nyuuzyou/wikis huggingface dataset:
Multiplication of Real Numbers is Right Distributive over Subtraction

== Theorem ==
That is, for any number $a$ and for any integers $m, n$:
:$m a - n a = \paren {m - n} a$
== Proof ==
Let two magnitudes $AB, CD$ be equimultiples of two magnitudes $E, F$.
Let $AG, CH$ subtracted from them be equimultiples of the same two $E, F$.
We need to show that the remainders $GB, HD$ are either equal to $E, F$ or are equimultiples of them.
:
First let $GB = E$.
Let $CK$ be made equal to $F$.
We have that $AG$ is the same multiple of $E$ that $CH$ is of $F$, while $GB = E$ and $KC = F$.
Therefore from the Distributive Laws of Arithmetic, $AB$ is the same multiple of $E$ that $KH$ is of $F$.
But , $AB$ is the same multiple of $E$ that $CD$ is of $F$.
Since then, each of the magnitudes $KH, CD$ is the same multiple of $F$.
Therefore $KH = CD$.
Let $CH$ be subtracted from each.
Therefore the remainder $KC$ equals the remainder $HD$.
But $F = KC$, so $HD = F$.
Hence, if $GB = E$ then $HD = F$.
Similarly we can prove that, even if $GB$ is a multiple of $E$, then $HD$ is also the same multiple of $F$.
== Also see ==
* Multiplication of Real Numbers is Left Distributive over Subtraction
* Distributive Laws of Arithmetic
== Sources ==
* : Book $\text{V}$. Propositions
Category:Real Multiplication
Category:Real Subtraction
Category:Examples of Distributive Operations
Category:Multiplication of Real Numbers Distributes over Subtraction
Multiplication of Real and Imaginary Parts

== Theorem ==
Let $w, z \in \C$ be complex numbers.
$(1)$ If $w$ is wholly real, then:
:$\map \Re {w z} = w \, \map \Re z$
and:
:$\map \Im {w z} = w \, \map \Im z$
$(2)$ If $w$ is wholly imaginary, then:
:$\map \Re {w z} = -\map \Im w \, \map \Im z$
and:
:$\map \Im {w z} = \map \Im w \, \map \Re z$
Here, $\map \Re z$ denotes the real part of $z$, and $\map \Im z$ denotes the imaginary part of $z$.
== Proof ==
Assume that $w$ is wholly real.
Then:
This equation shows that $\map \Re {w z} = w \, \map \Re z$, and $\map \Im {w z} = w \, \map \Im z$.
This proves $(1)$.
Now, assume that $w$ is wholly imaginary.
Then:
This equation shows that $\map \Re {w z} = -\map \Im w \, \map \Im z$, and $\map \Im {w c} = \map \Im w \, \map \Re z$.
This proves $(2)$.
Category:Complex Multiplication
Multiplication on 1-Based Natural Numbers is Cancellable

== Theorem ==
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $\times$ be multiplication on $\N_{>0}$.
Then:
:$\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a = b$
:$\forall a, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$
That is, $\times$ is cancellable on $\N_{>0}$.
== Proof ==
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
:$a = b$
:$a < b$
:$b < a$
Suppose $a < b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$a \times c < b \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c = b \times c$.
Similarly, suppose $b > a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$b \times c < a \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c = b \times c$.
The only other possibility is that $a = b$.
So 
:$\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a = b$
and so $\times$ is right cancellable on $\N_{>0}$.
From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
:$\forall , b, c \in \N_{>0}: a \times b = a \times c \implies b = c$
So $\times$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
== Sources ==
* : Introduction $\S 4$: The natural numbers: $\text{M} 3$
* : $2.2$: Theorem $2.13$
Category:Natural Numbers/1-Based
Category:Natural Number Multiplication
Multiplication on 1-Based Natural Numbers is Cancellable for Ordering

== Theorem ==
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $\times$ be multiplication on $\N_{>0}$.
Let $<$ be the strict ordering on $\N_{>0}$.
Then:
:$\forall a, b, c \in \N_{>0}: a \times c < b \times c \implies a < b$
:$\forall a, b, c \in \N_{>0}: a \times b < a \times c \implies b < c$
That is, $\times$ is cancellable on $\N_{>0}$ for $<$.
== Proof ==
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
:$a = b$
:$a < b$
:$b < a$
Let $a \times c < b \times c$.
Suppose $a = b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$a \times c = b \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c < b \times c$.
Similarly, suppose $b < a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$b \times c < a \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c < b \times c$.
The only other possibility is that $a < b$.
So 
:$\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a < b$
and so $\times$ is right cancellable on $\N_{>0}$ for $<$.
Let $a \times b < a \times c$.
Suppose $b = c$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$a \times b = a \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times b < a \times c$.
Similarly, suppose $c < b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
:$a \times c < a \times b$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times b < a \times c$.
The only other possibility is that $b < c$.
So 
:$\forall a, b, c \in \N_{>0}: a \times b < a \times c \implies b < c$
and so $\times$ is left cancellable on $\N_{>0}$ for $<$.
From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
:$\forall a, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$
So $+$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
== Sources ==
* : $2.2$: Theorem $2.14 \ \text{(ii)}$
Category:Natural Numbers/1-Based
Category:Natural Number Multiplication
Multiplication on Normed Algebra is Continuous

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Let $\struct {A^2, \norm {\, \cdot \,}_{A^2} }$ be the direct product of $A$ with itself equipped with the direct product norm.
Define $m : A^2 \to A$ by:
:$\map m {a, b} = a b$
for each $\tuple {a, b} \in A^2$. 
Then $m$ is continuous.
== Proof ==
Fix $\tuple {a, b} \in A^2$. 
Let $\tuple {a', b'} \in A^2$. 
Then we have:
Let $\epsilon > 0$.
Consider first the case of $b = {\mathbf 0}_A$. 
Take $\norm {a - a'} \le 1$. 
Then by the Reverse Triangle Inequality: Normed Vector Space, we have $\norm {a'} \le \norm a + 1$.
Hence, by taking: 
:$\ds \norm {b - b'} < \frac \epsilon {\norm a + 1}$ 
we have:
:$\norm b \norm {a - a'} + \norm {a'} \norm {b - b'} < \epsilon$
So by taking:
:$\ds \norm {\tuple {a, b} - \tuple {a', b'} }_{A^2} < \min \set {1, \frac \epsilon {\norm a + 1} }$
we have:
:$\norm {\map m {a, b} - \map m {a', b'} }_{A^2} < \epsilon$ 
So $m$ is continuous at $\tuple {a, {\mathbf 0}_A}$ for each $a \in A$.
Now consider $b \ne {\mathbf 0}_A$. 
If we take:
:$\ds \norm {a - a'} < \frac \epsilon {2 \norm b}$
then by the Reverse Triangle Inequality: Normed Vector Space, we have:
:$\ds \norm {a'} < \norm a + \frac \epsilon {2 \norm b}$
Then we can take:
:$\ds \norm {b - b'} < \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1}$
to obtain:
:$\ds \norm b \norm {a - a'} + \norm {a'} \norm {b - b'} < \norm b \paren {\frac \epsilon {2 \norm b} } + \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} } \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1} = \epsilon$
To conclude, by taking:
:$\ds \norm {\tuple {a, b} - \tuple {a', b'} }_{A^2} < \min \set {\frac \epsilon {2 \norm b}, \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1} }$
we have:
:$\norm {\map m {a, b} - \map m {a', b'} }_{A^2} < \epsilon$ 
So $m$ is continuous at $\tuple {a, b}$ for each $a \in A$, $b \ne {\mathbf 0}_A$.
Hence $m$ is continuous at $\tuple {a, b}$ for each $\tuple {a, b} \in A$.
Hence $m$ is continuous.
Category:Normed Algebras
Multiplication using Parabola

== Theorem ==
:
Let the parabola $P$ defined as $y = x^2$ be plotted on the Cartesian plane.
Let $A = \tuple {x_a, y_a}$ and $B = \tuple {x_b, y_b}$ be points on the curve $\map f x$ so that $x_a < x_b$.
Then the line segment joining $A B$ will cross the $y$-axis at $-x_a x_b$.
Thus $P$ can be used as a nomogram to calculate the product of two numbers $x_a$ and $x_b$, as follows:
:$(1) \quad$ Find the points $-x_a$ and $x_b$ on the $x$-axis.
:$(2) \quad$ Find the points $A$ and $B$ where the lines $x = -x_a$ and $x = x_b$ cut $P$.
:$(3) \quad$ Lay a straightedge on the straight line joining $A$ and $B$ and locate its $y$-intercept $c$.
Then $x_a x_b$ can be read off from the $y$-axis as the position of $c$.
== Proof ==
Let $\map f x = x^2$.
Then:
:$\map f {x_a} = x_a^2$
and:
:$\map f {x_b} = x_b^2$
Then the slope $m$ of the line segment joining $A B$ will be:
From Equation of Straight Line in Plane: Slope-Intercept Form:
:$y = \paren {x_b + x_a} x + c$
where $c$ denotes the $y$-intercept.
Substituting the coordinates of point $A = \tuple {x_a, x_a^2}$ for $\tuple {x, y}$:
== Sources ==
* : On Being your Own Teacher: Labour Saving Devices
Category:Multiplication
Category:Parabolas
Category:Nomograms
Multiplicative Auxiliary Relation iff Congruent

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below lattice.
Let $\RR$ be an auxiliary relation on $S$.
Then $\RR$ is multiplicative :
:$\forall a, b, x, y \in S: \tuple {a, x}, \tuple {b, y} \in \RR \implies \tuple {a \wedge b, x \wedge y} \in \RR$
That is $\RR$ is a congruence relation for $\wedge$.
== Proof ==
=== Sufficient Condition ===
Let $\RR$ be multiplicative.
Let $a, b, x, y \in S$ such that
:$\tuple {a, x}, \tuple {b, y} \in \RR$
By Meet Precedes Operands:
:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
By definition of reflexivity:
:$x \preceq x$ and $y \preceq y$
By definition of auxiliary relation:
:$\tuple {a \wedge b, x}, \tuple {a \wedge b, y} \in \RR$
Thus by definition of multiplicative relation:
:$\tuple {a \wedge b, x \wedge y} \in \RR$
=== Necessary Condition ===
Suppose:
:$\forall a, b, x, y \in S: \tuple {a, x}, \tuple {b, y} \in \RR \implies \tuple {a \wedge b, x \wedge y} \in \RR$
Let $a, x, y \in S$ such that
:$\tuple {a, x}, \tuple {a, y} \in \RR$
By Meet is Idempotent:
:$a \wedge a = a$
Thus :
:$\tuple {a, x \wedge y} \in \RR$
== Sources ==
* 
* 
Category:Order Theory
Category:Auxiliary Relations
Multiplicative Auxiliary Relation iff Images are Filtered

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below lattice.
Let $\RR$ be an auxiliary relation on $S$.
Then $\RR$ is multiplicative :
:for all $x \in S$: $\map \RR x$ is filtered
where $\map \RR x$ denotes the $\RR$-image of $x$.
== Proof ==
=== Sufficient Condition ===
Let $\RR$ be multiplicative.
Let $x \in S$.
Let $a, b \in \map \RR x$.
By definition of $\RR$-image of element:
:$\tuple {x, a}, \tuple {x, b} \in \RR$
By definition of multiplicative relation:
:$\tuple {x, a \wedge b} \in \RR$
By definition of $\RR$-image of element:
:$a \wedge b \in \map \RR x$
By Meet Precedes Operands:
:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
Thus
:$\exists c \in \map \RR x: c \preceq a \land c \preceq b$
Hence $\map \RR x$ is filtered.
=== Necessary Condition ===
Suppose that:
:for all $x \in S$: $\map \RR x$ is filtered
Let $a, x, y \in S$ such that:
:$\tuple {a, x}, \tuple {a, y} \in \RR$
By definition of $\RR$-image of element:
:$x, y \in \map \RR a$
By Auxiliary Relation Image of Element is Upper Section:
:$\map \RR a$ is an upper section.
By assumption:
:$\map \RR a$ is filtered.
By Filtered in Meet Semilattice:
:$x \wedge y \in \map \RR a$
Thus by definition of $\RR$-image of element:
:$\tuple {a, x \wedge y} \in \RR$
== Sources ==
* 
* 
Category:Order Theory
Category:Auxiliary Relations
Multiplicative Function that Converges to Zero on Prime Powers

== Theorem ==
Let $f$ be a multiplicative function such that:
:$\ds \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$
where $p^k$ runs though all prime powers.
Then:
:$\ds \lim_{n \mathop \to \infty} \map f n = 0$
where $n$ runs through the integers.
== Proof ==
By hypothesis, there exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > 1$.
Let $\ds A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.
Thus $A \ge 1$.
Let $0 < \dfrac \epsilon A$. 
There exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > \dfrac \epsilon A$.
Therefore there are only finitely many integers $n$ such that:
:$\size {\map f {p^k} } > \dfrac \epsilon A$
for every prime power $p^k$ that divides $n$.
Therefore if $n$ is sufficiently large there exists a prime power $p^k$ that divides $n$ such that:
:$\size {\map f {p^k} } < \dfrac \epsilon A$
Therefore $n$ can be written as:
:$\ds n = \prod_{i \mathop = 1}^ r p_i^{k_i} \prod_{i \mathop = r + 1}^{r + s} p_i^{k_i} \prod_{i \mathop = r + s + 1}^{r + s + t} p_i^{k_i}$
where $t \ge 1$ and:
Therefore:
This shows that $\map f n$ can be made arbitrarily small for sufficiently large $n$.
Category:Analytic Number Theory
Category:Multiplicative Functions
Multiplicative Group of Complex Numbers is not Direct Product of Reals with Reals

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then the direct product $\struct {\R_{\ne 0}, \times} \times \struct {\R_{\ne 0}, \times}$ is not isomorphic with $\struct {\C_{\ne 0}, \times}$.
== Proof ==
Let $\tuple {a, b}$ and $\tuple {c, d}$ be pairs of non-zero real numbers:
:$\tuple {a, b} \in \R_{\ne 0} \times \R_{\ne 0}$
:$\tuple {c, d} \in \R_{\ne 0} \times \R_{\ne 0}$
Then by definition of group direct product:
:$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c, b \times d}$
However, by interpreting $\tuple {a, b}$ and $\tuple {c, d}$ as complex numbers:
:$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c - b \times d, b \times c + a \times d}$
by definition of complex multiplication.
Hence the result.
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Multiplicative Group of Real Numbers
Category:Multiplicative Group of Complex Numbers
Category:Group Direct Products
Multiplicative Group of Complex Numbers is not Isomorphic to Additive Group of Complex Numbers

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\C_{\ne 0}, \times}$ is not isomorphic to $\struct {\C, +}$.
== Proof ==
A direct application of Additive Group and Multiplicative Group of Field are not Isomorphic.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.9$
Category:Examples of Group Isomorphisms
Category:Multiplicative Group of Complex Numbers
Category:Additive Group of Complex Numbers
Multiplicative Group of Complex Numbers is not Isomorphic to Additive Group of Real Numbers

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\C_{\ne 0}, \times}$ is not isomorphic to $\struct {\R, +}$.
== Proof ==
 $\struct {\C_{\ne 0}, \times}$ is isomorphic to $\struct {\R, +}$.
Let $\phi: \C_{\ne 0} \to \R$ be an isomorphism.
Note that $\order i = 4$ in $\struct {\C_{\ne 0}, \times}$.
By Group Isomorphism Preserves Order of Group Element:
:$\order {\map \phi i} = 4$ in $\struct {\R, +}$
Then $4 \map \phi i = 0$.
So $\map \phi i = 0$.
However $\order 0 = 1$.
Thus there is no order $4$ element in $\struct {\R, +}$, and thus $\map \phi i$ does not exist.
Therefore $\phi$ is not an isomorphism.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.9$
Category:Examples of Group Isomorphisms
Category:Multiplicative Group of Complex Numbers
Category:Additive Group of Real Numbers
Multiplicative Group of Complex Numbers is not Isomorphic to Multiplicative Group of Real Numbers

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then $\struct {\C_{\ne 0}, \times}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.
== Proof ==
 $\struct {\C_{\ne 0}, \times}$ is isomorphic to $\struct {\R_{\ne 0}, \times}$.
Let $\phi: \C_{\ne 0} \to \R_{\ne 0}$ be an isomorphism.
Note that $\order i = 4$ in $\struct {\C_{\ne 0}, \times}$.
By Group Isomorphism Preserves Order of Group Element:
:$\order {\map \phi i} = 4$ in $\struct {\R_{\ne 0}, \times}$
Then $\paren {\map \phi i}^4 = 1$.
So $\paren {\map \phi i}^2 = \pm 1$.
In $\R$ we must have $\paren {\map \phi i}^2 \ge 0$.
Hence $\paren {\map \phi i}^2 = 1$.
This gives $\map \phi i = \pm 1$.
However $\order 1 = 1$ and $\order {-1} = 2$ in $\struct {\R_{\ne 0}, \times}$.
Thus there is no order $4$ element in $\struct {\R_{\ne 0}, \times}$, and thus $\map \phi i$ does not exist.
Therefore $\phi$ is not an isomorphism.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.9$
Category:Examples of Group Isomorphisms
Category:Multiplicative Group of Complex Numbers
Category:Multiplicative Group of Real Numbers
Multiplicative Group of Complex Roots of Unity is Subgroup of Circle Group

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 0$.
Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity.
Let $\struct {K, \times}$ denote the circle group.
Then $\struct {U_n, \times}$ is a subgroup of $\struct {K, \times}$.
== Proof ==
By definition of the multiplicative group of complex $n$th roots of unity:
:$U_n := \set {z \in \C: z^n = 1}$
By definition of the circle group:
:$K = \set {z \in \C: \cmod z = 1}$
By Modulus of Complex Root of Unity equals 1:
:$\forall z \in U_n: \cmod z = 1$
Thus:
:$U_n \subseteq K$
We further have that the operation $\times$ on both $U_n$ and $K$ is complex multiplication.
Finally, from Roots of Unity under Multiplication form Cyclic Group, we have that $\struct {U_n, \times}$ is a group.
The result follows by definition of subgroup.
Category:Circle Group
Category:Multiplicative Groups of Complex Roots of Unity
Multiplicative Group of Field is Abelian Group

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Let $F^* := F \setminus \set 0$ be the set $F$ less its zero.
The algebraic structure $\struct {F^*, \times}$ is an abelian group.
== Proof 1 ==
== Proof 2 ==
Category:Field Theory
Category:Abelian Groups
Category:Multiplicative Group of Field is Abelian Group
Multiplicative Group of Field is Abelian Group/Proof 1

== Theorem ==
== Proof ==
From the field axioms:
Hence the result.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
Category:Multiplicative Group of Field is Abelian Group
Multiplicative Group of Galois Field is Cyclic

== Theorem ==
Let $\GF$ be a Galois field of order $q$.
Then its multiplicative group is cyclic of order $q-1$:
:$\GF^\times \cong C_{q - 1}$
== Proof ==
Follows immediately from Finite Multiplicative Subgroup of Field is Cyclic.
Category:Galois Fields
Multiplicative Group of Positive Rationals is Non-Cyclic

== Theorem ==
Let $\struct {\Q_{>0}, \times}$ be the multiplicative group of positive rational numbers.
Then $\struct {\Q_{>0}, \times}$ is not a cyclic group.
== Proof ==
 $\struct {\Q_{>0}, \times}$ is cyclic.
Then $\struct {\Q_{>0}, \times}$ has a generator $x$ such that $x > 1$.
It would follow that:
:$\Q = \set {\ldots, x^{-2}, x^{-1}, 1, x, x^2, \ldots}$
where the elements are arranged in ascending order.
But then consider $y \in \Q: y = \dfrac {1 + x} 2$.
So $1 < y < x$ and so $y \notin \Q$.
From this contradiction it is concluded that there can be no such single generator of $\struct {\Q_{>0}, \times}$.
Therefore, by definition, $\struct {\Q_{>0}, \times}$ is not a cyclic group.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $17$
Category:Rational Numbers
Multiplicative Group of Rationals is Normal Subgroup of Complex

== Theorem ==
Let $\struct {\Q, \times}$ be the multiplicative group of rational numbers.
Let $\struct {\C, \times}$ be the multiplicative group of complex numbers.
Then $\struct {\Q, \times}$ is a normal subgroup of $\struct {\C, \times}$.
== Proof ==
From Multiplicative Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, \times} \lhd \struct {\R, \times}$.
From Multiplicative Group of Reals is Normal Subgroup of Complex, $\struct {\R, \times} \lhd \struct {\C, \times}$.
Thus $\struct {\Q, \times} \le \struct {\C, \times}$.
From Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group, $\struct {\C, \times}$ is abelian.
From Subgroup of Abelian Group is Normal it follows that $\struct {\Q, \times} \lhd \struct {\C, \times}$.
Category:Multiplicative Group of Rational Numbers
Category:Multiplicative Group of Complex Numbers
Category:Examples of Normal Subgroups
Multiplicative Group of Rationals is Normal Subgroup of Reals

== Theorem ==
Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then $\struct {\Q_{\ne 0}, \times}$ is a normal subgroup of $\left({\R_{\ne 0}, \times}\right)$.
== Proof ==
From the definition of real numbers, it follows that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$.
As $\struct {\R_{\ne 0}, \times}$ is a group, and $\struct {\Q_{\ne 0}, \times}$ is a group, it follows from the definition of subgroup that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
As $\struct {\R_{\ne 0}, \times}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q_{\ne 0}, \times}$ is normal in $\struct {\R_{\ne 0}, \times}$.
== Sources ==
* : $\S 36$: Subgroups: Simple illustrations: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $4$: Subgroups: Example $4.3$
Category:Multiplicative Group of Rational Numbers
Category:Multiplicative Group of Real Numbers
Category:Examples of Normal Subgroups
Multiplicative Group of Reals is Normal Subgroup of Complex

== Theorem ==
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Then $\struct {\R_{\ne 0}, \times}$ is a normal subgroup of $\struct {\C_{\ne 0}, \times}$.
== Proof ==
Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are wholly real, we have that $x, y \in \R_{\ne 0}$.
Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also wholly real.
Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i$ which is also wholly real.
Thus by the Two-Step Subgroup Test, $\struct {\R_{\ne 0}, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.
From Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group, $\struct {\C_{\ne 0}, \times}$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $4$: Subgroups: Example $4.3$
Category:Multiplicative Group of Real Numbers
Category:Multiplicative Group of Complex Numbers
Category:Examples of Normal Subgroups