Right Distributive Law for Natural Numbers

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Theorem

The operation of multiplication is right distributive over addition on the set of natural numbers $\N_{> 0}$:

$\forall x, y, n \in \N_{> 0}: \paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$


Proof

Using the axiomatization:

\((\text A)\)   $:$     \(\ds \exists_1 1 \in \N_{> 0}:\) \(\ds a \times 1 = a = 1 \times a \)      
\((\text B)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \)      
\((\text C)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \)      
\((\text D)\)   $:$     \(\ds \forall a \in \N_{> 0}, a \ne 1:\) \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \)      
\((\text E)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds \)Exactly one of these three holds:\( \)      
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)      
\((\text F)\)   $:$     \(\ds \forall A \subseteq \N_{> 0}:\) \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)      


For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \paren {a + b} \times 1\) \(=\) \(\ds a + b\) Axiom $\text A$
\(\ds \) \(=\) \(\ds \paren {a \times 1} + \paren {b \times 1}\) Axiom $\text A$

and so $\map P 1$ 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:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$


Then we need to show:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$


Induction Step

This is our induction step:

\(\ds \paren {a + b} \times \paren {k + 1}\) \(=\) \(\ds \paren {\paren {a + b} \times k} + \paren {a + b}\) Axiom $\text B$
\(\ds \) \(=\) \(\ds \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}\) Induction hypothesis
\(\ds \) \(=\) \(\ds \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}\) Natural Number Addition is Commutative
\(\ds \) \(=\) \(\ds \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }\) Axiom $\text B$

The result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources

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