Re: An infinite number question

On Mon, 23 Jan 2006 10:46:08 -0500, "Robert J. Kolker"
<nowhere@xxxxxxxxxxx> wrote:

>r norman wrote:
>> On Sun, 22 Jan 2006 21:30:20 -0500, "Robert J. Kolker"
>> <nowhere@xxxxxxxxxxx> wrote:
>>>r norman wrote:

>>>>
>>>>Except that virtually all number sets are closed under multiplication,
>>>>but not under division, not every number has an inverse.
>>>>
>>>
>>>Every number but 0. If you exclude 0, the remaining numbers form a group
>>>under multiplication with 1 as the identity.
>>>
>>
>> But then you lose addition. Of course, in a group it doesn't matter
>> whether you call the one operator "addition" or "multiplication" or
>> "group operator".
>
>So? The point is that 0 has no multiplicative inverse and only 0. Get
>rid of it and the remaining set is a group under multiplication.
>Division is the inverse of multiplication.
>
>Now if you want a field that is different you need both addition (and
>its inverse) and multiplication (and its inverse).
>

We have been going around in circles mainly because of my carelessness
in expression. It is clear that I was quite wrong in asserting that
"virtually all number sets are closed under multiplication". But
there are a large number and variety of sets that really are closed
under multiplication where many elements beyond the additive identity
(zero) have no inverse and where division is not possible. Division,
when it exists, is defined as the inverse of multiplication. But
multiplication can exist without division.

A set with both addition and multiplication is a ring, provided that
addition obeys the usual rules. In particular, the set must be a
group under addition. However many rings exist closed under
multiplication but where no element whatsoever has an inverse; the
integers are an example. The even integers don't even have a
multiplicative identity.

If the set is a cyclic group under addition, as would be the infinite
set of integers or all integers multiples of a particular value or the
integers from 0 through n-1 where addition is mod(n), then you can
define multiplication as repeated addition so the set must be closed
under multiplication. If n is prime, then as you already pointed out,
you have a field and all elements except the additive identity have a
multiplicative inverse; division except by zero is possible. But if n
is composite, you don't.

Consider, for example, the integers 0, 1, 2, ..., 11 where addition
and multiplication are taken mod(12). Then 0, 3, 4, 5, 6, 8, and 10
have no multiplicative inverse. The only elements that do have one
are 1, 5, 7 = -5, and `11 = -1 and they are all their own reciprocal:
1*1 = 5*5 = 7*7 = 11*11 = 1. Division doesn't work.

I am sure you know all this -- it is elementary algebra (but rapidly
fading from memory). It is simply that in a news groups dealing with
creation and evolution, we tend to get a bit sloppy talking about
math.

In any event, none of this has any relation to the original (and quite
meaningless) question about infinities.


.

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