map f c = \map \varphi x \paren {x - c}$
:$(2): \quad \map \varphi c = \map {f'} c$
== Proof ==
=== Necessary Condition ===
Suppose $f$ is differentiable at $c$.
Then by definition $\map {f'} c$ exists.
So we can define $\varphi$ by:
:$\map \varphi x = \begin{cases}
\dfrac {\map f x - \map f c} {x - c} & : x \ne c, x \in I \\
\map {f'} c & : x = c
\end{cases}$
Condition $(2)$, that $\varphi$ is continuous at $c$, is satisfied, since:
Finally, condition $(1)$ is vacuous for $x = c$.
For $x \ne c$, it follows from the definition of $\varphi$ by dividing both sides of $(1)$ by $x - c$.
=== Sufficient Condition ===
Suppose a $\varphi$ as in the theorem statement exists.
Then for $x \ne c$, we have that:
:$\map \varphi x = \dfrac {\map f x - \map f c} {x - c}$
Since $\varphi$ is continuous at $c$:
:$\ds \map \varphi c = \lim_{x \mathop \to c} \map \varphi x = \lim_{x \mathop \to c} \frac {\map f x - \map f c} {x - c}$
That is, $f$ is differentiable at $c$, and $\map {f'} c = \map \varphi c$.
Category:Direct Proofs
Category:Real Analysis
Category:Differential Calculus
Carathéodory's Theorem (Convex Analysis)

== Theorem ==
Let $E \subset \R^l$.
Let $\mathbb x \in \map {\operatorname {conv} } E$, where $\map {\operatorname {conv} } E$ denotes the convex hull of $E$.
Then $\mathbf x$ is a convex combination of affinely independent points of $E$. 
In particular, $\mathbf x$ is a convex combination of at most $l + 1$ points of $E$.
== Proof ==
Since $\mathbb x \in \map {\operatorname {conv} } E$, $\mathbf x$ is a convex combination of points in $E$.
From the definition of convex combination:
:$\ds \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$
with:
:$\gamma_i \ge 0$
:$\ds \sum_{i \mathop = 1}^k \gamma_i = 1$
and:
:$\mathbf y_i \in E$
Let $K \subset \N$ be the set of all possible $k$ such that $\mathbf x$ is a convex combination of $k$ elements of $E$.
By the Well-Ordering Principle, $K$ is well-ordered. 
$K$ has a smallest element $k_s$ by the definition of well-ordered sets.
Suppose that the set $\set {\mathbf y_i: i \le k_s}$ is affinely dependent.
From the condition for affinely dependent set, there exists a set $\set {\alpha_i: i \le k_s}$ such that for some $\alpha_i > 0$:
:$\ds \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i = 0$
and:
:$\ds \sum_{i \mathop = 1}^{k_s} \alpha_i = 0$
Pick the smallest $\dfrac {\gamma_j} {\alpha_j} > 0$.
Then:
:$\ds \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} + 0 = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} - \paren {\frac {\gamma_j} {\alpha_j} \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i}$ 
Rearrange to get:
:$\ds \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \paren {\gamma_i - \frac {\gamma_j} {\alpha_j} \alpha_i} \mathbf y_i}$
Since $\gamma_j > 0$ is made the smallest and $\alpha_j > 0$ is made the greatest:
:$\dfrac {\alpha_i} {\alpha_j} \gamma_j \le \gamma_j \le \gamma_i$
for all $i \le k_s$.
Thus:
:$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} \ge 0$
For $i = j$:
:$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} = 0$
For $i \ne j$, let:
:$\gamma'_{i'} \equiv \gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i$
We can express:
:$\ds x = \sum_{i' \mathop = 1}^{k_s-1} \gamma'_{i'} \mathbf y_i$
which contradicts the assumption that $k_s = \min K$.
== Sources ==
* 
* : $\S 5.4.b$ Theorem $5.4.13$
* : $2.1$: Properties of convex sets
Category:Convex Analysis
Carathéodory's Theorem (Measure Theory)

== Theorem ==
Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a semi-ring of subsets of $X$.
Let $\mu: \SS \to \overline \R$ be a pre-measure on $\SS$.
Let $\map \sigma \SS$ be the $\sigma$-algebra generated by $\SS$.
Then $\mu$ extends to a measure $\mu^*$ on $\map \sigma \SS$.
=== Corollary ===
== Proof ==
== Sources ==
* : $6.1$
Category:Measure Theory
Carathéodory's Theorem (Measure Theory)/Corollary

== Corollary to Carathéodory's Theorem (Measure Theory) ==
Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a semi-ring of subsets of $X$.
Let $\mu: \SS \to \overline \R$ be a pre-measure on $\SS$.
Let $\map \sigma \SS$ be the $\sigma$-algebra generated by $\SS$.
Suppose there exists an exhausting sequence $\sequence {S_n}_{n \mathop \in \N} \uparrow X$ in $\SS$ such that:
:$\forall n \in \N: \map \mu {S_n} < +\infty$
Then the extension $\mu^*$ is unique.
== Proof ==
== Sources ==
* : $6.1$
Category:Measure Theory
Card Game with Bluffing is Completely Mixed Game

== Theorem ==
The game of cards with bluffing is a completely mixed game.
== Proof ==
From Solution to Card Game with Bluffing, the optimum strategy for the [[Definition:Card Game with Bluffing|card game with bluffing, is:
:For $A$, to use both pure strategies but favour $A_1$ by $2/3$ to $1/3$.
:For $B$, to use both pure strategies but favour $B_1$ by $2/3$ to $1/3$.
Hence the result by definition of completely mixed game.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Two-Person Zero-Sum Games
Cardano's Formula

== Theorem ==
Let $P$ be the cubic equation:
:$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
Then $P$ has solutions:
where:
where:
The expression $D = Q^3 + R^2$ is called the discriminant of the equation.
=== Real Coefficients ===
Let $a, b, c, d \in \R$.
Then:
=== Trigonometric Form of Solutions ===
== Proof ==
First the cubic is depressed, by using the Tschirnhaus Transformation:
:$x \to x + \dfrac b {3 a}$:
Now let:
:$y = x + \dfrac b {3 a}, Q = \dfrac {3 a c - b^2} {9 a^2}, R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$
Thus we have obtained the depressed cubic $y^3 + 3 Q y - 2 R = 0$.
Now let $y = u + v$ where $u v = -Q$.
Then:
Thus the resolvent equation is obtained.
This resolvent is seen to be a quadratic in $u^3$.
From Solution to Quadratic Equation:
:$u^3 = \dfrac {2 R \pm \sqrt {4 Q^3 + 4 R^2}} 2 = R \pm \sqrt {Q^3 + R^2}$
We have from above $u v = -Q$ and hence $v^3 = -\dfrac {Q^3} {u^3}$.
Let us try taking the positive root: $u^3 = R + \sqrt {Q^3 + R^2}$.
Then:
The same sort of thing happens if you start with $u^3 = R - \sqrt {Q^3 + R^2}$: we get $v^3 = R + \sqrt {Q^3 + R^2}$.
Thus we see that taking either square root we arrive at the same solution.
:
Let:
From Roots of Complex Number, we have the three cube roots of $u^3$ and $v^3$:
:$u = \begin{cases}
& S \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & S \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & S \\ 
\end{cases}$
:$v = \begin{cases}
& T \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & T \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & T \\ 
\end{cases}$
Because of our constraint $u v = -Q$, there are only three combinations of these which are possible such that $y = u + v$:
:$y = \begin{cases}
& S + T \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T} \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T} \\ 
\end{cases}$
As $y = x + \dfrac b {3 a}$, it follows that the three roots are therefore:
== Also presented as ==
== Also known as ==
== Examples ==
== Also see ==
* Tartaglia's Poem
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Cubic Equations: $3.8.2$
* : $\S 9$: Solutions of Algebraic Equations Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.3$
* : Introduction
* 
* : Historical Introduction: Polynomial Equations
* 
* : Chapter $4$: Lure of the Unknown: Cubic equations
* : $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.3.$
Category:Cubic Equations
Category:Cardano's Formula
Cardano's Formula/Also known as

== Cardano's Formula: Also known as ==
Cardano's Formula is also sometimes seen referred to as Cardan's Formula, from the English form of 's name, Jerome Cardan.
Some use the term Cardano's Method.
Some sources refer to it as the Tartaglia Formula, acknowledging the work of in its development.
== Sources ==
* 
* 
* 
* 
Category:Cardano's Formula
Cardano's Formula/Also presented as

== Cardano's Formula: Also presented as ==
Cardano's Formula can also be presented in the following form:
Let $P$ be the cubic equation:
:$x^3 + a_2 x^2 + a_1 x + a_0 = 0$
Then $P$ has solutions:
where:
where:
== Proof ==
From the main presentation of Cardano's formula, we have that:
Hence setting $a \gets 1$, $b \gets a_2$, $c \gets a_1$ and $d \gets a_0$:
after simplification.
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Cubic Equations: $3.8.2$
* : $\S 9$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.3$
* : Introduction
* : $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.3.$
Category:Cardano's Formula
Cardano's Formula/Examples/x^3 - 15x - 4

== Example of Use of Cardano's Formula ==
:$\ds x^3 - 15x - 4 = 0$
has solutions $x = 4$, $x = -2 + \sqrt 3$ and $x = -2 - \sqrt 3$.
== Proof ==
This is in the form:
:$a x^3 + b x^2 + c x + d = 0$
where:
:$a = 1$
:$b = 0$
:$c = -15$
:$d = -4$
From Cardano's Formula:
:$x = S + T$
where:
:$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
:$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$
where:
and:
Thus:
Using Cardano's Formula for Real Coefficients, we know from
:$D = Q^3 + R^2 = -121 < 0$
that all roots are real and unequal.
From Roots of Complex Number/Examples/Cube Roots of 2+11i, we have:
:$\sqrt [3] {\paren {2 + 11 i } } = \set {2 + i, -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3}}$
From Roots of Complex Number/Examples/Cube Roots of 2-11i, we have:
:$\sqrt [3] {\paren {2 - 11 i } } = \set {2 - i, -1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3}}$
We need to investigate $9$ sums and find solutions where the complex part vanishes.
Our $3$ solutions are:
Hence:
The result follows.
== Sources ==
* : Chapter $6$: Cardano and the Solution of the Cubic
Category:Cardano's Formula
Cardano's Formula/Historical Note

== Historical Note on Cardano's Formula ==
Cardano's Formula (in an incomplete form) was first published by in $1545$, in his .
He learned the technique from , who had sworn him to secrecy.
However, as learned in $1543$, the technique had in fact first been discovered by , so he no longer felt bound by his oath to .
The latter did not see the matter in the same light, and entered into a feud with that lasted a decade.
This technique was not actually analyzed in depth until the work of , who was the first one to solve the problem of what to do about the "imaginary numbers" that inevitably arose when using this formula.
The method detailed here is that given by , who delivered it some time around $1650$.
== Sources ==
* : Introduction
* : Chapter $4$: Lure of the Unknown: Cubic equations
Category:Historical Notes
Cardano's Formula/Real Coefficients

== Theorem ==
Let $P$ be the cubic equation:
:$a x^3 + b x^2 + c x + d = 0$
with $a \ne 0$.
Let $a, b, c, d \in \R$.
Let $D$ be the discriminant of $P$:
:$D := Q^3 + R^2$
where:
:$Q = \dfrac {3 a c - b^2} {9 a^2}$
:$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$
Then:
:$(1): \quad$ If $D > 0$, then one root is real and two are complex conjugates.
:$(2): \quad$ If $D = 0$, then all roots are real, and at least two are equal.
:$(3): \quad$ If $D < 0$, then all roots are real and unequal.
== Proof ==
From Cardano's Formula, the roots of $P$ are:
where:
=== Zero Discriminant ===
First the easy case: $D = 0$.
Hence $S = T = \sqrt [3] R$, and so $S + T = 2 \sqrt [3] R, S - T = 0$.
From the above, this gives us:
Thus the roots $x_2$ and $x_3$ are equal, and all three roots are real.
They are all equal when $R = 0$.
=== Positive Discriminant ===
Let $D = Q^3 + R^2 > 0$.
Then $S = R + \sqrt{Q^3 + R^2}$ and $T = R - \sqrt{Q^3 + R^2}$ are wholly real and distinct.
Therefore, so are $S + T$ and $S - T$.
Hence:
:$\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
and
:$\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$
are complex conjugates.
=== Negative Discriminant ===
Let $D = Q^3 + R^2 < 0$.
Then $\sqrt D = \pm i \left|{Q^3 + R^2}\right| = \pm i E$, say, where $E > 0$.
Thus $S^3 = R + i E, T^3 = R - i E$.
Let $\sqrt [3] {R + i E} = p + i q$, and so $\sqrt [3] {R - i E} = p - i q$.
Hence $S + T = 2 p, S - T = 2 i q$.
So:
Subtracting $\dfrac b {3 a}$ from the above, we obtain the three distinct real solutions:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Cubic Equations: $3.8.2$
* : $\S 9$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.3$
* 
* 
* : $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.3.$
Category:Cardano's Formula
Cardano's Formula/Trigonometric Form

== Theorem ==
Let $P$ be the cubic equation:
:$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
Let:
Let $a, b, c, d \in \R$.
Let the discriminant $D < 0$, where $D := Q^3 + R^2$.
Then the solutions of $P$ can be expressed as:
where:
:$\cos \theta = \dfrac R {\sqrt {-Q^3} }$
== Proof ==
From Cardano's Formula, the roots of $P$ are:
where:
Let $D = Q^3 + R^2 < 0$.
Then:
:$S^3 = R + i \sqrt {\size {Q^3 + R^2} }$
We can express this in polar form:
:$S^3 = r \paren {\cos \theta + i \sin \theta}$
where:
:$r = \sqrt {R^2 + \paren {\sqrt {Q^3 + R^2} }^2} = \sqrt {R^2 - \paren {Q^3 + R^2} } = \sqrt {-Q^3}$
:$\tan \theta = \dfrac {\sqrt {\size {Q^3 + R^2} } } R$
Then:
:$\cos \theta = \dfrac R {\sqrt {-Q^3} }$
Similarly for $T^3$.
The result:
follows after some algebra.
== Historical Note ==
== Sources ==
* : $\S 9$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.4$
* 
* 
* : $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.4.$
Category:Cardano's Formula
Cardinal Equal to Collection of All Dominated Ordinals

== Theorem ==
Let $S$ be a set.
Let $\preccurlyeq$ denote the dominance relation.
Let $\On$ denote the class of all ordinals.
Let $x = \set {y \in \On: y \preccurlyeq S}$
Then:
:$(1): \quad x$ is an element of the class of all cardinals.
:$(2): \quad$ There is no injection $f$ such that $f : x \to S$
== Proof ==
=== $x$ is an ordinal ===
$x$ is clearly a subset of the class of all ordinals.
Moreover, suppose $y \in x$ and $z \in y$.
Then $z \subseteq y$ by the fact that $y$ is an ordinal.
$y \in x$ means that $f : y \to S$ for some injective mapping $f$ by the definition of dominance.
But then, $f \restriction_z : z \to S$ is also an injection by Restriction of Injection is Injection.
Thus, $z \in x$ by the definition of $x$, so $x$ is transitive.
Therefore, $x$ is an ordinal, since it is a transitive subset of the ordinal class.
=== $x$ is a cardinal ===
 $x$ is not a cardinal.
It follows that $\size x \in x$ and $x \sim \set x$ by corollary to Cardinal of Cardinal Equal to Cardinal and Ordinal Number Equivalent to Cardinal Number.
Let $g: x \to \size x$ be a bijection.
Let $f: \size x \to S$ be an injection.
Then $f \circ g : x \to S$ is an injection by Composite of Injections is Injection.
It follows that $x \preccurlyeq S$, so $x \in x$.
This means that $x \subsetneq x$ by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction.
Therefore, $x$ must be an element of the class of all cardinals.
=== No injection $f: x \to S$ ===
 that there is an injection $f: x \to S$.
By the definition of $x$, it follows that $x \in x$.
This means that $x \subsetneq x$ by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction.
Therefore, no injection can exist.
== Sources ==
* : $\S 10.40$
Category:Cardinals
Cardinal Inequality implies Ordinal Inequality

== Theorem ==
Let $T$ be a set.
Let $\card T$ denote the cardinal number of $T$.
Let $x$ be an ordinal.
Then:
:$x < \card T \iff \card x < \card T$
== Proof ==
=== Sufficient Condition ===
By Cardinal Number Less than Ordinal, it follows that $\card x \le x$.
So if $x < \card T$, then $\card x < \card T$.
=== Necessary Condition ===
Suppose $\card T \le x$.
Then:
By the Rule of Transposition:
: $\neg \card T \le \card x \implies \neg \card T \le x$
By Ordinal Membership is Trichotomy:
:$x < \card T \iff \card x < \card T$
== Sources ==
* : $\S 10.31$
Category:Cardinals
Category:Ordinals
Cardinal Number Equivalence or Equal to Universe

== Theorem ==
Let $S$ be a set.
Let $\card S$ denote the cardinal number of $S$.
Let $\mathbb U$ denote the universal class.
Then:
:$S \sim \card S \lor \card S = \mathbb U$
== Proof ==
By Condition for Set Equivalent to Cardinal Number:
If $\exists x \in \On: S \sim x$, then:
:$S \sim \card S$
If $\neg \exists x \in \On: S \sim x$, then:
And thus by the definition of cardinal number:
:$\card S = \mathbb U$
== Sources ==
* : $\S 10.10$
Category:Cardinals
Cardinal Number Less than Ordinal

== Theorem ==
Let $S$ be a set.
Let $\card S$ denote the cardinal number of $S$.
Let $x$ be an ordinal such that $S \sim x$.
Then:
:$\card S \le x$
=== Corollary ===
== Proof ==
Since $S \sim x$, it follows that:
: $x \in \set {y \in \On : S \sim y}$
By Intersection is Subset: General Result, it follows that:
: $\ds \bigcap \set {y \in \On: S \sim y} \subseteq x$
Therefore $\card S \le x$ by the definition of cardinal number.
== Sources ==
* : $\S 10.12$
Category:Cardinals
Category:Ordinals
Cardinal Number Less than Ordinal/Corollary

== Corollary to Cardinal Number Less than Ordinal ==
Let $x$ be an ordinal.
Let $\card x$ denote the cardinal number of $x$.
Then:
:$\card x \le x$
== Proof ==
By Set Equivalence behaves like Equivalence Relation:
:$x \sim x$
By Cardinal Number Less than Ordinal:
:$\card x \le x$
== Sources ==
* : $\S 10.13$
Category:Cardinals
Category:Ordinals
Cardinal Number Plus One Less than Cardinal Product

== Theorem ==
Let $x$ be an ordinal such that $x > 1$.
Then:
:$\card {x + 1} \le \card {x \times x}$
Where $\times$ denotes the Cartesian product.
== Proof ==
Since $x > 1$, then $0 < x$ and $1 < x$.
Define the function $f: x + 1 \to x \times x$ as follows:
:$\map f y = \begin{cases}
\tuple {y, 0} &: y < x \\
\tuple {0, 1} &: y = x
\end{cases}$
If $\map f y = \map f z$, then $y = z$ by cases.
=== Case 1: $y = x$ ===
If $y = x$, then $\map f y = \tuple {0, 1}$ by the definition of $f$.
So if $z < x$, then $\map f y \ne \map f z$.
This contradicts the hypothesis, so $z = x$ and therefore $z = y$.
=== Case 2: $y < x$ ===
If $y < x$, then $\map f y = \tuple {y, 0}$ by the definition of $f$.
$z = x$ yields the contradictory statement $\tuple {y, 0} = \tuple {0, 1}$.
Therefore, $z < x$ and $\tuple {y, 0} = \tuple {z, 0}$.
By Equality of Ordered Pairs, it follows that $y = z$.
It follows that $f$ is an injection.
By Injection iff Cardinal Inequality, it follows that $\card {x + 1} \le \card {x \times x}$.
== Sources ==
* : $\S 10.32$
Category:Ordinals
Category:Cardinals
Cardinal Number is Ordinal

== Theorem ==
Let $S$ be a set such that $S \sim x$ for some ordinal $x$.
Let $\card S$ denote the cardinality of $S$.
Then:
:$\card S \in \On$
where $\On$ denotes the class of all ordinals.
== Proof ==
If $S \sim x$, then $\set {x \in \On: S \sim x}$ is a non-empty set of ordinals.
It follows that this set has a minimal element, its intersection.
This minimal element is the cardinal number of $S$, by the definition of cardinal number.
Thus, it is an ordinal.
== Sources ==
* : $\S 10.8$
Category:Cardinals
Cardinal One is Cancellable for Cardinal Sum

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$
where $\mathbf 1$ is (cardinal) one.
== Proof ==
Suppose $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1$.
Then from Condition for Existence of Cardinal Sum there exists some cardinal $\mathbf c$ such that 
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 = \mathbf c$
By definition of cardinal there exists a set $C$ such that $\mathbf c = \map \Card C$.
$C$ also has subsets $A$ and $B$ such that there also exist elements of $\alpha, \beta \in C$ such that:
:$\relcomp C A = \set \alpha$
:$\relcomp C B = \set \beta$
If $\alpha = \beta$ then $A = B$ and so $\mathbf a = \mathbf b$.
On the other hand, if $\alpha \ne \beta$ then $\alpha \in B$ and $\beta \in A$ and so:
:$\set \beta \cup \paren {A \cap B} = A$
:$\set \alpha \cup \paren {A \cap B} = B$
Since:
:$\set \beta \cap A \cap B = \O = \set \alpha \cap A \cap B$
it follows that:
:$\mathbf a = \map \Card A = \mathbf 1 + \map \Card {A \cap B} = \map \Card B = \mathbf b$
== Sources ==
* : $\S 8$: Theorem $8.9$
Category:Cardinals
Cardinal Product Distributes over Cardinal Sum

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a \paren {\mathbf b + \mathbf c} = \mathbf a \mathbf b + \mathbf a \mathbf c$
where:
: $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
: $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.
== Proof ==
Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
Let $B$ and $C$ be pairwise disjoint, that is:
:$B \cap C = \O$
Then we can define:
:$B \sqcup C := B \cup C$
where $B \sqcup C$ denotes the disjoint union of $B$ and $C$.
Then we have:
:$\mathbf b + \mathbf c = \card {B \sqcup C} = \card {B \cup C}$
Then:
Then:
Then:
== Sources ==
* : $\S 8$: Theorem $8.6$
Category:Cardinals