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 $\N$:

$\forall x, y, z \in \N:$
$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$
$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$


Proof

We are to show that:

$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

for all $x, y, z \in \N$.


From the definition of natural number multiplication, we have by definition that:

\(\displaystyle \forall m, n \in \N: \ \ \) \(\displaystyle m \times 0\) \(=\) \(\displaystyle 0\)
\(\displaystyle m \times n^+\) \(=\) \(\displaystyle \left({m \times n}\right) + m\)


Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $P \left({z}\right)$ be the proposition:

$\forall x, y \in \N: \left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\displaystyle \left({x + y}\right) \times 0\) \(=\) \(\displaystyle 0\) Definition of Natural Number Multiplication
\(\displaystyle \) \(=\) \(\displaystyle 0 + 0\) Definition of Natural Number Addition
\(\displaystyle \) \(=\) \(\displaystyle x \times 0 + y \times 0\) Definition of Natural Number Multiplication

and so $P \left({0}\right)$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.


So this is our induction hypothesis:

$\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$


Then we need to show:

$\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$


Induction Step

This is our induction step:


\(\displaystyle \left({x + y}\right) \times k^+\) \(=\) \(\displaystyle \left({x + y}\right) \times k + \left({x + y}\right)\) Definition of Natural Number Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times k}\right) + \left({y \times k}\right) + \left({x + y}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({x \times k}\right) + x}\right) + \left({\left({y \times k}\right) + y}\right)\) Natural Number Addition is Commutative and Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times k^+}\right) + \left({y \times k^+}\right)\) Definition of Natural Number Multiplication

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \left({x + y}\right) \times n = \left({x \times z}\right) + \left({y \times z}\right)$

$\Box$


Next we need to show that:

$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

for all $x, y, z \in \N$.

So:

\(\displaystyle z \times \left({x + y}\right)\) \(=\) \(\displaystyle \left({x + y}\right) \times z\) Natural Number Multiplication is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times z}\right) + \left({y \times z}\right)\) from above
\(\displaystyle \) \(=\) \(\displaystyle \left({z \times x}\right) + \left({z \times y}\right)\) Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

$\blacksquare$


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