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## Multiplication

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:

 by induction hypothesis

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:

 by induction hypothesis

Proposition 6.1.20 (Associative Law)   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:

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 :
• For , the property is obvious.
• Assume that . Then there exists a natural number such that and we have

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:

Proposition 6.1.26 (Cancellation)   For any triple of natural numbers,

 or

We leave the proof to the reader.

Next: Ordering Up: The natural numbers Previous: Addition   Contents
root 2002-06-10