Natural Number Multiplication is Associative
Theorem
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 1
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:
- $x \times \paren {z \times y} = x \times \paren {y \times z}$
$\blacksquare$
Proof 2
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:
\(\ds \forall m, n \in \N: \, \) | \(\ds m \times 0\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds m \times \paren {n + 1}\) | \(=\) | \(\ds \paren {m \times n} + m\) |
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:
\(\ds \paren {x \times y} \times 0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \times 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y \times 0}\) |
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:
\(\ds \paren {x \times y} \times \paren {k + 1}\) | \(=\) | \(\ds \paren {\paren {x \times y} \times k} + \paren {x \times y}\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times \paren {y \times k} } + \paren {x \times y}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times \paren {y \times k} }\) | Natural Number Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y + \paren {y \times k} }\) | Natural Number Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {\paren {y \times k} + y}\) | Natural Number Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y \times \paren {k + 1} }\) | Definition of Natural Number Multiplication |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$
Proof 3
In the Axiomatization of 1-Based Natural Numbers, this is rendered:
- $\forall x, y, z \in \N_{> 0}: \paren {x \times y} \times z = x \times \paren {y \times z}$
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:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
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:
\(\ds \paren {x \times y} \times 1\) | \(=\) | \(\ds x \times y\) | Axiom $\text A$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y \times 1}\) | 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 $\map P {k^+}$ 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:
\(\ds \paren {x \times y} \times \paren {k + 1}\) | \(=\) | \(\ds \paren {\paren {x \times y} \times k} + \paren {x \times y}\) | Axiom $\text B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times \paren {y \times k} } + \paren {x \times y}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \times y} + \paren {x \times \paren {y \times k} }\) | Natural Number Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y + \paren {y \times k} }\) | Natural Number Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {\paren {y \times k} + y}\) | Natural Number Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \times \paren {y \times \paren {k + 1} }\) | Definition of Natural Number Multiplication |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$
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: $\text{M} 1$
- 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 $1.1 \ \text {(f)}$