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Definition 6.1.17
We call
multiplication the operation in
denoted either by
or by
and defined by the two following conditions:
 (A1)

.
 (A2)

.
The number
is called the
product of
and
.
When there is no ambiguity, it is possible to denote instead of .
For example,
Proposition 6.1.18 (Distributive Law on the right)
For any triple
of natural numbers,
.
Proof.
By induction on
.
 For , the result is obvious.
 Assume that for some natural number , the property holds for every pair of natural numbers. Then we have:
Proposition 6.1.19 (Distributive Law on the left)
For any triple
of natural numbers,
.
Proof.
By induction on
.
 For , the result is obvious.
 Assume that for some natural number , the property holds for every pair of natural numbers. Then we have:
Proof.
By induction on
.
 For , the result is obvious.
 Assume that for some natural number , the property holds for every pair of natural numbers. Then we have:
Lemma 6.1.21
For any natural number
,
.
Proof.
By induction on
.
 If , the result is obvious.
 Assume that for some natural number we have . Then
As by definition,
, the commutative law is true in a particular case.
Lemma 6.1.22
For any natural number
,
.
Proof.
By induction on
.
 If , the result is obvious.
 Assume that for some natural number , the property is true. Then:
Proposition 6.1.23 (Commutative Law)
For every pair
of natural numbers, we have:
.
Proof.
By induction on
.
 If , the result is obvious.
 Assume that for some natural number , the property is true. Then:
Proposition 6.1.24
For any two natural numbers
and
we have:
or 

Proof.
The
is trivial. We prove
by induction on
:
Proposition 6.1.25
For any two natural numbers
and
we have:
Proof.
The
is trivial. In an unusual way, we do not use induction.
If
, then
and
, and there exist two natural numbers
and
such that
and
. Now we have:
i.e.
hence, by axiom (Pe1),
. By Proposition
1.16 it follows that
, and we have:
We have no division in
; nevertheless cancellation is generally possible in products:
We leave the proof to the reader.
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