Natural Number Multiplication Distributes over Addition/Proof 2
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
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:
- $\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic