Zero is Zero Element for Natural Number Multiplication
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Theorem
Let $\N$ be the natural numbers.
Then $0$ is a zero element for multiplication:
- $\forall n \in \N: 0 \times n = 0 = n \times 0$
Proof
Proof by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
- $0 \times n = 0 = n \times 0$
Basis for the Induction
By definition, we have:
- $0 \times 0 = 0 = 0 \times 0$
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, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis $\map P k$:
- $0 \times k = 0 = k \times 0$
Then we need to show that $\map P {k + 1}$ follows directly from $\map P k$:
- $0 \times \paren {k + 1} = 0 = \paren {k + 1} \times 0$
Induction Step
This is our induction step:
\(\ds 0 \times \paren {k + 1}\) | \(=\) | \(\ds \paren {0 \times k} + 0\) | Definition of Natural Number Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 + 0\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Definition of Natural Number Addition |
By definition:
- $\paren {k + 1} \times 0 = 0$
So $\map P k \implies \map P {k + 1}$, and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N: 0 \times n = 0 = n \times 0$
$\blacksquare$
Sources
- 1982: Alan G. Hamilton: Numbers, Sets and Axioms ... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(g)}$