On limits

[Organic]

This is the reason why I used the word "General".

Please show us how the limit concept is rigorous.

[matt grime]

Oh heck, not again. The limit concept is a definition, and therefore by, um, definition it is rigorous. Do you need to see the definition of limits?

[Organic]

Yes Matt,

Please write a formal definition of it, and then please explain it in plain English.

Thank you.

[matt grime]

Ok, for real numbers we say that a sequence x_n indexed by the natural numbers tends to a limit x if for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

A sequence that does not converge is said to diverge. In particular, if x_n is a sequence such that for every X in R x_n>X for all n sufficiently large then x_n is said to diverge to infinity. Note this does not tell you waht infinity is. It tells you what the phrase tends to infinity means.

This idea that infinity is a quantity larger than any real numbers is a short hand and unmathematical way of stating that fact about divergent series. The more rigorous way of saying 1/0 is infinity is to say that if x_n is any sequnce of (positive) real numbers converging to zero then 1/x_n diverges to infinity.

OK? Infinity isn't there, that is the easiest way of thinking about it, it isn't part of the real numbers.

[Organic]

for real numbers we say that a sequence x_n indexed by the natural numbers tends to a limit x if for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

Please tell me if this is a correct picture of this definition:


| |
- e - e
| |
- |x_m-x| <---> - m<n --> |x_n-x|
| | |
| - <---------'
| |
- 0 - 0

[matt grime]

Perhaps if by that you mean that if you plotted all the points |x_n-x| on the real axis, then all of the ones you plot after the m'th lie in the interval [0,e]

[Organic]

Perhaps if by that you mean that if you plotted all the points |x_n-x| on the real axis, then all of the ones you plot after the m'th lie in the interval [0,e]
Yes, |x_n-x| is for (what you call) all points and also |x_m-x| and e.


| |
- e - e
| |
- |x_m-x| <---> - m<n --> |x_n-x|
| | |
| - <---------'
| |
- 0 - 0


What a show is the invariant state that stands in the basis of the rigorous definition, or more to the point this invariant picture is the reason that we call it rigorous.

Do you agree?

[matt grime]

I cannot agree because I cannot decide what invariant state means. Seeing as this needs to be true for every e>0 and each time the m is different I see nothing very invariant. Moreover if you were to offer that as the definition of convergence of a sequence you'd be once more not providing enough information.

[Organic]

...for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

This is the invariant state that for any given n there is m<n in N.

Right?

[ShawnD]

Organic, the difference between a limit and normal numbers is that a limit is what the number gets close to.
the limit of x -->0 for 1/x is infinity. the value of 1/0 is undefined.

Thanks for answering the question guys.

[chroot]

The limit of 1/x with x from the right is infinity; from the left it's negative infinity.

- Warren

[Organic]

Hi ShawnD,

for real numbers we say that a sequence x_n indexed by the natural numbers tends to a limit x if for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

|x_n-x| is for (what you call) all points and also |x_m-x| and e.

| |
- e - e
| |
- |x_m-x| <---> - m<n --> |x_n-x|
| | |
| - <---------'
| |
- 0 - 0 is the limit x in this case

[matt grime]

This is the invariant state that for any given n there is m<n in N.

Right?


No, get your quantifiers in the right order: for any e>0, there is an m in N (dependent on e) such that for all n>m....

It is not the trivial true or false assertion (dependent on n being 1 or not) that for any given n there is m<n at all. That deosn't even mention x_i, x or e, does it?

[Organic]

Matt,

Please look at this:

http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html

If you translate newton's method to the limit of a sequence, do you get delta-epsilon definition?

[matt grime]

I don't understand what you want. If you mean: in order to make Newton-Raphson's iterative method rigorous in the sense that we wish to prove the iterated sequence converges must we use epsilons? Yes, and no. There are other equivalent definitions of convergence using nets and ultrafilters and topologies, but here I imagine one could prove that given certain constraints the algorithm converges in the epsilon definition. Generally N-R converges very rapidly as can generally be indicated graphically. One may prove for instance the that x^2-p yields a recurrence formual x_n = (p + x_(n-1)2)2/x_(n-1) or something which gives a monotone decreasing sequence bounded below which thus convrgese and to sqrt(p) without using an epsilon at all.

[Organic]

So, can we define a rigorous definition to a limit of a sequence without using an epsilon?

If yes, can you write a rigorous definition, which is based on N-R?


Generally N-R converges very rapidly as can generally be indicated graphically

It is a non-linear sequence of x1, x2, x3, ... that converges very slowly to the limit and run away from the limit (diverges to infinity) very rapidly, as we can see here: http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html

But in both directions no sequence (geometrical or arithmetical) can reach the limit, isn't it?

[matt grime]

Look, that is one particular example of N-R, you cannot decide how rapidly it converges in general from one example, and I'm not talking temporally to do with the animation anyway.

The rest of what you write is wrong: it is perfectly possible for a sequnce in N-R to attain the limit in a finite number of steps, nor does one reach infinity, in any sense because infinity isn't there, we cannot emphasize that enough to you.

I can see no way of using Newton-Raphson to define the notion of a limit of a sequence, or what it means for a sequence to converge.

[Organic]

The rest of what you write is wrong: it is perfectly possible for a sequnce in N-R to attain the limit in a finite number of steps

No, it is possible only if a curve become a straight line before the limit point, or changes its direction before or in the limit point.

I am am talking about a curve that does not become a sraight line or changes its direction before or in the limit point.

Therefore if we reach the limit then we make a phase transition that leave behind (ignore) N-R method.

If you don't think so, please prove That by N-R we are not approaching but reaching the limit in finitely many steps, where a curve does not become a sraight line or changes its direction before or in the limit point.

[matt grime]

A straight line is a curve, it has curvature zero at all points. Why is N-R not applicable to straight lines anyway? Moreover your assertion is not true, there are other curves that will reach their root in a finite number of steps (if your initial term is the root for instance), others may even cycle with period. The second sentence makes no sense to me. I don't know exactly what you're asking me to prove in the third step, but I think the trivial counter example you need is to take the initial term to be the root. That converges after 1 step, as well as the other trivial disproof that a straight line is a curve. You have shown a lack of belief of proofs in the past anyway, so what would it do for me to prove a result for you anyway? I've given you an example where it terminates in a finite number of steps, you've given an example of this, apparently, I've not checked the details, which you've dismissed for no good reason (a straight line is a curve, just not a very curvy one). Perhaps there are no other examples but I imagine you might be able to construct something that isn't technically a straight line (adjoin a non-straight line to it).

Ah you've now edited your post to remove the counter examples that you have NOW decided are not allowed for some reason, even though you didn't say they weren't allowed before! Naughty, naughty.

[Janitor]

Does this website own the copyright to the ongoing Grime-Organic debate? I'm smelling $ in the air on this one! How about publishing it in print?

[matt grime]

I want to stop, really I do, but I just can't leave ignorance alone. It's pointless, but some of us somehow appear to take it in turns to duke it out. Hurkyl did for a while and now he's stopped. This annoys me though because it is hijcaking another persons thread *yet* *again*. Hopefully it'll be locked again.

[Hurkyl]

So, can we define a rigorous definition to a limit of a sequence without using an epsilon?

How about: The sequence <x> converges to the limit L iff:

For every &rho; > 0 there exists a natural number N such that for all m > N:
|L - xm| < &rho;



There are certainly other ways to go about defining limits; they're all equivalent to this one, though, when applied to sequences of real numbers. A common one is to use the topological notions of neighborhoods or nearness.

[Organic]

Hi Hurkyl,

...neighborhoods or nearness.

When these words are used, do we can understand that no element in the sequence reaching the limit?

[Organic]

Matt,

The rest of what you write is wrong: it is perfectly possible for a sequnce in N-R to attain the limit in a finite number of steps


In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

Do you think that in this case we get infenitely many x_n values where abs(x_n-x)>0?

[matt grime]

Hi Hurkyl,

When these words are used, do we can understand that no element in the sequence reaching the limit?

It is perfectly possible in some convergent sequence x_n tending to x, for none, finitely many, infinitely many but not all, or all of the x_n to equal x. Why is this a problem?

As for N-R. What do you mean by 'change direction of curvature'?

At this rate you're going to say: ah but it's true except for all the counter examples.

Here's one for you

Consider a curve given by x^(1/2) for x positive, -(-x)^1/2 for x negative. Call this f(x). Pick some point x in (0,1) draw the tangent line in. Suppose just for the sake of the argument that I can tell exactly where that tangent line cuts curve again, call that point t. Then the curve g(x)= f(x)-t has a non-trivial starting point for a N-R sequence that converges in a finite number of steps. Does that have 'curvature that changes direction' though? Perhaps you mean sign?

It seems to me that you're just going to keep inventing restrictions each time to make any counter examples false. Why do you do that? Are those all the criteria you need? Are you sure? Not going to change your mind again?

[Hurkyl]

When these words are used, do we can understand that no element in the sequence reaching the limit?

No...

The normal topological definition says that L is the limit of <x> iff:

For every neighborhood U of L, there is a natural number N such that if m > N, then xm is in U.


Anyways, Organic, have you considered the sequence that is always 1? Shouldn't it's limit be 1?

[Organic]

What do you mean by 'change direction of curvature'?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

Do you think that in this case we get infenitely many x_n values where abs(x_n-x)>0?

[matt grime]

Then my counter example possibly doesn't cut it. However as the way to prove you claim has been merely to elminate all counter examples offered, why don't you offer a proof with all your many hypotheses, to demonstrate that the N-R method is only terminal for certain kinds of functions.

What that has to do with the defintion of limit is beyond a bear of little brain like me.

[Organic]

Let me ask this:

Is the case of abs(x_n-x)=0 (which means that x_n=x) exists, according to what I wrote?

[matt grime]

What you wrote when? If you'll excuse the ungrammatical paraphrasing.

Here's a reason why you won't get s generalized notion of seqences converging from N-R for a very good reason, and bearing in mind we're having to guess what it is that you mean, AGAIN.

The seqeunce x_n =1 n even, x_n=1+1/n n odd cannot arise from any Newton Raphson method given that if the iteration sends a to b, it always sends a to b, so there's no way 1 can go to 1+1/3 and 1+1/5 and 1+1/7 and....

[Organic]

What do you mean by 'change direction of curvature'?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

Let me ask this:

Is the case of abs(x_n-x)=0 (which means that x_n=x) exists, according to the curve that was described by me?

[matt grime]

I don't know. It's your conjecture, why dont you try to prove it in the light of all your artificial hypotheses which are solely to exclude counter examples that you didn't exclude originally?

If I may rephrase your question so that it makes sense to others who haven't read your stuff before - is it possible that there is some curve satisfying various criteria which is differentiable, and such that the Newton Raphson method does not converge to the exact root in a finite number of steps. The criteria are that the curve is nowhere linear, and that "the tangent line always stays on the same side of the curve". Note, that although he hasn't excluded it, one must presume that Organic also means that we do not choose the root as the initial input.

I don't know the answer, nor do I care really - my geometric intuition tells me it is probably true as one isn't allowed to do something bizarre like 1-sided derivatives.

Does that answer your query?

[Organic]

Matt,

What is "probably true"?

By your point of view x_n=x?

Please answer by yes or no.

Thank you.

[matt grime]

My view is that the conditions you've decided to impose on the type of curve probably imply that any non root initail choice for a N-R iteration will not reach the root in a finite number of steps. Probably means I can see how that proof would go (MVT or Rolle's theorem plus a little thinking).. I cannot answer yes or no because I have not proved it and have no proof in mind, only a possible proof. But it isn't my conjecture, it is yours, and I have no real desire to make it rigorous, try proving it yourself.

[Zurtex]

Matt,


In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

What?

You just said it was a curve with non-zero curvature at all points... therefore it is always changing its direction. How can anyone take you seriously when you make such obvious contradictions in your own argument?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.
Well why don't you actually say what you mean rather than just writing nonsense. Half the problems with your posts is you say one thing and mean something else all together, another problem is you don't actually seem to be proving anything...

[Organic]

Zurtex,

I gave this example:
http://phys23p.sl.psu.edu/~mrg3/mat..._I/newtons.html
before I wrote something about it.

This kind of a curve has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

By "does not change its direction" I mean that the tangent line stays in one and only one side of the curve, when N-R is used.

Shortly speaking, it is not switching sides.

If you can't understand all this by looking at the graphic example then it is your problem not mines.


another problem is you don't actually seem to be proving anything...

I do more than that, I make a paradigm shift in the infinity concept.

Please read this:
http://www.geocities.com/complementarytheory/NewDiagonalView.pdf

[matt grime]

Organic, how on earth can you expect people to know exactly what you mean just by posting one example without explanation. How can you not have realized yet that this is physicsforums.net not psychicforums.net. The confusion is entirely your causing by not explaining what you want. From one picture we are supposed to understand that you only want curves with THIS set of properties. Well, that curve in that link is also convex, should the curves only be convex? It only has one root if we carry it on in a naive smooth fashion, it cuts the x axis at positive x, must the curve always do this? The tangent has positive slope at all points, must this be true as well?

[Organic]

MVT or Rolle's theorem plus a little thinking

Examples of it can be found here:http://www.ies.co.jp/math/java/calc/rolhei/rolhei.html

But this is not the case that I show here:
http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html

I gave this example, and by this I mean that I am talking only about this example.

If instaed you want to speack about another types of curevs, then you are talking to yourself, not to me.

[matt grime]

Erm, thanks, Organic, I do know what the MVT and Rolle's theorem state, and, shockingly, I know how to prove them, amazing, someone might think I was a mathematician or something.

WHy didn't you say you only cared about that one example.

[Organic]

If I am not mistake we are in a thread that dealing with the limit problem, where there are infinitely many steps that cannot reach the limit point.

I gave this N-R example as something which is another example that is different from the epsilon-deltha method.

Is it not understood?

[matt grime]

You are making a mistake. This thread didn't start about that. You hijacked it so that you could talk about your interests again. And I sadly came along too.

Different from? As you've not proven anything to do with convergence that's a little rich.

[Organic]

If you look at the previous thread will see the you started this:

But is still makes no sense. x/0 is not a well-defined symbol in the real number system that one can manipulate like this.

[matt grime]

And what's that got to do with Newton-Raphson iteration? We explained what the more formal interpretation of your beloved Wolfram definition of infinity is.

[Organic]

x/0 deeply connected to the limit problem.

[HallsofIvy]

x/0 deeply connected to the limit problem.

Only in the minds of those who do not understand what a limit is.

"x/0" is "undefined" if x is a non-zero constant and "undetermined" if x is zero.

If you mean that x is a variable, then "x/0" makes no sense at all.

If you mean "lim as a-> 0 of x/a" then you should say that: the whole point of the theory of limits is that "x/0" will tell you nothing about the limit.

[Organic]

1/0 is the same as 1*oo and in both cases we are no longer in a finite system.

The whole idea of the interesting point of view of the limit concept is that no infinitely many elements can reach the limit itself.

This unclosed gap which is > 0 cannot be closed by infinitely many elements.

therefore the sum of .999... or the intervals of N-R is undefined by definition.

For example, Cauchy method only forcing the impossible to be possible by "raping" infinitely many ... to have a sum.

[matt grime]

You're wrong and veering off topic again with you own personal incorrect view of mathematics. Please stay on topic.

Counter examples: let x_n=0 if n is even 1/n n odd. this sequence converges to zero, adn reaches 0 infinitely often. OR let x_n^{M} be the sequence define to be 1 for n<M 0 other wise - this sequence converges to zero and is zero for all n>M. Demonstrate a non-zero real number between 0.9999.. and 1. Hint: can't be done. The infinite sum os defined. It is the limit of the partial sums. (N-R, or Newton Raphson, has no need to be here). I presume Chausy is Cauchy. I don't think you understand enough of the mathematics to be able to form an opinion about completions wrt norms. So, this is mathematics, in the real numbers in decimal notation 0.9999.. is the same as 1. It has been proven many times. If you're going to tell us we're wrong then please don't do so in this thread. Start another one and attempt to understand the answers that will be given. Don't hijack this one please - I've answered your post and told you where you're wrong conceptually as well as physically. If you don't accept that then you aren't using the mathematics correctly and you aren't adding to this thread's worth. Start one in TD say, but this topic has been done to death and that you cannot accept the PROOF is a reflection on you not the mathematics.

[Organic]

Another example:

PI exact plece in the real line is unknown, because in any representation method of it we have to use infinitely many elements to define it.

Demonstrate a non-zero real number between 0.9999.. and 1

Demonstrate a zero gap between 0.999... and 1

[matt grime]

Again? There are at least 2 proofs of this fact in this thread alone. Let x_n be the n'th partial sum 0f 0.9+.009+.0009...

|1-x_n|= 1/10^n

0.999.. =lim x_n

hence |1-0.999...| =0 as the difference with the limit tends to zero, ie can be made of arbitrarily small absolute value.

If you disagree with that then you are disagreeing with the definition of the real numbers. Got it? If you want to work in a different number system then start a different thread or something.

Just realized this isn't in the thread I thought it was in (new page carry error, bane of mathematics) so rant away in your own private langauge at will.

[Zurtex]

Another example:

PI exact plece in the real line is unknown, because in any representation method of it we have to use infinitely many elements to define it.

Demonstrate a zero gap between 0.999... and 1
How do you mean it is unknown?

I'm fairly sure it is at \pi... If you let \pi 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 \pi in terms of decimals?

[matt grime]

Here's a little thing you need that you don't seem to know, Organic.

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

[Organic]

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

proof:

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

Let d be the difference.

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

therefore non-zero/2 > 0.

[matt grime]

But your hypothesis is false: if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e. You do understand what the quantifier for all means?

[matt grime]

Actually that 'proof' of yours should go down in history: you assume a and b are distinct numbers, make a false claim about them and use that false claim to prove that a and b a different, which is part of the hypothesis... fantastic

[matt grime]

What are you trying to prove there anyway, now you've edited it? cos looking at it you can't really tell.

[Organic]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

let e=d>2,

This in not the case because e>d always.

[matt grime]

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]

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]

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]

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]

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]

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...

[matt grime]

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

so you are saying that for any real positive number |a-b|<e Got it? That's what that quantifier for all means. Notice that you desired conclusion is part of the hypothesis. Note also that by the proof immediately preceding the first appearance of this claim, there are no pairs of real numbers satisftying both hypotheses, and thus the statement is vacuous ((AandB)=>A is true tautologically as well)

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

What's th point of this line above? there is an 'if' , but no 'then'

proof:

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

Let d be the difference.

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

therefore non-zero/2 > 0.


but that doesn't show anything other than your errors. In fact you said that e = d/2 wasn't allowed a couple of posts ago. And now it is? You've just shown d<d/2 ie d is negative. And you've proved d>0 allegedly after assuming d =|a-b| which was already greater than zero by assumption. So the best you#ve done is prove that if you assume X you can deduce X from that assumption. Howeever the logic contains so many errors that even that is in doubt. (X=>X is tautologically true again)

This makes no sense, Organic.

[Organic]

Behind any rigorous agreement there is a meaning.

During the time, people forgetting the meaning and using only the technical tools of the agreement.

When this is happens, it means that we are dealing with a dieing system.

The soul of Math is based on philosophy, is body is it’s the rigorous agreement.

We need both of them to keep Math alive, no more no less.

[Organic]

I used your original proof, also a corrected it:

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]

As you evidently don't know what or where the body is (see the above expanded repudiation of your 'proof') how do you even know there is a soul?

[Organic]

I used your original proof, also I corrected it:

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 are both > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.

[pig]

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

i am having trouble understanding this, could you please rephrase it in a clearer way?

[matt grime]

But my proof that you thoughtfully corrected (ha!) wasn't of the statement you made. I showed that if, for all e>0 |a-b|<e then a=b.

The statement you've made above:

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

are you attempting to say that if e>|a-b| and a and b are distinct that e is greater than zero? But that is trivially true and doesn't require a proof, and doesn't even require that a and b are distinct.

As it stands your initial statement does'nt even have an obvious conclusion, it appears there is nothing to prove. It doesn't make sense.

It looks vaguely mathematical but there's nothing in what you've just written.

You cannot let e=d/2 since e is strictly greater than d, you even said I wasn't allowed to let e=d/2 in an earlier post.

[Organic]

Matt,

First let us write your rigorous proof:

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

You clime that if for any e>0 we cen find |a-b|=d<e then a=b.

Your proof is:

1) a is not b

2) |a-b|=d>0

3) e=d/2, but since we know that |a-b|=d<e, then e=d/2 breaking our on rules, because |a-b|=d<e does not exist anymore.

Instead we have to use now |a-b|=d/2<e.

Shortly seaking, this part:

let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

breaking the rules, therefore this proof doesn't hold because e=<e.

[matt grime]

I will remove the 'then' that Organic objects to so that instead of being a statement in inverted commas (ie incorrect) it is a quote. It doesn't alter the meaning of the sentence though


"You clime that if for any e>0 we cen find |a-b|=d<e then a=b."

That is not what I claim.

a and b are given to you at the start. You do not "find" an a and b with |a-b|=d

Now d is a non-zero positive number,and thus so is d/2, so given the hypothesis that |a-b| <e for any e>0 it must be that it is true if I let e = d/2
thus d<d/2 which is impossible so my assumption that a is not equal to b is incorrect, hence a=b.

0.99999... and 1 satisfy the hypotheses of the lemma, hence they are equal.

Note it should be a-b is non zero not a=b is nonzero in the statement of the lemma

[Organic]

Matt don't add words that I did not use (for example: "then")!

Here it is again, and now give your answer step by step, according to what I write:

First let us write your rigorous proof:

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

You clime that if for any e>0 we have |a-b|=d<e then a=b.

Your proof is:

1) a is not b

2) |a-b|=d>0

3) e=d/2, but since we know that |a-b|=d<e, then e=d/2 breaking our on rules, because |a-b|=d<e does not exist anymore.

Instead we have to use now |a-b|=d/2<e.

Shortly seaking, this part:

let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

breaking the rules, therefore this proof doesn't hold because e=<e.

The problem is in e=<e, therefore your proof does't hold water.

[pig]

if |a-b|=d>0, d/2 is still > 0 so e=d/2 doesn't break the rule that e>0.

the contradiction which appears is not the result of the fallacy of the proof, it is the whole point of the proof.

[matt grime]

Do you understand what proof be contradiction means? By the very fact that we are have the non-sensical assertion that d<d/2 (which is "allowed" by hypothesis on a and b) we have shown that the assumption that a does not equal b is incorrect, and thus a=b.

As I now see Pig has eloquently stated too.

[Organic]

But it breacks the rule that |a-b|=d<e, which means that e always MUST be greater than d.

When you write e=d/2, you break the rules.

[matt grime]

And this is the contradiction that shows the assumption that a=/=b is false, hence a=b as we were require to prove. We have two "rules" that by assumption must both be satisfied, yet this is nonsensical so it must be that our assumption is incorrect. That is what proof by contradiction does.

[pig]

i will try to restate this in a way you will understand it.

if |a-b| is smaller than ANY number larger than 0, then |a-b| cannot be larger than 0, or it would by definition "smaller than any number > 0" have to be smaller than itself.

if |a-b| < any number larger than 0, then

a!=b -> |a-b|>0 -> |a-b|<|a-b|

[Organic]

Matt,

What are you talking about?

You start with this statment |a-b|=d<e

Then you contradict your own statment by writing that e=d/2.

The result is e=<e, which is a contradiction.

Therefore e is logicaly meaningless and you can't use it to prove anything about |a-b|.

Edit: The result is e not= e, which is a contradiction.

[Organic]

|a-b| is smaller than ANY number larger than 0

|a-b| is a general notation to say that the gap between d and 0 is always d and not 0.

[matt grime]

I think the easiest way to point out that you haven't got a clue is to state

e<=e is not a contradiction.

any real number is less than or equal to itself.

I have a property that a and b satisfy, if a and b are distinct then I can show there is a strictly positive number d that satisfies d<d/2 (in fact d<d)

If you dont like me setting e=d/2 then how about let e be any non-zero positive number less than d, then by hypothesis |a-b|<e and simultaneously e<|a-b| contradiction, hence a=b.

You do see the word contradiction don't you? The only rule the e must satisfy is that it is greater than zero. It is not that e satisfies the rule that it must be greater then |a-b| but that |a-b| must be less than e, that is subtly different, this is not a result about e, which is allowed to be any positive real number, but about a and b.

Look at where the quantifiers come in the construction of the proposition

[Organic]

let e be any non-zero positive number less than d

It does not change anything because, when you say |a-b|=d < e then e is alway greater than d.

Threfore:

"let e be any non-zero positive number less than d" --> e < d

"|a-b|=d < e" AND "e < d" --> false, and cannot be used in your proof.

I made a miskate in my pervious post, because by your proof we get:

e not= e which is a contradiction, therefore e is meaningless.

[matt grime]

No, no, no, no. Look at which points the quantifiers and the thier referents occur.

let a and b be two real numbers. Suppose for all e>o that |a-b|<e, then this implies a=b.

IF |a-b| is not zero then we have a contradiction, because |a-b|= d is a real number >0, thus a and be cannot satisfy the hypothesis since no real number satisfies d<d. Is that better for you? I've not said e=d/2 or anything else that can't be true. Thus we can safely say the proposition is true.

[pig]

this is my last attempt. i think you are confused with what the number e represents. let's try this way:

1. let a and b be real numbers.
2. let S be a set of all real numbers > 0.
3. let |a-b| < all members of that set.

if a!=b, then |a-b|>0 therefore |a-b| is a member of S* and therefore |a-b|<|a-b|*. this is impossible.

if a=b, |a-b|=0 therefore |a-b| is not a member of S*.

notice that the contradiction doesn't result from wrong reasoning, but from the impossibility of a!=b if the first 3 statements are true.

[Organic]

let a and b be two real numbers. Suppose for all e>o that |a-b|<e, then this implies a=b.

Yes,yes,yes,yes.

I am talking about abs(a-b)=d therefore d is always a positive value greater than 0.

When you say: abs(a-b)=d < e, it means that no matter how d is small, e is always bigger than d.

Threfore:

"let e be any non-zero positive number less than d" --> e < d

"|a-b|=d < e" AND "e < d" --> false, and cannot be used in your proof.

e not= e which is a contradiction, therefore e is meaningless.

[matt grime]

I'm afraid saying things like 'all' when the set is infinite is only likely to provoke Organic to even greater heights of crankiness.

[Organic]

let |a-b| < all members of that set

It means that S is not complete, no more no less.

Shortly speaking, statment 2 is meaninless.

[matt grime]

So you agree the proposition is true, but that the proof is not correct? Well what's your proof then? (I can think of at least two more) But the problem isn't the result it's the idea of proof by contradiction, isn't it?

I know that the "and" is false ok, so on of the 'inputs' is false in some way agreed the only way for that to happen is if d=0 which leads us to the conclusion that a=b.


Suppose I prove sqrt(2) is irrational, by saying suppose it's p/q p and q both not even.... and then get a contradiction thus sqrt(2) is not rational. But the first step was to assume that it's rational, therefore as it's irrational my proof can't be valid because I've got two mutually exclusive things happening!

This is how contradiction works. I am not saying e must be greater than d, but that if the hypotheses are true that we get a contradiction. Anyway, I've rewritten the proof to omit entirely the mention of e a couple of posts back does that ease your worried mind?

[Organic]

if the hypotheses are true that we get a contradiction

Why?

1) a not= b

2) abs(a-b)=d < e > 0

So, where is the proof by contradiction?

[matt grime]

If a and b satisfied |a-b|<e for all e>0 then it impossible for a not be equal to b since if a were not eqaul to b then |a-b| is some positive number and we would have to have BY HYPOTHESIS that |a-b|<|a-b| which is impossible hence if the hypotheses are true it implies that a=b (as we've seen that the assumption otherwise invalidates the hypothesis ***A CONTRADICTION***).

[Organic]

and we would have to have BY HYPOTHESIS that |a-b|<|a-b|

You never get to this HYPOTHESIS because:

1) a not= b

2) abs(a-b)=d < e > 0

e cannot be both litte than and greater than d, and you have no hypothesis that shows the contradiction that we get from |a-b|<|a-b|.

[matt grime]

Do you know what proof by contradiction means? We entertain a silly idea temporarily putting aside our intuition and show that there is a genuine contradiction.

What we do here is to prove A=>B is to show that not(B) => not(A)

we've now omitted the step you find objectionable entirely by proving that it is not possible.

So let us take what I had in my previous post: would have to have. Do you understand what that means?

We are showing that |a-b| not zero implies that (since |a-b| <|a-b| is impossible) that they hypothesis is impossible. That is what we want, because it shows the negation of the conclusion implies the negation of the hypotheses, thus the hypotheses imply the conclusion.

This is very simple logic and every time you post indicating you don't understand the proof you are weakening yet more your stance as being some ultimate arbiter of mathematical correctness.

I wonder, how long do you actually take to try and understand the answers you get?

[Organic]

|a-b| <|a-b|


1) a not= b

2) abs(a-b)=d < e > 0 < d

Please show how by (1) and (2) you can prove by contradiction that d = 0

[pig]

if d is smaller than any possible number > 0, then d cannot be > 0 because it would have to be smaller than itself. what is so hard for you to understand here?

[Organic]

if d is smaller than any number > 0, then d cannot be > 0 because it would have to be smaller than itself

No, abs(a-b)=d < e > 0 is smaller then any number accsept 0, because d approaching but never reaching 0.

I you can't understand that simple thing?

[matt grime]

Occam's razor: two seemingly eloquent expositors of mathematics say you are wrong, you insist you are right without being able to explain yourself coherently... hmm, wonder how that's going to work out.

We've omitted any mention of e, and e<d and all the things you found unacceptable so what else are we supposed to do?

We are saying that it is impossible for any real strictly positive number to be less than every other real strictly positive number, that's all.

It's not very hard, and it doesn't even mention e or d.

[pig]

No, abs(a-b)=d < e > 0 is smaller then any number accsept 0, because d approaching but never reaching 0.

I you can't understand that simple thing?

wait a minute. you say it is smaller than any number except 0, but is "never reaching 0". what is it then? santa claus? i am wasting my time here.

[matt grime]

but d isn't approaching anything because d is fixed, organic.

[Organic]

Matt,

You are looking on your impossible d from the wrong side.

Look at d from 0 side and see how d always > 0, because there are infinitely many of them between any given d and 0.

Shortly speaking, if d=0 then we have the smallest d, and as we know, this collection of infenitely many elements > 0 has no smallest element.

[pig]

please demonstrate this on the example of a=2, b=2. what is there between d and 0?

[matt grime]

that could be a goody...

d is |a-b| and a and b are fixed not mention the fact thta card((0,d)) isn't important in the slightest, really (not that we expect you to realize that and not that you even know what card means either).

[Organic]

please demonstrate this on the example of a=2, b=2

a not= b, this is the first rule.

[pig]

a not= b, this is the first rule.

sorry, i didn't realize we were still stuck there, remind me what exactly are you trying to prove? just please be clear and exact.

i mean, i now have a feeling you're trying to prove that if d>0, then d cannot be 0, which is self-evident.

[Hurkyl]

Since we've left the domain of mathematics, we're also leaving the domain of the mathematics forum.

[matt grime]

were we (well organic) ever in the realm of mathematics?

[Organic]

Matt,

not that we expect you to realize that and not that you even know what card means either


This is the whole idea.

a not= b --> 1=|{d}|

a = b --> 0=|{}|

So as you see to reach 0, d has a phase transition form 1(exists) to 0(does not exist).

By this phase transition we are breaking the rules by eliminating d existence.

And why we are breaking the rules? because our case is the Math universe where a not= b.

When a=b we made a phase transition to another Math universe where d does not exist, therefore cannot be used in our proof as the impossible d<d.

if d is smaller than any possible number > 0, then d cannot be > 0 because it would have to be smaller than itself. what is so hard for you to understand here?

1) a not= b

2) abs(a-b)=d < e > 0

d is always smaller than e where e>0, and we never comparing d to itself but only to e.

Therefore we can never conclude that d<d.

[matt grime]

crank the crackpot index by a couple more points.

[pig]

Therefore we can never conclude that d<d.

we conclude that a=b, because otherwise d would be < d which is impossible. what do you find wrong with my statement that you quoted? i am trying to avoid mentioning "e" because you clearly do not understand what it represents from the very beginning.

[Organic]

if d is smaller than any possible number > 0, then d cannot be > 0 because it would have to be smaller than itself. what is so hard for you to understand here?

1) a not= b

2) abs(a-b)=d < e > 0

d is always smaller than e where e>0, and we never comparing d to itself but only to e.

Therefore we can never conclude that the impossibility of d<d --> a=b.

[matt grime]

But that isn't what we conlcude at all

[Organic]

Therefore we can never conclude that the impossibility of d<d --> a=b.

[matt grime]

If we show you a stick, are you guaranteed to get the wrong end of it?

[pig]

organic, here is a simpler but related problem written in a way that should be familiar to you.

Little Johnny is the best mathematician in his class. One morning, during the break, the teacher decides to test his logical skills. She knows Johnny often mentions how he likes all numbers which are larger than 0, and only those. When she waves at Johnny, a candy bar in her hand, he runs up to her desk within seconds. "Johnny, dear, I am thinking about a number. You wouldn't like this number, it is too small, but it is not negative. If you guess the number, the candy bar is yours. If you guess wrong, the fat kid in the third row gets it." Johnny's answer was zero.

Who got the candy bar?

A) Johnny
B) The fat kid in the third row
C) Eric Cartman
D) All of them approach, but never reach the candy bar
E) The candy bar was phase-transfered to another candy bar universe because it broke the rules

[Organic]

Pig, it is a beautiful story!

I am not so good in English, but I'll try my best.

One morning the farmer said to his last pig in the farm:

Here is a sand clock.

After the last grain of sand felled down you will be eaten.

Pig loved life very much, so he prayed to the god of mercy

The god of mercy heard his pray and came to help him.

He looked on the sand clock and said:

"Do not be afraid little pig, I have here another sand clock, that from outside looks exactly like the farmer's sand clock but inside of it time never ends, because this is the send clock of eternity of gods.

But little pig, you have to find by yourself how it works, otherwise I cannot help you, all I can tell you is that one and only one grain is falling down to the bottom of the sand clock each time ."

The pig knew Math and he loved very much irrational numbers, so he thought to himself:

What if each grain of sand is like some irrational number that its right side is always open, and it means that it has no exact value.

It means that I can use this open side to find how the sand clock of eternity works.

So he said to the god of mercy:

"My answer is this:

Each time when a one grain of sand is falling down to the bottom of the sand clock, each grain in the top side of the clock is divided to two pieces, and the width of the entrance between the top and the bottom sides of the send clock, becomes 1/2 of its previous width.

In this case our sand clock is the clock of eternity."

And the little pig got his life back and enjoyed every piece of grain of it, until his last day.

And before he died he discoverd the secret of the clock of eternity to his sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons, and they discovered it to their sons ...


(By the way, from my point of view 0 is not positive AND also not negative).

[matt grime]

presumes real space and so on is infinitely divisible, which it isn't. See what happens when you think that physical inutuition and mathematical intuition are the same thing?

[Organic]

Yes Matt, my intuition is the theoretical one, your is the physical one where infinitely many elements has fixed properties, and we can use words like "all" or "complete" that related to them.

On the contrary, the theoretical infinity (what I call potential infinity) is an open collection, which incompleteness is one of its fundamental properties.

And the reason that in your world 0 is a posivite number, based on the excluded middle idea of the Boolean or Fuzzy Logics.

My intuitions are based on Complementary Logic:

http://www.geocities.com/complementarytheory/BFC.pdf

Boolean or Fuzzy Logics are proper sub-systems of Complementary Logic, therefore They are going to get off stage as main logical systems.

[matt grime]

you aren't making sense. In fact you've entirely switched your view that I am a thoerist just interested in definitions and it was you that was dealing with the acutal reality of mathematics as a physical entity.

Your second paragraph starts as though you are going to explain why I think something, and then doesn't explain anything. (and whether or not you declare 0 to be positive or negative or neither or both is personal preference)

[Organic]

and it was you that was dealing with the acutal reality of mathematics as a physical entity.

What I showed is that actual infinity is too strong or too weak to be modulated by using Math language.

More details can be found here:
http://www.geocities.com/complementarytheory/Theory.pdf

Your second paragraph starts as though you are going to explain why I think something, and then doesn't explain anything.

Between us, please show me a one case where you tried to understand me.

[matt grime]

Well you could take that second paragraph and rewrite it so that it makes sense, in some sense. For instance if you replaced the comma with 'is' it might make some sense, but would still be inaccurate.

Why don't you demonstrate that at some point you've attempted to understand any of the very basic mathematics upon which you've seen fit to pronounce judgement despite obviously being incredibly ignorant and idiotically stupid?

[Organic]

Even if math originated solely from our brain it would still remain an underlying principle of nature.

Do you think that our brain totally-included, half-included or totally not-included to the underlying principle of nature?

Can you demonstrate an abstract thought which is totally not influenced by reality?

Do you think that professional mathematicians can develop Math in one hand but on the other hand they say: "We don't care about reality when we develop our definitions"?

For example:


Speaking up for (some of ) the mathematicians: we don't care. If we did we'd be doing philosophy. Would the martians have derived that equation? Perhaps, perhaps not - they almost certianly wouldn't have devised the same way of presenting it, and we couldn't tell if they'd picked i or -i as their square root of -1, which they may have called something else anyway. That answer has a superficial and a non-superficial part to it.


There is a "philosophy" section to Physics forum and this probably belongs there.


Can we ignore our abilities to develop Math language by saying that our abilities to develop Math is not mathematical but a philosophical question?

Please be aware that not some of but most of the professional mathematicians have Matt's opinion on these questions, and the reason is very simple, beside learning Math in the universities they also learn from their teachers that the logical realm of "pure" Math has no connections with the real world.

And when we say: "the logical realm of "pure" Math has no connections with the real world", don't we use philosophy?

As you can see Matt, I understand very well your game, and because I understand it I have the motivation to develop a better game which philosophy, our cognitive abilities and reality influence are legitimate parts of it.

[pig]

impressive use of enlarged letters, organic. the blue is also a nice touch.

good luck with developing a "better game". i am looking forward to a rigorous, logically consistent system which will be more efficient in describing real world problems than math is :)

[matt grime]

Organic, you also wanted me to show that I have understood you, well, in this thread we've pointed out that if you're so against proof by contradiction one can remove that part of the argument, but you've not commented upon that nicety.

Did you miss it?

Let X be the property "is a positive real number is smaller than any other positive real number"
Let Y be "the number is zero"

if d, a positive real number satisfies not(Y), then as d<d/2 is clearly not true, it follows that not(X) is true

ie not(Y)=>not(X), ie X=>Y

see, no assumption of mutually exclusive events.

Also I pointed out that given two statements, say A, B, that AandB false, which is what happens, is not meaningless, but tells you one of the two statements A and B is false.

DId you take time out to smell the roses and realize that any of those things had actually happened?

[Organic]

Invariant state 1) a not= b --> abs(a-b)=d > 0
Invariant state 2) d=d < e > 0

Therefore by (1) and (2) for any e > d, d > 0


Please show us the proof by contradiction, according what I wrote.



Also this time please show how x_n = limit-point in this particular curve:

http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html

[matt grime]

What's an invariant state. And why would I wish to demonstrate something for that example that is probably not true, and is impossible to verify becuase it;s just a diagram, after all, if we look close enough, because of the limits of the diagram, that curve will have to be a straight line, possibly one pixel large long at the current distance. That curve doesn't have an equation describing it, you can't possibly know how it behaves if we 'zoom' in. How can I prove the impossible? I offered the observation that, in general, there is nothing to stop N-R iteration converging after a finite number of steps, I said nothing about that particular example, it was only your imprecise language that allowed that mistake to propogate.

[Organic]

Invariant state means: no matter how you zoom in, always you will find that:

1) a not= b --> abs(a-b)=d > 0
2) d=d < e > 0

Therefore by (1) and (2) for any e > d, d > 0

[matt grime]

but how on earth does one 'zoom into' a proposition? how can i prove things using definitions that do not make sense to me. I have proved the assertion that if two real numbers, a and b, satisfy the property that for all e>0 |a-b|<e then a=b. I did it by contradiction, understood that you disbelieved it because you don't see how that method of proof works so I rewrote it without contradiction (ie using the contrapositive). I cannot help you learn how to understand proof by contradiction because it is part of your dogmatic and religious approach that you refuse to be open minded to it (you refusing to attempt to learn how it works). All of your reasons for not believing it are fanciful and illogical, and you can't teach someone to be logical, just show them how it works, and hope the use it. We've explained how it works, why don't you learn about it rather than automatically gain-saying it because someone like me, who you distrust automatically, told you about it? You cannot be hypocritical and hope not to lose respect.

[Organic]

who you distrust automatically

I am like a marsian that do not see infinity as you see it, and the way I see infinity your logical proposition does not hold.

The reason is this:

1) a not= b

Matt: By my way (1) is an hypothesis.

Organic: By my way (1) is an invariant state.

2) abs(a-b)=d < e > 0

Matt: a) By my way you compare d to a set S that includes in it all R members > 0 in this case d<d is impossible; therefore d must be = 0 --> a=b

Matt: b) Another version of my way is to say that e=d/2 but then |a-b|=d AND |a-b|<d/2 which is impossible; therefore a=b.

Organic: e and d relation remaining unchanged in any arbitrary scale that you choose, which means: d is always smaller then e but greater than 0, it means that e=d/2 is impossible because e > d/n > 0.

Organic: S is an open collection (has infinitely many elements) therefore cannot be completed by definition. Only finite collection can be a complete collection. Therefore there is no such thing like S which includes all r > 0.

Matt: e and d are fixed values.

Organic: e and d are variables, and both of them always greater 0.

[matt grime]

could you do everyone a favour? when you're about to switch and start talking about mathematics as only you consider it and not as the rest of the world understands it could you give us a signal? we prove things in context. if you're just going to artificially change the context and the rules at least have the decency to let us know so we can just file it under 'organic goes off on one again' for instance we state d is fixed (it is |a-b| and they are fixed), so if you're going to use another incompatible definition of it give us a heads up.

[Organic]

'organic goes off on one again'

I think that the one who goes on an off is you Matt, for example:

By your method, infinitely many elements have the same property like finitely many elements, and this is the reason that you have no problem to use words like 'all' and 'complete' together with infinitely many elements.

But sometimes you use the open interval Idea, and then no infinitely many elements can reach the limit.

Trough my point of view, a collection of infinitely many elements cannot be completed by definition, therefore I call it an open collection.

[matt grime]

Then you aren't using any of those words with the definition attached to them that other people give them. Therefore it is no wonder you are constantly saying things that are wrong on appearance, as everyone keeps telling you, you are using words with a different meaning from everyone else, and hte fact that what they say is 'wrong' if given your meaning to the words is not a particularly relevant issue because of your misuse of the terms.

[Organic]

Have you heard about the words "Paradigm Shift"?

Paradigm shift is like a mutation where mutation changes the system from within and not just adding another pretty thing to the existing system without changing any fundamental concept of it.

For example my organic numbers:

Let x be a general notation for a singleton.

When a finite collection of singletons have the same color, we mean that all singletons are identical, or have the maximum symmetry-degree.

When each singleton has its own unique color, we mean that each singleton in the finite collection is unique, or the collection has the minimum symmetry-degree.

Multiplication can be operated only among identical singletons, where addition is operated among unique singletons.

Each natural number is used as some given quantity, where in this given quantity we can order several different sets, that have the same quantity of singletons, but they are different by their symmetrical degrees.

In more formal way, within the same quantity we can define all possible degrees, which existing between a multiset and a "normal" set, where the complete multiset and the complete "normal" set are included too.

If we give an example of transformations between multisetes and "normal" sets, we can see that the internal structure of n+1 > 1 ordered forms, constructed by using all previous n >= 1 forms:

1
(+1) = {x}

2
(1*2) = {x,x}
((+1)+1) = {{x},x}

3
(1*3) = {x,x,x}
((1*2)+1) = {{x,x},x}
(((+1)+1)+1) = {{{x},x},x}

4
(1*4) = {x,x,x,x} <------------- Maximum symmetry-degree,
((1*2)+1*2) = {{x,x},x,x} Minimum information's
(((+1)+1)+1*2) = {{{x},x},x,x} clarity-degree
((1*2)+(1*2)) = {{x,x},{x,x}} (no uniqueness)
(((+1)+1)+(1*2)) = {{{x},x},{x,x}}
(((+1)+1)+((+1)+1)) = {{{x},x},{{x},x}}
((1*3)+1) = {{x,x,x},x}
(((1*2)+1)+1) = {{{x,x},x},x}
((((+1)+1)+1)+1) = {{{{x},x},x},x} <------ Minimum symmetry-degree,
Maximum information's
clarity-degree
(uniqueness)
5
...


Can someone give me an address of some mathematical brach that researches these kind of relations between multisets and "normal" sets?

Thank you,

Organic

[matt grime]

Ah, paradigm shifts, automatic 40 point penalty on the crackpot index. Organic, I have forgotten more about this than you will ever learn, stop patronizing please.

[Organic]

automatic 40 point penalty on the crackpot index

1) patronizing? Who gives the points here, me?

2) Please show us a mathematical branch that define organic numbers as I show in my previous post.

3) Once you asked me to explain how N is not a complete collection by show n which is not in N.

My answer is very simple: natural numbers do not exists because of the existence of N, but because of the axioms that define them, N is only the name of the container that its content is infinitely many elements that can never be completed.

[matt grime]

How can we demonstrate organic numbers, as you've never defined them?

[Organic]

Here they are, in colors, to help you to understand their structures:

Let x be a general notation for a singleton.

When a finite collection of singletons have the same color, we mean that all singletons are identical, or have the maximum symmetry-degree.

When each singleton has its own unique color, we mean that each singleton in the finite collection is unique, or the collection has the minimum symmetry-degree.

Multiplication can be operated only among identical singletons, where addition is operated among unique singletons.

Each natural number is used as some given quantity, where in this given quantity we can order several different sets, that have the same quantity of singletons, but they are different by their symmetrical degrees.

In more formal way, within the same quantity we can define all possible degrees, which existing between a multiset and a "normal" set, where the complete multiset and the complete "normal" set are included too.

If we give an example of transformations between multisetes and "normal" sets, we can see that the internal structure of n+1 > 1 ordered forms, constructed by using all previous n >= 1 forms:

1
(+1) = {x}

2
(1*2) = {x,x}
((+1)+1) = {{x},x}

3
(1*3) = {x,x,x}
((1*2)+1) = {{x,x},x}
(((+1)+1)+1) = {{{x},x},x}

4
(1*4) = {x,x,x,x} <------------- Maximum symmetry-degree,
((1*2)+1*2) = {{x,x},x,x} Minimum information's
(((+1)+1)+1*2) = {{{x},x},x,x} clarity-degree
((1*2)+(1*2)) = {{x,x},{x,x}} (no uniqueness)
(((+1)+1)+(1*2)) = {{{x},x},{x,x}}
(((+1)+1)+((+1)+1)) = {{{x},x},{{x},x}}
((1*3)+1) = {{x,x,x},x}
(((1*2)+1)+1) = {{{x,x},x},x}
((((+1)+1)+1)+1) = {{{{x},x},x},x} <------ Minimum symmetry-degree,
Maximum information's
clarity-degree
(uniqueness)
5
...


Can you give me an address of some mathematical brach that researches these kind of relations between multisets and "normal" sets?

Thank you,

Organic

[matt grime]

You're just defining certain types of partition functions. Why should we do your research? Do I patronize you? Tell you things you already know? Strange you don't seem to know lots of things. I'm certainly not polite to you, but that's because you don't earn respect from me, not that you'd want to obviously.

[Organic]

You're just defining certain types of partition functions

Have you seen before any use of these partition functions as I do?

[matt grime]

Yes and no. I have not seen people assign words to things as you do, but then as you never explain what any of those words mean that is highly irrelevant. You don't actually do anything with the things you write down as you admit yourself (and that scores you 50 more points on the crackpot index as well).

[Organic]

You don't actually do anything with the things you write down as you admit yourself

Please look here:

http://www.geocities.com/complementarytheory/CATpage.html

and show us an example of how I don't actually do anything with the things I write down.