# Multiplication of integers

1. Feb 27, 2005

### C0nfused

Hi everybody,
We define multiplication as an operation with these properties :
a(b+c)=ab+ac and (a+b)c=ac+bc ,a*0=0 and a*1=a with a,b,c natural numbers and of course the two properties Zurtex mentioned ab=ba and a(bc)=(ab)c-I "forgot" to mention them because I didn't use them in what is included in this post . Also the naturals are produced from the function S(n)=n+1, meaning that for every n=natural, S(n)=natural. So starting from 0 and 1, we can "create" any natural.

Question: the definition of multiplication as repeatitive addition,
m*n=m+m+...+m (n m's are added) (n natural) is a result of the above definitions? I mean, the fact that every natural is equal to the sum of its previous natural and 1 shows that every natural n is an addition of n 1's. So
n=1+1+...1 (n 1's are added). Using this result and the generalisation of the distributive property we get that m*n=m*(1+1+...+1)=m+m+...+m (n times) (with m ending up as any quantity)

Are these correct? If so, do they have any actual meaning?

Thanks

Last edited: Feb 27, 2005
2. Feb 27, 2005

### Zurtex

For multiplication of integers you have the commutative law:

a*b = b*a

The associative law:

(a*b)*c = a*(b*c)

And the multiplicative identity:

a*1=a

When you include addition you get the distributive law:

a*(b + c) = a*b + a*c

As well as the various law of addition:

a + b = b + a

a + 0 = 0

a + (-a) = 0 (where -a is the additive inverse and it has yet to be proved that -1*a = -a)

From these laws (and a few more) you can construct all the laws of the integers entirely.

3. Feb 27, 2005

### C0nfused

My question still remains because I have seen multiplication defined both ways: with the properties and with the term "repeatitive(?i am not sure if it's called this way) addition". Is the latter just a result of these properties? Is what I have written the correct way to reach this result?
Thanks again

4. Feb 27, 2005

### Hurkyl

Staff Emeritus
What you mean by "repetative addition" is this recursive definition of multiplication on the natural numbers:

0*y = 0
(x+1)*y = x*y + y

It's a useful exercise to prove that this definition (and that of addition) satisfy the axioms that Zurtex posted. (except for the last, which is not true for the natural numbers)

5. Feb 27, 2005

### chronon

1) Does this follow from the standard axioms of arithmetic?
Yes. (At least you can prove m*2=m+m, m*3=m+m+m and so on)

2) Wouldn't this be just as good a way to define multiplication in the axioms?
The problem here is with the ..., which isn't really defined. So you end up having to say what you mean by m+m+...m (n times), which just gets you back where you started.

6. Feb 27, 2005

### Zurtex

I've never included 0 in the set of Natural Numbers but as I clearly specify in my post I am talking about integers. I wasn't sure what rules to put up but as I was writing it the title thread "Multiplication of integers" was the only thing that really caught my attention.

7. Feb 27, 2005

### strid

are you sure about that? :tongue:

hehe...

8. Feb 27, 2005

### Zurtex

Lame .... :shy: