Natural Number Multiplication Distributes over Addition
Theorem
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 1
\(\ds \paren {x + y} \times z\) | \(=\) | \(\ds +^z \paren {x + y}\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {+^z x} + \paren {+^z y}\) | Power of Product of Commuting Elements in Semigroup equals Product of Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times z + y \times z\) |
$\Box$
\(\ds z \times \paren {x + y}\) | \(=\) | \(\ds +^{x + y} z\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {+^x z} + \paren {+^y z}\) | Index Laws for Semigroup: Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds z \times x + z \times y\) |
$\blacksquare$
Proof 2
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:
\(\ds \forall m, n \in \N: \, \) | \(\ds m \times 0\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds m \times n^+\) | \(=\) | \(\ds \paren {m \times n} + m\) |
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:
\(\ds \paren {x + y} \times 0\) | \(=\) | \(\ds 0\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 + 0\) | Definition of Natural Number Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times 0 + y \times 0\) | Definition of Natural Number Multiplication |
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:
\(\ds \paren {x + y} \times k^+\) | \(=\) | \(\ds \paren {x + y} \times k + \paren {x + y}\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times k} + \paren {y \times k} + \paren {x + y}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {x \times k} + x} + \paren {\paren {y \times k} + y}\) | Natural Number Addition is Commutative and Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times k^+} + \paren {y \times k^+}\) | Definition of Natural Number Multiplication |
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:
- $\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$
$\Box$
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:
\(\ds z \times \paren {x + y}\) | \(=\) | \(\ds \paren {x + y} \times z\) | Natural Number Multiplication is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times z} + \paren {y \times z}\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {z \times x} + \paren {z \times y}\) | Natural Number Multiplication is Commutative |
Thus we have proved:
- $\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
$\blacksquare$
Proof 3
In the Axiomatization of 1-Based Natural Numbers, this is rendered:
- $\forall x, y, z \in \N_{> 0}:$
- $\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
- $z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
Using the following axioms:
\((\text A)\) | $:$ | \(\ds \exists_1 1 \in \N_{> 0}:\) | \(\ds a \times 1 = a = 1 \times a \) | ||||||
\((\text B)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \) | ||||||
\((\text C)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \) | ||||||
\((\text D)\) | $:$ | \(\ds \forall a \in \N_{> 0}, a \ne 1:\) | \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \) | ||||||
\((\text E)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds \)Exactly one of these three holds:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
Left Distributive Law for Natural Numbers
First we show that:
- $n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$
Let us cast the proposition in the form:
- $\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds a \times \paren {b + 1}\) | \(=\) | \(\ds \paren {a \times b} + a\) | Axiom $\text B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times b} + \paren {a \times 1}\) | Axiom $\text A$ |
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:
- $\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$
Then we need to show:
- $\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$
Induction Step
This is our induction step:
\(\ds a \times \paren {b + \paren {k + 1} }\) | \(=\) | \(\ds a \times \paren {\paren {b + k} + 1}\) | Axiom $\text C$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times \paren {b + k} } + a\) | Axiom $\text B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {a \times b} + \paren {a \times k} } + a\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times b} + \paren {\paren {a \times k} + a}\) | Natural Number Addition is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times b} + \paren {a \times \paren {k + 1} }\) | Axiom $\text B$ |
The result follows by the Principle of Mathematical Induction.
$\Box$
Right Distributive Law for Natural Numbers
Then we show that:
- $\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \paren {a + b} \times 1\) | \(=\) | \(\ds a + b\) | Axiom $\text A$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times 1} + \paren {b \times 1}\) | Axiom $\text A$ |
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:
- $\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$
Then we need to show:
- $\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$
Induction Step
This is our induction step:
\(\ds \paren {a + b} \times \paren {k + 1}\) | \(=\) | \(\ds \paren {\paren {a + b} \times k} + \paren {a + b}\) | Axiom $\text B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}\) | Natural Number Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }\) | Axiom $\text B$ |
The result follows by the Principle of Mathematical Induction.
$\Box$
The result follows.
$\blacksquare$
Sources
- 1937: Richard Courant: Differential and Integral Calculus: Volume $\text { I }$ (2nd ed.) ... (previous) ... (next): Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: $\text{D}$
- 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 4$: Number systems $\text{I}$: Peano's Axioms
- 1982: Alan G. Hamilton: Numbers, Sets and Axioms ... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(f)}$