nyuuzyou/wikis huggingface dataset:
De Polignac's Formula

== Theorem ==
Let $n!$ be the factorial of $n$.
Let $p$ be a prime number.
Then $p^\mu$ is a divisor of $n!$, and $p^{\mu + 1}$ is not, where:
:$\ds \mu = \sum_{k \mathop > 0} \floor {\frac n {p^k} }$
where $\floor {\, \cdot \,}$ denotes the floor function.
=== Technique ===
== Proof ==
Note that although the summation given in the statement of the theorem is given as an infinite sum, in fact it terminates after a finite number of terms (because when $p^k > n$ we have $0 < n/p^k < 1$).
From Number of Multiples less than Given Number, we have that $\floor{\dfrac n {p^k} }$ is the number of integers $m$ such that $0 < m \le n$ which are multiples of $p^k$.
We look more closely at $n!$:
:$n! = 1 \times 2 \times \ldots \times \paren {n - 1} \times n$
We see that any integer $m$ such that $0 < m \le n$ which is divisible by $p^j$ and not $p^{j + 1}$ must be counted exactly $j$ times.
That is:
:once in $\floor {\dfrac n p}$
:once in $\floor {\dfrac n {p^2} }$
$\ldots$
:once in $\floor {\dfrac n {p^j} }$
And that is all the occurrences of $p$ as a factor of $n!$.
Thus:
:$\mu = \floor {\dfrac n p} + \floor {\dfrac n {p^2} } + \dotsb + \floor {\dfrac n {p^j} }$
Hence the result.
== Examples ==
== Also known as ==
De Polignac's Formula is also known as Legendre's Formula for .
== Sources ==
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $5$
* : $\S 1.2.5$: Permutations and Factorials: $(8)$
* : $\S 3.6$
Category:Factorials
Category:Prime Decompositions
Category:De Polignac's Formula
De Polignac's Formula/Examples

== Examples of Use of De Polignac's Formula ==
=== Prime Factors of $1000!$ ===
=== Multiplicity of $2$ in $1000!$ ===
=== Multiplicity of $3$ in $1000!$ ===
=== Multiplicity of $5$ in $1000!$ ===
=== Multiplicity of $7$ in $1000!$ ===
=== Multiplicity of $11$ in $1000!$ ===
=== Multiplicity of $13$ in $1000!$ ===
=== Multiplicity of $2$ in $720!$ ===
=== Multiplicity of $3$ in $720!$ ===
=== Multiplicity of $5$ in $720!$ ===
=== Prime Factors of $20!$ ===
Category:De Polignac's Formula
De Polignac's Formula/Examples/11 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $11$ appears in $1000!$ to the power of $98$.
That is:
:$11^{98} \divides 1000!$
but:
:$11^{99} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $11$ which divides $1000!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/13 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $13$ appears in $1000!$ to the power of $81$.
That is:
:$13^{81} \divides 1000!$
but:
:$13^{82} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $13$ which divides $1000!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/2 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $2$ appears in $1000!$ to the power of $994$.
That is:
:$2^{994} \divides 1000!$
but:
:$2^{995} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $2$ which divides $1000!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/2 in 720 Factorial

== Example of Use of De Polignac's Formula ==
The prime factor $2$ appears in $720!$ to the power of $716$.
That is:
:$2^{716} \divides 720!$
but:
:$2^{717} \nmid 720!$
== Proof ==
Let $\mu$ denote the power of $2$ which divides $720!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/3 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $3$ appears in $1000!$ to the power of $498$.
That is:
:$3^{498} \divides 1000!$
but:
:$3^{499} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $3$ which divides $1000!$
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
Category:De Polignac's Formula
De Polignac's Formula/Examples/3 in 720 Factorial

== Example of Use of De Polignac's Formula ==
The prime factor $3$ appears in $720!$ to the power of $356$.
That is:
:$3^{356} \divides 720!$
but:
:$3^{357} \nmid 720!$
== Proof ==
Let $\mu$ denote the power of $3$ which divides $720!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/5 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $5$ appears in $1000!$ to the power of $249$.
That is:
:$5^{249} \divides 1000!$
but:
:$5^{250} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $5$ which divides $1000!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/5 in 720 Factorial

== Example of Use of De Polignac's Formula ==
The prime factor $5$ appears in $720!$ to the power of $178$.
That is:
:$5^{178} \divides 720!$
but:
:$5^{179} \nmid 720!$
== Proof ==
Let $\mu$ denote the power of $5$ which divides $720!$
Category:De Polignac's Formula
De Polignac's Formula/Examples/7 in 1000

== Example of Use of De Polignac's Formula ==
The prime factor $7$ appears in $1000!$ to the power of $164$.
That is:
:$7^{164} \divides 1000!$
but:
:$7^{165} \nmid 1000!$
== Proof ==
Let $\mu$ denote the power of $7$ which divides $1000!$
Category:De Polignac's Formula
De Polignac's Formula/Technique

== Theorem ==
Let $n!$ be the factorial of $n$.
Let $p$ be a prime number.
Let $\mu$ be defined as:
:$\ds \mu = \sum_{k \mathop > 0} \floor {\frac n {p^k} }$
When calculating $\mu$, the easiest way to calculate the next term is simply to divide the previous term by $p$ and discard the remainder:
:$\floor {\dfrac n {p^{k + 1} } } = \floor {\floor {\dfrac n {p^k} } / p}$
== Proof ==
From Floor of $\dfrac {x + m} n$: Corollary:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
which is valid for all integers $m, n$ such that $n > 0$.
In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(8)$
Category:De Polignac's Formula
De Rham Cohomology of Sphere

== Theorem ==
Let $S^n$ denote the $n$-Sphere.
Then the de Rham Cohomology of $S^n$ are:
:$\map {H^k} {S^n} = \begin {cases} \Z^2 & : k = 0, n = 0 \\ \Z & : k = 0 \text { or } k = n, n > 0 \\ 0 & : 0 < k < n \end {cases}$
and Higher de Rham Cohomology Vanishes.
== Proof ==
Category:de Rham Cohomology
Decay Equation

== Theorem ==
The first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = k \paren {y_a - y}$
where $k \in \R: k > 0$
has the general solution:
:$y = y_a + C e^{-k x}$
where $C$ is an arbitrary constant.
If $y = y_0$ at $x = 0$, then:
:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
This differential equation is known as the decay equation.
== Proof ==
This is our general solution.
Suppose we have the initial condition:
:$y = y_0$ when $x = 0$
Then:
:$y_0 = y_a + C e^{-k \cdot 0} = y_a + C$
and so:
:$C = y_0 - y_a$
Hence the solution:
:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
== Also see ==
* Newton's Law of Cooling
* First-Order Reaction
Category:Decay Equation
Category:Examples of Solutions to Separable Differential Equation
Category:Examples of Linear First Order ODEs
Decimal Expansion/Examples

== Examples of Decimal Expansions ==
=== Decimal Number $234 \cdotp 568$ ===
=== Decimal Number $0.207$ ===
=== Decimal Number $23.23$ ===
=== Decimal Number $6735 \cdotp 249$ ===
=== Decimal Expansion of $17 / 10$ ===
=== Decimal Expansion of $9 / 100$ ===
=== Decimal Expansion of $1 / 6$ ===
=== $1 \cdotp 23999 \ldots$ is not a Decimal Expansion ===
Category:Examples of Decimal Expansions
Decimal Expansion/Examples/1.23999...

== Example of Decimal Expansion ==
:$1 \cdotp 23999 \ldots$
is not a decimal expansion as defined on .
This is because it ends in an infinite sequence of $9$s.
The number $1 \cdotp 23999 \ldots$ is equal to, and is best expressed as, $1 \cdotp 24$.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $2$
Category:Decimal Notation
Decimal Expansion of Fourth Power of Pi

== Theorem ==
The decimal expansion of the $4$th power of $\pi$ begins:
:$97 \cdotp 40909 \, 10340 \, 0 \ldots$
=== Continued Fraction Expansion ===
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $97 \cdotp 40909 \, 10340 \, 0 \ldots$
Category:Pi
Category:Fourth Powers
Decimal Notation/Examples/234.568

== Example of Decimal Expansion ==
The number:
:$234 \cdotp 568$
is effectively shorthand for:
:$2$ hundreds
:$3$ tens
:$4$ units
plus:
:$5$ tenths
:$6$ hundredths
:$8$ thousandths
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals
Category:Examples of Decimal Expansions
Decimal Notation/Examples/6735.249

== Example of Decimal Notation ==
The number:
:$6735 \cdotp 249$
is shorthand for:
:$\paren {6 \times 10^3} + \paren {7 \times 10^2} + \paren {3 \times 10^1} + \paren {5 \times 10^0} + \paren {2 \times 10^{-1} } + \paren {4 \times 10^{-2} } + \paren {9 \times 10^{-3} }$
== Sources ==
* 
* 
Category:Examples of Decimal Notation
Decision Theory/Examples/Two-Player Game

== Examples of Applications of Decision Theory ==
Consider a two-player game with an entrepreneur as one player and the competition (for example: the world, the environment, or a market) as the other player.
The entrepreneur may adopt any of a finite number of strategies, but knows only the probabilities that the competition will adopt each of its own strategies.
The payoff matrix gives the consequences (either gain or loss) to the entrepreneur for each action made, correlated with each action made by the competition.
Hence, knowing or estimating the probability of choosing a particular action made by the competition, the entrepreneur may calculate the expected payoff from a particular action.
== Sources ==
* 
* 
Category:Examples of Decision Theory
Decision Tree/Examples/Arbitrary Example 1

== Example of Decision Tree ==
Consider the case of a builder who is allowed to tender for only one of two contracts.
If he tenders for contract $A$, there is a $70 \%$ probability of winning the contract, and thereby making a profit of $\$10\,000$.
If he tenders for contract $B$, there is only a $20 \%$ probability of winning the contract, but his profit will be $\$30\,000$.
If he fails to win either contract, he will do other work instead and make a profit of $\$5000$.
The decision tree for making the decision as to which contract to bid for is shown here:
:
Here, the expected gains for contracts $A$ and $B$ are calculated as follows:
:For contract $A$: $\$10\,000 \times 0.7 + \$5000 \times 0.3 = \$8500$
:For contract $B$: $\$30\,000 \times 0.2 + \$5000 \times 0.8 = \$10\,000$
Hence, for a maximum expected gain, the builder should tender for contract $B$.
== Sources ==
* 
* 
Category:Examples of Decision Trees
Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles

== Theorem ==
Let $D$ be a deck of $52$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then after $52$ such shuffles, the cards of $D$ will be in the same order they started in.
== Proof 1 ==
== Proof 2 ==
Category:Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Category:52
Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles/Proof 1

== Theorem ==
== Proof ==
From Position of Card after n Modified Perfect Faro Shuffles, after $n$ shuffles a card in position $x$ will be in position $2^n x \pmod {53}$.
So for all $52$ cards in $D$, we need to find $n$ such that:
:$2^n x \equiv x \pmod {53}$
We have that $53$ is a prime number, so $x$ can be cancelled from either side:
:$2^n \equiv 1 \pmod {53}$
By Fermat's Little Theorem:
:$2^{52} \equiv 1 \pmod {53}$
So the cards of $D$ will return to their original order after $52$ modified perfect faro shuffles.
== Sources ==
* : $\text {4-3}$ Riffling
Category:Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles
Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles/Proof 2

== Theorem ==
== Proof ==
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod {53}$
By Fermat's Little Theorem:
:$2^{52} \equiv 1 \pmod {53}$
So the cards of $D$ will return to their original order after $52$ modified perfect faro shuffles.
== Sources ==
* : $\text {4-3}$ Riffling
Category:Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles
Decoding Received Word using Coset Decoding Table

== Theorem ==
Let $C$ be a linear code.
Let $v$ be a received word, which may have transmission errors.
To find out the transmitted codeword $u$ corresponding to $v$:
:$(1): \quad$ Find $v$ in the coset decoding table for $C$.
:$(2): \quad$ The corresponding transmitted codeword $u$ will be found at the top of the column where $v$ can be found.
== Proof ==
== Sources ==
* : Chapter $6$: Error-correcting codes
Category:Linear Codes
Decomposition into Even-Odd Integers is not always Unique

== Theorem ==
For every even integer $n$ such that $n > 1$, if $n$ can be expressed as the product of one or more even-times odd integers, it is not necessarily the case that this product is unique.
== Proof ==
Let $n \in \Z$ be of the form $2^2 p q$ where $p$ and $q$ are odd primes.
Then:
:$n = \paren {2 p} \times \paren {2 q} = 2 \times \paren {2 p q}$
A specific example that can be cited is $n = 60$:
:$60 = 6 \times 10$
and:
:$60 = 2 \times 30$.
Each of $2, 6, 10, 30$ are even-times odd integers:
Every $n \in \Z$ which has a divisor in that same form $2^2 p q$ can similarly be decomposed non-uniquely into even-times odd integers.
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $1$
Category:Even Integers
Decomposition into Product of Power of 2 and Odd Integer is Unique

== Theorem ==
Let $n \in \Z$ be an integer.
Then $n$ can be decomposed into the product of a power of $2$ and an odd integer.
== Proof ==
 there exists $n \in \Z$ which can be decomposed into a power of $2$ and an odd integer in more than one way.
That is:
:$n = 2^a r$
and:
:$n = 2^b s$
where:
:$a, b \in \Z_{\ge 0}$
:$r$ and $s$ are odd integers.
:either $a \ne b$ or $r \ne s$.
Suppose $r = s$.
Then:
:$\dfrac n r = \dfrac n s = 2^a = 2^b$
which leads to:
:$a = b$
which contradicts our supposition that either $r \ne s$ or $a \ne b$.
Similarly, suppose $a = b$.
Then:
:$\dfrac n {2^a} = \dfrac n {2^b} = r = s$
which also contradicts our supposition that either $r \ne s$ or $a \ne b$.
 
Thus both $r \ne s$ and $a \ne b$.
, let $a > b$.
Then $2^{a - b} = 2^c$ where $c \ge 1$.
Thus:
:$\dfrac n {2^b} = 2^c r = s$
which means $s = 2 u$ for some $u \in \Z$.
Thus $s$ is an even integer.
This contradicts our supposition that $r$ and $s$ are both odd.
All possibilities have been investigated:
:$a = b, r \ne s$
:$a \ne b, r = s$
:$a \ne b, r \ne s$
All have been shown to lead to a contradiction.
Thus by Proof by Contradiction it can be seen that an decomposition of $n$ into the product of a power of $2$ and an odd integer is unique.
Hence the result.
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $2$
Category:Integers
Decomposition of 2-Form on 4-Dimensional Riemannian Manifold

== Theorem ==
Let $M$ be a $4$-dimensional oriented Riemannian manifold.
Let $\omega$ be a $2$-form on $M$.
Then $\omega$ can be written uniquely as a sum of a self-dual $2$-form and an anti-self-dual $2$-form.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Decomposition of Complex Measure into Finite Signed Measures

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Then there exists unique finite signed measures $\mu_R$ and $\mu_I$ such that: 
:$\mu = \mu_R + i \mu_I$
== Proof ==
=== Existence ===
For each $A \in \Sigma$ define the function $\mu_R : X \to \R$ by: 
:$\map {\mu_R} A = \map \Re {\map \mu A}$
Similarly, for each $A \in \Sigma$ define the function $\mu_I : X \to \R$ by: 
:$\map {\mu_I} A = \map \Im {\map \mu A}$
Clearly we have: 
for each $A \in \Sigma$.
So:
:$\mu = \mu_R + \mu_I$
It remains to show that $\mu_R$ and $\mu_I$ are finite signed measures.
We verify both of the conditions for a signed measure in turn. 
We have: 
:$\map \mu \O = 0$
so:
:$\map {\mu_R} A = \map \Re 0 = 0$
and:
:$\map {\mu_I} A = \map \Im 0 = 0$
verifying $(1)$ for $\mu_R$ and $\mu_I$. 
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets. 
Then: 
and:
verifying $(2)$ for $\mu_R$ and $\mu_I$.
So $\mu_R$ and $\mu_I$ are signed measures. 
Since $\mu$ is a complex-valued function, we have: 
:$\cmod {\map \mu X} < \infty$
We then have, from Modulus Larger than Real Part: 
:$\size {\map \Re {\map \mu X} } < \infty$ 
so:
:$\size {\map {\mu_R} X} < \infty$
Similarly, from Modulus Larger than Imaginary Part:
:$\size {\map \Im {\map \mu X} } < \infty$
so:
:$\size {\map {\mu_I} X} < \infty$
So:
:$\mu_R$ and $\mu_I$ are finite.
=== Uniqueness ===
Suppose that $\mu_R^{(1)}$, $\mu_R^{(2)}$, $\mu_I^{(1)}$, $\mu_I^{(2)}$ are finite signed measures with: 
:$\mu = \mu_R^{(1)} + i \mu_I^{(1)} = \mu_R^{(2)} + i \mu_I^{(2)}$
Then for each $A \in \Sigma$, we have: 
:$\map {\mu_R^{(1)} } A - \map {\mu_R^{(2)} } A = i \paren {\map {\mu_I^{(2)} } A - \map {\mu_I^{(1)} } A}$
The is real, while the is imaginary, so: 
:$\map {\mu_R^{(1)} } A - \map {\mu_R^{(2)} } A = i \paren {\map {\mu_I^{(2)} } A - \map {\mu_I^{(1)} } A} = 0$
so:
:$\map {\mu_R^{(1)} } A = \map {\mu_R^{(2)} } A$
and:
:$\map {\mu_I^{(1)} } A = \map {\mu_I^{(2)} } A$
Since $A \in \Sigma$ was arbitrary, we have: 
:$\mu_R^{(1)} = \mu_R^{(2)}$
and:
:$\mu_I^{(1)} = \mu_I^{(2)}$
so we obtain the desired uniqueness.
== Sources ==
* : $4.1$: Signed and Complex Measures
Category:Complex Measures
Decomposition of Distribution Function of Finite Signed Borel Measure

== Theorem ==
Let $\mu$ be a finite signed Borel measure on $\R$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then: 
:$F_\mu = F_{\mu^+} - F_{\mu^-}$
where:
:$F_\mu$ is the distribution function of $\mu$
:$F_{\mu^+}$ and $F_{\mu^-}$ are the distribution functions of $\mu^+$ and $\mu^-$ respectively.
== Proof ==
For each $x \in \R$ we have:
Category:Distribution Function of Finite Signed Borel Measure
Decomposition of Field Extension as Separable Extension followed by Purely Inseparable

== Theorem ==
Let $E / F$ be an algebraic field extension.
Then the relative separable closure $K = F^{sep}$ in $E$ is the unique intermediate field with the following properties:
:$K / F$ is separable
:$E / K$ is purely inseparable.
== Proof ==
Let $K = F^{sep}$ be the set of elements of $E$ which are separable over $F$.
This is a subextension by Separable Elements Form Field.
Then elements of $E \setminus K$ are not separable over $F$, since all elements that are separable over $F$ are in $K$.
Then elements of $E \setminus K$ are not separable over $K$, since if an element of $E$ is separable over $K$, then it is separable over $F$, by Transitivity of Separable Field Extensions.
By definition of purely inseparable field extension, $E/K$ is purely inseparable.
== Source ==
* : Chapter $\text V$: $\S 6$: Inseparable Extensions: Proposition $6.6$
Category:Field Extensions
Decomposition of Matrix Exponential

== Theorem ==
Let $\mathbf A$ be a square matrix of order $m$ for some $m \in \Z_{\ge 1}$.
Let $t \in \R$ be a real number.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Let $\mathbf P$ be a nonsingular matrix of order $m$.
Then:
:$e^{\mathbf P \mathbf A \mathbf P^{-1} } = \mathbf P e^{\mathbf A} \mathbf P^{-1}$
== Proof ==
$\paren {\mathbf P \mathbf A \mathbf P^{-1} }^n = \mathbf P \mathbf A^n \mathbf P^{-1}$ by induction.
Category:Matrix Exponential
Decomposition of Mean Squared Error

== Theorem ==
<onlyinclude>
Let $\theta$ be a population parameter of some stochastic model.
Let $\hat \theta$ be an estimator of $\theta$.
We then have:
:$\map {\operatorname {MSE} } {\hat \theta} = \var {\hat \theta} + \paren {\map {\operatorname {bias}} {\hat \theta} }^2 $
where:
:$\map {\operatorname {MSE} } {\hat \theta}$ denotes the Mean Squared Error of $\hat \theta$.
:$\var {\hat \theta}$ denotes the variance of $\hat \theta$.
:$\map {\operatorname {bias} } {\hat \theta}$ denotes the bias of $\hat \theta$.
== Proof ==
Let $\delta = \hat \theta - \theta$.
By :
:$\expect {\delta ^2} = \map {\operatorname {MSE} } {\hat \theta}$
and:
Therefore:
Or, equivalently:
:$\map {\operatorname {MSE} } {\hat \theta} = \var {\hat \theta} + \paren {\map {\operatorname {bias} } {\hat \theta} }^2$
Category:Inductive Statistics
Category:Variance
Dedekind's Theorem

== Theorem ==
Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.
Then there exists a unique real number which is a producer of $\tuple {L, R}$.
Thus it is proved that the totally ordered set $\R$ is Dedekind complete, and that is why it is referred to as the continuum.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
Dedekind's Theorem is also known as the completeness theorem for the real numbers.
== Also see ==
* Axiom:Axiom of Continuity
* Continuum Property
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.32$. Theorem (Dedekind)
Category:Real Analysis
Category:Dedekind Cuts
Category:Dedekind's Theorem
Dedekind's Theorem/Corollary

== Theorem ==
Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.
Then either $L$ contains a largest number or $R$ contains a smallest number.
== Proof ==
From Dedekind's Theorem, there exists a unique real number such that:
:$l \le \gamma$ for all $l \in L$
:$\gamma \le r$ for all $r \in R$.
Let $\gamma \in L$.
Then by definition $\gamma$ is the largest number in $L$.
Let $\gamma \in R$.
Then by definition $\gamma$ is the smallest number in $R$.
By the definition of Dedekind cut, $\tuple {L, R}$ is a partition of $\R$.
Hence $\gamma$ is either in $L$ or $R$, but not both.
That is, $\gamma$ is either:
:the largest number in $L$
or:
:the smallest number in $R$.
Hence the result.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.32$. Corollary
Category:Real Analysis
Category:Dedekind's Theorem
Dedekind's Theorem/Proof 1

== Theorem ==
== Proof ==
Suppose $P$ and $Q$ are two properties which are mutually exclusive.
Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$.
Suppose that any number having $P$ is less than any which have $Q$.
Let us call the numbers with $P$ the left hand set $L$, and the ones with $Q$ the right hand set $R$.
There are two possibilities, as follows.
* $L$ has a greatest element, or
* $R$ has a least element.
It is not possible that both of the above can happen.
Because suppose $l$ is the greatest element of $L$ and $r$ is the least element of $R$.
Then the number $\dfrac {l + r} 2$ is greater than $l$ and less than $r$, so it could not be in either class.
However, one of the above must occur.
Because, suppose the following.
Let $L_1$ and $R_1$ be the subsets of $L$ and $R$ respectively consisting of only the rational numbers in $L$ and $R$.
Then $L_1$ and $R_1$ form a section of the set of rational numbers $\Q$.
There are two cases to think about:
Maybe $L_1$ has a greatest element $\alpha$.
In this case, $\alpha$ must also be the greatest element of $L$.
Because if not, then there's a greater one, which we can call $\beta$.
There are always rational numbers between $\alpha$ and $\beta$ from Rational Numbers are Densely Ordered.
These are less than $\beta$ and thus belong to $L$ and (because they're rational) also to $L_1$.
This is a contradiction, so if $\alpha$ is the greatest element of $L_1$, it's also the greatest element of $L$.
On the other hand, $L_1$ may not have a greatest element.
In this case, the section of the rational numbers formed by $L_1$ and $R_1$ is a real number $\alpha$.
It must belong to either $L$ or $R$.
If it belongs to $L$ we can show, like we did before, that it is the greatest element of $L$.
Similarly, if it belongs to $R$ we can show it is the least element of $R$.
So in any case, either $L$ has a greatest element or $R$ has a least element.
Thus, any section of the real numbers corresponds to a real number.
Category:Dedekind's Theorem