nyuuzyou/wikis huggingface dataset:
Real and Imaginary Parts of Holomorphic Function are Harmonic

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a holomorphic complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
Then $u$ and $v$ are harmonic functions.
== Proof ==
By Cauchy-Riemann Equations, $u$ and $v$ satisfy:
Consider the partial derivative of $(1)$ $x$, and the partial derivative of $(2)$ $y$:
By Clairaut's Theorem:
:$\dfrac {\partial^2 v} {\partial x \partial y} = \dfrac {\partial^2 v} {\partial y \partial x}$
Thus $(3) + (4)$ yields:
:$\dfrac {\partial^2 u} {\partial x^2} + \dfrac {\partial^2 u} {\partial y^2} = 0$
So $u$ is a harmonic function by definition.
Now consider the partial derivative of $(1)$ $y$, and the partial derivative of $(2)$ $x$:
By Clairaut's Theorem :
:$\dfrac {\partial^2 u} {\partial x \partial y} = \dfrac {\partial^2 u} {\partial y \partial x}$
Thus $(5) - (6)$ yields:
:$0 = \dfrac {\partial^2 v} {\partial y^2} + \dfrac {\partial^2 v} {\partial x^2}$
So $v$ is a harmonic function by definition.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Laplace's Equation: $3.7.32$
Category:Holomorphic Functions
Category:Harmonic Functions
Real and Imaginary Parts of Integer Power of Complex Number are Harmonic

== Theorem ==
Let $z \in \C$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n$ denote $z$ raised to the $n$th power.
Then both the real part $\map \Re {z^n}$ and the imaginary part $\map \Im {z^n}$ of $z^n$ are harmonic polynomials.
== Proof ==
=== Real Part of Integer Power of Complex Number is Harmonic ===
=== Imaginary Part of Integer Power of Complex Number is Harmonic ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.23$
Category:Harmonic Polynomials
Category:Complex Powers
Realification of Normed Dual is Isometrically Isomorphic to the Normed Dual of Realification

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\C$. 
Let $\struct {X_\R, \norm {\, \cdot \,} }$ be the realification of $\struct {X, \norm {\, \cdot \,} }$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,} }$.
Let $\struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast_\R} }$ be the normed dual of $\struct {X_\R, \norm {\, \cdot \,} }$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R$ be the realification of $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
Then $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R$ and $\struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast} }$ are isometrically isomorphic.
Further $T : \struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R \to \struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast} }$ defined by: 
:$T f = \map \Re f$
is an isometric isomorphism. 
== Proof ==
We first need to show that if $f \in X^\ast$, then $T f = \map \Re f \in X^\ast_\R$.
From Real Part of Linear Functional is Linear Functional, we have that $\map \Re f : X \to \R$ is $\R$-linear.
We just need to show that $\map \Re f$ is bounded. 
We have for each $x \in X$:
So we have that: 
:$\map \Re f$ is bounded
with: 
:$\norm {\map \Re f}_{X^\ast_\R} \le \norm f_{X^\ast}$
We want to show that: 
:$\norm {\map \Re f}_{X^\ast_\R} = \norm f_{X^\ast}$
for each $f \in X^\ast$. 
Let $f \in X^\ast$.
Let $\sequence {u_n}_{n \in \N}$ be a sequence with $\norm {u_n} = 1$ and:
:$\ds \norm f_{X^\ast} - \frac 1 n \le \cmod {\map f {u_n} } \le \norm f_{X^\ast}$ 
so that $\cmod {\map f {u_n} } \to \norm f_{X^\ast}$.
Pick $\alpha_n \in \C$ with $\cmod {\alpha_n} = 1$ such that $\cmod {\map f {u_n} } = \alpha_n \map f {u_n}$.
Since $f : X \to \C$ is $\C$-linear, we have $\alpha_n \map f {u_n} = \map f {\alpha_n u_n}$. 
So we have $\map \Re {\map f {\alpha_n u_n} } = \cmod {\map f {u_n} }$.
Then: 
:$\map \Re {\map f {\alpha_n u_n} } \to \norm f_{X^\ast}$
while: 
:$\norm {\alpha_n u_n} = 1$ for each $n \in \N$.
This gives: 
:$\norm f_{X^\ast} \le \norm {\map \Re f}_{X^\ast_\R}$
so that: 
:$\norm {\map \Re f}_{X^\ast_\R} = \norm f_{X^\ast}$
Finally, we show that $T$ is surjective.
This is given by Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional. 
Category:Realifications of Complex Vector Spaces
Category:Normed Dual Spaces
Category:Isometric Isomorphisms (Normed Vector Spaces)
Realification of Topological Vector Space is Topological Vector Space

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space over $\C$.
Let $X_\R$ be the realification of $X$.
Then $\struct {X_\R, \tau}$ is a topological vector space over $\R$. 
Further, $\struct {X, \tau}$ is a Hausdorff topological vector space $\struct {X_\R, \tau}$ is a Hausdorff topological vector space. 
== Proof ==
Since $\struct {X, \tau}$ is a topological vector space, the vector addition map $+_X : X \times X \to X$ is continuous.
Note that the underlying sets and topologies of $X_\R$ and $X$ are identical, we are only restricting the scalar field.
So the vector addition map $+_{X_\R} : X_\R \times X_\R \to X_\R$ is also continuous.
Again since $\struct {X, \tau}$ is a topological vector space, the scalar multiplication map $\circ_X : \C \times X \to X$.
By the definition of realification, the scalar multiplication map $\circ_{X_\R} : \R \times X \to X$ on the realification is precisely the restriction of $\circ_X$ to $\R \times X$.
From Restriction of Continuous Mapping is Continuous, $\circ_{X_\R}$ is therefore continuous. 
So $\struct {X_\R, \tau}$ is a topological vector space over $\R$.
The topology on $X$ and $X_{\R}$ are the same, so $\struct {X, \tau}$ is a Hausdorff topological vector space $\struct {X_\R, \tau}$ is a Hausdorff topological vector space.
Category:Topological Vector Spaces
Category:Realifications of Complex Vector Spaces
Realization of Stochastic Process/Examples/Batch Process

== Example of Realization of Stochastic Process ==
Consider a manufacturing plant which outputs a product at some variable rate.
Let the output generate a discrete time series $T$, obtained by accumulation.
The measurement of the volume of the output at the timestamps $T$ are the realizations of the underlying stochastic process governing that output.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Stochastic Process
Category:Examples of Stochastic Processes
Reals are Isomorphic to Dedekind Cuts

== Theorem ==
Let $\mathscr D$ be the set of all Dedekind cuts of the totally ordered set $\struct {\Q, \le}$.
Define a mapping $f: \R \to \mathscr D$ as:
:$\forall x \in \R: \map f x = \set {y \in \Q: y < x}$
Then $f$ is a bijection.
== Proof ==
First, we will prove that:
:$\forall x \in \R: \map f x \in \mathscr D$
Let $x \in \R$.
It is to be proved that $\map f x$ is a proper subset of $\Q$ such that:
:$(1): \quad \forall z \in \map f x: \forall y \in \Q: y < z \implies y \in \map f x$
:$(2): \quad \forall z \in \map f x: \exists y \in \map f x: z < y$
We have that:
:$x \notin \map f x$
Therefore by definition $\map f x$ is a proper subset of $\Q$.
$(1): \quad$ Let $z \in \map f x, y \in \Q$ such that:
:$y < z$
By definition of $\map f x$:
:$z < x$
Then:
:$y < x$
Thus by definition of $\map f x$:
:$y \in \map f x$
$(2): \quad$ Let $z \in \map f x$.
By definition of $\map f x$:
:$z < x$
By Between two Real Numbers exists Rational Number:
:$\exists r \in \Q: z < r < x$
Then by definition of $\map f x$:
:$r \in \map f x$
Thus:
:$\exists r \in \map f x: z < r$
By definition of bijection it suffices to prove that $f$ is an injection and a surjection.
We will show by definition that $f: \R \to \mathscr D$ is an injection.
Let $x_1, x_2 \in \R$ such that
:$\map f {x_1} = \map f {x_2}$
 $x_1 \ne x_2$.
 suppose $x_1 < x_2$.
By Between two Real Numbers exists Rational Number:
:$\exists r \in \Q: x_1 < r < x_2$
Then by definition of $\map f x$:
:$r \notin \map f {x_1}$
and
:$r \in \map f {x_2}$
This contradicts $\map f {x_1} = \map f {x_2}$.
We will prove by definition that $f: \R \to \mathscr D$ is a surjection.
Let $L \in \mathscr D$.
By definition of Dedekind cut:
:$L$ is a proper subset of $\Q$.
By definition of proper subset:
:$\exists r \in \Q: r \notin L$
By definition of Dedekind cut:
:$(3): \quad \forall x \in L: \forall y \in \Q: y < x \implies y \in L$
Then
:$\forall x \in L: r \not < x \land r \ne x$
Hence
:$\forall x \in L: r > x$
Then $L$ is bounded above by definition.
By definition of supremum:
:$\map \sup L \le r$
Hence:
:$\map \sup L \in \R$
By definition of supremum:
:$\map \sup L$ is an upper bound of $L$.
Then by definition of upper bound:
:$\forall x \in L: x < \map \sup L$
We will prove that:
:$\forall x \in \Q: x < \map \sup L \implies x \in L$
Let $x \in \Q$ such that:
:$x < \map \sup L$
 $x \notin L$.
By $(3)$:
:$\forall x \in L: r \ge x$
By definition:
:$r$ is an upper bound of $L$.
By definition of supremum:
:$r \ge \map \sup L$
This contradicts $x < \map \sup L$.
Thus:
:$L = \map f {\map \sup L}$
Category:Real Numbers
Category:Dedekind Cuts
Rearrangement Operation as Replacement Operations

== Example of use of Replacement Operation ==
Let $\left({a, b, c, d}\right)$ be an ordered quadruple consisting of four variables whose values are to be rearranged into the order $\left({b, c, d, a}\right)$.
This can be implemented using replacement operations.
Let $t$ be a new variable which has been established for this purpose.
Then the sequence of replacement operations:
:$t \gets a$
:$a \gets b$
:$b \gets c$
:$c \gets d$
:$d \gets t$
performs the task.
== Proof ==
Observing the values of the variables after each replacement operation:
Operation 	$a$	$b$	$c$	$d$	$t$
$t \gets a$	$a$	$b$	$c$	$d$	$a$
$a \gets b$	$b$	$b$	$c$	$d$	$a$
$b \gets c$	$b$	$c$	$c$	$d$	$a$
$c \gets d$	$b$	$c$	$d$	$d$	$a$
$d \gets t$	$b$	$c$	$d$	$a$	$a$
Hence the result.
Notice how the sequence:
:$a \gets b, b \gets c, c \gets d, d \gets a$
does not do the job.
This is because, when $d \gets a$ is performed, $a$ no longer contains its original value, and the resulting ordered quadruple is $\left({b, c, d, b}\right)$.
== Sources ==
* : $\S 1.1$: Algorithms: Exercise $1$
Category:Algorithms
Rearrangement of Variables in Total Differential Equation

== Theorem ==
In a total differential equation, any one of the variables can be regarded as an independent variable, while the remainder can be treated as dependent variables.
== Proof by Example ==
Consider the total differential equation:
:$u \rd x + v \rd y + w \rd z = 0$
This may be written:
:$u + v \dfrac {\d y} {\d x} + w \dfrac {\d z} {\d x} = 0$
or another independent variable $t$ may be introduced:
:$u \dfrac {\d x} {\d t} + v \dfrac {\d y} {\d t} + w \dfrac {\d z} {\d t} = 0$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Total Differential Equations
Reciprocal/Examples/Euler's Number

== Example of Reciprocal ==
The reciprocal of Euler's Number $e$ is approximately:
:$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $0 \cdotp 367 \, 879 \, 441 \, 171 \, 442 \, 321 \, 595 \, 523 \, 770 \, 161 \, 460 \, 867 \, 445 \, 811 \, 131 \, 031$
* : $0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031$
Category:Examples of Reciprocals
Category:Euler's Number
Reciprocal/Examples/Pi

== Example of Reciprocal ==
The reciprocal of $\pi$ (pi) is approximately:
:$\dfrac 1 \pi \approx 0 \cdotp 31830 \, 98861 \, 83790 \, 67153 \, 77675 \, 26745 \, 02872 \, 40689 \, 19291 \, 480 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $0 \cdotp 301 \, 029 \, 995 \, 663 \, 981 \ldots$
* : $0 \cdotp 31830 \, 98861 \, 83790 \, 67153 \, 77675 \, 26745 \, 02872 \, 40689 \, 19291 \, 480$
Category:Examples of Reciprocals
Category:Pi
Reciprocal Curve/Examples/Arbitrary Example 1

== Examples of Reciprocal Curves ==
Let $\CC$ be the curve described by the equation:
:$y = 2 x$
The reciprocal curve of $\CC$ is the curve described by the equation:
:$y = \dfrac 1 {2 x}$
and .
== Sources ==
* 
* 
Category:Examples of Reciprocal Curves
Reciprocal Equation/Examples/Arbitrary Example 1

== Example of Reciprocal Equation ==
The equation:
:$x^2 + 1 = 0$
is an example of a reciprocal equation.
== Proof ==
Replacing $x$ with $\dfrac 1 x$ gives:
:$\dfrac 1 {x^2} + 1 = 0$
which, on multiplying through by $x^2$, gives:
:$1 + x^2 = 0$
== Sources ==
* 
* 
Category:Examples of Reciprocal Equations
Reciprocal Function is Continuous on Real Numbers without Zero

== Theorem ==
Let $\R_{\ne 0}$ denote the real numbers excluding $0$:
:$\R_{\ne 0} := \R \setminus \set 0$.
Let $f: \R_{\ne 0} \to \R$ denote the reciprocal function:
:$\forall x \in \R_{\ne 0}: \map f x = \dfrac 1 x$
Then $f$ is continuous on all real intervals which do not include $0$.
== Proof ==
From Identity Mapping is Continuous, the real function $g$ defined as:
:$\forall x \in \R: \map g x = x$
is continuous on $\R$.
From Constant Mapping is Continuous, the real function $h$ defined as:
:$\forall x \in \R: \map x h = 1$
We note that $\map g 0 = 0$.
The result then follows from Quotient Rule for Continuous Real Functions:
:$\map f x = \dfrac {\map h x} {\map g x}$
is continuous wherever $\map g x \ne 0$.
Category:Reciprocals
Category:Examples of Continuous Real Functions
Reciprocal Function is Discontinuous at Zero

== Theorem ==
Let $f$ be the real function defined as:
:$\map f x = \dfrac 1 x$
for $x \ne 0$
Then $\map f x$ cannot be extended to a real function $g$ on $\R$ such that $\map g x$ is continuous on $\openint {-1} 1$.
== Proof ==
It is apparent that $f$ is itself not continuous at $x = 0$.
We need to show that whatever we define $\map g 0$ to be, $\ds \lim_{x \mathop \to 0} \map f x \ne \map f 0$.
Let $\map g 0 = c$.
Let $\epsilon$ be selected.
Let $x$ be chosen such that $x < \dfrac 1 c$.
Then $\dfrac 1 x > c$.
It follows that
:$\ds \lim_{x \mathop \to 0^+} \map f x > c$
and so:
$\ds \lim_{x \mathop \to 0} \map f x \ne \map f 0$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity: Example $\text D$
Category:Reciprocals
Reciprocal Function is Strictly Decreasing

== Theorem ==
The reciprocal function:
:$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
is strictly decreasing:
:on the open interval $\openint 0 \to$
:on the open interval $\openint \gets 0$
== Proof 1 ==
== Proof 2 ==
== Warning ==
Though the reciprocal function is decreasing on $\openint \gets 0$ and on $\openint 0 \to$, it is not decreasing on $\openint \gets 0 \cup \openint 0 \to$.
This is because there is a nonremovable discontinuity at the origin.
== Also see ==
* Reciprocal Sequence is Strictly Decreasing
* Harmonic Series is Divergent
* Existence of Euler-Mascheroni Constant
Category:Real Analysis
Category:Reciprocals
Category:Reciprocal Function is Strictly Decreasing
Reciprocal Function is Strictly Decreasing/Proof 1

== Theorem ==
== Proof ==
Let $x \in \openint 0 {+\infty}$.
By the definition of negative powers:
:$\dfrac 1 x = x^{-1}$
From Power Rule for Derivatives:
From Square of Real Number is Non-Negative:
: $-x^{-2} < 0$
for all $x$ within the domain.
Thus from Derivative of Monotone Function, $\operatorname{recip}$ is strictly decreasing.
The proof for $x \in \openint {-\infty} 0$ is similar.
Category:Reciprocal Function is Strictly Decreasing
Reciprocal Function is Strictly Decreasing/Proof 2

== Theorem ==
== Proof ==
=== Strictly Increasing on $\openint 0 \to$ ===
Let $a, b \in \Dom {\operatorname {recip} }$ such that $a$ and $b$ are both (strictly) positive.
Let $a < b$.
Then $0 < a < b$.
By Properties of Totally Ordered Field:
:$0 < b^{-1} < a^{-1}$
That is:
:$\dfrac 1 b < \dfrac 1 a$
=== Strictly Increasing on $\openint \gets 0$ ===
Let $a, b \in \Dom {\operatorname {recip} }$ such that $a$ and $b$ are both (strictly) negative.
Let $a < b$.
Then $a < b < 0$.
By Inversion Mapping Reverses Ordering in Ordered Group:
:$0 < -b < -a$
By Properties of Totally Ordered Field:
:$0 < \paren {-a}^{-1} < \paren {-b}^{-1}$
By Negative of Product Inverse:
:$\paren {-b}^{-1} = -b^{-1}$
:$\paren {-a}^{-1} = -a^{-1}$
Thus:
:$0 < -a^{-1} < -b^{-1}$
By Inversion Mapping Reverses Ordering in Ordered Group:
:$-\paren {-b^{-1} } < -\paren {-a^{-1} } < 0$
By Inverse of Group Inverse:
:$b^{-1} < a^{-1} < 0$
Thus in particular:
:$\dfrac 1 b < \dfrac 1 a$
Category:Reciprocal Function is Strictly Decreasing
Reciprocal Function is Unbounded on Open Unit Interval

== Theorem ==
Let $A = \openint 0 1$ denote the open unit interval.
Let $f: A \to \R$ be the reciprocal function:
:$\forall x \in A: \map f x := \dfrac 1 x$
Then $f$ is unbounded.
== Proof ==
Let $K \in \R_{>0}$.
Then:
:$\exists x \in \R: 0 < x < \dfrac 1 K$ such that $x < 1$.
Then we have:
:$\map f x = \dfrac 1 x > K$
So whatever $K$ may be, it can never be large enough to be a bound of $f$ on $\openint 0 1$.
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Example $5.1.2$
Category:Reciprocals
Category:Examples of Unbounded Real-Valued Functions
Reciprocal Property of Pole and Polar

== Theorem ==
Let $\CC$ be a circle.
Let $P$ and $Q$ be points in the plane of $\CC$.
Let $\PP$ and $\QQ$ be the polars of $P$ and $Q$ with respect to $\CC$ respectively.
Let $Q$ lie on $\PP$ with respect to $\CC$.
Then $P$ lies on $\QQ$.
That is:
:if $P$ lies on the polar of $Q$, then $Q$ lies on the polar of $P$
:if the pole of $\PP$ lies on $\QQ$, then the pole of $\QQ$ lies on $\PP$
== Proof ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin of a Cartesian plane.
Let $P = \tuple {x_0, y_0}$.
Let $Q = \tuple {x_1, y_1}$.
The polar of $P$ is given by:
:$x x_0 + y y_0 = r^2$
The polar of $Q$ is given by:
:$x x_1 + y y_1 = r^2$
Let $Q$ lie on the polar of $P$.
Then $Q$ satisfies the equation:
:$x_0 x_1 + y_0 y_1 = r^2$
which is exactly the same as the condition for $P$ to lie on $Q$.
== Also see ==
* Definition:Conjugate Points (Geometry)
* Definition:Conjugate Lines
== Sources ==
* : Chapter $\text {III}$. The Circle: $8$. Reciprocal property of pole and polar
Category:Polars of Points
Category:Conjugate Points
Category:Conjugate Lines
Reciprocal Sequence is Strictly Decreasing

== Theorem ==
The reciprocal sequence:
:$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$
is strictly decreasing.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Harmonic Series is Divergent
* Existence of Euler-Mascheroni Constant
Category:Reciprocal Sequence is Strictly Decreasing
Category:Sequences
Category:Real Analysis
Category:Reciprocals
Reciprocal as Summation of Binomial Coefficients of Reciprocals

== Theorem ==
:$\forall n \in \Z_{>0}: \dfrac 1 n = \ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}$
where $\dbinom {n - 1} k$ denotes a binomial coefficient.
That is, for example:
== Proof ==
== Sources ==
* : $35$
* : $35$
Category:Binomial Coefficients
Category:Reciprocals
Reciprocal of 101

== Theorem ==
The decimal expansion of the reciprocal of $101$ is as follows:
:$\dfrac 1 {101} = 0 \cdotp \dot 009 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.00990099...
 --------------
101)1.00000000...
 909
 ---
 910
 909
 ---
 1000
 909
 ----
 910
 909
 ---
 ...
Category:101
Category:Examples of Reciprocals
Reciprocal of 103

== Theorem ==
The decimal expansion of the reciprocal of $103$ is as follows:
:$\dfrac 1 {103} = 0 \cdotp \dot 00970 \, 87378 \, 64077 \, 66990 \, 29126 \, 21359 \, 223 \dot 3$
== Proof ==
Performing the calculation using long division:
 0.0097087378640776699029126213592233009...
 --------------------------------------------
103)1.0000000000000000000000000000000000000...
 927 618 927 103 309
 --- --- --- --- ---
 730 420 300 370 1000
 721 412 206 309 927
 --- --- --- --- ----
 900 800 940 610 ....
 824 721 927 515
 --- --- --- ---
 760 790 130 950
 721 721 103 927
 --- --- --- ---
 390 690 270 230
 309 618 206 206
 --- --- --- ---
 810 720 640 240
 721 618 618 206
 --- --- --- ---
 890 1020 220 340
 824 927 206 309
 --- ---- --- ---
 660 930 140 310
 618 927 103 309
Category:103
Category:Examples of Reciprocals
Reciprocal of 1089

== Theorem ==
The reciprocal of $1089$ is:
:$\dfrac 1 {1089} = 0 \cdotp \dot 00091 \, 82736 \, 45546 \, 37281 \, 9 \dot 1$
Note that it consists mainly of the unit multiples of $9$ from $1$ to $9$ concatenated.
== Proof ==
Performing the calculation using long division:
 0.00091827364554637281910009...
 --------------------------------------------
1089)1.00000000000000000000000000...
 9801 5445 1089
 ---- ---- ----
 1990 5950 9910
 1089 5445 9801
 ---- ---- ----
 9010 5050 1090
 8712 4356 1089
 ---- ---- ----
 2980 6940 10000
 2178 6534 9801
 ---- ---- ----
 8020 4060 ....
 7623 3267
 ---- ----
 3970 7930
 3267 7623
 ---- ----
 7030 3070
 6534 2178
 ---- ----
 4960 8920
 4356 8712
 ---- ----
 6040 2080
 5445 1089 
== Historical Note ==
== Sources ==
* : $1089$
* : $1089$
Category:1089
Category:Examples of Reciprocals
Reciprocal of 11

== Theorem ==
The decimal expansion of the reciprocal of $11$ is as follows:
:$\dfrac 1 {11} = 0 \cdotp \dot 0 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.0909...
 ------------
11)1.000000000
 99
 ----
 100
 99
 ---
 ...
Category:11
Category:Examples of Reciprocals
Reciprocal of 121

== Theorem ==
The decimal expansion of the reciprocal of $121$ is as follows:
:$\dfrac 1 {121} = 0 \cdotp \dot 008 \, 264 \, 462 \, 809 \, 917 \, 355 \, 371 \, \dot 9$
== Proof ==
== Also see ==
* Reciprocal of Square of 1 More than Number Base
Category:121
Category:Examples of Reciprocals
Reciprocal of 13

== Theorem ==
The decimal expansion of the reciprocal of $13$ is as follows:
:$\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3 \ldots$
== Proof ==
Performing the calculation using long division:
 0.07692307...
 -------------
13)1.00000000...
 91 
 ---- 
 90
 78
 --
 120
 117
 ---
 30
 26
 --
 40
 39
 --
 100
 91
 ---
 ...
Category:13
Category:Examples of Reciprocals