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Multiplication

Definition 6.1.17   We call multiplication the operation in $ \mathbb{N}$ denoted either by $ \times$ or by $ \cdot$ and defined by the two following conditions:
(A1)
$ \forall n \in \mathbb{N}, \; n+0=0$.
(A2)
$ \forall m \in \mathbb{N}, \; \forall n \in \mathbb{N}, \; m \cdot n'=(m \cdot n) + m$.
The number $ m \cdot n$ is called the product of $ m$ and $ n$.

When there is no ambiguity, it is possible to denote $ mn$ instead of $ m \cdot n$. For example,

$\displaystyle m \cdot 1 = m \cdot 0' = (m \cdot 0) +m = 0 + m =m.$    

Proposition 6.1.18 (Distributive Law on the right)   For any triple $ m,n,p$ of natural numbers, $ (m+n)p=mp+np$.

Proof. By induction on $ p$. $ \qedsymbol$

Proposition 6.1.19 (Distributive Law on the left)   For any triple $ m,n,p$ of natural numbers, $ m(n+p)=mn+mp$.

Proof. By induction on $ p$. $ \qedsymbol$

Proposition 6.1.20 (Associative Law)   For any triple $ m,n,p$ of natural numbers, $ (mn)p=m(np)$.

Proof. By induction on $ p$. $ \qedsymbol$

Lemma 6.1.21   For any natural number $ n$, $ 0n=0$.

Proof. By induction on $ n$. $ \qedsymbol$

As by definition, $ n \cdot 0 =0$, the commutative law is true in a particular case.

Lemma 6.1.22   For any natural number $ n$, $ 1 \cdot n= n \cdot 1$.

Proof. By induction on $ n$. $ \qedsymbol$

Proposition 6.1.23 (Commutative Law)   For every pair $ (m,n)$ of natural numbers, we have: $ mn=nm$.

Proof. By induction on $ n$. $ \qedsymbol$

Proposition 6.1.24   For any two natural numbers $ m$ and $ n$ we have:

$\displaystyle mn=0 \Longleftrightarrow m=0$   or $\displaystyle n=0.$    

Proof. The $ \Longleftarrow$ is trivial. We prove $ \Longrightarrow$ by induction on $ n$: $ \qedsymbol$

Proposition 6.1.25   For any two natural numbers $ m$ and $ n$ we have:

$\displaystyle mn=1 \Longleftrightarrow m=n=1.$    

Proof. The $ \Longleftarrow$ is trivial. In an unusual way, we do not use induction. If $ mn=1$, then $ m \neq 0$ and $ n \neq 0$, and there exist two natural numbers $ p$ and $ q$ such that $ m=p'$ and $ n=q'$. Now we have:

$\displaystyle mn=1 \Longleftrightarrow p'q'=0'$    

i.e.

$\displaystyle 0'=p'q+p'=(p'q+p)'$    

hence, by axiom (Pe1), $ p'q+p=0$. By Proposition 1.16 it follows that $ p'q=p=0$, and we have:

\begin{displaymath}\begin{cases}
 p'q=0 \\  p=0
 \end{cases}
 \Longrightarrow
 \begin{cases}
 q=0 \\  p=0
 \end{cases}
 \Longrightarrow
 m=n=1.\end{displaymath}    

$ \qedsymbol$

We have no division in $ \mathbb{N}$; nevertheless cancellation is generally possible in products:

Proposition 6.1.26 (Cancellation)   For any triple $ m,n,p$ of natural numbers,

$\displaystyle mp=np \Longleftrightarrow p=0$    or $\displaystyle m=n.$    

We leave the proof to the reader.
next up previous contents
Next: Ordering Up: The natural numbers Previous: Addition   Contents
root 2002-06-10