nyuuzyou/wikis huggingface dataset:
Non-Zero Integer has Finite Number of Divisors

== Theorem ==
Let $n \in \Z_{\ne 0}$ be a non-zero integer.
Then $n$ has a finite number of divisors.
== Proof ==
Let $S$ be the set of all divisors of $n$.
Then from Absolute Value of Integer is not less than Divisors:
:$\forall m \in S: -n \le m \le n$
Thus $S$ is finite.
Category:Number Theory
Category:Divisors
Non-Zero Integer has Unique Positive Integer Associate

== Theorem ==
Let $a \in \Z$ be an integer such that $a \ne 0$.
Then $a$ has a unique associate $b \in \Z_{>0}$.
== Proof ==
Let $a \in \Z_{\ne 0}$.
By Integer Divides its Absolute Value:
:$a \divides \size a$ and $\size a \divides a$
Hence $\size a$ is an associate of $a$.
Now we prove its uniqueness.
Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$.
Let $a \sim b$ and $a \sim c$ where $\sim$ denotes the relation of associatehood.
By definition of associatehood:
:$a \divides b$ and $b \divides a$
and:
:$a \divides c$ and $c \divides a$
From Divisor Relation is Antisymmetric/Corollary/Proof 2:
:$a = \pm b$
and
:$a = \pm c$
That is:
:$\pm b = \pm c$
which means:
:$b = c$ or $b = -c$
But as both $b > 0$ and $c > 0$:
:$b = c$
Hence the result.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
Category:Integers
Non-Zero Integers Closed under Multiplication

== Theorem ==
The set of non-zero integers is closed under multiplication.
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ as:
:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
Integer multiplication is defined as:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$
From Integer Multiplication is Closed, we have that $x, y \in \Z \implies x y \in \Z$.
From Ring of Integers has no Zero Divisors, we have that $x, y \in \Z: x, y \ne 0 \implies x y \ne 0$.
Therefore multiplication on the non-zero integers is closed.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$
Category:Integer Multiplication
Category:Algebraic Closure
Non-Zero Integers are Cancellable for Multiplication

== Theorem ==
Every non-zero integer is cancellable for multiplication.
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
Some sources give this as the cancellation law, but this term is already in use in the context of a group.
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Integers
Category:Non-Zero Integers are Cancellable for Multiplication
Non-Zero Integers are Cancellable for Multiplication/Proof 1

== Theorem ==
== Proof ==
Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.
By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication.
Let $x < 0$.
We know that the Integers form Integral Domain and are thus a ring.
Then $-x > 0$ and so:
So whatever non-zero value $x$ takes, it is cancellable for multiplication.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.10$
Category:Non-Zero Integers are Cancellable for Multiplication
Non-Zero Integers are Cancellable for Multiplication/Proof 3

== Theorem ==
== Proof ==
Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.
By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication.
{{qed|lemma}
Let $x < 0$.
We know that the Integers form Integral Domain and are thus a ring.
Then $-x > 0$ and so:
So whatever non-zero value $x$ takes, it is cancellable for multiplication.
Category:Non-Zero Integers are Cancellable for Multiplication
Non-Zero Integers under Multiplication are not Subgroup of Reals

== Theorem ==
Let $\struct {\Z_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero integers under multiplication.
Let $\struct {\R_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero real numbers under multiplication.
Then, while $\struct {\Z_{\ne 0}, \times}$ is closed, it is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
== Proof ==
We have that Non-Zero Real Numbers under Multiplication form Group.
We also have that the set of non-zero integers $\Z_{\ne 0}$ form a subset of $\R_{\ne 0}$.
From Non-Zero Integers Closed under Multiplication:
:$\forall a, b \in \Z_{\ne 0}: a \times b \in \Z_{\ne 0}$
We have that:
:$\forall x \in \Z_{\ne 0}: 1 \times x= x = x \times 1$
and so $1$ is the identity of $\struct {\Z_{\ne 0}, \times}$.
But for $x \in \Z_{\ne 0}: x \ne 1$, there exists no $y \in \Z_{\ne 0}: x \times y = 1$.
Thus $\struct {\Z_{\ne 0}, \times}$ does not have inverses for all $x \in \Z_{\ne 0}$.
Thus, by definition, $\struct {\Z_{\ne 0}, \times}$ is not a group.
It follows that $\struct {\Z_{\ne 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $4$
Category:Integer Multiplication
Category:Real Multiplication
Non-Zero Modulo Numbers Closed under Multiplication then Modulo is Prime

== Theorem ==
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$ for $m > 1$.
Let $\Z'_m$ be the set of non-zero integers modulo $m$.
Let $\struct {\Z_m, \times_m}$ be closed under modulo multiplication.
Then $m$ is prime.
== Proof ==
Suppose $m$ is not prime.
Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$.
So $r, s \in \Z'_m$.
But:
:$r \times_m s \equiv 0 \pmod m$
and so $r \times_m s \notin \Z'_m$.
So if $m$ is not prime, $\struct {\Z_m, \times_m}$ is not closed.
The result follows from the Rule of Transposition.
== Also see ==
* Ring of Integers Modulo m is Ring
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$
Category:Modulo Arithmetic
Category:Ring of Integers Modulo m
Category:Prime Numbers
Non-Zero Multiple of Closed Linear Transformation is Closed

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces over $\C$.
Let $\map D T$ be a linear subspace of $X$.
Let $T : \map D T \to Y$ be a closed linear transformation.
Let $\alpha \in \C \setminus \set 0$. 
Then $\alpha T : \map D T \to Y$ is closed.
== Proof ==
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $\struct {X, \norm {\, \cdot \,}_X}$ valued in $\map D T$ such that $\sequence {\alpha T x_n}_{n \in \N}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$.
Suppose that $x_n \to x$ and $\alpha T x_n \to y$. 
Then from Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent, $\alpha x_n \to \alpha x$. 
From Sequential Characterization of Closed Linear Transformation and since $T$ is linear, we have that $\alpha x \in \map D T$ and $y = \map T {\alpha x} = \alpha T x$.
Since $\map D T$ is a linear subspace of $X$ and $\alpha \ne 0$, we have $x \in \map D T$. 
Applying Sequential Characterization of Closed Linear Transformation again, we have that $\alpha T$ is closed.
Category:Closed Linear Transformations
Non-Zero Natural Numbers under Addition do not form Monoid

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$.
The structure $\struct {\N_{>0}, +}$ does not form a monoid.
== Proof ==
From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ forms a commutative monoid.
From Natural Numbers Bounded Below under Addition form Commutative Semigroup, $\struct {\N_{>0}, +}$ forms a commutative semigroup.
From Identity Element of Natural Number Addition is Zero, $0$ is the identity of $\struct {\N, +}$ 
From Natural Number Addition is Cancellable, all elements of $\struct {\N, +}$ are cancellable.
 $\struct {\N_{>0}, +}$ is a monoid.
Then by definition $\struct {\N_{>0}, +}$ has an identity $e$.
From Identity of Cancellable Monoid is Identity of Submonoid, $e$ the same as the identity of $\struct {\N, +}$.
That is:
:$e = 0$
But $0 \notin \N_{>0}$.
Thus it follows that $\struct {\N_{>0}, +}$ has no identity.
Hence by Proof by Contradiction it follows that $\struct {\N_{>0}, +}$ is not a monoid.
== Sources ==
* : $\S 3.1$: Monoids
Category:Natural Number Addition
Category:Examples of Monoids
Non-Zero Natural Numbers under Addition form Semigroup

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero, that is:
:$\N_{>0} = \N \setminus \set 0$
Let $+$ denote natural number addition.
The structure $\struct {\N_{>0}, +}$ forms a semigroup.
== Proof ==
This is a specific instance of Natural Numbers Bounded Below under Addition form Commutative Semigroup.
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $1$
Category:Natural Number Addition
Category:Examples of Semigroups
Non-Zero Natural Numbers under Multiplication form Commutative Monoid

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$.
The structure $\struct{\N_{>0}, \times}$ forms a commutative monoid.
== Proof ==
From Non-Zero Natural Numbers under Multiplication form Commutative Semigroup, $\struct {\N_{>0}, \times}$ forms a commutative semigroup.
From Identity Element of Natural Number Multiplication is One, $\struct {\N_{>0}, \times}$ has an identity element which is $1$.
Hence the result, by definition of commutative monoid.
== Sources ==
* : $\S 3.1$: Monoids
Category:Natural Number Multiplication
Category:Examples of Commutative Monoids
Non-Zero Natural Numbers under Multiplication form Commutative Semigroup

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero, that is, $\N_{>0} = \N \setminus \set 0$.
The structure $\struct {\N_{>0}, \times}$ forms an infinite commutative semigroup.
== Proof ==
=== ===
From Natural Numbers have No Proper Zero Divisors:
:$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$
It follows that:
:$\forall m, n \in \N_{>0}: m \times n \ne 0$
and so:
:$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$
That is, $\struct {\N_{>0}, \times}$ is closed.
=== ===
Natural Number Multiplication is Associative.
=== Commutativity ===
	
Natural Number Multiplication is Commutative.
=== Infinite ===
We have that the Natural Numbers are Infinite.
Then we have that Infinite if Injection from Natural Numbers.
The mapping $s: \N \to \N_{>0}: \map s n = n + 1$ is such an injection.
Hence $\N_{>0}$ is infinite.
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $2$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
Category:Natural Number Multiplication
Category:Examples of Commutative Semigroups
Non-Zero Natural Numbers under Multiplication with Divisibility forms Ordered Semigroup

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero, that is, $\N_{>0} = \N \setminus \set 0$.
Let $\divides$ denote the divisibility relation on $\N_{>0}$:
:$\forall a, b \in \N_{>0}: a \divides b \iff \exists k \in \Z: k \times a = b$
where $\times$ denotes conventional integer multiplication.
The ordered structure $\struct {\N_{>0}, \times, \divides}$ forms an ordered semigroup.
== Proof ==
First we note that from Non-Zero Natural Numbers under Multiplication form Commutative Semigroup, $\struct {\N_{>0}, \times}$ is a semigroup.
From Divisor Relation on Positive Integers is Partial Ordering, $\struct {\N_{>0}, \divides}$ is an ordered set.
It remains to be shown that $\divides$ is compatible with $\times$.
Let $a, b \in \N_{>0}$ such that $a \divides b$.
We have:
Similarly:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.9$
Category:Examples of Ordered Semigroups
Category:Natural Number Multiplication
Category:Divisibility
Non-Zero Rational Numbers Closed under Multiplication

== Theorem ==
The set of non-zero rational numbers is closed under multiplication.
== Proof ==
Recall that Rational Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\Q_{\ne 0}, \times}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$
Category:Rational Multiplication
Category:Algebraic Closure
Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group

== Theorem ==
Let $\Q_{\ne 0}$ be the set of non-zero rational numbers:
:$\Q_{\ne 0} = \Q \setminus \set 0$
The structure $\struct {\Q_{\ne 0}, \times}$ is a countably infinite abelian group.
== Proof ==
From the definition of rational numbers, the structure $\struct {\Q, + \times}$ is constructed as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence from Multiplicative Group of Field is Abelian Group, $\struct {\Q_{\ne 0}, \times}$ is an abelian group.
From Rational Numbers are Countably Infinite, we have that $\struct {\Q_{\ne 0}, \times}$ is a countably infinite group.
== Also see ==
* Definition:Multiplicative Group of Rational Numbers
* Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
* : $\S 1$: Some examples of groups: Example $1.05$
* : Chapter $2$: The Definition of Group Structure: $\S 29 \alpha \ (2)$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $1$: Definitions and Examples: Example $1.5$
* : $\S 1.1$: Example $2$
Category:Rational Multiplication
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Non-Zero Real Numbers Closed under Multiplication

== Theorem ==
The set of non-zero real numbers is closed under multiplication:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$
Category:Real Multiplication
Category:Algebraic Closure
Category:Non-Zero Real Numbers Closed under Multiplication
Non-Zero Real Numbers Closed under Multiplication/Proof 1

== Theorem ==
== Proof ==
Recall that Real Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\R_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\R_{\ne 0}, \times}$.
Category:Non-Zero Real Numbers Closed under Multiplication
Non-Zero Real Numbers Closed under Multiplication/Proof 2

== Theorem ==
== Proof ==
Let $x \times y = 0$.
, suppose that $x \ne 0$.
Then:
Thus:
:$x \times y = 0, x \ne 0 \implies y = 0$
Mutatis mutandis
:$x \times y = 0, y \ne 0 \implies x = 0$
and so:
:$x \times y = 0 \implies y = 0 \lor x = 0$
So:
The result follows by the Rule of Transposition.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(l)}$
Category:Non-Zero Real Numbers Closed under Multiplication
Non-Zero Real Numbers under Multiplication form Abelian Group

== Theorem ==
Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
* : $\S 1$: Some examples of groups: Example $1.06$
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $1$: Definitions and Examples: Example $1.5$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Real Multiplication
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Category:Non-Zero Real Numbers under Multiplication form Abelian Group
Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 1

== Theorem ==
== Proof ==
Taking the group axioms in turn:
=== ===
From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication.
Note that proof 2 needs to be used specifically here, as proof 1 rests on this result.
=== ===
Real Multiplication is Associative.
=== : Real Multiplication Identity is One ===
=== : Inverse for Real Multiplication ===
=== $\text C$: Commutativity ===
	
Real Multiplication is Commutative.
=== Infinite ===
Real Numbers are Uncountably Infinite.
== Sources ==
* : $\S 4.5$. Examples of groups: Example $82$
Category:Non-Zero Real Numbers under Multiplication form Abelian Group
Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 2

== Theorem ==
== Proof ==
We have Real Numbers under Multiplication form Monoid.
From Inverse for Real Multiplication, the non-zero numbers are exactly the invertible elements of real multiplication.
Thus from Invertible Elements of Monoid form Subgroup of Cancellable Elements, the non-zero real numbers under multiplication form a group.
From:
:Real Multiplication is Commutative
:Subset Product within Commutative Structure is Commutative
it follows that this group is also Abelian.
Category:Non-Zero Real Numbers under Multiplication form Abelian Group
Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 3

== Theorem ==
== Proof ==
From Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ forms a group.
From Real Multiplication is Commutative it follows that $\struct {\R_{\ne 0}, \times}$ is abelian.
From Real Numbers are Uncountably Infinite it follows that $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.
Category:Non-Zero Real Numbers under Multiplication form Abelian Group
Non-Zero Real Numbers under Multiplication form Group

== Theorem ==
Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ forms a group.
== Proof ==
<onlyinclude>
Taking the group axioms in turn:
=== ===
From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication.
Note that proof 2 needs to be used specifically here, as proof 1 rests on this result.
=== ===
Real Multiplication is Associative.
=== : Real Multiplication Identity is One ===
=== : Inverse for Real Multiplication ===
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $5$
* : $\S 33$. The definition of a group
Category:Real Multiplication
Category:Examples of Groups
Non-Zero Subspace of Topological Vector Space is not von Neumann-Bounded

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$. 
Let $Y \ne \set {\mathbf 0_X}$ be a non-trivial subspace of $X$. 
Then $Y$ is not von Neumann-bounded.
== Proof ==
Let $x \in Y \setminus \set {\mathbf 0_X}$.
Then $t x \in Y$ for $t > 0$. 
Since $X$ is Hausdorff, there exists an open neighborhood $V$ of $x$ that does not contain $\mathbf 0_X$.
Then $t x \not \in t V$ for each $t > 0$. 
So we do not have $Y \subseteq t V$ for any $t > 0$. 
So $Y$ is not von Neumann-bounded.
== Sources ==
* : $1.29$: Bounded sets
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Non-Zero Value of Continuous Real-Valued Function has Neighborhood not including Zero

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $f: M \to \R$ be a continuous real-valued function.
Let $\map f a > 0$ for some $a \in M$.
Then there exists $\delta \in \R_{>0}$ such that:
:$\forall x \in \map {B_\delta} a$
where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$ in $M$.
== Proof ==
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \map f a$.
Let $\map {N_\epsilon} {\map f a} = \openint {\map f a - \epsilon} {\map f a + \epsilon}$ be the $\epsilon$-neighborhood of $\map f a$.
Then $\map {N_\epsilon} {\map f a}$ is an open interval of $\map f a$ such that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
We are given that $f$ is continuous.
So, by definition:
:$\exists U \subseteq A: \forall x \in U: \map f x \in \map {N_\epsilon} {\map f a}$
where $U$ is an open set of $M$.
As $a \map {N_\epsilon} {\map f a}$, it follows that:
:$a \in U$
By definition of open set:
:$\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$
Hence:
:$\forall x \in \map {B_\delta} a: \map f x \in \map {N_\epsilon} {\map f a}$
As we have already noted that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
the result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 13$
Category:Continuous Mappings on Metric Spaces
Non-Zero Vectors are Orthogonal iff Perpendicular

== Theorem ==
Let $\mathbf u$, $\mathbf v$ be non-zero vectors in the real Euclidean space $\R^n$.
Then $\mathbf u$ and $\mathbf v$ are orthogonal they are perpendicular.
== Proof ==
=== Necessary Condition ===
When $\theta$ denotes the angle between $\mathbf u$ and $\mathbf v$ measured in radians, we have:
where $\norm {\,\cdot\,}$ denotes the Euclidean norm.
As $\mathbf u, \mathbf v$ are non-zero vectors, it follows from the norm axiom of positive definiteness that $\norm {\mathbf u} \ne 0$, and $\norm {\mathbf v} \ne 0$.
Then $\cos \theta = 0$.
It follows from Zeroes of Sine and Cosine that $\theta = \dfrac \pi 2$, given that $\theta \in \closedint 0 \pi$ by the convention of the definition of angle between vectors.
Then $\mathbf u$ and $\mathbf v$ are perpendicular by the measurement of a right angle.
=== Sufficient Condition ===
If $\mathbf u$ and $\mathbf v$ are perpendicular, then the angle between them measures $\dfrac \pi 2$ in radians.
Then $\cos \theta = 0$, and the calculations above show that $\mathbf u \cdot \mathbf v = 0$.
Hence, $\mathbf u$ and $\mathbf v$ are orthogonal.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Linear Algebra
Category:Analytic Geometry