nyuuzyou/wikis huggingface dataset:
Real Numbers under Addition Modulo 1 form Group

== Theorem ==
Let $S = \set {x \in \R: 0 \le x < 1}$.
Let $\circ: S \times S \to S$ be the operation defined as:
:$x \circ y = x + y - \floor {x + y}$
That is, $\circ$ is defined as addition modulo $1$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
First note that Modulo Addition is Well-Defined.
Taking the group axioms in turn:
=== ===
In Real Number minus Floor it is demonstrated that:
:$\forall x, y \in S: x \circ y \in S$
Thus $\struct {S, \circ}$ is closed.
=== ===
The associativity of $\circ$ follows from that of the sum of real numbers.
Thus $\circ$ is associative on $S$.
=== ===
By definition of $S$:
:$0 \in S$
Let $x \in S$.
We have that:
:$0 \le x < 1$
and so by definition of floor function:
:$\floor x = 0$
So:
Hence $\struct {S, \circ}$ has $0$ as an identity element.
=== ===
Let $x \in S$.
First let $x = 0$.
We have that $0$ is the identity of $\struct {S, \circ}$:
:$0 \circ 0 = 0$
and so from Inverse of Identity Element is Itself, $0$ is its own inverse.
Now let $x \ne 0$.
By definition of $S$:
:$0 \le x < 1$
Hence:
:$1 - x \in S$
(Note that because $1 - 0 \notin S$, the above is not true for $x = 0$, which is why it has been treated as a special case.)
Therefore:
From above, $0$ is the identity of $\struct {S, \circ}$.
Thus every element $x$ of $\struct {S, \circ}$ has an inverse:
:$\begin {cases} 1 - x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
Category:Examples of Groups
Real Numbers under Addition form Group

== Theorem ==
Let $\R$ be the set of real numbers.
The structure $\struct {\R, +}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
Real Addition is Closed.
=== ===
Real Addition is Associative.
=== ===
From Real Addition Identity is Zero, we have that the identity element of $\struct {\R, +}$ is the real number $0$.
=== ===
From Inverse for Real Addition, we have that the inverse of $x \in \struct {\R, +}$ is $-x$.
== Sources ==
* : $\S 1$: Some examples of groups: Example $1.03$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $5$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Additive Group of Real Numbers
Real Numbers under Addition form Infinite Abelian Group

== Theorem ==
Let $\R$ be the set of real numbers.
The structure $\struct {\R, +}$ is an infinite abelian group.
== Proof ==
From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
Then:
:Real Addition is Commutative
and:
:Real Numbers are Uncountably Infinite.
== Sources ==
* : $\S 4.5$. Examples of groups: Example $81$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : Chapter $\text{II}$: Groups: The Group Property
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(i)}$
* : Chapter $1$: Definitions and Examples: Example $1.4$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Additive Group of Real Numbers
Real Numbers under Addition form Monoid

== Theorem ==
The set of real numbers under addition $\struct {\R, +}$ forms a monoid.
== Proof ==
Taking the monoid axioms in turn:
=== ===
Real Addition is Closed.
=== ===
Real Addition is Associative.
=== ===
From Real Addition Identity is Zero, we have that the identity element of $\struct {\R, +}$ is the real number $0$.
Hence the result.
== Sources ==
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
* : $\S 3.1$: Monoids
Category:Real Addition
Category:Examples of Monoids
Real Numbers under Multiplication do not form Group

== Theorem ==
The algebraic structure $\struct {\R, \times}$ consisting of the set of real numbers $\R$ under multiplication $\times$ is not a group.
== Proof ==
 that $\struct {\R, \times}$ is a group.
By the definition of the number $0 \in \R$:
:$\forall x \in \R: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\R, \times}$ is the trivial group.
But $\R$ contains other elements besides $0$.
From this contradiction it follows that $\struct {\R, \times}$ is not a group.
== Also see ==
* Non-Zero Real Numbers under Multiplication form Abelian Group
== Sources ==
* : $\S 4.5$. Examples of groups: Example $82$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
* : $\S 1$: Some examples of groups
Category:Real Multiplication
Category:Examples of Groups
Real Numbers under Multiplication form Monoid

== Theorem ==
The set of real numbers under multiplication $\struct {\R, \times}$ forms a monoid.
== Proof ==
Taking the monoid axioms in turn:
=== ===
Real Multiplication is Closed.
=== ===
Real Multiplication is Associative.
=== ===
Real Multiplication Identity is $1$.
Hence the result.
== Sources ==
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
* : $\S 3.1$: Monoids
Category:Real Multiplication
Category:Examples of Monoids
Real Numbers under Subtraction do not form Semigroup

== Theorem ==
The set of real numbers under subtraction $\struct {\R, -}$ does not form a semigroup.
== Proof ==
We have that Subtraction on Numbers is Not Associative.
Hence $\struct {\R, -}$ is not a semigroup by definition.
== Sources ==
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
Category:Real Subtraction
Category:Semigroups
Real Numbers with Absolute Value form Normed Vector Space

== Theorem ==
Let $\R$ be the set of real numbers.
Let $\size {\, \cdot \,}$ be the absolute value.
Then $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.
== Proof ==
We have that:
:Real Numbers form Vector Space
:Absolute Value is Norm
By definition, $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed Spaces
Category:Examples of Normed Vector Spaces
Real Ordering is not Compatible with Division

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$\dfrac a c > \dfrac b d$
That is, the usual ordering is not compatible with division.
== Proof ==
Proof by Counterexample:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $\dfrac a c = 1 \frac 1 4$ while $\dfrac b d = 3$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (4) \ \text{(ii)}$
Category:Real Division
Category:Algebra
Real Ordering is not Compatible with Subtraction

== Theorem ==
Let $a, b, c, d \in R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$a - c > b - d$
That is, the usual ordering is not compatible with subtraction.
== Proof ==
Proof by Counterexample:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $a - c = 1$ while $b - d = 2$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (4) \ \text{(i)}$
Category:Real Subtraction
Category:Algebra
Real Part as Mapping is Endomorphism for Complex Addition

== Theorem ==
Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $f: \C \to \R$ be the mapping from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Re z$
where $\map \Re z$ denotes the real part of $z$.
Then $f: \struct {\C, +} \to \struct {\R, +}$ is a group epimorphism.
Its kernel is the set:
:$\map \ker f = \set {i x: x \in \R}$
of wholly imaginary numbers.
== Proof ==
From Real Part as Mapping is Surjection, $f$ is a surjection.
Let $z_1, z_2 \in \C$.
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
So $f$ is a group homomorphism.
Thus $f$ is a surjective group homomorphism and therefore by definition a group epimorphism.
Finally:
:$\forall y \in \R: \map \Re {0 + i y} = 0$
It follows from Complex Addition Identity is Zero that: 
:$\map \ker f = \set {i x: x \in \R}$
== Also see ==
* Imaginary Part as Mapping is Endomorphism for Complex Addition
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Example $12.3$
* : Chapter $\text{II}$: Groups: Direct Products
Category:Additive Group of Complex Numbers
Category:Additive Group of Real Numbers
Category:Real Parts
Category:Complex Addition
Category:Group Endomorphisms
Real Part as Mapping is Surjection

== Theorem ==
Let $f: \C \to \R$ be the projection from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Re z$
where $\map \Re z$ denotes the real part of $z$.
Then $f$ is a surjection.
== Proof ==
Let $x \in \R$ be a real number.
Let $y \in \R$ be an arbitrary real number.
Let $z \in \C$ be the complex number defined as:
:$z = x + i y$
Then we have:
:$\map \Re z = x$
That is:
:$\exists z \in \C: \map f z = x$
The result follows by definition of surjection.
== Also see ==
* Imaginary Part as Mapping is Surjection
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Real Parts
Category:Examples of Surjections
Real Part of Complex Exponential Function

== Theorem ==
Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\exp z$ denote the complex exponential function.
Then:
:$\map \Re {\exp z} = e^x \cos y$
where:
:$\Re z$ denotes the real part of a complex number $z$
:$e^x$ denotes the real exponential function of $x$
:$\cos y$ denotes the real cosine function of $y$.
== Proof ==
From the definition of the complex exponential function:
:$\exp z := e^x \paren {\cos y + i \sin y}$
The result follows by definition of the real part of a complex number.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Exponential Function
Real Part of Complex Product

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\map \Re {z_1 z_2} = \map \Re {z_1} \map \Re {z_2} - \map \Im {z_1} \map \Im {z_2}$
== Proof ==
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
By definition of complex multiplication:
:$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $166 \ \text{(a)}$
Category:Complex Multiplication
Real Part of Element of *-Algebra is Real Linear

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $x, y \in A$ and $\lambda \in \R$.
Then we have:
:$\map \Re {x + \lambda y} = \map \Re x + \lambda \map \Re y$
where $\Re$ denotes real part. 
== Proof ==
We have:
Category:*-Algebras
Real Part of Imaginary Unit times Element of *-Algebra

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $x \in A$.
Then:
:$\map \Re {i x} = -\map \Im x$
where $\Re$ and $\Im$ denote the real and imaginary parts of $A$.
== Proof ==
We have:
Category:*-Algebras
Real Part of Integer Power of Complex Number is 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 the real part $\map \Re {z^n}$ of $z^n$ is a harmonic polynomial.
== Proof ==
Let $z = x + i y$.
Then:
Then we have:
and:
It remains to be shown that for $j \in \set {0, 1, \ldots, n - 2}$ that:
:$\paren {-1}^{j / 2} \dbinom n j \paren {n - j} \paren {n - j - 1} = -\paren {-1}^{\paren {j + 2} / 2} \dbinom n {j + 2} \paren {j + 2} \paren {j + 1}$
First we have that:
Hence it remains to be shown that:
:$\dbinom n j \paren {n - j} \paren {n - j - 1} = \dbinom n {j + 2} \paren {j + 2} \paren {j + 1}$
We have:
and:
Hence the result.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.23$
Category:Real Parts
Category:Harmonic Polynomials
Category:Complex Powers
Real Part of Linear Functional is Linear Functional

== Theorem ==
Let $X$ be a vector space over $\C$. 
Let $f : X \to \C$ be a linear functional. 
Define $g : X \to \R$ by: 
:$\map g x = \map \Re {\map f x}$
for each $x \in X$.
Then $f$ is $\R$-linear. 
== Proof ==
Let $x, y \in X$ and $\lambda, \mu \in \R$. 
Then:
Category:Linear Functionals
Real Part of Sine of Complex Number

== Theorem ==
Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\sin z$ denote the complex sine function.
Then:
:$\map \Re {\sin z} = \sin x \cosh y$
where:
:$\map \Re z$ denotes the real part of a complex number $z$
:$\sin$ denotes the sine function (real and complex)
:$\cosh$ denotes the hyperbolic cosine function.
== Proof ==
From Sine of Complex Number:
:$\map \sin {x + i y} = \sin x \cosh y + i \cos x \sinh y$
The result follows by definition of the real part of a complex number.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Sine Function
Real Plus Epsilon

== Theorem ==
Let $a, b \in \R$, such that:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
where $\R_{>0}$ is the set of strictly positive real numbers.
That is:
:$\epsilon > 0$
Then:
:$a \le b$
== Proof ==
 $a > b$.
Then:
:$a - b > 0$
By hypothesis, we have:
:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
Let $\epsilon = a - b$.
Then:
:$a < b + \paren {a - b} \implies a < a$
The result follows by Proof by Contradiction.
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.7$: Example
Category:Real Analysis
Category:Inequalities
Real Polynomial Function is Continuous

== Theorem ==
A (real) polynomial function is continuous at every point.
Thus a (real) polynomial function is continuous on every interval of $\R$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.4$: Continuity
* : $\S 9.2$
* 
* 
* 
Category:Real Polynomial Function is Continuous
Category:Continuous Real Functions
Category:Real Polynomial Functions
Real Polynomial Function is Continuous/Proof 1

== Theorem ==
== Proof ==
Let $f_n$ be an arbitrary real polynomial function of degree $n$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$f_n$ is continuous on $\R$.
$\map P 0$ is the case $f_0$, where $f_0$ is of zero degree.
Such a real polynomial function is a constant function.
It follows from Constant Real Function is Continuous that $f_0$ is continuous on $\R$.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
:$f_1$ is continuous on $\R$.
Such a real polynomial function is a linear function.
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:An arbitrary real polynomial function of degree $k$ is continuous on $\R$
from which it is to be shown that:
:An arbitrary real polynomial function of degree $k + 1$ is continuous on $\R$.
=== Induction Step ===
This is the induction step:
:$\forall x \in \R: \map {f_{k + 1} } x = a_{k + 1} x^{k + 1} + \map {f_k} x$
for some real polynomial function of degree $k$, where $a_{k + 1} \in \R$
From the induction hypothesis $f_k$ is continuous on $\R$.
From Real Power Function for Positive Integer Power is Continuous‎, the real function $x \mapsto x^{k + 1}$ is continuous.
Hence from the Combined Sum Rule for Continuous Real Functions:
:$f_{k + 1}$ is continuous.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: f_n$ is continuous on $\R$.
Category:Real Polynomial Function is Continuous
Real Polynomial Function is Continuous/Proof 2

== Theorem ==
== Proof ==
From Linear Function is Continuous‎, setting $\alpha = 1$ and $\beta = 0$, we have that:
:$\ds \lim_{x \mathop \to c} x = c$
Repeated application of the Product Rule for Limits of Real Functions shows us that:
:$\ds \forall k \in \N: \lim_{x \mathop \to c} x^k = c^k$
Now let $\map P x = a_n x^N + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$.
By repeated application of the Combined Sum Rule for Limits of Real Functions, we find that:
:$\ds \lim_{x \mathop \to c} \map P x = \map P c$
So whatever value we choose for $c$, we have that $\map P x$ is continuous at $c$.
From the definition of continuity on an interval, the second assertion follows.
== Sources ==
* : $\S 8.13$
Category:Real Polynomial Function is Continuous
Real Polynomial Function is Differentiable

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $f_n: \R \to \R$ be a real polynomial function.
Then $f_n$ is differentiable over $\R$.
== Proof ==
Let $f_n$ be an arbitrary real polynomial function of degree $n$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$f_n$ is differentiable over $\R$.
$\map P 0$ is the case $f_0$, where $f_0$ is of zero degree.
Such a real polynomial function is a constant function.
From Derivative of Constant we have that $f_0$ is differentiable over $\R$.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
From Derivative of Identity Function: Corollary we have that $f_1$ is differentiable over $\R$.
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:An arbitrary real polynomial function of degree $k$ is differentiable over $\R$
from which it is to be shown that:
:An arbitrary real polynomial function of degree $k + 1$ is differentiable over $\R$.
=== Induction Step ===
This is the induction step:
:$\forall x \in \R: \map {f_{k + 1} } x = a_{k + 1} x^{k + 1} + \map {f_k} x$
for some real polynomial function of degree $k$, where $a_{k + 1} \in \R$
From the induction hypothesis $f_k$ is differentiable over $\R$.
From Power Rule for Derivatives, the real function $x \mapsto x^{k + 1}$ is differentiable over $\R$.
Hence from the Linear Combination of Derivatives:
:$f_{k + 1}$ is differentiable over $\R$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: f_n$ is differentiable over $\R$.
Category:Real Polynomial Functions
Category:Differentiable Real Functions
Real Power Function for Positive Integer Power is Continuous

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map {f_n} x = x^n$
Then $f_n$ is continuous on $\R$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall x \in \R: f_n$ is continuous on $\R$.
$\map P 0$ is the case:
:$\forall x \in \R: \map {f_0} x = x^0 = 1$
Thus it is seen that $f_0$ is the constant mapping.
It follows from Constant Real Function is Continuous that $f_0$ is continuous on $\R$.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\forall x \in \R: \map {f_1} x = x^1 = x$
It follows from Linear Function is Continuous that $f_1$ is continuous on $\R$.
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\forall x \in \R: f_k$ is continuous on $\R$.
from which it is to be shown that:
:$\forall x \in \R: f_{k + 1}$ is continuous on $\R$.
=== Induction Step ===
This is the induction step:
From the basis for the induction:
:$f_1$ is continuous on $\R$.
From the induction hypothesis:
:$f_k$ is continuous on $\R$.
It follows from the Product Rule for Continuous Real Functions that $f_{k + 1}$ is continuous on $\R$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: f_n$ is continuous on $\R$.
Category:Continuous Real Functions
Category:Integer Powers
Real Power is of Exponential Order Epsilon

== Theorem == 
Let: 
:$f: \hointr 0 \to \to \R: t \mapsto t^r$
be $t$ to the power of $r$, for $r \in \R, r > -1$.
Then $f$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
== Proof ==
For $t > 0$, $t^r$ is continuous.
At $t = 0$, defining $0^r = 0$, the function is continuous from the right:
Fix $t > 1$.
By the Axiom of Archimedes, there is a natural number $n$ such that $n > r$.
By Real Power Function on Base Greater than One is Strictly Increasing, $t^r < t^n$.
Recall from Polynomial is of Exponential Order Epsilon, $t^n < K e^{at}$ for any $a > 0$, arbitrarily small in magnitude.
Therefore the inequality $t^r < K e^{at}$ has solutions of the same nature.
Category:Exponential Order
Real Power of Strictly Positive Real Number is Strictly Positive

== Theorem ==
Let $x$ be a strictly positive real number.
Let $y$ be a real number.
Then:
:$x^y > 0$
where $x^y$ denotes $x$ raised to the $y$th power.
== Proof ==
From the definition of power:
:$x^y = \exp \left({y \ln x}\right)$
From Exponential of Real Number is Strictly Positive:
:$x^y = \exp \left({y \ln x}\right) > 0$
Category:Real Analysis
Real Projective Plane is 2-Manifold

== Theorem ==
The real projective plane is a $2$-manifold.
The real projective plane can be modelled as a square $ABCD$ whose opposite sides are identified after a half-twist:
:
That is:
:$AB$ is identified with $CD$
:$DA$ is identified with $BC$
== Proof ==
== Sources ==
* 
Category:Real Projective Plane
Category:Topological Manifolds
Real Rational Function is Continuous

== Theorem ==
A real rational function is continuous at every point at which it is defined.
Thus a real rational function is continuous on every interval of $\R$ not containing a root of the denominator of the function.
== Proof ==
Let:
:$\map R x = \dfrac {\map P x} {\map Q x}$
be a real rational function, defined at all points of $\R$ at which $\map Q x \ne 0$.
Let $c \in \R$.
From Real Polynomial Function is Continuous‎:
:$\ds \lim_{x \mathop \to c} \map P x = \map P c$
and:
:$\ds \lim_{x \mathop \to c} \map Q x = \map Q c$
Thus by Quotient Rule for Limits of Real Functions:
:$\ds \lim_{x \mathop \to c} \map R x = \lim_{x \mathop \to c} \frac {\map P x} {\map Q x} = \frac {\map P c} {\map Q c}$
whenever $\map Q c \ne 0$.
So whatever value we choose for $c$ such that $\map Q c \ne 0$, we have that $\map P x$ is continuous at $c$.
From the definition of continuity on an interval, the second assertion follows.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.4$: Continuity
* : $\S 8.13$
* 
Category:Real Analysis
Category:Continuous Real Functions
Real Semi-Inner Product Space is Complex Semi-Inner Product Space

== Theorem ==
Let $\struct{V, \innerprod \cdot \cdot}$ be a real semi-inner product space.
Then $\struct{V, \innerprod \cdot \cdot}$ is a complex semi-inner product Space.
== Proof ==
This follows immediately from:
* 
* 
* Real Semi-Inner Product is Complex Semi-Inner Product
== Also see ==
* Real Inner Product Space is Complex Inner Product Space
Category:Semi-Inner Product Spaces
Real Semi-Inner Product is Complex Semi-Inner Product

== Theorem ==
Let $V$ be a vector space over a real subfield $\GF$.
Let $\innerprod \cdot \cdot: V \times V \to \GF$ be a real semi-inner product.
Then $\innerprod \cdot \cdot$ is a complex semi-inner product
== Proof ==
This follows immediately from:
* 
* 
* Complex Number equals Conjugate iff Wholly Real
== Also see ==
* Real Semi-Inner Product Space is Complex Semi-Inner Product Space
* Real Inner Product is Complex Inner Product
Category:Semi-Inner Product Spaces
Real Sequence/Examples

== Examples of Real Sequences ==
=== Example: $\sequence {\paren {-1}^n}$ ===
=== Example: $\sequence {\dfrac {\paren {-1}^{n + 1} } n}$ ===
=== Example: $\sequence {n^{-1} }$ ===
=== Example: $\sequence 1$ ===
=== Example: $\sequence {2^n}$ ===
=== Example: $\sequence {x^n}$ ===
=== Example: $\sequence {n^s}$ ===
=== Example: $\sequence {\dfrac 1 2 \paren {x_{n - 1} + \dfrac 2 {x_{n - 1} } } }_{n \mathop \ge 2}$ ===
=== Example: $\sequence {\dfrac n {n + 1} }$ ===
=== Example: $\sequence {\paren {-1}^n 2^{-n} }$ ===
=== Example: $\sequence {\dfrac {x^n} {n!} }$ ===
=== Example: Arbitrary Sequence $1$ ===
=== Example: $n$ ===
=== Example: $\dfrac {1 + \paren {-1}^n} 2$ ===
=== Example: $x^n$ for $0 < x < 1$ ===
=== Example: Arbitrary Sequence $2$ ===
=== Example: $\sqrt {2 + \sqrt {x_n} }$ ===
Category:Examples of Real Sequences