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**Ricky****Moderator**- Registered: 2005-12-04
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George has been questioning infinity and it's properties, and I want to dedicate this thread to a rigorous mathematical understanding of infinity. I shall start with sequences for now, hopefully it will spark some interest. This is one of the cases where it may not be best to start at the beginning, so I will be posting this assuming the concepts dealing with the natural numbers and countability are well-defined and go back to defining it if asked.

Many mathematicians use notation of a sequence of numbers such as:

The "..." meaning to infinity. This is of course not rigorous, but this is one of the times where rigor can be laborious. Things get much worse when you start dealing with uncountable sets, you should know this if you've ever seen the proof of Zorn's Lemma in the uncountable chain case (if you don't know that means, don't worry). However in general, when mathematicians write a sequence as above, it means that such a sequence is countable. To write the sequence rigorously, we would write in general:

In this specific case:

The notation above may make it seem like we have now lost the order of our sequence. After all, what I wrote was just a set, right? And sets don't have any order. Of course this is all correct, but the order of the sequence can be recovered by defining:

So we can now say that a_1 < a_5 or that a_3 > a_2, not talking about comparing the elements, but comparing their position. We say that "x appears in a sequence a_n" (where a_n is the set defined as above) if:

We now prove a rather trivial theorem.

Let x be an element in some set S, and let

be a sequence. Then x appears in the sequence if and only if x = a_n for some natural number n.Proof: Let x = a_n. Then it is clear that x appears in the sequence. For the forward direction, let x appear in the sequence. Then

To be an element of a set means to equal one of its elements. All elements of S are of the form a_n, and so if x is not equal to a_n for some n, then we would have that x is not in S. As x is in S, we conclude that x must be equal to some a_n. Thus, x = a_n, ending the proof.

Now George has said:

1/10= (1/8)(1/2+3/10) = 1/16+(1/8)(1/2)(1/2+1/10) = 1/16+1/32+(1/16)(1/10) =...

By this algorithm, 1/10 in binary can be approximated by

0.00011 0011 0011 recurring

And frankly this is what you can get from strict logic proof of induction, Ricky.

Concluding that the decimal digit 0.1 is in the set of binary digits:

{};0;

{1};0.1;

{2}, {1,2}; 0.01,0.11;

{3}, {1,3}, {2,3}, {1,2,3};0.001,0.101,0.011,0.111;

...; ...

Now it a fairly common fact (and easily proven) that 0.1 (decimal) can not be expressed by a finite amount of binary digits. As George has not seemed to contest this, I will not prove it at least for now. What I will show is that using this fact, we may prove that 0.1 (decimal) is not in the sequence George has described above.

Proof: As 0.1 (decimal) has no finite binary expansion, it suffices to show that all of the elements in the sequence above have finite binary expansions. For if we have equality between 0.1 (decimal) and an element of that sequence, we have a finite binary expansion of 0.1 (decimal), which is a contradiction.

We begin by defining the set above. Let n be an nonnegative integer, and define:

Where "minimal binary expansion" means that there is no shorter way to define the binary expansion of a. Note that this is well-defined by the well-ordering of the naturals. Also note that every element of S_n has a finite binary expansion. We now define our sequence:

Since S is composed of elements from S_n for some n, and each of these elements has finite binary expansion, we conclude that S only contains real numbers between 0 and 1 that have finite binary expansions. Therefore, we conclude that 0.1 is not in S.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**George,Y****Member**- Registered: 2006-03-12
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Ricky, just imagine we have already count over 2[sup]**N**[/sup] of the binaries between 0 and 1, j in the similiar way that you imagined you count over all the infinite digits of 0.111... . My set is well ordered and also infinite.

In fact, it has 2[sup]**No**[/sup] of elements as Cohen proved in Cantor's Continuous Assumption(Theorem), where **No** means the infinite amount of all natural numbers (primary infinity). He did state all the binaries between 0 and 1 represents all real numbers between 0 and 1, and his idea is to switch on and off each digits to show there should be 2^the length of the digits (primary infinity) of combinations. Just suppose you have infinite bulbs in a line and how many combinations you would have.

What I did, is just hold all the digits after the first off at 0, and gradually each by each, allowing them to be turned on as 1, and by this way I am sure to order all the 2[sup]**N**[/sup] elements, just as Cantor did to order all the **N*****N** elements of fractions (rationals).

Here I shall stress the symbol "**N**" or "**No**" here stands for Hebrew letter "alpha", it doesn't mean any natural number you think of, it represents the primary infinity you would tag at

1,2,3, ...->∞

or the amount of all natural numbers. (They call it cardinality, same thing)

*Last edited by George,Y (2009-02-07 09:28:46)*

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**Ricky****Moderator**- Registered: 2005-12-04
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George, you have completely ignored my post above. You also have not found fault in my proof. I'm not sure why I should be so nice as to reply to yours, but I suppose being a moderator here means I should hold myself up to a higher standard.

What I did, is just hold all the digits after the first off at 0, and gradually each by each, allowing them to be turned on as 1, and by this way I am sure to order all the 2N elements, just as Cantor did to order all the N*N elements of fractions (rationals).

In this way, no element of yours will ever have more than a finite amount of 1's. However, there are binary numbers in [0,1] which can only be expressed with an infinite amount of 1's.

And this all comes back to sequences as I tried to rigorously define above. Every element in a sequence has a finite position from the starting element. There is no element in a sequence that is of "infinite distance" to the start. And this is why you will never get more than a finite number of ones in any element in your sequence.

But if you believe you have put the binary numbers between [0,1] can be put in 1-1 correspondence with the natural numbers, then I offer you this challenge. What position of your sequence contains the decimal digit 0.1?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**George,Y****Member**- Registered: 2006-03-12
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2^No

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**George,Y****Member**- Registered: 2006-03-12
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let me see

0.111... is on 2^No

and

1/10 or

1/1001=0.00011 0011 0011 recurring

is somewhere between

2^(No-1) and 2^No

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**George,Y****Member**- Registered: 2006-03-12
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The reason why I ignored your post is simple, you continuously misunderstand potential infinity and realized infinity, which I have tried to distinct in the original post.

You think you have got infinite amount of digitals afterwards, but in fact you are getting only more and more yet finite amount of digitals. If you think you have got them whole, without bother to state the infiniteth digit and instead pretend you have only digits mapped to all natural numbers, you are ignoring the 0.999... post.

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**Ricky****Moderator**- Registered: 2005-12-04
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The reason why I ignored your post is simple, you continuously misunderstand potential infinity and realized infinity, which I have tried to distinct in the original post.

George, this is not a reason to ignore my post. This is a reason to discuss the differences.

and

1/10 or

1/1001=0.00011 0011 0011 recurring

is somewhere between

2^(No-1) and 2^No

Well, you attempted to answer my question, thank you. For now, I'll let you pretend that No-1 even makes sense.

However, we need to start on basic definitions which is what I was attempting to do in my first post. To make sure we have the same starting point, please answer me these two questions:

1. Is every natural number a finite distance away from 0?

2. Is every term in a sequence a finite distance from the starting term?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**George,Y****Member**- Registered: 2006-03-12
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1 No

2 No

The reason why? Right-end paradox I used in 0.999... post.

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**Ricky****Moderator**- Registered: 2005-12-04
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I would like to discuss why it is you think there is a natural number of infinite distance from zero. I really don't understand this at all. I would bring out an axiomatic discussion of the natural numbers, stemming from ZFC and the Peano postulates, but I don't think you care too much about that.

Instead, let me define the following set:

Do you believe that set to be well-defined, George? Does it contain any numbers of infinite distance from 0? How many numbers does it contain, finite or infinite?

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**TheDude****Member**- Registered: 2007-10-23
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George,Y wrote:

1 No

2 No

The reason why? Right-end paradox I used in 0.999... post.

Would you mind reposting your argument? I don't want to sift through nearly 40 pages of the 0.999... post.

Although I think what we're really getting at is something fundamental to mathematics. It appears that George either disagrees with one or more axioms related to set theory, limits, or whatever other fields of math are involved in this discussion, or believes that one or more of the theorems arising from those axioms is wrong. Would you mind explaining which it is George?

Wrap it in bacon

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**George,Y****Member**- Registered: 2006-03-12
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Well Ricky, I don't expect everyone can accept my new idea. The guy who discovered square root of 2 was thrown to the sea by his teacher. Although I argue he was deceived by seeing is believing, the reason he was rejected was not he was wrong, but people *are* simply reluctant to accept new ideas.

I will go on with my inquiry. Today I discovered this counting method help solve another important problem. But I guess I would like to leave it to some journal to decide whether I am right or wrong.

For Dude, you have a good curiosity, this is nice. And the next post is for you. Hope you find it make sense.

*Last edited by George,Y (2009-02-10 12:18:58)*

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**George,Y****Member**- Registered: 2006-03-12
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mrphysics, i used the similar method to locate the last 9 in 0.999...

you see, the new 0.9999... has exactly one more 9 in it so we wanna find where it is by compare and contrast

0.999...

0.9999...

and through a proof I posted in previous thread the one 9 previous 0.999... lacks has no suceeding 9 after it or it would be ...09

in other words, it has a last 9

Start from here,

and think it this way,

try to sum up 9's from the last 9 backwards, you will get 0, or infinitesimal, which doesn't matter. And no matter how many or much you have summed up, you will never be able to come up with an even small number larger than 0.

Is it a proof

0.9+0.09+0.009+...+last9=0?

This is my paradox

I know Ricky you would argue the cardinality (only a quible name for amount when it comes to infinity) of the digits of 0.999... is N0. But I know how this Hebrew alpha 0 is defined by Cantor, and I have a similar disproof for so called mapping integers to evens.

BTW, the "proof" of ∞=∞+1 is fake, or two proofs, one is mapping, the other is stating larger than finite.

for the former one

0.999...->0.0999... every 9 has already done one'o'one mapping

and +0.9 strictly adds in one more 9 to the latter

the only reason now to equate the 0.999... and the new 0.9999...

is by their similarity or unexaughstive counting, by simply looking at

0.9 0.99 0.999 ...

0.9 0.99 0.999 ...

and say since first second and third 9's are the same, and the examine process can go on beyond our human's ability to count all over

the 9's in the two are the same amount.

such finite "examination" is plausible because this method itself is defected since it is not done at all. (Have it examined "all" digits for either 0.999...?) You cannot force us to testify someone innocent by stressing s/he was not seen in the crime scene. And another piece of evidence might be sufficient to nail down the case.

The only exhaustive method is by wholistic mapping

when you do the division by 10 to 0.999...

every digit is already assumed moving rightwards by one digit

(or if you try to move the 9's one by one, you never can finish the division in your life

and you should not claim you can do the division in your life if you are rigorous),

I want to mention here not a single one more 9 is created or less 9 is diminished in this division

otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)

then add in one more 9

yes just the one more 9 to compose 0.9+0.999.../10

and now we have already created ∞+1>∞

and then you can use my proof to just locate where the one more 9 is

yes the one more 9 is 0.9

but taking it at a different angle, suppose the 9's representing the same 9*10^-r play the same role and be treated the same in both 0.999...'s and from previous knowledge we know there is one 9's difference, one not and one has, the former can be filled with 0.

And this pair of 0 and 9 will not have a succeeding 9 'n' 9 pair to compose the paradox

...09

...99

Logically it follows this pair is on the right end. A 9 has no 9's on its right is clearly the right end of a bunch of 9's

Simple,huh? And I did not use countive mapping to try to conclude not enough evidence for the last 9. Additional evidence has already been provided to pin it down.

appendix

"I want to mention here not a single one more 9 is created or less 9 is diminished in this division

otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)"

Now I relax it, but could anyone tell me which 9 can be diminished? The 9 at the right end?

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**Ricky****Moderator**- Registered: 2005-12-04
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Well Ricky, I don't expect everyone can accept my new idea. The guy who discovered square root of 2 was thrown to the sea by his teacher.

The man who won't defend his idea is throwing himself overboard.

You come here every time the moon is full (or so it seems) to discuss your ideas on infinity, I know you want further discussion. Why will you not discuss them any longer? I asked you three specific questions about the set I defined, they would have taken a few seconds to answer. If you don't wish to discuss it any further, that's fine, but I can't help but feeling this is because my set creates a paradox in your view of infinity. Specifically, I believe you think that any infinite set of numbers must contain a member of infinite distance from 0, and my set S demonstrates how this idea is flawed.

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**George,Y****Member**- Registered: 2006-03-12
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To post 9

You have to first distinguish what your infinite means, Ricky.

We are in a circle now. You keep imagining induction guarantees you all the natural numbers, but it only gives you a growing yet FINITE amont of numbers at every given STEP. If you imagine you have got them all, you have to have gone through infinite steps, which means you have included infinity in your natuaral number set.

Step1 {1}

Step2 {1,2}

Step3 {1,2,3}

....

Tell me Ricky, how many steps you have gone through so far and how large your ending natuaral number is at your latest step?

I remembered a book called

1,2,3 and then infinity

Ha, such casual argument should belongs to poets only.

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**Ricky****Moderator**- Registered: 2005-12-04
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You have to first distinguish what your infinite means, Ricky.

Infinite means not finite. I ask again for you to at least try to answer the three questions I posed in post #9.

You keep imagining induction guarantees you all the natural numbers, but it only gives you a growing yet FINITE amont of numbers at every given STEP. If you imagine you have got them all, you have to have gone through infinite steps, which means you have included infinity in your natuaral number set.

George, you are so close you can almost taste it. Induction will give you every number that you * can* reach in a finite amount of steps. And indeed, we can reach an infinite amount of numbers where each individual number is reached in a finite number of steps. But here is the kicker, and I think this is where you are misunderstanding induction arguments. I believe you see them as:

There exists an N such that for all integers K we will reach K in N or less steps.

But induction arguments really go:

For all integers K, there exists an N such that we will reach K in N or less steps.

Do you see the difference? For every different integer K, you are allowed to choose a different integer N.

It is best to see this with an example. By induction, we shall prove that for all n >= 1, n is an integer.

Base case: n = 1. 1 is an integer, done.

Inductive assumption: n is an integer, we will prove that n+1 is an integer. Since both n and 1 are integers, and integers are closed under addition (i.e. integer + integer is an integer), we have that n+1 is also an integer.

So we can take this very simple example George an analyze it. You say that induction will only prove it for finitely many numbers. At what point does this proof cease to work? 10^10? 10^1000? What number does this proof not work for? If it does indeed work for all integers, then wouldn't it work for an infinite (meaning not finite) amount of them?

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**George,Y****Member**- Registered: 2006-03-12
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Induction only gives you finite numbers increasing, how can you get them all, Ricky?

Put it in other words, if you have got them all, you have already passed the induction phase.

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**Ricky****Moderator**- Registered: 2005-12-04
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I would still like to know about the set I posted in #9, George.

Put it in other words, if you have got them all, you have already passed the induction phase.

Given any number k, it is either true or false that k has a property p. There is no concept of "using induction to get them". What is true was always and will always be true, whether we know it or not. Induction is just a method to know if it is true.

So what we're saying is: I know I can prove it for 1, and I can also prove that if it holds for k-1, then it must also hold for k. Every integer can be reached in a finite number of steps using this, and so we conclude that there are no integers that we can't prove it for. This of course is just saying that it isn't false for every integer, or rather it is true for every integer.

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**George,Y****Member**- Registered: 2006-03-12
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Ricky, if you trust induction algorithm is all that entails infinity, my inductive algorithm can produce decimals any long thus is infinity as well.

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**Ricky****Moderator**- Registered: 2005-12-04
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I would still like to know about the set I posted in #9, George.

if you trust induction algorithm is all that entails infinity

There may be a translation problem, but that doesn't make any sense.

my inductive algorithm can produce decimals any long thus is infinity as well

No, you can't George. Induction only works when the number may be reached in a finite amount of steps. It may very well be that this finite amount is arbitrarily large, but it says nothing about what happens at infinity.

Let's say I have proven that proposition P holds for the value 1, and I know that if it holds for a number n, then it holds for n+1. Please answer me these questions:

1. Does such a proposition hold for the number 2?

2. What integer may proposition P not hold for?

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**George,Y****Member**- Registered: 2006-03-12
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Sorry for leaving for so long.

For post 9

Could I define a pair of numbers, each number larger than the other, Ricky?

The major misconception of your definition is that you know it well within simple cases, and you assume you know *all* the cases.

Let me ask you a question,

When you *try* to define such a set, have you considered the possibility that it does not exist?

Let us assume it has a steady amount of elements in it.

Would you tell me how large it is?

If you assume not steady, please tell me why you are so confident to define a set without knowing either the total amount of it or whether it has a total amount at all?

Or simply you don't define a set, you define a process or an algorithm and you mistake it as a set?

Hope you figure it out.

*Last edited by George,Y (2009-07-04 14:53:58)*

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**Ricky****Moderator**- Registered: 2005-12-04
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Hope you figure it out.

My question was not posed so that *I* could understand this set, indeed I know exactly what this set is. My question was posed to you so that I could understand where your views on mathematics differ from what is accepted in standard mathematics.

When you try to define such a set, have you considered the possibility that it does not exist?

It exists because the integers exist. It is an element of the power set of integers, just like any other set of integers.

Would you tell me how large it is?

This is what I want you to tell me. But I don't care about how large it is, only if it is finite or infinite.

Let us assume it has a steady amount of elements in it... If you assume not steady...

You're going to have to define "steady", because I can't find it anywhere.

please tell me why you are so confident to define a set without knowing either the total amount of it or whether it has a total amount at all?

Either an integer is a finite distance away from zero, or it is an infinite distance away. I am taking all the integers that are a finite distance away, and I call this my set that I posted in #9. What is wrong with this?

Also, what the heck does "total amount" mean?

Or simply you don't define a set, you define a process or an algorithm and you mistake it as a set?

A choosing function (what you mistakenly call an algorithm... it's not) is a perfectly valid mathematical tool. Indeed, we lose a ton of mathematics if we aren't allowed to use them (e.g. all of number theory).

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**George,Y****Member**- Registered: 2006-03-12
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"Either an integer is a finite distance away from zero, or it is an infinite distance away. I am taking all the integers that are a finite distance away, and I call this my set that I posted in #9. What is wrong with this?"

The problem is, Ricky, the conception of integers.

When people think of positive integers, they think of 1, 2, 3, and *so on*

And the "so on" is the problem

Is it finite or infinite?

It has a finite side that every integer is finite, just as you said.

Yet it appears infinite that the growing process might go on and on, without a boundary. (and this is what you want me to admit on infinte set)

This is the most wide-spreading misconception of infinity, I have to say.

Aristotle name it as *potential infinity*, he referred to time as evidence

Let's examine more carefully the two sides.

The former is experimental, the latter is imagination.

When you observe time or days, you do see the pattern

1, 2, 3, 4, 5, and might go on.

But actually, the cardinals 1, 2, 3, 4, 5 and perhaps more don't represent absolute quantity.

The fact is, you cannot put 5 days together, shuffle them, seperate them, group them as you do with 5 apples.

Are 5 days of 1, 2, 3, 4, 5 the same as the 5 days of 1, 3, 5, 7, 9? Just as you arbitarily choose 5 apples with almost the same quality?

We all know time changes, it cannot be the same.

The major difference is, 5 apples are 5 apples, they are independent, true 5 individuals, free of consequential or consecutive relations.

whereas 5 days, are actually 1 day mutated 5 times, each following day is the previous day evolved.

When you try to group {Day1, Day2, Day3, Day4, Day5, ...} and think you are forming a good example of integer set,

you are mistaken.

You are like forming {you in yesterday, you at today, you at tommorrow, ...} but not {1 of you, duplication of you, triple of you...} or even {1 days, 2 days, 3 days, 4 days, ...}

The latter is a list of quantities, whereas the former is a list of cardinals. Cardinals doesn't necessarily represent quantities.

It is really an illusion to form a list of integers by forming a list of cardinals like these, since they represent only one thing changing.

Since cardinals are not good example or proof of existence, we can think of quantities

when you believe you can form a set as

{1 apples, 2 apples, 3 apples, ...} , which includes *all* the possibilities of amount of apples

You have included infinite apples at the right end, which I have proved.

Please don't tell me you haven't finished listing the amount of apples imaginarily

Only cardinals like date have this right not to be shown at the same time, since they are actually only one, since changing the appearance needs time. But true quantities can be gathered at the same time, all at now.

*Last edited by George,Y (2009-07-05 00:19:18)*

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**Ricky****Moderator**- Registered: 2005-12-04
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George, I must say with the greatest sincerity that that was entirely incomprehensible. You seem to have this idea that it's important what it is your counting, but I haven't the faintest clue as to why. I suppose you are missing the whole point about the abstractness that numbers provide, but even of that I'm not really sure. I'm thoroughly convinced that there is no point in further discussion.

To end on a sad note, it is depressing to see your mathematical imagination blocked by what we can or can't do. You are missing out on the beauties of a universe, even if it isn't our own.

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**George,Y****Member**- Registered: 2006-03-12
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Ricky wrote:

George, I must say with the greatest sincerity that that was entirely incomprehensible. You seem to have this idea that it's important what it is your counting, but I haven't the faintest clue as to why. I suppose you are missing the whole point about the abstractness that numbers provide, but even of that I'm not really sure. I'm thoroughly convinced that there is no point in further discussion.

To end on a sad note, it is depressing to see your mathematical imagination blocked by what we can or can't do. You are missing out on the beauties of a universe, even if it isn't our own.

I tried to explain it to you before in a simple way that how mathematicians like you misinterprete words, and simbols. And I show this simple logic disproof of your infinite set again:

Regardless of what particular infinite set you define

Try to shift all elements one position rightwards by some pratical algorithm (in this case +1)

Then add in the starting element lost in the shifting process (in this case 1)

And then it is undeniable logic truth that I have just created an infinite set with one element more than the old one

And now it is time to reveal what the old set lacks - the right end

And the right end is the infiniteth set revealed

sometimes infinity, sometimes infinitesimal

so the paradox is discovered.

"Let's say I have proven that proposition P holds for the value 1, and I know that if it holds for a number n, then it holds for n+1. "

This inductive model seems to stretch out the infinite set to you,

and you keep saying

Can you find any error in my set by this approach?

insead of asking

What other approach might prove my set wrong?

Ricky, have you encountered some classmate asking you the question:

"don't tell me your way and how you get it correct, tell me why my way is wrong try to work my way out to get the correct answer"

I heard these questions numerous times, and I have to admit, they are always much harder questions.

Nice as I am, I have to give you an answer, even though you might not be satisfied with it.

The misconception is

you mistake induction as infinite

Yes many had mistaken it,

But where is it mistaken?

You add in its infinite potential to imagine it as infinite

If we step firmly,

we should say inductive sets can never qualify infinite set

you have 1 element in it at first,

then by inductive process, you have 2,

you have 3,

you have 4,

just like day 1, day 2, day 3, day 4

but come on

when can you have the inductive set or sets to capture its ever changing nature

countless?

Every stage it is an Finite set.

This truth looks so boring

So most mathematicians like you say

Imagine(define) a set to include all element an inductive process can have, or the union of *all* inductive sets possible

like the

{1}U{1,2}U{1,2,3}U...

thing

In this way so can you *create* an infinite set without the dynamic inductive process and its each finite stage

You wanna chop off the process and leave the result

Fine,

just after you have done that, your set claiming to include "all positive integers" in it no longer adds in new elements like the inductive set, right?

Otherwise how can it be the union of *all* inductive set?

Then we are good to go

Go to four paragraphs above

And reread

"Regardless of what particular infinite set you define

Try to shift all elements one position rightwards by some pratical algorithm, in this case +1 ..."

You would discover why it is an illusion to you

that your infinite set doesn't contain *the infiniteth* element or *the right end*

Your only defense would be going back to the inductive process, back from the beginning

{1} "I haven't seen an infinite number!"

{1,2} "I haven't seen an infinite number, not yet!"

{1,2,3} "not yet at all"

n, n+1 "they are defined as integers, not infinite numbers"

Ricky, your way and my way get different answers, and I know one part of your reasoning is flawed

That if you use n and n+1 *only*

You are burried in the "infinite" inductive loop

If you have a computer program or anything like that to run it

You are in the "dead loop"

{1}

{1,2}

{1,2,3}

...

{1,2,3,...,n}

not infinite yet?

...

{1,2,3,...,n,...,m}

still not infinite?

...

{1,2,3,...,n,...,m,...,q}

not infinite, no way!

And there is *nowhen* you can achieve your ideal infinite set

which contains *all* inductive sets *possible* (really?)

To get away from stucked by dead loop

You make a jump-

You just claim you can get or define such an infinite set,

which in fact is *never*(literally) a product of inductive growing

And it has *now* invited **the infiniteth element demon** into it, (as I proved)

which innocent n and n+1 algorithm had never ever and done and will never ever do. (indeed you won't find "infinity" in n 'n' n+1)

And here is where your way (or the way chosen by mathematics community) is flawed, Ricky.

*Last edited by George,Y (2009-07-08 02:25:15)*

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,365

Ricky wrote:

George, I must say with the greatest sincerity that that was entirely incomprehensible. You seem to have this idea that it's important what it is your counting, but I haven't the faintest clue as to why.

Because maths derives from reality, the concept "all" "natural numbers" "fraction" "set" comes from abstraction on reality. And maths does not belong to only mathematicians. If you educate anyone on a street 2 isn't 1 plus another 1, but instead some set of any rationals smaller than 2, or some "infinite" Cauchy sequence, or some field, there is no way that the "abstract beautiful mathematic concept" win over common sense, or simple truth. And if mathematicians force people to accept their own idea in instituations like university, in the name of abstractness, in the name of beauty or in the name of intellect.

To end on a sad note, it is depressing to see your mathematical imagination blocked by what we can or can't do. You are missing out on the beauties of a universe, even if it isn't our own.

Imaginition isn't the only thing that matters.

Beyond it there are logic and clear awareness of what you say.

**X'(y-Xβ)=0**

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