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Addition

Definition 6.1.7   We call addition the operation in $ \mathbb{N}$ denoted by $ +$ and defined by the two following conditions:
(A1)
$ \forall n \in \mathbb{N}, \; n+0=n$.
(A2)
$ \forall m \in \mathbb{N}, \; \forall n \in \mathbb{N}, \; m+n'=(m+n)'$.
The number $ m+n$ is called the sum of $ m$ and $ n$.

Proposition 6.1.8   For any natural number $ m$, $ m'=m+1$.

Proof. For any natural number $ m$, $ m+1=m+0'=(m+0)'=m'$. $ \qedsymbol$

So we have: $ 1+1=1+0'=(1+0)'=1'=2$. It was worth to learn all the stuff until this point!

Proposition 6.1.9   The addition is well-defined.

Proof. The proposition means that the sum $ m+n$ is defined for every pair $ (m,n)$ of natural numbers, and that there exists a unique operation verifying the con ditions of definition 1.7. Fix the number $ m$ and consider the set $ E$ of the natural numbers $ n$ for which the sum $ m+n$ is computable. By the Induction Principle, $ M=\mathbb{N}$. $ \qedsymbol$

Proposition 6.1.10 (Associative Law)   For any triple $ m,n,p$ of natural numbers, $ (m+n)+p=m+(n+p)$.

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

Lemma 6.1.11   For any natural number $ n$, we have $ n+0=0+n$.

We prove the lemma by induction on $ n$:

Lemma 6.1.12   For any natural number $ n$, we have $ n+1=1+n$.

We prove the lemma by induction on $ n$: Now we can prove the commutative law:

Proposition 6.1.13 (Commutative Law)   For every pair $ (m,n)$ of natural numbers, we have: $ m+n=n+m$.

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

A direct consequence of the definition of the addition and of the commutative law is as follows:

Corollary 6.1.14 (Neutral element)   For any natural number $ n$, $ n+0=0+n=n$.

Proposition 6.1.15 (Cancellation)   For any triple $ m,n,p$ of natural numbers,

$\displaystyle m+p=n+p \Longleftrightarrow m=n.$    

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

Proposition 1.15 is particularly important: we cannot use substraction for cancelling $ p$ in both sides of the equation $ m+p=n+p$, because we have not defined such an operation. In fact, it is impossible to define substraction in $ \mathbb{N}$; the definition of the substraction has to wait for the construction of the set $ \mathbb{Z}$ of the integers and of the operations in $ \mathbb{Z}$ (v.i. 2).

Proposition 6.1.16   For any pair $ (m,n)$ of natural numbers,

$\displaystyle m+n=0 \Longleftrightarrow m=n=0.$    

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


next up previous contents
Next: Multiplication Up: The natural numbers Previous: The natural numbers   Contents
root 2002-06-10