Natural Number Multiplication Distributes over Addition/Proof 2

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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

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