nyuuzyou/wikis huggingface dataset:
Replicative Function of x minus Floor of x is Replicative/Lemma

== Theorem ==
Let $x \in \R$.
Suppose $x - \floor x < \dfrac 1 n$.
Then:
:$\floor {x + \dfrac k n} = \dfrac {\floor {n x} } n$
for any $0 \le k \le n - 1$.
== Proof ==
We have $n x < n \floor x + 1$.
By Number less than Integer iff Floor less than Integer:
:$\floor {n x} < n \floor x + 1$
Thus $\floor {n x} \le n \floor x$.
From definition of floor function:
:$n x \ge n \floor x$
By Number not less than Integer iff Floor not less than Integer:
:$\floor {n x} \ge n \floor x$
Therefore $\floor {n x} = n \floor x$ and thus $\floor x = \dfrac {\floor {n x} } n$.
Now we show $\floor {x + \dfrac k n} = \floor x$ for $0 \le k \le n - 1$.
We have:
:$\floor x \le \floor {x + \dfrac k n} \le \floor {x + \dfrac {n - 1} n}$
We also have:
:$x + \dfrac {n - 1} n < \floor x + \dfrac 1 n + \dfrac {n - 1} n = \floor x + 1$
By Number less than Integer iff Floor less than Integer:
:$\floor {x + \dfrac {n - 1} n} < \floor x + 1$
This gives:
:$\floor {x + \dfrac {n - 1} n} \le \floor x$
Combining the above:
:$\floor {x + \dfrac k n} = \floor x$
Therefore $\floor {x + \dfrac k n} = \floor x = \dfrac {\floor {n x} } n$.
Category:Replicative Function of x minus Floor of x is Replicative
Representation of 1 as Sum of n Unit Fractions

== Theorem ==
Let $U \left({n}\right)$ denote the number of different ways of representing $1$ as the sum of $n$ unit fractions.
Then for various $n$, $U \left({n}\right)$ is given by the following table:
:
$n$	$U \left({n}\right)$
$1$	$1$
$2$	$1$
$3$	$3$
$4$	$14$
$5$	$147$
$6$	$3462$
== Proof ==
Trivially:
:$1 = \dfrac 1 1$
and it follows that:
$U \left({1}\right) = 1$
Also trivially:
:$1 = \dfrac 1 2 + \dfrac 1 2$
and it follows that:
$U \left({2}\right) = 1$
From Sum of 3 Unit Fractions that equals 1:
:$U \left({3}\right) = 3$
From Sum of 4 Unit Fractions that equals 1:
:$U \left({4}\right) = 14$
From Sum of 5 Unit Fractions that equals 1:
:$U \left({5}\right) = 147$
== Historical Note ==
== Sources ==
* 
* : $147$
Category:Unit Fractions
Representation of Degree One is Irreducible

== Theorem ==
Let $\struct {G, \cdot}$ be a finite group.
Let $\rho: G \to \GL V$ be a linear representation of $G$ on $V$ of degree $1$. 
Then $\rho$ is an irreducible linear representation.
== Proof ==
By the definition of degree of a linear representation, it is known that $\dim V = 1$.
Let $W$ be a proper vector subspace of $V$.
It follows from Dimension of Proper Subspace is Less Than its Superspace that:
:$\dim W < 1$
and hence $\dim W = 0$.
Now from Trivial Vector Space iff Zero Dimension, it follows that: 
:$W = \set {\mathbf 0}$
But this is not a non-trivial proper subspace of $V$.
Thus $V$ has no non-trivial proper vector subspaces.
Hence, by definition, $\rho$ is an irreducible linear representation.
Category:Representation Theory
Representation of Integers in Balanced Ternary

== Theorem ==
Let $n \in \Z$ be an integer.
$n$ can be represented uniquely in balanced ternary:
:$\ds n = \sum_{j \mathop = 0}^m r_j 3^j$
:$\sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}$
such that:
where:
:$m \in \Z_{>0}$ is a strictly positive integer such that $3^m < \size {2 n} < 3^{m + 1}$
:all the $r_j$ are such that $r_j \in \set {\underline 1, 0, 1}$, where $\underline 1 := -1$.
== Proof ==
Let $n \in \Z$.
Let $m \in \Z_{\ge 0}$ be such that:
:$3^m + 1 \le \size {2 n} \le 3^{m + 1} - 1$
where $\size {2 n}$ denotes the absolute value of $2 n$.
As $2 n$ is even, this is always possible, because $3^r$ is always an odd integer for non-negative $r$.
Let $d = \dfrac {3^{m + 1} - 1} 2$.
Let $k = n + d$.
We have that:
Let $k = n + d \in \Z$ be represented in ternary notation:
:$k = \ds \sum_{j \mathop = 0}^m s_j 3^j$
where $s_j \in \set {0, 1, 2}$.
By the Basis Representation Theorem, this expression for $k$ is unique.
Now we have:
Hence we see:
Hence $n$ has a representation in balanced ternary.
The representation for $k$ in ternary notation is unique, as established.
Hence the representation in balanced ternary for $n$ is also unique.
== Sources ==
* : $\text {1-2}$ The Basis Representation Theorem: Exercise $4$
Category:Balanced Ternary Representation
Representation of Integers in Golden Mean Number System

== Theorem ==
The positive integers $n$ are represented in the golden mean number system in their simplest form $S_n$ as follows:
:
$n$ 	$S_n$
$1$ 	$1$
$2$ 	$10 \cdotp 01$
$3$ 	$100 \cdotp 01$
$4$ 	$101 \cdotp 01$
$5$ 	$1000 \cdotp 1001$
$6$ 	$1010 \cdotp 0001$
$7$ 	$10000 \cdotp 0001$
$8$ 	$10001 \cdotp 0001$
$9$ 	$10010 \cdotp 0101$
$10$	$10100 \cdotp 0101$
$11$	$10101 \cdotp 0101$
$12$	$100000 \cdotp 101001$
$13$	$100010 \cdotp 001001$
$14$	$100100 \cdotp 001001$
$15$	$100101 \cdotp 001001$
$16$	$101000 \cdotp 100001$
== Proof ==
:$1$ is represented by $\left[{1}\right]_\phi = \phi^0$.
From there, the algorithm for Addition of 1 in Golden Mean Number System is run.
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $35$
Category:Golden Mean Number System
Representation of Number Base in that Base

== Theorem ==
Let $b \in \Z$ be an integer such that $b > 1$.
Then $b$ is expressed in base $b$ as $10$.
== Proof ==
By the Basis Representation Theorem, $b$ can be expressed uniquely in the form:
:$\ds b = \sum_{j \mathop = 0}^m r_j b^j$
where:
:$m$ is such that $b^m \le n < b^{m + 1}$
:all the $r_j$ are such that $0 \le r_j < b$.
As $b = b^1$, we have that:
:$b = 1 \times b^1 + 0 \times b^0$
That is, by definition of base $b$:
:$b = \sqbrk {1 0}_b$
Hence the result.
== Examples ==
== Sources ==
* 
* 
Category:Representation of Number Base in that Base
Category:Number Bases
Representation of Ternary Expansions

== Theorem ==
Let $x \in \R$ be a real number.
Let $x$ be represented in base $3$ notation.
While it may be possible for $x$ to have two different such representations, for example:
:$\dfrac 1 3 = 0.100000 \ldots_3 = 0.022222 \ldots_3$
it is not possible for $x$ be written in more than one way without using the digit $1$.
== Proof ==
It is sufficient to show that two distinct representations represents two distinct numbers.
Let $a$ and $b$ two real numbers representable as the form above.
Their signs are easy to distinguish, so we consider $\size a$ and $\size b$.
There is some $n$ such that:
:$\size a, \size b < 3^n$
In that case, $\dfrac {\size a} {3^n}$ can be represented as:
:$0.a_1 a_2 a_3 \ldots$
and $\dfrac {\size b} {3^n}$ can be represented as:
:$0.b_1 b_2 b_3 \ldots$
where $a_i, b_i$ are either $0$ or $2$.
Let $N$ be the smallest integer such that $a_N \ne b_N$.
 assume that $a_N = 2$ and $b_N = 0$.
We have:
and thus $\size a$ and $\size b$ are distinct.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set
Category:Ternary Notation
Representations for 1 in Golden Mean Number System

== Theorem ==
Then there are infinitely many ways to express the number $1$ in the golden mean number system.
== Proof ==
We have that:
:$\phi^0 = 1$
and so $1$ has the representation $S_1$ as:
:$S_1 := \left[{1 \cdotp 00}\right]_\phi$
By inspection it is seen that $S_1$ is the simplest form of $1$.
Expanding $S_1$:
:$S_2 := \left[{0 \cdotp 11}\right]_\phi$
which can then be expressed as:
:$S_2 := \left[{0 \cdotp 1100}\right]_\phi$
Expanding $S_2$:
:$S_3 := \left[{0 \cdotp 1011}\right]_\phi$
and so:
:$S_4 := \left[{0 \cdotp 101011}\right]_\phi$
Continuing in this way we obtain an infinite sequence $\left\langle{S_n}\right\rangle$ all of which are a representation of $1$.
== Examples ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $35$
Category:Golden Mean Number System
Category:Representations for 1 in Golden Mean Number System
Representations for 1 in Golden Mean Number System/Examples/0.01111...

== Example of Representations for 1 in Golden Mean Number System ==
$1$ can be represented in the golden mean number system as $\left[{0 \cdotp 01111 \ldots}\right]_\phi$.
== Proof ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $35$
Category:Representations for 1 in Golden Mean Number System
Representations for 1 in Golden Mean Number System/Examples/0.11

== Example of Representations for 1 in Golden Mean Number System ==
$1$ can be represented in the golden mean number system as $\sqbrk {0 \cdotp 11}_\phi$.
== Proof ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $35$
Category:Representations for 1 in Golden Mean Number System
Representative of P-adic Number is Representative of Equivalence Class

== Theorem ==
Let $p$ be any prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $x \in \Q_p$.
Then for any sequence $\sequence{y_n}$ of the rational numbers $\Q$:
:$\sequence{y_n}$ is a representative of the $p$-adic number $x$ $\sequence{y_n}$ is a representative of the equivalence class $x$.
== Proof ==
By definition of the $p$-adic numbers:
:$\Q_p$ is quotient ring
By definition of a quotient ring:
:$\Q_p$ is a coset space
By definition of a coset space:
:Every $p$-adic number $x$ is an equivalence class
By definitions of a representative of a $p$-adic number and a representative of an equivalence class:
:for any sequence $\sequence{y_n}$ of the rational numbers $\Q$: 
::$\sequence{y_n}$ is a representative of the $p$-adic number $x$ $\sequence{y_n}$ is a representative of the equivalence class $x$.
Category:P-adic Number Theory
Representatives of same P-adic Number iff Difference is Null Sequence

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rational numbers $\Q$.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence{\alpha_n}$ and $\sequence{\beta_n}$ be Cauchy sequences in $\struct {Q, \norm {\,\cdot\,}_p}$.
Then: 
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are representatives of the same $p$-adic number in $\Q_p$
:
:the sequence $\sequence {\alpha_n - \beta_n}$ is a null sequence.
== Proof ==
Let $\CC$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\NN$ be the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
Then $\Q_p$ is the quotient ring $\CC \, \big / \NN$ by definition of the $p$-adic numbers.
Hence:
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are representatives of the same $p$-adic number in $\Q_p$ 
:
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are elements of the same left coset of $\NN$ in $\CC$.
From Element in Left Coset iff Product with Inverse in Subgroup:
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are elements of the same left coset of $\NN$ in $\CC$ 
: 
:$\sequence {\alpha_n} - \sequence {\beta_n} = \sequence {\alpha_n - \beta_n} \in \NN$.
The result follows.
Category:P-adic Number Theory
Repunit 19 is Unique Period Prime with Period 19

== Theorem ==
The repunit prime $R_{19}$ is a unique period prime whose reciprocal has a period of $19$:
:$\dfrac 1 {1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111} = 0 \cdotp \dot 00000 \, 00000 \, 00000 \, 000 \dot 9$
== Proof ==
The reciprocal of a repunit $R_n$ is of the form:
:$\dfrac 1 {R_n} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{n - 1} \dot 9$
Thus $\dfrac 1 {R_{19} }$ has a period of $19$.
From Period of Reciprocal of Prime, for prime numbers such that:
:$p \nmid 10$
we have that the period of such a prime is the order of $10$ modulo $p$.
That is, the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
The only other possible primes $p$ whose reciprocals might have a period of $19$ must also satisfy:
:$10^{19} \equiv 1 \pmod p$
that is:
:$p \divides \paren {10^{19} - 1} = 9 \times R_{19}$
Therefore the only other possible prime whose reciprocal might have a period of $19$ is $3$.
Trivially:
:$\dfrac 1 3 = 0 \cdotp \dot 3$
which has a period of $1$.
Hence the result.
Category:1,111,111,111,111,111,111
Category:Examples of Reciprocals
Category:Examples of Unique Period Primes
Repunit Expressed using Power of 10

== Theorem ==
The repunit number $R_n$ can be expressed as:
:$R_n = \dfrac {10^n - 1} 9$
== Proof ==
The result follows from the Basis Representation Theorem.
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunits
Repunit Integer as Product of Base - 1 by Increasing Digit Integer

== Theorem ==
That is:
:$\ds 9 \sum_{j \mathop = 0}^n \paren {n - j} 10^j + n + 1 = \sum_{j \mathop = 0}^n 10^j$
=== General Result ===
== Proof ==
A specific instance of the general result:
where $b = 10$.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $20 \ \text({a})$
Category:Repunits
Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result

== Theorem ==
Let $b \in \Z_{>1}$.
Then:
:$\ds \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1 = \sum_{j \mathop = 0}^n b^j$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $20 \ \text({b, c})$
Category:Recreational Mathematics
Repunit Prime/Sequence

== Sequence of Repunit Primes ==
The sequence of repunit primes (base $10$) begins:
:$11, 1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111, 11 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111, \ldots$
=== Repunit Primes by Indices ===
== Linguistic Note ==
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunit Primes
Repunit Prime/Sequence/Indices

== Sequence of Repunit Primes by Indices ==
Expressing a repunit prime of $n$ digits by $R_n$, the complete sequence of known repunit primes (base $10$) can be expressed as:
:$R_2, R_{19}, R_{23}, R_{317}, R_{1031}, R_{49 \, 081}, R_{86 \, 453}, R_{109 \, 297}, R_{270 \, 343}, R_{5 \, 794 \, 777}, R_{8 \, 177 \, 207}$
According to , only up to $R_{86 \, 453}$ are known to be prime.
The rest are only probably prime.
== Linguistic Note ==
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
* 
* 
Category:Repunit Primes
Repunit cannot be Square

== Theorem ==
A repunit (apart from the trivial $1$) cannot be a square.
== Proof ==
Let $m$ be a repunit with $r$ digits such that $r > 1$.
By definition, $m$ is odd.
Thus from Square Modulo 4, if $m$ were square it would be of the form:
:$m \equiv 1 \pmod 4$.
$m$ is of the form $\ds \sum_{k \mathop = 0}^{r - 1} 10^k$ where $r$ is the number of digits.
Thus for $r \ge 2$:
Hence:
:$m \equiv 3 \pmod 4$
and so cannot be square.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $7$
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunits
Category:Square Numbers
Repunit in Base 9 is Triangular

== Theorem ==
Let $m$ be a repunit base $9$.
Then $m$ is a triangular number.
== Proof ==
Let $m$ be a repunit base $9$ with $n$ digits.
We have:
The result follows from Closed Form for Triangular Numbers.
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunits
Category:Triangular Numbers
Repunit is Zuckerman Number

== Theorem ==
Let $n$ be a repunit.
Then $n$ is also a Zuckerman number.
== Proof ==
The digits of a repunit are by definition all $1$.
Thus the product of the digits of a repunit is $1$.
By One Divides all Integers, $1$ is a divisor of $n$.
Hence the result, by definition of Zuckerman number.
== Sources ==
* : $11$
Category:Repunits
Category:Zuckerman Numbers
Requirement for Connected Domain to be Simply Connected Domain

== Theorem ==
Let $D \subseteq \C$ be a connected domain.
Then $D$ is simply connected all paths in $D$ with the same initial points and end points are path-homotopic.
That is, given two paths in $D$:
:$\gamma: \closedint a b \to D$
:$\sigma: \closedint c d \to D$
with $\map \gamma a = \map \sigma c$ and $\map \gamma b = \map \sigma d$, there exists a continuous function $H: \closedint 0 1 \times \closedint 0 1 \to D$ such that:
:$\map H {s, 0} = \map \gamma s$ for all $s \in \closedint 0 1$
:$\map H {s, 1} = \map \sigma s$ for all $s \in \closedint 0 1$
:$\map H {t, 0} = \map \gamma t$ for all $t \in \closedint 0 1$
:$\map H {t, 1} = \map \sigma t$ for all $t \in \closedint 0 1$
== Proof ==
From Open Domain is Connected iff it is Path-Connected, it follows that $D$ is path-connected.
The results now follows by the definition of simply connected sets.
== Sources ==
* : $\S 3.1$
Category:Complex Analysis
Category:Simply Connected Spaces
Rescaling is Linear Transformation

== Theorem ==
Let $\struct {R, +, \cdot}$ be a commutative ring.
Let $\struct {V, +, \circ}_R$ be an $R$-module.
Then for any $r \in R$, the rescaling:
:$m_r: V \to V, v \mapsto r \circ v$
is a linear transformation.
== Proof ==
Let $v \in V$ and $s \in R$.
Then:
Next, for $v, w \in V$:
It follows that $m_r$ is a linear transformation.
Category:Linear Transformations
Residual/Examples/Simple Linear Regression

== Examples of Residuals ==
Consider a simple linear regression in which the following model is specified:
:$y_i = \alpha + \beta x_i + \epsilon_i$
where:
:$\alpha$, $\beta$ and $\epsilon_i$ are unknown
:$\epsilon_i$ is the realized value of a random variable $\epsilon$, assumed to be normally distributed with expectation $0$ and variance (usually unknown) $\sigma^2$.
Let the least-squares estimators of $\alpha$ and $\beta$ be $a$ and $b$ respectively.
Then:
:$e_i = y_i - a - b x_i$
and the $e_i$ are correlated, even though the $\epsilon_i$ are independent.
== Sources ==
* 
* 
Category:Examples of Residuals
Residual Variation/Examples/Arbitrary Example 1

== Example of Residual Variation ==
Consider the set of data $\tuple {x_i, y_i}$ for $i = 1, 2, \ldots, n$.
Let a linear regression $y = a + b x$ be determined from this set of data.
Let the estimate of $y_i$ from this model be denoted $\hat y_i$.
Then the residual is given as $e_i = y_i - \hat y_i = y_i - a - b x_i$
The error mean square has the value:
:$\ds \dfrac 1 {n - 2} \sum_{i \mathop = 1} n {e_i}^2$
where the divisor $n - 2$ represents the number of degrees of freedom.
== Sources ==
* 
* 
Category:Examples of Residual Variation
Residue (Number Theory)/Examples/Arbitrary Example 1

== Example of Residue in context of Number Theory ==
Consider the congruence:
:$x^3 \equiv 1 \pmod 9$
This has a solution:
:$x = 4$
so $1$ is a residue of $9$ of order $3$.
== Sources ==
* 
* 
Category:Examples of Residues (Number Theory)
Residue Class/Examples/Modulo 7

== Example of Residue Class ==
The residue class of $2$ modulo $7$ on the integers is:
:$\eqclass 2 7 = \set {\ldots, -19, -12, -5, 2, 9, 16, \ldots}$
== Sources ==
* 
* 
Category:Examples of Residue Classes
Residue Classes form Partition of Integers

== Theorem ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\Z_m$ be the set of residue classes modulo $m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \dotsc, \eqclass {m - 1} m}$
Then $\Z_m$ forms a partition of $\Z$.
== Proof ==
By definition of the set of residue classes modulo $m$, $\Z_m$ is the quotient set of congruence modulo $m$:
:$\Z_m = \dfrac \Z {\RR_m}$
where $\RR_m$ is the congruence relation modulo $m$ on the set of all $a, b \in \Z$:
:$\RR_m = \set {\tuple {a, b} \in \Z \times \Z: \exists k \in \Z: a = b + k m}$
By the Fundamental Theorem on Equivalence Relations, $\Z_m$ is a partition of $\Z$.
== Sources ==
* : $\S 18.2$: Congruence classes
Category:Residue Classes
Residue Field of P-adic Norm on Rationals

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is isomorphic to the field $\F_p$ of integers modulo $p$.
== Proof ==
By Valuation Ring of P-adic Norm on Rationals:
:$\Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b}$
is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.
By Valuation Ideal of P-adic Norm on Rationals:
:$p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
is the induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$.
By definition, the induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the quotient ring $\Z_{\ideal p} / p \Z_{\ideal p}$.
By Quotient Ring of Integers with Principal Ideal, $\F_p$ is isomorphic to $\Z / p \Z$, where $p \Z$ is the principal ideal of $\Z$ generated by $p$.
To complete the proof it is sufficient to show that $\Z / p \Z$ is isomorphic to $\Z_{\ideal p} / p \Z_{\ideal p}$.
By Integers form Subring of Valuation Ring of P-adic Norm on Rationals then $\Z$ is a subring of $\Z_{\ideal p}$.
Let $\phi : \Z \to \Z_{\ideal p} / p \Z_{\ideal p}$ be the mapping defined by:
:$\forall a \in \Z: \map \phi a = a + p \Z_{\ideal p}$
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
Hence $\phi$ is a ring epimorphism with:
:$p \Z = \map \ker \phi$
By Quotient Ring of Kernel of Ring Epimorphism then $\Z / p \Z$ is isomorphic to $\Z_{\ideal p} / p \Z_{\ideal p}$
The result follows.
== Sources ==
* : $\S 2.4$ Algebra: Proposition $2.4.3$
Category:P-adic Number Theory
Category:Residue Field of P-adic Norm on Rationals
Residue Field of P-adic Norm on Rationals/Lemma 1

== Theorem ==
Let $\norm {\, \cdot \,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\Z_{\paren p}$ be the induced valuation ring on $\struct {\Q, \norm {\, \cdot \,}_p}$.
Let $p \, \Z_{\paren p}$ be the induced valuation ideal on $\struct {\Q, \norm {\, \cdot \,}_p}$.
Let $\phi: \Z \to \Z_{\paren p} / p \, \Z_{\paren p}$ be the mapping defined by:
:$\forall a \in \Z: \map \phi a = a + p \, \Z_{\paren p}$
Then:
:$\phi$ is a homomorphism.
== Proof ==
Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \Z_{\paren p}$.
Hence $\Z \subset \Z_{\paren p}$ is a subring of $\Z_{\paren p}$.
Let $i: \Z \to \Z_{\paren p}$ be the inclusion mapping defined by:
:$\map i a = a$
By Inclusion Mapping is Monomorphism then $i$ is a ring monomorphism.
Let $q: \Z_{\paren p} \to \Z_{\paren p} / p \, \Z_{\paren p}$ be the quotient ring epimorphism from $\Z_{\paren p}$ to $\Z_{\paren p} / p \, \Z_{\paren p}$.
By Quotient Ring Epimorphism is Epimorphism, $q$ is a epimorphism.
By definition, $\phi = q \circ i$ is the composition of $i$ with $q$.
By Composition of Ring Homomorphisms is Ring Homomorphism, $\phi$ is a homomorphism.
== Sources ==
* : $\S 2.4$ Algebra, Proposition $2.4.3$
Category:Residue Field of P-adic Norm on Rationals
Residue Field of P-adic Norm on Rationals/Lemma 2

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\Z_{\ideal p}$ be the induced valuation ring on $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $p \Z_{\ideal p}$ be the induced valuation ideal on $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\phi : \Z \to \Z_{\ideal p} / p \Z_{\ideal p}$ be the mapping defined by:
:$\forall a \in \Z: \map \phi a = a + p \Z_{\ideal p}$
Then:
:$p \Z = \map \ker \phi$
== Proof ==
Let $\map \ker \phi$ denote the kernel of $\phi$.
Then:
Hence:
:$p \Z = \map \ker \phi$
== Sources ==
* : $\S 2.4$ Algebra, Proposition $2.4.3$
Category:Residue Field of P-adic Norm on Rationals
Residue Field of P-adic Norm on Rationals/Lemma 3

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\Z_{\paren p}$ be the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.
Let $p\Z_{\paren p}$ be the induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$.
Let $\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ be the mapping defined by:
:$\forall a \in \Z: \map \phi a = a + p \Z_{\paren p}$
Then:
:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a surjection.
== Proof ==
Let $a / b \in \Z_{\paren p}$, where $a / b$ are in canonical form.
Then $p \nmid b$
Let $\F_p$ be the field of integers modulo $p$.
By the definition of a field:
:$\exists b' \in \Z: b b' \equiv 1 \pmod p$
By the definition of congruence modulo $p$:
:$p \divides b b' - 1$
By Divisor Divides Multiple:
:$\forall a \in \Z: p \divides a b b' - a$
By Valuation Ideal of P-adic Norm on Rationals then:
:$a b' - \dfrac a b = \dfrac {a b b' - a} b \in p \Z_{\paren p}$
By Element in Left Coset iff Product with Inverse in Subgroup:
:$\map \phi {a b'} = a b' + p \Z_{\paren p} = a / b + p \Z_{\paren p}$
It follows that:
:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a surjection.
== Sources ==
* : $\S 2.4$ Algebra, Proposition $2.4.34$
Category:Residue Field of P-adic Norm on Rationals