nyuuzyou/wikis huggingface dataset:
Multiplication by Power of 10 by Moving Decimal Point

== Theorem ==
Let $n \in \R$ be a real number.
Let $n$ be expressed in decimal notation.
Let $10^d$ denote a power of $10$ for some integer $d$
Then $n \times 10^d$ can be expressed in decimal notation by shifting the decimal point $d$ places to the right.
Thus, if $d$ is negative, and so $10^d = 10^{-e}$ for some $e \in \Z_{>0}$, $n \times 10^d$ can be expressed in decimal notation by shifting the decimal point $e$ places to the left.
== Proof ==
Let $n$ be expressed in decimal notation as:
:$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0 \cdotp a_{-1} a_{-2} \dotso a_{-s} a_{-s - 1} \dotso}$
That is:
:$n = \ds \sum_{k \mathop \in \Z} a_k 10^k$
Then:
The effect of presenting digit $a_{k - d}$ in position $k$ of $n$ is the same as what you get having moved the decimal point from between $a_0$ and $a_{-1}$ to between $a_{-d}$ and $a_{-d - 1}$.
Thus:
:if $d$ is positive, that is equivalent to moving the decimal point $d$ places to the right
and:
:if $d$ is negative, that is equivalent to moving the decimal point $d$ places to the left.
== Sources ==
* : Chapter $1$: Scientific Notation
Category:Decimal Notation
Multiplication is Arithmetically Definable

== Theorem ==
Let $f: \N^2 \to \N$ be defined as:
:$\map f {x, y} = x \times y$
Then there exists a WFF $\map \phi {z, x, y}$ of $3$ free variables and containing no quantifiers such that:
:$z = \map f {x, y} \iff \N \models \map \phi {\sqbrk z, \sqbrk x, \sqbrk y}$
where $\sqbrk a$ is the unary representation of $a \in \N$.
== Proof ==
Define:
:$\map \phi {z, x, y} := z = x \times y$
Correctness is trivial.
Category:Natural Number Multiplication
Multiplication is Primitive Recursive

== Theorem ==
The function $\operatorname{mult}: \N^2 \to \N$, defined as:
:$\map \Mult {n, m} = n \times m$
is primitive recursive‎.
== Proof ==
We observe that:
:$\map \Mult {n, 0} = n \times 0 = 0$
and that
:$\map \Mult {n, m + 1} = n \times \paren {m + 1} = \paren {n \times m} + n = \map \Add {\map \Mult {n, m}, n}$.
We are to show that $\Mult$ is obtained by primitive recursion from known primitive recursive‎ functions.
First we note that:
:$\map \Mult {n, 0} = 0 = \map \Zero n$
where $\map \Zero n$ is the zero function, which is basic primitive recursive‎.
Next we need to find a primitive recursive‎ function $g: \N^3 \to \N$ such that:
:$\map \Mult {n, m + 1} = \map g {n, m, \map \Mult {n, m} }$
Because $\map \Mult {n, m + 1} = \map \Add {\map \Mult {n, m}, n}$, we see that a suitable function for $g$ is:
:$\map g {n_1, n_2, n_3} = \map \Add {n_3, n_1}$
Using projection functions, we can write this as:
:$\map g {n_1, n_2, n_3} = \map \Add {\map {\pr_3^3} {n_1, n_2, n_3}, \map {\pr_1^3} {n_1, n_2, n_3} }$.
This shows that $g$ is obtained by substitution from $\Add$, which is primitive recursive, and the basic primitive recursive‎ functions $\pr_3^3$ and $\pr_1^3$.
So $g$ is primitive recursive‎.
So $\Mult$ is obtained by primitive recursion from the primitive recursive functions‎ $g$ and $\Zero$, and so is primitive recursive‎.
Category:Primitive Recursive Functions
Multiplication is Superfunction

== Theorem ==
The function $f: \C \to \C$, defined as:
:$\map f z = z \times c$
is a superfunction for any complex number $c$.
== Proof ==
Define $h: \C \to \C$ by $\map h z = z + c$.
Then:
Thus $\map f z = z \times c$ is a superfunction and $\map h z = z + c$ is the corresponding transfer function.
Category:Superfunctions
Multiplication of Complex Number by -1 is Equivalent to Rotation through Two Right Angles

== Theorem ==
Let $z \in \C$ be a complex number.
Let $z$ be interpreted as a vector in the complex plane.
Let $w \in \C$ be the complex number defined as $z$ multiplied by $-1$:
:$w = \left({-1}\right) z$
Then $w$ can be interpreted as the vector $z$ after being rotated through two right angles.
The direction of rotation is usually interpreted as being anticlockwise, but a rotated through two right angles is the same whichever direction the rotation is performed.
== Proof ==
:
By definition of the imaginary unit:
:$-1 = i^2$
and so:
:$-1 \times z = i \paren {i z}$
From Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle, multiplication by $i$ is equivalent to rotation through a right angle, in an anticlockwise direction.
So multiplying by $i^2$ is equivalent to rotation through two right angles in an anticlockwise direction.
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Multiplication
Category:Geometry of Complex Plane
Multiplication of Cross-Relation Equivalence Classes on Natural Numbers is Well-Defined

== Theorem ==
Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
Let $\otimes$ be the binary operation defined on these equivalence classes as:
:$\forall \eqclass {a, b} {}, \eqclass {c, d} {} \in \N \times \N: \eqclass {a, b} {} \otimes \eqclass {c, d} {} = \eqclass {\tuple {a \cdot c} + \tuple {b \cdot d}, \tuple {a \cdot d} + \tuple {b \cdot c} } {}$
where $a \cdot c$ denotes natural number multiplication between $a$ and $c$.
The operation $\otimes$ on these equivalence classes is well-defined, in the sense that:
== Proof ==
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
Both $+$ and $\times$ are commutative and associative on $\N$. Thus:
So we have $a_1 \cdot c_1 + b_1 \cdot d_1 + a_2 \cdot d_2 + b_2 \cdot c_2 = a_1 \cdot d_1 + b_1 \cdot c_1 + a_2 \cdot c_2 + b_2 \cdot d_2$ and so, by the definition of $\boxtimes$, we have:
:$\eqclass {a_1 \cdot c_1 + b_1 \cdot d_1, a_1 \cdot d_1 + b_1 \cdot c_1} {} = \eqclass {a_2 \cdot c_2 + b_2 \cdot d_2, a_2 \cdot d_2 + b_2 \cdot c_2} {}$
So, by the definition of integer multiplication, this leads to:
:$\eqclass {a_1, b_1} {} \otimes \eqclass {c_1, d_1} {} = \eqclass {a_2, b_2} {} \otimes \eqclass {c_2, d_2} {}$
Thus integer multiplication has been shown to be well-defined.
== Sources ==
* : $\S 2.5$: Theorem $2.22$
Category:Integers
Category:Multiplication
Category:Examples of Well-Defined Mappings‎
Multiplication of Cuts Distributes over Addition

== Theorem ==
Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
:$\alpha + \beta$ denote the sum of $\alpha$ and $\beta$.
:$\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.
$\alpha \paren {\beta + \gamma}$ is the set of all rational numbers $s$ of the form:
:$s = p \paren {q + r}$
such that:
:$s < 0$
or:
:$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
$\alpha \beta + \alpha \gamma$ is the set of all rational numbers $s$ of the form:
:$s = p q + p r$
such that:
:$s < 0$
or:
:$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
From Rational Multiplication Distributes over Addition:
$p \paren {q + r} = p q + p r$
and the result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(c)}$
Category:Cuts
Category:Addition
Category:Multiplication
Category:Examples of Distributive Operations
Multiplication of Cuts is Associative

== Theorem ==
Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let $\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.
$\paren {\alpha \beta} \gamma$ is the set of all rational numbers $s$ of the form $s = \paren {p q} r$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
Similarly, $\alpha \paren {\beta \gamma}$ is the set of all rational numbers $s$ of the form $s = p \paren {q r}$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
From Rational Multiplication is Associative we have that:
::$\paren {p q} r = p \paren {q r}$
Thus we have that:
:$\size \alpha \paren {\size \beta \, \size \gamma} = \paren {\size \alpha \, \size \beta} \size \gamma$
and the result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(b)}$
Category:Cuts
Category:Multiplication
Category:Examples of Associative Operations‎
Multiplication of Cuts is Commutative

== Theorem ==
Let $\alpha$ and $\beta$ be cuts.
Let the $\alpha \beta$ be the product of $\alpha$ and $\beta$.
Then:
:$\alpha \beta = \beta \alpha$
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$ and $\beta \ge 0^*$.
:$\gamma := \alpha \beta$
where $\gamma$ is the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p q$
$\alpha \beta$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = p q$ such that $p \in \alpha$ and $q \in \beta$.
Similarly, $\beta \alpha$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = q p$ such that $p \in \alpha$ and $q \in \beta$.
From Rational Multiplication is Commutative we have that:
:$p q = q p$
Thus we have that:
:$\size \alpha \, \size \beta = \size \beta \, \size \alpha$
and the result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(a)}$
Category:Cuts
Category:Multiplication
Category:Examples of Commutative Operations
Multiplication of Distribution induced by Locally Integrable Function by Smooth Function

== Theorem ==
Let $f \in \map {L^1_{loc} } {\R^d}$ be a locally integrable function.
Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth function.
Let $T_f \in \map {\DD'} {\R^d}$ be a distribution induced by $f$.
Then in the distributional sense it holds that:
:$\alpha T_f = T_{\alpha f}$
== Proof ==
Let $\Omega \subseteq \R^d$ be a compact subset.
Then for all $\mathbf x \in \Omega$ we have that $\map \alpha {\mathbf x}$ is bounded.
Hence, $\alpha f$ is locally integrable.
Let $\phi \in \map \DD {\R^d}$ be a test function.
Then:
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Distributions
Multiplication of Fractions

== Theorem ==
Let $a, b, c, d \in \Z$ such that $b d \ne 0$.
Then:
:$\dfrac a b \times \dfrac c d = \dfrac {a c} {b d}$
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:Multiplication of Fractions
Category:Fractions
Category:Multiplication
Multiplication of Integers is Primitive Recursive

== Theorem ==
Let $t : \N^2 \to \N$ be defined as:
:$\map t {m, n} = p$
where:
:$m$ is the code number for the integer $k$
:$n$ is the code number for the integer $\ell$
:$p$ is the code number for the integer $k \times \ell$
Then $t$ is a primitive recursive function.
== Proof ==
We have:
:$\size {k \times \ell} = \size k \times \size \ell$
By:
* Multiplication is Primitive Recursive
* Absolute Value of Integer is Primitive Recursive
we have that:
:$\map a {m, n} = \size {k \times \ell}$
is primitive recursive.
Additionally, we have that:
:$k \times \ell > 0$
 either:
:$k > 0 \land \ell > 0$
or:
:$k < 0 \land \ell < 0$
By
* Set of Strictly Positive Integers is Primitive Recursive
* Set of Strictly Negative Integers is Primitive Recursive
* Set Operations on Primitive Recursive Relations
the above statement is a primitive recursive relation.
Let $c^+ : \N \to \N$ be defined as:
:$\map {c^+} n = k_{\mathop + n}$
where $k_{\mathop + n}$ is the code number for the integer $+n : \Z$.
Let $c^- : \N \to \N$ be defined as:
:$\map {c^-} n = k_{\mathop - n}$
where $k_{\mathop - n}$ is the code number for the integer $-n : \Z$.
By
* Code Number for Non-Negative Integer is Primitive Recursive
* Code Number for Non-Positive Integer is Primitive Recursive
both $c^+$ and $c^-$ are primitive recursive.
We now have:
:$\map t {m, n} = \begin{cases}
\map {c^+} {\map a {m, n}} & : \paren {k > 0 \land \ell > 0} \lor \paren {k < 0 \land \ell < 0} \\
\map {c^-} {\map a {m, n}} & : \text{otherwise}
\end{cases}$
which is primitive recursive by:
:Definition by Cases is Primitive Recursive/Corollary
Category:Primitive Recursive Functions
Multiplication of Natural Numbers is Provable

== Theorem ==
Let $x, y \in \N$ be natural numbers.
Then there exists a formal proof of:
:$\sqbrk x \times \sqbrk y = \sqbrk {x \times y}$
in minimal arithmetic, where $\sqbrk a$ is the unary representation of $a$.
== Proof ==
By Unary Representation of Natural Number, let $\sqbrk a$ denote the term $\map s {\dots \map s 0}$, where there are $a$ applications of the successor mapping to the constant $0$.
Proceed by induction on $y$.
=== Base Case ===
Let $y = 0$.
Then $\sqbrk y = 0$.
The following is a formal proof:
But $x \times 0 = 0$.
Therefore, $\sqbrk {x \times 0} = \sqbrk 0$.
Thus, the proof above is a formal proof of $\sqbrk x \times 0 = \sqbrk {x \times 0}$.
=== Induction Step ===
Let there exist a formal proof of $\sqbrk x \times \sqbrk y = \sqbrk {x \times y}$.
Then the following is a formal proof:
Therefore, there exists a formal proof of $\sqbrk x \times \map s {\sqbrk y} = \sqbrk {\paren {x \times y} + x}$.
But $\sqbrk y$ consists of $y$ applications of the successor mapping to the constant $0$.
Thus, $\map s {\sqbrk y}$ consists of $y + 1 = \map s y$ mappings of $s$ to $0$.
As that is the unary representation of $\map s y$ it follows that:
:$\map s {\sqbrk y} = \sqbrk {\map s y}$
By definition of multiplication:
:$x \times \map s y = \paren {x \times y} + x$
Therefore, the above proof also proves:
:$\sqbrk x \times \sqbrk {\map s y} = \sqbrk {x \times \map s y}$
as the WFFs are syntactically identical.
Thus, by the Principle of Mathematical Induction, there exists such a formal proof for every $y \in \N$.
As $x$ was arbitrary, there exists such a formal proof for every $x, y \in \N$.
Category:Natural Number Multiplication
Category:Proofs by Induction
Multiplication of Numbers is Left Distributive over Addition

== Theorem ==
That is, if $m a$, $m b$, $m c$ etc. be any equimultiples of $a$, $b$, $c$ etc., then:
:$m a + m b + m c + \cdots = m \paren {a + b + c + \cdots }$
== Proof ==
Let any number of magnitudes whatever $AB, CD$ be respectively equimultiples of any magnitudes $E, F$ equal in multitude.
Then we are to show that whatever multiple $AB$ is of $E$, then that multiple will $AB + CD$ be of $E + F$.
:
Since $AB$ is the same multiple of $E$ that $CD$ is of $F$, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $CD$ equal to $F$.
Let $AB$ be divided into the magnitudes $AG, GB$ equal to $E$, and $CH, HD$ equal to $F$.
Then the multitude of the magnitudes $AG, GB$ will be equal to the multitude of the magnitudes $CH, HD$.
Since $AG = E$ and $CH = F$ it follows that $AG = E$ and $AG + CH = E + F$.
For the same reason, $GB = E$ and $GB + HD = E + F$.
Therefore, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $AB + CD$ equal to $E + F$.
== Also see ==
* Real Multiplication Distributes over Addition
* Multiplication of Numbers is Right Distributive over Addition
== Sources ==
* : Book $\text{V}$. Propositions
* : Chapter $2$: The Logic of Shape: The golden mean
Category:Multiplication
Category:Addition
Category:Examples of Distributive Operations
Category:Distributive Laws of Arithmetic
Multiplication of Numbers is Right Distributive over Addition

== Theorem ==
That is:
:$m a + n a + p a + \cdots = \paren {m + n + p + \cdots} a$
== Proof ==
Let a first magnitude, $AB$, be the same multiple of a second, $C$, that a third, $DE$, is of a fourth, $F$.
Let a fifth, $BG$, be the same multiple of $C$ that a sixth, $EH$, is of $F$.
:
We need to show that $AG = AB + BG$ is the same multiple of $C$ that $DH = DE + EH$ is of $F$.
We have that $AB$ is the same multiple of $C$ that $DE$ is of $F$.
It follows that as many magnitudes as there are in $AB$ equal to $C$, so many also are there in $DE$ equal to $F$.
For the same reason, as many as there are in $BG$ equal to $C$, so many also are there in $EH$ equal to $F$.
So as many as there are in the whole $AG$ equal to $C$, so many also are there in the whole $DH$ equal to $F$.
Therefore the sum of the first and fifth, $AG$, is the same multiple of the second, $C$, that the sum of the third and sixth, $DH$ is of the fourth, $F$.
== Also see ==
* Real Multiplication Distributes over Addition
* Multiplication of Numbers is Left Distributive over Addition
== Sources ==
* : Book $\text{V}$. Propositions
Category:Multiplication
Category:Examples of Distributive Operations
Category:Distributive Laws of Arithmetic
Multiplication of Polynomials Distributes over Addition

== Theorem ==
Multiplication of polynomials is left- and right- distributive over addition.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
:$\ds h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
be arbitrary polynomials in the indeterminates $\set{X_j: j \in J}$ over $R$.
By Multiplication of Polynomials is Commutative, it is sufficient to prove that $\circ$ is left distributive over addition only.
Then
 
 
 
Therefore, $f \circ \paren {g + h} = f \circ g + f \circ h$ for all polynomials $f, g, h$.
Therefore, polynomial multiplication is distributive over addition.
Category:Polynomial Theory
Category:Examples of Distributive Operations
Multiplication of Polynomials is Associative

== Theorem ==
Multiplication of polynomials is associative.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity with zero $0_R$.
To improve readability of the expressions used, we will write the ring product $\circ$ in multiplicative notation.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
:$\ds h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then it follows from Polynomials Closed under Ring Product that:
:$\ds f \circ \paren {g \circ h} = \sum_{k \mathop \in Z} m_k \mathbf X^k$
for some $m_k \in R$, and:
:$\ds \paren {f \circ g} \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$
for some $n_k \in R$.
To establish associativity of $\circ$ we compute $m_k$ and $n_k$, and check that they are equal. 
We have:
 
 
Similarly we compute:
 
 
Since $p$, $q$, $r$ and $s$ are all dummy variables, it follows that $m_k = n_k$ for all $k \in Z$.
Therefore, $f \circ \paren {g \circ h} = \paren {f \circ g} \circ h$ for all polynomials $f$, $g$ and $h$.
Hence multiplication of polynomials is associative.
Category:Polynomial Theory
Multiplication of Polynomials is Commutative

== Theorem ==
Multiplication of polynomials is commutative.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then:
 
 
Therefore, $f \circ g = g \circ f$ for all polynomials $f$ and $g$.
Therefore, polynomial multiplication is commutative.
Category:Polynomial Theory
Multiplication of Positive Cuts preserves Ordering

== Theorem ==
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Let $\alpha$, $\beta$ and $\gamma$ be cuts such that:
:$0^* < \alpha < \beta$
:$0^* < \gamma$
where $<$ denotes the strict ordering on cuts.
Then
:$\alpha \gamma < \beta \gamma$
where $\alpha \gamma$ denotes the product of $\alpha$ and $\gamma$.
== Proof ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(g)}$
Category:Cuts
Category:Multiplication
Multiplication of Real Numbers is Left Distributive over Subtraction

== Theorem ==
That is, for any numbers $a, b$ and for any integer $m$:
:$m a - m b = m \paren {a - b}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Multiplication of Real Numbers is Right Distributive over Subtraction
* Distributive Laws of Arithmetic
Category:Real Multiplication
Category:Real Subtraction
Category:Examples of Distributive Operations
Category:Multiplication of Real Numbers Distributes over Subtraction
Category:Multiplication of Real Numbers is Left Distributive over Subtraction
Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 1

== Theorem ==
== Proof ==
Let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the part $AE$ subtracted is of the part $CF$ subtracted.
We need to show that the remainder $EB$ is also the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:
Whatever multiple $AE$ is of $CF$, let $EB$ be made that multiple of $CG$.
We have that $AE$ is the same multiple of $CF$ that $AB$ is of $GC$.
So from Distributive Property, $AE$ is the same multiple of $CF$ that $AB$ is of $GF$.
By by assumption, $AE$ is the same multiple of $CF$ that $AB$ is of $CD$.
Therefore $AB$ is the same multiple of each of the magnitudes $GF, CD$.
Therefore $GF = CD$.
Let $CF$ be subtracted from each.
Then the remainder $GC$ is equal to the remainder $FD$.
Since:
: $AE$ is the same multiple of $CF$ that $EB$ is of $GC$
: $GC = DF$
it follows that $AE$ is the same multiple of $CF$ that $EB$ is of $CD$.
That is, the remainder $EB$ will be the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Multiplication of Real Numbers is Left Distributive over Subtraction