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On limits

by Organic
Tags: limits
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matt grime
#55
Apr7-04, 09:18 AM
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What are you trying to prove there anyway, now you've edited it? cos looking at it you can't really tell.
Organic
#56
Apr7-04, 09:21 AM
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How do you mean it is unknown?

I'm fairly sure it is at ... If you let be your base unit then it is really easy to mark it on.

Or do you just mean there is no given ratio between 1 and in terms of decimals?
What I say is very simple: Pi is a notation an element, which its exact place in the real line is unknown.

More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example:

3/9 place is well-known 3/10 place is unknown.
matt grime
#57
Apr7-04, 09:26 AM
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Why is 1/3's place known? How do you know where 1 is? Or zero? The real numbers aren't actually physically a line, Organic. You are confusing the representation of something with the something... Oh, no, you're going to talk about x and model(x) again aren't you?

Actually the statement above is trivially true because it is of the form A=>B whre A is false....
Organic
#58
Apr7-04, 09:27 AM
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if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e.
The two different a and b are both < e.

Therefore |a-b| = d < e, but both d and e > 0.
Zurtex
#59
Apr7-04, 09:28 AM
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Quote Quote by Organic
What I say is very simple: Pi is a notation an element, which its exact place in the real line is unknown.

More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example:

3/9 place is well-known 3/10 place is unknown.
What are you on about?

Do you know what a number line is? It is not something physical...
matt grime
#60
Apr7-04, 09:31 AM
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Quote Quote by Organic
The two different a and b are both < e.

Therefore |a-b| = d < e, but both d and e > 0.

but that isn't deducible from your hypothesis: just because |a-b|<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption.

Try writing out the statement of the lemma again, and its proof making sure all the hypotheses are written correctly and that it is not vacuous (which it was first time)

and seeing as the statement was for all e, then you've just shown a=b=0
Organic
#61
Apr7-04, 09:33 AM
P: 1,210
Oh, no, you're going to talk about x and model(x) again aren't you?
Yes exactly, Math is only a theory therefore x-itself does not exist is its scope, only x-model can be used by Math language.
matt grime
#62
Apr7-04, 09:34 AM
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good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.
Organic
#63
Apr7-04, 09:46 AM
P: 1,210
but that isn't deducible from your hypothesis: just because |a-b|<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption.
Thank you for this correction you are right.

When we writing |a-b| < e we mean that d < e.
if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e
If a and b are distinct real numbers then for any e > |a-b| = d > 0.
Organic
#64
Apr7-04, 09:49 AM
P: 1,210
good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.
No, cauchy sequences do not prove that 0.999... = 1 without breaking infinitely many elements to become finitely many elements, by reaching the limit.
matt grime
#65
Apr7-04, 09:52 AM
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Quote Quote by Organic
Thank you for this correction you are right.

When we writing |a-b| < e we mean that d < e.

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

let e=d>2, then you have

d/2>d and d>0. now think for a second.

d/2>d => d>2d => 0>d ,yet d>0.

Want to rethink that at all.
Organic
#66
Apr7-04, 09:57 AM
P: 1,210
let e=d>2,
This in not the case because e>d always.
matt grime
#67
Apr7-04, 09:59 AM
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Quote Quote by Organic
No, cauchy sequences do not prove that 0.999... = 1 without breaking infinitely many elements to become finitely many elements, by reaching the limit.
As real numbers are defined... oh look, circles. Tell you what, why don't you tell us what you think the real numbers are? Since your definition must be equivalent to the one using cauchy sequences where 0.9999 =1 by definition you are in trouble. I think this is because when mathematicians speak of a model, in the sense of something satisfying the axioms, an example, they don't mean what you think the mean. ie a model in the sense of a model of turbulence, or something, which is only an approximation (at the moment). There is no approximation; you are confusing the concrete and the abstract. The Cauchy sequence argument is not some "best approximation" mathematically to the "physical" real numbers, they are the real numbers, in and of themselves, it is the things that you draw on the page using axes that are the approximation, not the other way round.

breaking, infinitely many, finite, that's you wishing something to be true that isn't, you are thinking unmathematically (perhaps intuiitive and physically in your opinion, but that isnt' mathematics).
matt grime
#68
Apr7-04, 10:04 AM
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Quote Quote by Organic
This in not the case because e>d always.
then your initial quantifier, for all e>0, is not correct is it?


Here is your initial post:

Suppose a and b are different real numbers and for any e>0 we know |a-b|<e then a not= b.


No restriction on e>d. at all and it says for any e>0 doesn't it?

Once more you move the goal posts half way through your argument when someone points out where it's gone wrong.

So want to start from the beginning and clearly write out what it is you are trying to prove again?

Because so far you're not doing very well. I mean what was the point of it anyway?
Organic
#69
Apr7-04, 10:10 AM
P: 1,210
you are thinking unmathematically
There is no objective thing like mathematics that we can compare our way of thoughts to it.

Math language is only a rigorous agreement between people, no more no less.

I have found that the current agreement includes lot of weak point in it, where one of them is the infinity concept.
Organic
#70
Apr7-04, 10:17 AM
P: 1,210
Suppose a and b are distinct real numbers and for any e>0 we know |a-b|<e then a not= b.

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

If a is not b then |a-b|>0.

Let d be the difference.

Let e = d/2 then |a-b|=d or |a-b| = d/2 > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.
matt grime
#71
Apr7-04, 10:17 AM
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No, the idea of infinity is well understood, but apparently not by you. Neither, it seems, is the idea of axioms and definition.

All of the 'problems' you've come across have been because of your own refusal to accept the definitions that are there. (Cantor, Natural numbers, axiom of infinity, convergence, real numbers).


There are some deep and troubling issues in mathematics that we don't understand and have to live with. They cause no practical problems. Your findings aren't these, though. If you want to say things like 'there is no objective thing like maths' at least take the time to learn some of it, you might, well, learn something.
Zurtex
#72
Apr7-04, 10:18 AM
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Quote Quote by Organic
There is no objective thing like mathematics that we can compare our way of thoughts to it.

Math language is only a rigorous agreement between people, no more no less.
What you think of maths is not relevant to how maths works. Maybe you should go and argue about the philosophy of maths rather than fail to attempt to argue in maths...


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