Is free-will an illusion?

@ughaibu,

Quote:

Then you can't reverse the ******* order, can you?

You are right I can't reverse the order. I put "reservse" in quotations because I knew that. I meant only that I would start with infinitely long natural numbers instead of finite natural numbers. This was done to produce the same format of endless digits that the decimal numbers were in. Now that this is cleared up. Maybe you can better understand the proof.

From my proof, if you are caught up on starting out with 9999... as the first natural number replace it with any other natural number it does not matter. Start with 1000... or 1010.... or 87633222... Which number is bigger? Who knows? It does not matter since they are all elements of the set of natural numbers.

2 Replies

@Fil Albuquerque,

Quote:

so they’re not the same size; and in every attempt at a bijection, you can cover all of the integers, but you’ll always miss some reals – so the set of the reals is larger than the set of the integers.

The way they come to the conclusion there is no bijection between the integers and the reals is from the diagonal argument. If the diagonal argument is a false argument then a lack of one-to-one correspondence cannot be supported (which is what bijection means.)

So this is what is at stake. Give my argument a second look. And see if you can find any problems with it. If it is not clear what I have done, I will try to explain it better. I just really feel it is a solid proof.

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@tomr,

...contain or is contained would define bigger and smaller, what other definition we need ?

1 Reply

@tomr,

tomr wrote:

I meant only that I would start with infinitely long natural numbers instead of finite natural numbers.

There are no infinitely long natural numbers.

1 Reply

@ughaibu,

Quote:

There are no infinitely long natural numbers.

Infinity means endless. For every n there is an n + 1. What is to stop me from unendingly writing 9's such as 9999.... You have got to be joking here.

1 Reply

@tomr,

tomr wrote:

What is to stop me from unendingly writing 9's such as 9999..

Yes, you can write it, in principle, but it isn't a natural number.

1 Reply

@Fil Albuquerque,

Quote:

...contain or is contained would define bigger and smaller, what other definition we need ?

I am not certain what you are refering to. If one set contains the another I would say the first is equal to or greater than the second. Yet if there is a bijection or a one-to-one matching between the decimal reals and the natural numbers than the cardinality of both is the same. So neither is bigger or smaller. I may be off base with your intented comment. What do you think of the proof I made, I really value your feedback.

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@tomr,

...to my view the fact remains that you can describe a number which is not on that infinite list...while I think what you said is interesting and has a certain appeal I am not yet convinced diagolanization is a wrong method...what bottom line you seam to be saying is that whatever real number we can come up with in the natural numbers list you can always ad one to infinity to make the match...but in principle you are forgetting that list is already complete...you seam to just be postponing the problem...

1 Reply

@ughaibu,

A natural number is defined to be the unending positive integers. 0,1,2,3,etc... So any number that follows the pattern 1,2,3, ... n ... n+1 is a natural number. Therefore I can endlessly add numbers to any given value and produce unending numbers. n= 1110, n+1 = 1111, etc... n=9999, n + 1111 = 10000 and so forth until unending numbers are created like ...999 = 999...

2 Replies

@tomr,

tomr wrote:

A natural number is defined to be the unending positive integers. 0,1,2,3,etc... So any number that follows the pattern 1,2,3, ... n ... n+1 is a natural number. Therefore I can endlessly add numbers to any given value and produce unending numbers. n= 1110, n+1 = 1111, etc... n=9999, n + 1111 = 10000 and so forth until unending numbers are created like ...999 = 999...

Wrong. For any natural number greater than one, there is a second smaller natural number, and by successively subtracting the second number from the first, we will eventually be left with a natural number less than or equal to the second number. Your number doesn't meet this requirement, so it is not a natural number.

1 Reply

@Fil Albuquerque,

Quote:

what bottom line you seam to be saying is that whatever real number we can come up with in the natural numbers list you can always ad one to infinity to make the match...but in principle you are forgetting that list is already complete...you seam to just be postponing the problem...

I am not saying this anymore. I avoid the problems of diagonalization by writting out the natural numbers as infinitely long numbers. That way diagonalization can be applied to both sets. You may have not seen the most recent proof heres what I am refering to now.

Quote:

n = 1 or n = 99999999... <--> .00000000.....
n = 2 or n = 98989898... <--> .010101010.....
n = 3 or n = 98898898... <--> .01101101....
n = 4 or n = 99988899... <--> .00011100...
.
.
.

Now both sets of numbers can be shown to correspond one to one with each other. For every anti-diagonal in the natural numbers there is one in the decimal reals. Problem solved. The decimal reals are countable.

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@tomr,

...the fact that a natural number can't ever be reached by a real number sequence should in turn be the alarming bell here...after all definitions are conventions...a finite number containing an infinity of numbers makes us wonder what a natural number really is...

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@ughaibu,

Quote:

Wrong. For any natural number greater than one, there is a second smaller natural number, and by successively subtracting the second number from the first, we will eventually be left with a natural number less than or equal to the second number. Your number doesn't meet this requirement, so it is not a natural number.

If you can endlessly add numbers to produce an endlessly long number then you can endlessly subtract to get the number you are talking about. This is a simple result of the inverse nature of the operations of addition and subtraction.

1 Reply

@tomr,

tomr wrote:

If you can endlessly add numbers to produce an endless number then you can endlessly subtract to get the number you are talking about. This a simple result of the inverse nature of the operations of addition and subtraction.

For fucks' sake! Do a Google search. It should take you less than five minutes to confirm that you're mistaken. As I have already spent much longer than that on this futile endeavour, you are yet again on your own.

1 Reply

@ughaibu,

Quote:

For fucks' sake! Do a Google search. It should take you less than five minutes to confirm that you're mistaken. As I have already spent much longer than that on this futile endeavour, you are yet again on your own.

You can find anything you want on the internet. If you have a real argument you can express it right here.

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@tomr,

BTW, if you can add or multiply, you can also subtract or divide; that's the wonder of math.

Isn't that "free will?" LOL

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@tomr,

tomr wrote:

If you have a real argument you can express it right here.

But I can't educate you, can I? That's your responsibility. You don't understand that subtraction from an infinite number leaves an infinite remainder, unless the number subtracted is the same as the number from which it's subtracted. Otherwise you would understand that your number is not a natural number. In short, you do not understand the basics of the notions that you're attempting to employ in your proof. There is nothing interesting for me in repeatedly pointing out your basic mistakes, and it's an utter waste of my time if you're not going to educate yourself and stop insisting that you're correct when you're almost unexceptionally mistaken.

1 Reply

@ughaibu,

Quote:

You don't understand that subtraction from an infinite number leaves an infinite remainder, unless the number subtracted is the same as the number from which it's subtracted.

Then what does an infinite series of subtractions leave. If I can produce by endless addition the infinitely long number a1 + a2 + a3 ... = bbbbb.... . Then by the that same process i can create an infinitely large negative number -a1 + -a2 + -a3 ... = -bbbbbb..... That would be the number you are referring to when you say "unless the number subtracted is the same as the number from which it's subtracted". An infinite subtraction process is equivalent to an infinite addition process of negative numbers.

You are trying to apply addition/subtraction rules for finite numbers onto numbers that are endless and therefore incomplete. So not everything will work like it does with finite numbers.

There are only two possible options. Either you claim that we can have endlessly long natural numbers or you have bounded the set and made it finite.

1 Reply

@tomr,

tomr wrote:

Quote:
You don't understand that subtraction from an infinite number leaves an infinite remainder, unless the number subtracted is the same as the number from which it's subtracted.

Then what does an infinite series of subtractions leave. If I can produce by endless addition the infinitely long number a1 + a2 + a3 ... = bbbbb.... . Then by the that same process i can create an infinitely large negative number -a1 + -a2 + -a3 ... = -bbbbbb..... That would be the number you are referring to when you say "unless the number subtracted is the same as the number from which it's subtracted". An infinite subtraction process is equivalent to an infinite addition process of negative numbers.

You are trying to apply addition/subtraction rules for finite numbers onto numbers that are endless and therefore incomplete. So not everything will work like it does with finite numbers.

There are only two possible options. Either you claim that we can have endlessly long natural numbers or you have bounded the set and made it finite.

Every natural number is the successor of a natural number. If your number is a natural number, then subtracting 1 from it leaves a natural number. What is that number?
In short, there are no infinite natural numbers.
Now will you please get your finger out of your arse, Google this stuff and spend some time studying it and taking it in.

1 Reply

@ughaibu,

Quote:

Every natural number is the successor of a natural number. If your number is a natural number, then subtracting 1 from it leaves a natural number. What is that number?
In short, there are no infinite natural numbers.
Now will you please get your finger out of your arse, Google this stuff and spend some time studying it and taking it in.

Okay. I binged it:
http://wiki.answers.com/Q/Show_that_the_natural_set_of_numbers_is_an_unbounded_set

The natural numbers are unbounded. So you must be able to write them as endless strings. End of story.

Quote:

Every natural number is the successor of a natural number. If your number is a natural number, then subtracting 1 from it leaves a natural number. What is that number?

As I have shown an infinite subtraction operation must undo an infinite addition operation. Because infinite numbers are endless, subtracting a finite number is not distiguishable since it would come from the "end" of the number. Oh wait I could show this:

.....9999 - 1 = ......9998

But if you want to say that an endless string of whole numbers is not a natural number then you have put an upper limit on the set. Which nine must be the one I stop at? You tell me? The 1000th the 1000000th:

999999999999999999999999999999999999999999999.....

Where is the finite point where this is not a natural number? If you give one then you are saying you have bounded the set. This could not be any more obvious.

1 Reply

Is free-will an illusion?

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