• Security incident: ISF was recently accessed by intruders. Please change your password, and change it anywhere else you used it. Read more

Cont: Deeper than primes - Continuation 2

Status
Not open for further replies.
EDIT:

You claimed it was "place value method". Does each digit position have a place value or not?

If not, than this isn't "place value method".
It is an infinite extension of the place value method to the "left" side of the redix point.

Moreover, since you are also claiming behavior different from aleph0, your sequence of 17 symbols doesn't represent the same thing as aleph0. It does not represent the cardinality of the set of natural numbers.
It replaces the notion of infinite cardinality as represented by aleph0.

By the standard notion of the set of natural numbers, this set is a complete set (no one of its members is missing), which means that no new members can be added to it.

By using the diagonalization argument

[qimg]https://upload.wikimedia.org/wikipedia/commons/thumb/b/b7/Diagonal_argument_01_svg.svg/250px-Diagonal_argument_01_svg.svg.png[/qimg]

it is shown that there are more real numbers than natural numbers, but it is also shown that the set of natural numbers is actually incomplete, since we always enable to add new members to it (and in this example this ability is represented by the s member when 10111010011... is added, without loss of generality).

Once again, the notion at the basis of aleph0 can't be used in order to show this mathematical fact since, for example, aleph0+1 = aleph0.

This is not the case about the notion at the basis of, for example, 1,000,000,... < 1+1,000,000,... by 1, which is used to show this mathematical fact (as observed above) and may help us to refine our understanding about the sizes of infinite sets (http://www.internationalskeptics.com/forums/showpost.php?p=11242445&postcount=1036 is a possible example of such mathematical research).
 
Last edited:
It is an infinite extension of the place value method to the "left" side of the redix point.

Then what is the place value of the left-most digit? What is the place value of the position to it's immediate right? What is the place value of the next position to the right?

Either each and every position has a place value, or it is not anything resembling the "place value method".

It replaces the notion of infinite cardinality as represented by aleph0.

And yet you have it behaving differently.

By the standard notion of the set of natural numbers, this set is a complete set (no one of its members is missing), which means that no new members can be added to it.

That is a somewhat twisted view. The idea is simply that there is a set of all the natural numbers. The set contains all the natural numbers. It is complete in that sense.

By using the diagonalization argument...it is shown that there are more real numbers than natural numbers

Yes.

...but it is also shown that the set of natural numbers is actually incomplete

No.

...since we always enable to add new members to it (and in this example this ability is represented by the s member when 10111010011... is added, without loss of generality).

No. Emphatically, no. There is nowhere in the proof that numbers are added to any existing set. The proof simply exhibits a real number (not a natural number) not in some hypothetical list of real numbers.
 
Last edited:
Then what is the place value of the left-most digit?
1,000,000,000,...

What is the place value of the position to it's immediate right?
100,000,000,...

What is the place value of the next position to the right?
10,000,000,...

And yet you have it behaving differently.
Right, for example, aleph0+1 = aleph0, where 1+1,000,000,000,... > 1,000,000,000,... by 1.


That is a somewhat twisted view. The idea is simply that there is a set of all the natural numbers. The set contains all the natural numbers. It is complete in that sense.
Well, this no more than the traditional view of this set.

I do not use this view.


No. Emphatically, no. There is nowhere in the proof that numbers are added to any existing set. The proof simply exhibits a real number (not a natural number) not in some hypothetical list of real numbers.
My use of the diagonalization argument, which clearly shows that one can add new members to set of natural numbers (http://www.internationalskeptics.com/forums/showpost.php?p=11245518&postcount=1061) is not restricted to the traditional view and the proof that is involved with the diagonalization argument.
 
Last edited:
1

0


0

Here they are in case that you have missed them:

1,000,000,000,...


Place values. You know, like ones, tens, millions, ten trillion, ....

But I suspect you actually knew that and instead opted for what you thought a clever insult. If that is the level of discourse you've selected for this, your latest misunderstanding of mathematics, enjoy yourself (emphasis on that final word).
 
Place values. You know, like ones, tens, millions, ten trillion, ....
It was my mistake, I first thought that you are asking about the values in each place.

When I realized that this is not you have asked, I corrected my reply about this questions.

So once again the answer to your questions is:

Code:
1,000,000,000,...
  100,000,000,...
   10,000,000,...

But I suspect ...
There is no basis to your suspicion.

Please reply to the rest of http://www.internationalskeptics.com/forums/showpost.php?p=11245795&postcount=1063.
 
Last edited:
It was my mistake, I first thought that you are asking about the values in each place.

When I realized that this is not you have asked, I corrected my reply about this questions.

So once again the answer to your questions is:

Code:
1,000,000,000,...
  100,000,000,...
   10,000,000,...

Attempting to define the meaning of a term by simply repeating the term accomplishes nothing. Your "place value method" notation remains without definition and therefore meaningless.
 
You claimed it was "place value method". Does each digit position have a place value or not?

If not, than this isn't "place value method".

Moreover, since you are also claiming behavior different from aleph0, your sequence of 17 symbols doesn't represent the same thing as aleph0. It does not represent the cardinality of the set of natural numbers.

It was my mistake, I first thought that you are asking about the values in each place.

When I realized that this is not you have asked, I corrected my reply about this questions.

So once again the answer to your questions is:

Code:
1,000,000,000,...
  100,000,000,...
   10,000,000,...


There is no basis to your suspicion.

Please reply to the rest of http://www.internationalskeptics.com/forums/showpost.php?p=11245795&postcount=1063.

With the number 123, the digit 1 is in the hundreds position, the digit 2 is in the tens position and the digit 3 is in the ones position.

What is the name of the position of the digit 1 in your notation of 1,000,000,000... ?
 
Attempting to define the meaning of a term by simply repeating the term accomplishes nothing. Your "place value method" notation remains without definition and therefore meaningless.
jsfisher, please tell me, what is the name of each "place value" position in the following number (please define the infinitely many names of these positions, otherwise this number is not well defined ("and therefore meaningless" , according to your own restrictions) ? :

3.141592653589793238462643383279502884097169399375105820974944592307816406286...10
 
Last edited:
jsfisher, please tell me, what is the name of each "place value" position in the following number (please define the infinitely many names of these positions, otherwise this number is not well defined ("and therefore meaningless" , according to your own restrictions) ? :

3.141592653589793238462643383279502884097169399375105820974944592307816406286...10

For reference purposes, identify the digit positions in your number, 3.1415926..., as the sequence, di, where d0 is 3, d1 is 1, d2 is 4, d3 is 1, and so on.

The place value for di is R^(-i), where R is the radix. For the case at hand, the radix is 10, so the place value for di is 10^(-i).
 
For reference purposes, identify the digit positions in your number, 3.1415926..., as the sequence, di, where d0 is 3, d1 is 1, d2 is 4, d3 is 1, and so on.

The place value for di is R^(-i), where R is the radix. For the case at hand, the radix is 10, so the place value for di is 10^(-i).
You did not define the infinitely many place values of number 3.141592653589793238462643383279502884097169399375 105820974944592307816406286...10, so, according to your own restrictions it is not well defined "and therefore meaningless".

In other words, by following your own restrictions the digonalization argument is meaningless in the first place, yet traditional mathematicians like you use it.
 
You did not define the infinitely many place values of number 3.141592653589793238462643383279502884097169399375 105820974944592307816406286...10, so, according to your own restrictions it is not well defined "and therefore meaningless".

Oh really? Which one did I leave out? di speaks for each and everyone of them.
 
So once again the answer to your questions is:

Code:
1,000,000,000,...
  100,000,000,...
   10,000,000,...

Aren't all three numbers a 1 followed infinitely many zeros?

Yes, <...>

Then those 'numbers' are the same and they can't be the answer to jfisher's question.

The notation you use is meaningless. I'm reminded of people who think 0.999... is not equal to 1 because 0.999....5 is in between.
 
Last edited:
Oh really? Which one did I leave out? di speaks for each and everyone of them.

There are infinitely many place values but the index i has only finite values, so infinitely many places are always left out, or in other words, a number like 3.141592653589793238462643383279502884097169399375 105820974944592307816406286...10, according to your own restrictions is not well defined "and therefore meaningless", yet you are using it, for example, in the digonalization argument.
 
There are infinitely many place values but the index i has only finite values, so infinitely many places are always left out

Index i may take on any finite value, and there are countable-infinitely many of them. Nothing is left out.

Basic stuff, Doronshadmi, and basic stuff you already know.

It would seem that you have reduced your position to just empty quibbles and petty distractions. If that's all you have, this conversation is over. As much as you may try, you cannot take an infinite sequence of digits and put a number at the end of it (be it to the right or left).
 
Index i may take on any finite value, and there are countable-infinitely many of them. Nothing is left out.
Really? please define some infinite place value position of such index.

If you can't do that, then your i is actually restricted only to finitely many place value positions.

Basic stuff, jsfisher, and basic stuff you already know.
 
As much as you may try, you cannot take an infinite sequence of digits and put a number at the end of it (be it to the right or left).

It can be done at both left and right sides of the redix point, by using the diagonalization argument, which clearly shows that one can add new members to set of natural numbers (http://www.internationalskeptics.com/forums/showpost.php?p=11245518&postcount=1061).

This ability is not restricted to the traditional notion and the proof that is involved with the diagonalization argument.

If that's all you have, this conversation is over.
If all you have is the traditional notion about this subject, then indeed this conversation with you is over.

--------------------

Again, if one reads (by not being restricted only to the traditional view of the considered subjects) this thread from post http://www.internationalskeptics.com/forums/showpost.php?p=11242445&postcount=1036 forward, one can easily realizes that the following numbers
Code:
1,000,000,000,...
  100,000,000,...
   10,000,000,...
are not the same, and they are also define their own place values positions, exactly as done among finite numbers.
 
Last edited:
Really? please define some infinite place value position of such index.

There is no infinite position. Every digit of pi sits at some finite position. There are infinitely many such positions, but each of which corresponds to an integer index.
 
Status
Not open for further replies.

ISF - Join now!

Every member here is approved by hand. No bots, no spam, just people who care about evidence and honest debate.

Membership is free!

Create your free account

Back
Top Bottom