Natural Number Multiplication is Associative/Proof 2
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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
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.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic