avewright/proofwiki-math huggingface dataset:
Theorem: Natural Number Addition is Commutative

The operation of addition on the set of natural numbers $\N$ is commutative:
:$\forall m, n \in \N: m + n = n + m$

Proof:
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$x + n = n + x$
=== Basis for the Induction ===
From Natural Number Commutes with 1 under Addition, we have that:
:$\forall x \in \N_{> 0}: x + 1 = 1 + x$
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$x + k = k + x$
Then we need to show:
:$x + \paren {k + 1} = \paren {k + 1} + x$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = x + \paren {k + 1}
      | r = \paren {x + k} + 1
      | c = Natural Number Addition is Associative
}}
{{eqn | r = \paren {k + x} + 1
      | c = Induction Hypothesis
}}
{{eqn | r = k + \paren {x + 1}
      | c = Natural Number Addition is Associative
}}
{{eqn | r = k + \paren {1 + x}
      | c = Basis for the Induction
}}
{{eqn | r = \paren {k + 1} + x
      | c = Natural Number Addition is Associative
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Natural Number Addition is Associative

The operation of addition on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$

Proof:
Consider the natural numbers defined as a naturally ordered semigroup.
By  definition, the operation in a semigroup is associative.
Hence the result.
{{qed}}
Theorem: Natural Number Addition is Associative

The operation of addition on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$

Proof:
Consider the von Neumann construction of natural numbers $\N$, as elements of the minimally inductive set $\omega$.
We are to show that:
:$\paren {x + y} + n = x + \paren {y + n}$
for all $x, y, n \in \N$.
From the definition of addition, we have that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
      | l = m + 0
      | r = m
      | c = 
}}
{{eqn | l = m + n^+
      | r = \paren {m + n}^+
      | c = 
}}
{{end-eqn}}
Let $x, y \in \N$ be arbitrary.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {x + y} + n = x + \paren {y + n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {x + y} + 0
      | r = x + y
      | c = 
}}
{{eqn | r = x + \paren {y + 0}
      | c = 
}}
{{end-eqn}}
and so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\paren {x + y} + k = x + \paren {y + k}$
Then we need to show:
:$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \paren {x + y} + k^+
      | r = \paren {\paren {x + y} + k}^+
      | c = {{Defof|Addition in Minimally Inductive Set}}
}}
{{eqn | r = \paren {x + \paren {y + k} }^+
      | c = induction Hypothesis
}}
{{eqn | r = x + \paren {\paren {y + k}^+}
      | c = {{Defof|Addition in Minimally Inductive Set}}
}}
{{eqn | r = x + \paren {y + k^+}
      | c = {{Defof|Addition in Minimally Inductive Set}}
}}
{{end-eqn}}
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Natural Number Addition is Associative

The operation of addition on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$

Proof:
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Let $x, y \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {x + y} + n = x + \paren {y + n}$
=== Basis for the Induction ===
From Axiom $\text C$, we have by definition that:
:$\forall x, y \in \N_{> 0}: \paren {x + y} + 1 = x + \paren {y + 1}$
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\paren {x + y} + k = x + \paren {y + k}$
Then we need to show:
:$\paren {x + y} + \paren {k + 1} = x + \paren {y + \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \paren {x + y} + \paren {k + 1}
      | r = \paren {\paren {x + y} + k} + 1
      | c = Basis for the Induction
}}
{{eqn | r = \paren {x + \paren {y + k} } + 1
      | c = Induction Hypothesis
}}
{{eqn | r = x + \paren {\paren {y + k} + 1}
      | c = Basis for the Induction
}}
{{eqn | r = x + \paren {y + \paren {k + 1} }
      | c = Basis for the Induction
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Rational Multiplication is Commutative

The operation of multiplication on the set of rational numbers $\Q$ is commutative:
:$\forall x, y \in \Q: x \times y = y \times x$

Proof:
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is commutative on $\Q$.
{{qed}}
Theorem: Rational Multiplication is Associative

The operation of multiplication on the set of rational numbers $\Q$ is associative:
:$\forall x, y, z \in \Q: x \times \paren {y \times z} = \paren {x \times y} \times z$

Proof:
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is associative on $\Q$.
{{qed}}
Theorem: Natural Number Multiplication is Commutative

The operation of multiplication on the set of natural numbers $\N$ is commutative:
:$\forall x, y \in \N: x \times y = y \times x$

Proof:
Let $A, B$ be two (natural) numbers, and let $A$ by multiplying $B$ make $C$, and $B$ by multiplying $A$ make $D$.
We need to show that $C = D$.
:350px
We have that $A \times B = C$.
So $B$ measures $C$ according to the units of $A$.
But the unit $E$ also measures $A$ according to the units in it.
So $E$ measures $A$ the same number of times that $B$ measures $C$.
Therefore from {{EuclidPropLink|book = VII|prop = 15|title = Alternate Ratios of Multiples}} $E$ measures $B$ the same number of times that $A$ measures $C$.
We also have that $A$ measures $D$ according to the units of $B$ since $B \times A = D$.
But the unit $E$ also measures $B$ according to the units in it.
Therefore $E$ measures $B$ the same number of times that $A$ measures $D$.
But we also have that $E$ measures $B$ the same number of times that $A$ measures $C$.
So $A$ measures $C$ and $D$ the same number of times.
Therefore $C = D$.
{{qed}}
Theorem: Natural Number Multiplication is Commutative

The operation of multiplication on the set of natural numbers $\N$ is commutative:
:$\forall x, y \in \N: x \times y = y \times x$

Proof:
Natural number multiplication is recursively defined as:
:$\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \paren {n + 1} & = m \times n + m \end{cases}$
From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition; that is, such that:
:$\forall n \in \N: \begin{cases} \map f 0 = 0 \\ \map f {n + 1} = \map f n + m \end{cases}$
Consider now $f'$ defined as $\map {f'} n = n \times m$.
Then by Zero is Zero Element for Natural Number Multiplication:
:$\map {f'} 0 = 0 \times m = 0$
Furthermore:
{{begin-eqn}}
{{eqn | l = \map {f'} {n + 1}
      | r = \paren {n + 1} \times m
}}
{{eqn | r = n \times m + m
      | c = Natural Number Multiplication Distributes over Addition
}}
{{eqn | r = \map {f'} n + m
}}
{{end-eqn}}
showing that $f'$ also satisfies the definition of $m \times n$.
By the Principle of Recursive Definition it follows that:
:$m \times n = n \times m$
{{qed}}
Theorem: Natural Number Multiplication is Commutative

The operation of multiplication on the set of natural numbers $\N$ is commutative:
:$\forall x, y \in \N: x \times y = y \times x$

Proof:
Proof by induction:
From the definition of natural number multiplication, we have that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
      | l = m \times 0
      | r = 0
}}
{{eqn | l = m \times n^+
      | r = \paren {m \times n} + m
}}
{{end-eqn}}
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \N: m \times n = n \times m$
=== Basis for the Induction ===
From Zero is Zero Element for Natural Number Multiplication:
:$\forall m \in \N: m \times 0 = 0 = 0 \times m$
Thus $\map P 0$ is seen to be true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis $\map P k$:
:$\forall m \in \N: m \times k = k \times m$
Then we need to show that $\map P {k^+}$ follows from $\map P k$:
:$\forall m \in \N: m \times k^+ = k^+ \times m$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = m \times k^+
      | r = \paren {m \times k} + m
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = m + \paren {m \times k}
      | c = Natural Number Addition is Commutative
}}
{{eqn | r = m + \paren {k \times m}
      | c = Induction Hypothesis
}}
{{eqn | r = k^+ \times m
      | c = Natural Number Multiplication Distributes over Addition
}}
{{end-eqn}}
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n \in \N: m \times n = n \times m$
{{qed}}
Theorem: Natural Number Multiplication is Commutative

The operation of multiplication on the set of natural numbers $\N$ is commutative:
:$\forall x, y \in \N: x \times y = y \times x$

Proof:
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a \in \N_{> 0}: a \times n = n \times a$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = a \times 1
      | r = a
      | c = Axiom $\text A$
}}
{{eqn | r = 1 \times a
      | c = Axiom $\text A$
}}
{{end-eqn}}
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall a \in \N: a \times k = k \times a$
Then we need to show:
:$\forall a \in \N: a \times \paren {k + 1} = \paren {k + 1} \times a$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = a \times \paren {k + 1}
      | r = \paren {a \times k} + \paren {a \times 1}
      | c = Left Distributive Law for Natural Numbers
}}
{{eqn | r = \paren {k \times a} + \paren {a \times 1}
      | c = Induction hypothesis
}}
{{eqn | r = \paren {k \times a} + \paren {1 \times a}
      | c = Basis for the Induction
}}
{{eqn | r = \paren {k + 1} \times a
      | c = Right Distributive Law for Natural Numbers
}}
{{end-eqn}}
The result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Natural Number Multiplication is Associative

The operation of multiplication on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$

Proof:
From Index Laws for Semigroup: Product of Indices we have:
:$+^{z \times y} x = \map {+^z} {+^y x}$
By definition of multiplication, this amounts to:
:$x \times \paren {z \times y} = \paren {x \times y} \times z$
From Natural Number Multiplication is Commutative, we have:
:$x \times \paren {z \times y} = x \times \paren {y \times z}$
{{qed}}
Theorem: Natural Number Multiplication is Associative

The operation of multiplication on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$

Proof:
We are to show that:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
for all $x, y, n \in \N$.
From the definition of natural number multiplication, we have that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
      | l = m \times 0
      | r = 0
      | c = 
}}
{{eqn | l = m \times \paren {n + 1}
      | r = \paren {m \times n} + m
      | c = 
}}
{{end-eqn}}
Let $x, y \in \N$ be arbitrary.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {x \times y} \times 0
      | r = 0
}}
{{eqn | r = x \times 0
}}
{{eqn | r = x \times \paren {y \times 0}
}}
{{end-eqn}}
and so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {x \times y} \times k = x \times \paren {y \times k}$
Then we need to show:
:$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \paren {x \times y} \times \paren {k + 1}
      | r = \paren {\paren {x \times y} \times k} + \paren {x \times y}
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = \paren {x \times \paren {y \times k} } + \paren {x \times y}
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {x \times y} + \paren {x \times \paren {y \times k} }
      | c = Natural Number Addition is Commutative
}}
{{eqn | r = x \times \paren {y + \paren {y \times k} }
      | c = Natural Number Multiplication Distributes over Addition
}}
{{eqn | r = x \times \paren {\paren {y \times k} + y}
      | c = Natural Number Addition is Commutative
}}
{{eqn | r = x \times \paren {y \times \paren {k + 1} }
      | c = {{Defof|Natural Number Multiplication}}
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Natural Number Multiplication is Associative

The operation of multiplication on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$

Proof:
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Let $x, y \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {x \times y} \times 1
      | r = x \times y
      | c = Axiom $\text A$
}}
{{eqn | r = x \times \paren {y \times 1}
      | c = Axiom $\text A$
}}
{{end-eqn}}
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\paren {x \times y} \times k = x \times \paren {y \times k}$
Then we need to show:
:$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \paren {x \times y} \times \paren {k + 1}
      | r = \paren {\paren {x \times y} \times k} + \paren {x \times y}
      | c = Axiom $\text B$
}}
{{eqn | r = \paren {x \times \paren {y \times k} } + \paren {x \times y}
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {x \times y} + \paren {x \times \paren {y \times k} }
      | c = Natural Number Addition is Commutative
}}
{{eqn | r = x \times \paren {y + \paren {y \times k} }
      | c = Natural Number Multiplication Distributes over Addition
}}
{{eqn | r = x \times \paren {\paren {y \times k} + y}
      | c = Natural Number Addition is Commutative
}}
{{eqn | r = x \times \paren {y \times \paren {k + 1} }
      | c = {{Defof|Natural Number Multiplication}}
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Natural Number Multiplication Distributes over Addition

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, z \in \N:$
::$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
::$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + y} \times z
      | r = +^z \paren {x + y}
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = \paren {+^z x} + \paren {+^z y}
      | c = Power of Product of Commuting Elements in Semigroup equals Product of Powers
}}
{{eqn | r = x \times z + y \times z
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = z \times \paren {x + y}
      | r = +^{x + y} z
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = \paren {+^x z} + \paren {+^y z}
      | c = Index Laws for Semigroup: Sum of Indices
}}
{{eqn | r = z \times x + z \times y
}}
{{end-eqn}}
{{qed}}
Theorem: Natural Number Multiplication Distributes over Addition

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, z \in \N:$
::$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
::$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

Proof:
We are to show that:
:$\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
From the definition of natural number multiplication, we have by definition that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
      | l = m \times 0
      | r = 0
}}
{{eqn | l = m \times n^+
      | r = \paren {m \times n} + m
}}
{{end-eqn}}
Let $x, y \in \N$ be arbitrary.
For all $z \in \N$, let $\map P z$ be the proposition:
:$\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {x + y} \times 0
      | r = 0
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = 0 + 0
      | c = {{Defof|Natural Number Addition}}
}}
{{eqn | r = x \times 0 + y \times 0
      | c = {{Defof|Natural Number Multiplication}}
}}
{{end-eqn}}
and so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\forall x, y \in \N: \paren {x + y} \times k = \paren {x \times k} + \paren {y \times k}$
Then we need to show:
:$\forall x, y \in \N: \paren {x + y} \times k^+ = \paren {x \times k^+} + \paren {y \times k^+}$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \paren {x + y} \times k^+
      | r = \paren {x + y} \times k + \paren {x + y}
      | c = {{Defof|Natural Number Multiplication}}
}}
{{eqn | r = \paren {x \times k} + \paren {y \times k} + \paren {x + y}
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {\paren {x \times k} + x} + \paren {\paren {y \times k} + y}
      | c = Natural Number Addition is Commutative and Associative
}}
{{eqn | r = \paren {x \times k^+} + \paren {y \times k^+}
      | c = {{Defof|Natural Number Multiplication}}
}}
{{end-eqn}}
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:
:$\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$
{{qed|lemma}}
Next we need to show that:
:$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
So:
{{begin-eqn}}
{{eqn | l = z \times \paren {x + y}
      | r = \paren {x + y} \times z
      | c = Natural Number Multiplication is Commutative
}}
{{eqn | r = \paren {x \times z} + \paren {y \times z}
      | c = from above
}}
{{eqn | r = \paren {z \times x} + \paren {z \times y}
      | c = Natural Number Multiplication is Commutative
}}
{{end-eqn}}
Thus we have proved:
:$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
{{qed}}
Theorem: Natural Number Multiplication Distributes over Addition

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, z \in \N:$
::$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
::$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

Proof:
Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
=== Left Distributive Law for Natural Numbers ===
First we show that:
:$n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$
{{:Left Distributive Law for Natural Numbers}}{{qed|lemma}}
=== Right Distributive Law for Natural Numbers ===
Then we show that:
:$\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$
{{:Right Distributive Law for Natural Numbers}}{{qed|lemma}}
The result follows.
{{qed}}
Theorem: Distributive Laws/Arithmetic

On all the number systems:
:natural numbers $\N$
:integers $\Z$
:rational numbers $\Q$
:real numbers $\R$
:complex numbers $\C$
the operation of multiplication is distributive over addition:
:$m \paren {n + p} = m n + m p$
:$\paren {m + n} p = m p + n p$

Proof:
This is demonstrated in these pages:
:Natural Number Multiplication Distributes over Addition
:Integer Multiplication Distributes over Addition
:Rational Multiplication Distributes over Addition
:Real Multiplication Distributes over Addition
:Complex Multiplication Distributes over Addition
{{qed}}
Theorem: Rational Multiplication Distributes over Addition

The operation of multiplication on the set of rational numbers $\Q$ is distributive over addition:
:$\forall x, y, z \in \Q: x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
:$\forall x, y, z \in \Q: \paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$

Proof:
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore {{afortiori}} $\times$ is distributive over $+$ on $\Q$.
{{qed}}
Category:Rational Addition
Category:Rational Multiplication
Category:Examples of Distributive Operations
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Theorem: Associative Law of Multiplication

:$\forall x, y, z \in \mathbb F: x \times \paren {y \times z} = \paren {x \times y} \times z$
That is, the operation of multiplication on the standard number sets is associative.

Proof:
Let a first magnitude $A$ be the same multiple of a second $B$ that a third $C$ is of a fourth $D$.
Let equimultiples $EF, GH$ be taken of $A, C$.
We need to show that $EF$ is the same multiple of $B$ that $GH$ is of $D$.
We have that $EF$ is the same multiple of $A$ that $GH$ is of $C$.
Therefore as many magnitudes as there are in $EF$ equal to $A$, so many also are there in $GH$ equal to $C$.
Let $EF$ be divided into the magnitudes $EK, KF$ equal to $A$, and $GH$ into the magnitudes $GL, LH$ equal to $C$.
Then the multitude of the magnitudes $EK, KF$ will be equal to the multitude of the magnitudes $GL, LH$.
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We have that $A$ is the same multiple of $B$ that $C$ is of $D$, while $EK = A$ and $GL = C$.
So $EK$ is the same multiple of $B$ that $GL$ is of $D$.
For the same reason, $KF$ is the same multiple of $B$ that $LH$ is of $D$.
So we have that:
:a first magnitude $EK$ is the same multiple of a second $B$ that a third $GL$ is of a fourth $D$
:a fifth $KF$ is also of the same multiple of the second $B$ that a sixth $LH$ is of the fourth $D$.
Therefore the sum of the first and fifth, $EF$, is also the same multiple of the second $B$ that the sum of the third and sixth, $GH$ is of the fourth $D$.
{{qed}}