Re: Question


zuhair wrote:
Arturo Magidin wrote:
In article <1146169646.101883.284550@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
zuhair <zaljohar@xxxxxxxxx> wrote:


Just because you don't like the answer does not mean it is not an
answer. Your question was the logical equivalent of "what kind of duck
is a fox terrier?" It would seem you would consider the answer "a fox
terrier is not a duck" to be "an escape rather than an answer."

A sequence is, BY DEFINITION, a map whose domain is the natural
numbers.

Fair enough, this mean that the biggest number of terms in a sequence
would be Omega.

You really need to stop using standard mathematical words with your
own private meaning.

"Omega" is not a "quantity"; you do not speak about there being "omega
things". Omega is an ordinal. It corresponds to a well-ordering type.

There should be some expression which can describe the quantity of
terms in a sequance
Since a sequence is a function who's domain is the natural numbers.
Then it should have
Aleph-0 of terms in it, because the multiplicity of terms in the
natural numbers set is descriped
by Aleph-0.

So 0.9999......... if it is to represent a real number then it should
be a cauchy sequence
and accordingly it should have Aleph-0 of nines in it.

Is that Correct?


And, no. It means EXACTLY that a sequence is a function whose domain
is the natural numbers. There are ->exactly<- one term for each
element of omega. Not "the biggest number of terms". There is exactly
one term for each element of omega.

Or in other words a sequence either contains n terms when n is finite,
or Omega of terms if the number of terms in it is infinite.

Will you ever learn to read?

A sequence is a function whose DOMAIN is the natural numbers. Not
whose domain is a subset of the natural numbers. There is always one
term for each natural number.

If you want to talk about "finite sequences", whose domains are the
sets I called "n" (where the set 0 is the empty set, and the set "n+1"
is the set {0, 1, ..., n}), then call them finite sequences.

Therefore the decimal expansion of 0.99999......... has Omega of 9 in
it.

No. Omega is not a quantity. omega is an ordinal.


and I repeat it would be lesser than 1 by 1/10^Omega.

And it does not matter how many times you repeat it, it is still
nonsense. The decimal expansion represents a sequence. The sequence is
equivalent to the constant sequence 1, and therefore, since "1" and
"0.999..." are two distinct representations of the same equivalence
class, they are the SAME real number, by definition.

Perhaps this is non sense. I don't know.

Yes, you do know. Because you have been told this many times. You just
ignore it and repeat it over and over and over. No matter how often
you repeat it, it is still nonsense. The symbol "10^omega" does not
represent a real number. You can define addition and multiplication of
ordinals, as well as exponentiation; under the standard definitions,
10^omega is the ordinal omega. And there is no standard definition of
"division of ordinals", so no matter how many times you repeat the
nonsense of writing "1/10^omega", it is still a symbol that lack any
meaning. It is nonsense.

Just because you can write down a symbol it does not, ipso facto,
grant it sense.


However you succeeded in illustrating to me that the symbole
0.9999.......9 is not a sequence

simply because the last 9 though a finite number but it is at a
transfinite position that is Omega+1

Actually, no. It would be in position omega, not position omega+1. The
order type of the digits is Omega+1; that is, your symbol represents a
function with DOMAIN omega+1, which has value 9 at each natural number
and value 9 at omega (recall that, BY DEFINITION,
omega+1 = omega \/ {omega} = {0, 1, 2, ..., n, ...} \/ {omega}


and therefore it is not a sequence.

well according to what you are saying then 1/10^Omega =0

No. It still has no meaning whatsoever.

according to
standard mathematics.

No. According to standard mathematics, the symbol you insist on
repeating is still complete and utter nonsense.


which something very strange as I see.

That is quite plain.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

I wonder why subtraction and division are not defined for Transfinite
Ordinal numbers?

For example : n + Omega = Omega

but Omega + n > Omega

Now we can define two operators of subtraction , on is left subtraction
denoted as " L -" and the other is right subtraction denoted as "-R" as
followes

Define:

Omega L- n = Omega
Omega -R n < Omega

n is a finite number

Also about division we can also define two division operators in a
similar way L / and /R as below

Omega L / n= x is x: n.x = Omega
Omega /R n = z is z : z.n = Omega

For example Omega L / 3 = ( 1L / 3 ) Omega < Omega, because
3.Omega>Omega

While Omega / R 3 = Omega. 3 = 3+3+3+3+.......... = Omega

Now 1 L / Omega < Omega = infentismal = Left one omegath

but 1 / R Omega = ???

Any how if these are valid both right and left division operators
cannot be zero.

because even if the right division operator is not defined yet it
cannot be zero because

Omega of zeros is a zero and not one.

It is the left divion operator that might be interesting

1 L / Omega = 0 means 0.Omega =1 means Omega L - Omega = 1

now Omega L - Omega would be and indeterminate number like 0/0 .

While Omega -R Omega = Zero.

However all the above might seem to be loonatic to professional
mathematicians.

Anyhow I am still conviniced that we can form numbers like 1/2^Omega or
1/2^Aleph-x

And they cannot be zeros, but of coarse we should specify the division
operator.

Zuhair

.

Relevant Pages

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